problem_id int64 1 978 | question stringlengths 86 2.11k | source stringlengths 19 76 | solution stringlengths 94 14.7k | asymptote_code stringlengths 44 17.8k | solution_image_url stringclasses 16
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226 | Point $D$ lies on side $\overline{BC}$ of $\triangle ABC$ so that $\overline{AD}$ bisects $\angle BAC.$ The perpendicular bisector of $\overline{AD}$ intersects the bisectors of $\angle ABC$ and $\angle ACB$ in points $E$ and $F,$ respectively. Given that $AB=4,BC=5,$ and $CA=6,$ the area of $\triangle AEF$ can be written as $\tfrac{m\sqrt{n}}p,$ where $m$ and $p$ are relatively prime positive integers, and $n$ is a positive integer not divisible by the square of any prime. Find $m+n+p.$ | 2020 AIME I Problem 13 | Let $B=(0,0)$ and $BC$ be the line $y=0$.
We compute that $\cos{\angle{ABC}}=\frac{1}{8}$, so $\tan{\angle{ABC}}=3\sqrt{7}$.
Thus, $A$ lies on the line $y=3x\sqrt{7}$. The length of $AB$ at a point $x$ is $8x$, so $x=\frac{1}{2}$.
We now have the coordinates $A=\left(\frac{1}{2},\frac{3\sqrt{7}}{2}\right)$, $B=(0,0)$ and $C=(5,0)$.
We also have $D=(2,0)$ by the angle-bisector theorem and $M=\left(\frac{5}{4},\frac{3\sqrt{7}}{4}\right)$ by taking the midpoint.
We have that because $\cos{\angle{ABC}}=\frac{1}{8}$, $\cos{\frac{\angle{ABC}}{2}}=\frac{3}{4}$ by half angle formula.
We also compute $\cos{\angle{ACB}}=\frac{3}{4}$, so $\cos{\frac{\angle{ACB}}{2}}=\frac{\sqrt{14}}{4}$.
Now, $AD$ has slope $-\frac{\frac{3\sqrt{7}}{2}}{2-\frac{1}{2}}=-\sqrt{7}$, so it's perpendicular bisector has slope $\frac{\sqrt{7}}{7}$ and goes through $\left(\frac{5}{4},\frac{3\sqrt{7}}{4}\right)$.
We find that this line has equation $y=\frac{\sqrt{7}}{7}x+\frac{4\sqrt{7}}{7}$.
As $\cos{\angle{CBI}}=\frac{3}{4}$, we have that line $BI$ has form $y=\frac{\sqrt{7}}{3}x$.
Solving for the intersection point of these two lines, we get $x=3$ and thus $E=\left(3, \sqrt{7}\right)$
We also have that because $\cos{\angle{ICB}}=\frac{\sqrt{14}}{4}$, $CI$ has form $y=-\frac{x\sqrt{7}}{7}+\frac{5\sqrt{7}}{7}$.
Intersecting the line $CI$ and the perpendicular bisector of $AD$ yields $-\frac{x\sqrt{7}}{7}+\frac{5\sqrt{7}}{7}=\frac{x\sqrt{7}}{7}+\frac{4\sqrt{7}}{7}$.
Solving this, we get $x=\frac{1}{2}$ and so $F=\left(\frac{1}{2},\frac{9\sqrt{7}}{14}\right)$.
We now compute $EF=\sqrt{\left(\frac{5}{2}\right)^2+\left(\frac{5\sqrt{7}}{14}\right)^2}=\frac{5\sqrt{14}}{7}$.
We also have $MA=\sqrt{\left(\frac{3}{4}\right)^2+\left(\frac{3\sqrt{7}}{4}\right)^2}=\frac{3\sqrt{2}}{2}$.
As ${MA}\perp{EF}$, we have $[\triangle{AEF}]=\frac{1}{2}\left(\frac{3\sqrt{2}}{2}\times\frac{5\sqrt{14}}{7}\right)=\frac{15\sqrt{7}}{14}$.
The desired answer is $15+7+14=\boxed{036}$ ~Imayormaynotknowcalculus | // Block 1
size(8cm); defaultpen(fontsize(10pt));
pair A,B,C,I,D,M,T,Y,Z,EE,F;
A=(0,3sqrt(7));
B=(-1,0);
C=(9,0);
I=incenter(A,B,C);
D=extension(A,I,B,C);
M=(A+D)/2;
draw(B--EE,gray+dashed);
draw(C--F,gray+dashed);
draw(A--B--C--A);
draw(A--D);
draw(B--(5,sqrt(28)));
draw(M--(5,sqrt(28)));
draw(C--(0,9sqrt(7)/7));
draw(M--(0,9sqrt(7)/7));
dot("$A$",A,NW);
dot("$B$",B,SW);
dot("$C$",C,SE);
dot("$D$",D,S);
dot("$E$",(5,sqrt(28)),N);
dot("$M$",M,dir(70));
dot("$F$",(0,9sqrt(7)/7),N);
label("$2$",B--D,S);
label("$3$",D--C,S);
label("$6$",A--C,N);
label("$4$",A--B,W);
// Block 2
size(8cm); defaultpen(fontsize(10pt)); pair A,B,C,I,D,M,T,Y,Z,EE,F; A=(0,3sqrt(7)); B=(-1,0); C=(9,0); I=incenter(A,B,C); D=extension(A,I,B,C); M=(A+D)/2; draw(B--EE,gray+dashed); draw(C--F,gray+dashed); draw(A--B--C--A); draw(A--D); draw(B--(5,sqrt(28))); draw(M--(5,sqrt(28))); draw(C--(0,9sqrt(7)/7)); draw(M--(0,9sqrt(7)/7)); dot("$A$",A,NW); dot("$B$",B,SW); dot("$C$",C,SE); dot("$D$",D,S); dot("$E$",(5,sqrt(28)),N); dot("$M$",M,dir(70)); dot("$F$",(0,9sqrt(7)/7),N); label("$2$",B--D,S); label("$3$",D--C,S); label("$6$",A--C,N); label("$4$",A--B,W); | [] |
226 | Point $D$ lies on side $\overline{BC}$ of $\triangle ABC$ so that $\overline{AD}$ bisects $\angle BAC.$ The perpendicular bisector of $\overline{AD}$ intersects the bisectors of $\angle ABC$ and $\angle ACB$ in points $E$ and $F,$ respectively. Given that $AB=4,BC=5,$ and $CA=6,$ the area of $\triangle AEF$ can be written as $\tfrac{m\sqrt{n}}p,$ where $m$ and $p$ are relatively prime positive integers, and $n$ is a positive integer not divisible by the square of any prime. Find $m+n+p.$ | 2020 AIME I Problem 13 | As usual, we will use homogenized barycentric coordinates.
We have that $AD$ will have form $3z=2y$. Similarly, $CF$ has form $5y=6x$ and $BE$ has form $5z=4x$.
Since $A=(1,0,0)$ and $D=\left(0,\frac{3}{5},\frac{2}{5}\right)$, we also have $M=\left(\frac{1}{2},\frac{3}{10},\frac{1}{5}\right)$.
It remains to determine the equation of the line formed by the perpendicular bisector of $AD$.
This can be found using EFFT. Let a point $T$ on $EF$ have coordinates $(x, y, z)$.
We then have that the displacement vector $\overrightarrow{AD}=\left(-1, \frac{3}{5}, \frac{2}{5}\right)$ and that the displacement vector $\overrightarrow{TM}$ has form $\left(x-\frac{1}{2},y-\frac{3}{10},z-\frac{1}{5}\right)$.
Now, by EFFT, we have $5^2\left(\frac{3}{5}\times\left(z-\frac{1}{5}\right)+\frac{2}{5}\times\left(y-\frac{3}{10}\right)\right)+6^2\left(-1\times\left(z-\frac{1}{5}\right)+\frac{2}{5}\times\left(x-\frac{1}{2}\right)\right)+4^2\left(-1\times\left(y-\frac{3}{10}\right)+\frac{3}{5}\times\left(x-\frac{1}{2}\right)\right)=0$.
This equates to $8x-2y-7z=2$.
Now, intersecting this with $BE$, we have $5z=4x$, $8x-2y-7z=2$, and $x+y+z=1$.
This yields $x=\frac{2}{3}$, $y=-\frac{1}{5}$, and $z=\frac{8}{15}$, or $E=\left(\frac{2}{3},-\frac{1}{5},\frac{8}{15}\right)$.
Similarly, intersecting this with $CF$, we have $5y=6x$, $8x-2y-7z=2$, and $x+y+z=1$.
Solving this, we obtain $x=\frac{3}{7}$, $y=\frac{18}{35}$, and $z=\frac{2}{35}$, or $F=\left(\frac{3}{7},\frac{18}{35},\frac{2}{35}\right)$.
We finish by invoking the Barycentric Distance Formula twice; our first displacement vector being $\overrightarrow{FE}=\left(\frac{5}{21},-\frac{5}{7},\frac{10}{21}\right)$.
We then have $FE^2=-25\left(-\frac{5}{7}\cdot\frac{10}{21}\right)-36\left(\frac{5}{21}\cdot\frac{10}{21}\right)-16\left(\frac{5}{21}\cdot-\frac{5}{7}\right)=\frac{50}{7}$, thus $FE=\frac{5\sqrt{14}}{7}$.
Our second displacement vector is $\overrightarrow{AM}=\left(-\frac{1}{2},\frac{3}{10},\frac{1}{5}\right)$.
As a result, $AM^2=-25\left(\frac{3}{10}\cdot\frac{1}{5}\right)-36\left(-\frac{1}{2}\cdot\frac{1}{5}\right)-16\left(-\frac{1}{2}\cdot\frac{3}{10}\right)=\frac{9}{2}$, so $AM=\frac{3\sqrt{2}}{2}$.
As ${AM}\perp{EF}$, the desired area is $\frac{\frac{5\sqrt{14}}{7}\times\frac{3\sqrt{2}}{2}}{2}={\frac{15\sqrt{7}}{14}}\implies{m+n+p=\boxed{036}}$. ~Imayormaynotknowcalculus
Remark: The area of $\triangle{AEF}$ can also be computed using the Barycentric Area Formula, although it may increase the risk of computational errors; there are also many different ways to proceed once the coordinates are determined. | // Block 1
size(8cm); defaultpen(fontsize(10pt));
pair A,B,C,I,D,M,T,Y,Z,EE,F;
A=(0,3sqrt(7));
B=(-1,0);
C=(9,0);
I=incenter(A,B,C);
D=extension(A,I,B,C);
M=(A+D)/2;
draw(B--EE,gray+dashed);
draw(C--F,gray+dashed);
draw(A--B--C--A);
draw(A--D);
draw(B--(5,sqrt(28)));
draw(M--(5,sqrt(28)));
draw(C--(0,9sqrt(7)/7));
draw(M--(0,9sqrt(7)/7));
dot("$A$",A,NW);
dot("$B$",B,SW);
dot("$C$",C,SE);
dot("$D$",D,S);
dot("$E$",(5,sqrt(28)),N);
dot("$M$",M,dir(70));
dot("$F$",(0,9sqrt(7)/7),N);
label("$2$",B--D,S);
label("$3$",D--C,S);
label("$6$",A--C,N);
label("$4$",A--B,W);
// Block 2
size(8cm); defaultpen(fontsize(10pt)); pair A,B,C,I,D,M,T,Y,Z,EE,F; A=(0,3sqrt(7)); B=(-1,0); C=(9,0); I=incenter(A,B,C); D=extension(A,I,B,C); M=(A+D)/2; draw(B--EE,gray+dashed); draw(C--F,gray+dashed); draw(A--B--C--A); draw(A--D); draw(B--(5,sqrt(28))); draw(M--(5,sqrt(28))); draw(C--(0,9sqrt(7)/7)); draw(M--(0,9sqrt(7)/7)); dot("$A$",A,NW); dot("$B$",B,SW); dot("$C$",C,SE); dot("$D$",D,S); dot("$E$",(5,sqrt(28)),N); dot("$M$",M,dir(70)); dot("$F$",(0,9sqrt(7)/7),N); label("$2$",B--D,S); label("$3$",D--C,S); label("$6$",A--C,N); label("$4$",A--B,W); | [] |
226 | Point $D$ lies on side $\overline{BC}$ of $\triangle ABC$ so that $\overline{AD}$ bisects $\angle BAC.$ The perpendicular bisector of $\overline{AD}$ intersects the bisectors of $\angle ABC$ and $\angle ACB$ in points $E$ and $F,$ respectively. Given that $AB=4,BC=5,$ and $CA=6,$ the area of $\triangle AEF$ can be written as $\tfrac{m\sqrt{n}}p,$ where $m$ and $p$ are relatively prime positive integers, and $n$ is a positive integer not divisible by the square of any prime. Find $m+n+p.$ | 2020 AIME I Problem 13 | To get the area of $\triangle AEF$, we try to find $AM$ and $\angle EAF$.
Since $AD$ is the angle bisector, we can get that $BD=2$ and $CD=3$. By applying Stewart's Theorem, we can get that $AD=3\sqrt{2}$. Therefore $AM=\frac{3\sqrt{2}}{2}$.
Since $EF$ is the perpendicular bisector of $AD$, we know that $AE = DE$. Since $BE$ is the angle bisector of $\angle BAC$,
we know that $\angle ABE = \angle DBE$. By applying the Law of Sines to $\triangle ABE$ and $\triangle DBE$, we know that
$\sin \angle BAE = \sin \angle BDE$. Since $BD$ is not equal to $AB$ and therefore these two triangles are not congruent, we know that $\angle BAE$ and $\angle BDE$ are supplementary. Then we know that $\angle ABD$ and $\angle AED$ are also supplementary. Given that $AE=DE$, we can get that $\angle DAE$ is half of $\angle ABC$. Similarly, we have $\angle DAF$ is half of $\angle ACB$.
By applying the Law of Cosines, we get $\cos \angle ABC = \frac{1}{8}$, and then $\sin \angle ABC = \frac{3\sqrt{7}}{8}$. Similarly, we can get $\cos \angle ACB = \frac{3}{4}$ and $\sin \angle ACB = \frac{\sqrt{7}}{4}$. Based on some trig identities, we can compute that $\tan \angle DAE = \frac{\sin \angle ABC}{1 + \cos \angle ABC} = \frac{\sqrt{7}}{3}$, and $\tan \angle DAF = \frac{\sqrt{7}}{7}$.
Finally, the area of $\triangle AEF$ equals $\frac{1}{2}AM^2(\tan \angle DAE + \tan \angle DAF)=\frac{15\sqrt{7}}{14}$. Therefore, the final answer is $15+7+14=\boxed{036}$. ~xamydad
Remark: I didn't figure out how to add segments $AF$, $AE$, $DF$ and $DE$. Can someone please help add these segments?
(Added :) ~Math_Genius_164) | // Block 1
size(8cm); defaultpen(fontsize(10pt));
pair A,B,C,I,D,M,T,Y,Z,EE,F;
A=(0,3sqrt(7));
B=(-1,0);
C=(9,0);
I=incenter(A,B,C);
D=extension(A,I,B,C);
M=(A+D)/2;
draw(B--EE,gray+dashed);
draw(C--F,gray+dashed);
draw(A--B--C--A);
draw(A--D);
draw(A--(5,sqrt(28)));
draw(A--(0,9sqrt(7)/7));
draw(D--(0,9sqrt(7)/7));
draw(D--(5,sqrt(28)));
draw(B--(5,sqrt(28)));
draw(M--(5,sqrt(28)));
draw(C--(0,9sqrt(7)/7));
draw(M--(0,9sqrt(7)/7));
dot("$A$",A,NW);
dot("$B$",B,SW);
dot("$C$",C,SE);
dot("$D$",D,S);
dot("$E$",(5,sqrt(28)),N);
dot("$M$",M,dir(70));
dot("$F$",(0,9sqrt(7)/7),N);
label("$2$",B--D,S);
label("$3$",D--C,S);
label("$6$",A--C,N);
label("$4$",A--B,W);
// Block 2
size(8cm); defaultpen(fontsize(10pt)); pair A,B,C,I,D,M,T,Y,Z,EE,F; A=(0,3sqrt(7)); B=(-1,0); C=(9,0); I=incenter(A,B,C); D=extension(A,I,B,C); M=(A+D)/2; draw(B--EE,gray+dashed); draw(C--F,gray+dashed); draw(A--B--C--A); draw(A--D); draw(A--(5,sqrt(28))); draw(A--(0,9sqrt(7)/7)); draw(D--(0,9sqrt(7)/7)); draw(D--(5,sqrt(28))); draw(B--(5,sqrt(28))); draw(M--(5,sqrt(28))); draw(C--(0,9sqrt(7)/7)); draw(M--(0,9sqrt(7)/7)); dot("$A$",A,NW); dot("$B$",B,SW); dot("$C$",C,SE); dot("$D$",D,S); dot("$E$",(5,sqrt(28)),N); dot("$M$",M,dir(70)); dot("$F$",(0,9sqrt(7)/7),N); label("$2$",B--D,S); label("$3$",D--C,S); label("$6$",A--C,N); label("$4$",A--B,W); | [] |
226 | Point $D$ lies on side $\overline{BC}$ of $\triangle ABC$ so that $\overline{AD}$ bisects $\angle BAC.$ The perpendicular bisector of $\overline{AD}$ intersects the bisectors of $\angle ABC$ and $\angle ACB$ in points $E$ and $F,$ respectively. Given that $AB=4,BC=5,$ and $CA=6,$ the area of $\triangle AEF$ can be written as $\tfrac{m\sqrt{n}}p,$ where $m$ and $p$ are relatively prime positive integers, and $n$ is a positive integer not divisible by the square of any prime. Find $m+n+p.$ | 2020 AIME I Problem 13 | Let $x = \angle BAD = \angle CAD$, $y = \angle CBE = \angle ABE$, and $z = \angle BCF = \angle ACF$. Notice that $x+y+z = 90^\circ$.
In $\triangle ABD$, segment $\overline{BE}$ is the bisector of $\angle ABD$, and $E$ lies on the perpendicular bisector of side $\overline{AD}$. Therefore $E$ is the midpoint of arc $\stackrel{\textstyle\frown}{AD}$ on the circumcircle of $\triangle ABD$. It follows that $\angle BED = \angle BAD = x$ and $\angle EDA = \angle EBA = y$. Likewise, $ACDF$ is cyclic, $\angle CFD = \angle CAD = x$, and $\angle FDA = \angle FCA = z$. Because $\overline{EF}$ is the perpendicular bisector of $\overline{AD}$, triangles $AEF$ and $DEF$ are congruent, implying that
\begin{align*} [\triangle AEF] = [\triangle DEF] &= \frac{DE\cdot DF\cdot\sin(\angle EDF)}{2}\\ &= \frac{DE\cdot DF\cdot\sin(y+z)}{2} = \frac{DE\cdot DF\cdot\cos x}{2}. \end{align*}
Applying the Law of Sines to $\triangle BED$ and $\triangle CFD$ gives\[DE = BD\cdot\frac{\sin y}{\sin x}\text{~ and ~} DF = CD\cdot\frac{\sin z}{\sin x}.\]By the Angle Bisector Theorem, $BD = 2$ and $CD = 3$. Combining the above information yields
\[[\triangle AEF] = \frac{3\sin y\cdot\sin z\cdot\cos x}{\sin^2 x}.\]Applying the Law of Cosines to $\triangle ABC$ gives $\cos 2x = \frac9{16}$, $\cos 2y = \frac1{8}$, and $\cos 2z = \frac34$. By the Half Angle Formulas,\[\sin^2x = \frac7{32},~~ \cos x = \sqrt{\frac{25}{32}},~~ \sin y = \sqrt{\frac7{16}}, \text{~ and ~} \sin z = \sqrt{\frac18}.\]Therefore
\[[\triangle AEF] = \frac{3\cdot\sqrt{\frac{7}{16}}\cdot\sqrt{\frac{1}{8}}\cdot\sqrt{\frac{25}{32}}} {\frac{7}{32}} = \frac{15\sqrt{7}}{14}.\]The requested sum is $15+7+14 = 36$. | // Block 1
unitsize(0.8 cm);
pair A, B, C, D, E, F, I;
real angleC = aCos(3/4);
C = (0,0);
A = 6*dir(270 - angleC/2);
B = 5*dir(270 + angleC/2);
I = incenter(A,B,C);
D = extension(A, I, B, C);
E = extension((A + D)/2, (A + D)/2 + rotate(90)*(A - D), B, I);
F = extension((A + D)/2, (A + D)/2 + rotate(90)*(A - D), C, I);
draw(A--B--C--cycle);
draw(A--interp(A,D,1.4));
draw(A--F--D--E--cycle);
draw(E--F);
draw(circumcircle(A,B,D));
draw(B--interp(B,E,1.5));
draw(C--interp(C,F,1.5));
dot("$A$", A, SW);
dot("$B$", B, dir(0));
dot("$C$", C, N);
dot("$D$", D, dir(0));
dot("$E$", E, N);
dot("$F$", F, dir(0));
// Block 2
unitsize(0.8 cm); pair A, B, C, D, E, F, I; real angleC = aCos(3/4); C = (0,0); A = 6*dir(270 - angleC/2); B = 5*dir(270 + angleC/2); I = incenter(A,B,C); D = extension(A, I, B, C); E = extension((A + D)/2, (A + D)/2 + rotate(90)*(A - D), B, I); F = extension((A + D)/2, (A + D)/2 + rotate(90)*(A - D), C, I); draw(A--B--C--cycle); draw(A--interp(A,D,1.4)); draw(A--F--D--E--cycle); draw(E--F); draw(circumcircle(A,B,D)); draw(B--interp(B,E,1.5)); draw(C--interp(C,F,1.5)); dot("$A$", A, SW); dot("$B$", B, dir(0)); dot("$C$", C, N); dot("$D$", D, dir(0)); dot("$E$", E, N); dot("$F$", F, dir(0)); | [] |
226 | Point $D$ lies on side $\overline{BC}$ of $\triangle ABC$ so that $\overline{AD}$ bisects $\angle BAC.$ The perpendicular bisector of $\overline{AD}$ intersects the bisectors of $\angle ABC$ and $\angle ACB$ in points $E$ and $F,$ respectively. Given that $AB=4,BC=5,$ and $CA=6,$ the area of $\triangle AEF$ can be written as $\tfrac{m\sqrt{n}}p,$ where $m$ and $p$ are relatively prime positive integers, and $n$ is a positive integer not divisible by the square of any prime. Find $m+n+p.$ | 2020 AIME I Problem 13 | Let the point $M$ be the midpoint of $\overline{AD}$, let $I$ be the incenter of $\triangle ABC$ which is the common point of lines $AD$, $BE$, and $CF$, and let $r$ be the inradius of $\triangle ABC$. The semiperimeter of $\triangle ABC$ is\[s = \frac{AB + BC + CA}2 = \frac{15}2,\]and Heron's Formula gives the area of $\triangle ABC$ as\[\sqrt{s(s-AB)(s-BC)(s-CA)} = \frac{15\sqrt7}4.\]This area is also $rs$ implying that $r = \frac{\sqrt7}2$. Stewart's Theorem gives $AD =3\sqrt2$. Because the ratio of the areas of $\triangle IBC$ and $\triangle ABC$ is $\frac{ID}{AD},$ it follows that\[ID = AD\cdot\frac{\frac{r\cdot BC}2}{\frac{15\sqrt7}4} = \sqrt2.\]Thus $IM = MD - ID = \frac{\sqrt2}2$.
Note that $\angle EFI = 90^{\circ} - \angle FIA = 90^{\circ} - \angle CID = 90^{\circ} - \frac{\angle A}{2} - \frac{\angle C}{2} = \frac{\angle B}{2} = \angle IBC$. Thus
$\triangle IBC \sim \triangle IFE$. The height of $\triangle IBC$ to $I$ is $r=\frac{\sqrt7}2$, and the height of $\triangle IFE$ to $I$ is $IM=\frac{\sqrt2}2$, so $EF = BC\cdot \frac{IM}r = \frac{5\sqrt{14}}{7}$. The needed area of $\triangle AEF$ is $\frac12\cdot EF\cdot \frac{AD}2 = \frac{15\sqrt7}{14}$, as above. | // Block 1
unitsize(0.8 cm);
pair A, B, C, D, E, F, I, M;
real angleC = aCos(3/4);
C = (0,0);
A = 6*dir(270 - angleC/2);
B = 5*dir(270 + angleC/2);
I = incenter(A,B,C);
D = extension(A, I, B, C);
E = extension((A + D)/2, (A + D)/2 + rotate(90)*(A - D), B, I);
F = extension((A + D)/2, (A + D)/2 + rotate(90)*(A - D), C, I);
M = (A + D)/2;
draw(A--B--C--cycle);
draw(A--interp(A,D,1.4));
draw(A--F--D--E--cycle);
draw(E--F);
draw(B--interp(B,E,1.5));
draw(C--interp(C,F,1.5));
dot("$A$", A, SW);
dot("$B$", B, dir(0));
dot("$C$", C, N);
dot("$D$", D, dir(0));
dot("$E$", E, N);
dot("$F$", F, dir(0));
dot("$I$", I, dir(120));
dot("$M$", M, W);
// Block 2
unitsize(0.8 cm); pair A, B, C, D, E, F, I, M; real angleC = aCos(3/4); C = (0,0); A = 6*dir(270 - angleC/2); B = 5*dir(270 + angleC/2); I = incenter(A,B,C); D = extension(A, I, B, C); E = extension((A + D)/2, (A + D)/2 + rotate(90)*(A - D), B, I); F = extension((A + D)/2, (A + D)/2 + rotate(90)*(A - D), C, I); M = (A + D)/2; draw(A--B--C--cycle); draw(A--interp(A,D,1.4)); draw(A--F--D--E--cycle); draw(E--F); draw(B--interp(B,E,1.5)); draw(C--interp(C,F,1.5)); dot("$A$", A, SW); dot("$B$", B, dir(0)); dot("$C$", C, N); dot("$D$", D, dir(0)); dot("$E$", E, N); dot("$F$", F, dir(0)); dot("$I$", I, dir(120)); dot("$M$", M, W); | [] |
227 | Let $\triangle ABC$ be an acute triangle with circumcircle $\omega,$ and let $H$ be the intersection of the altitudes of $\triangle ABC.$ Suppose the tangent to the circumcircle of $\triangle HBC$ at $H$ intersects $\omega$ at points $X$ and $Y$ with $HA=3,HX=2,$ and $HY=6.$ The area of $\triangle ABC$ can be written in the form $m\sqrt{n},$ where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n.$ | 2020 AIME I Problem 15 | The following is a power of a point solution to this menace of a problem:
Let points be what they appear as in the diagram below. Note that $3HX = HY$ is not insignificant; from here, we set $XH = HE = \frac{1}{2} EY = HL = 2$ by PoP and trivial construction. Now, $D$ is the reflection of $A$ over $H$. Note $AO \perp XY$, and therefore by Pythagorean theorem we have $AE = XD = \sqrt{5}$. Consider $HD = 3$. We have that $\triangle HXD \cong HLK$, and therefore we are ready to PoP with respect to $(BHC)$. Setting $BL = x, LC = y$, we obtain $xy = 10$ by PoP on $(ABC)$, and furthermore, we have $KH^2 = 9 = (KL - x)(KL + y) = (\sqrt{5} - x)(\sqrt{5} + y)$. Now, we get $4 = \sqrt{5}(y - x) - xy$, and from $xy = 10$ we take \[\frac{14}{\sqrt{5}} = y - x.\] However, squaring and manipulating with $xy = 10$ yields that $(x + y)^2 = \frac{396}{5}$ and from here, since $AL = 5$ we get the area to be $3\sqrt{55} \implies \boxed{058}$. ~awang11's sol | // Block 1
defaultpen(fontsize(12)+0.6); size(250);
pen p=fontsize(10)+gray+0.4;
var phi=75.5, theta=130, r=4.8;
pair A=r*dir(270-phi-57), C=r*dir(270+phi-57), B=r*dir(theta-57), H=extension(A,foot(A,B,C),B,foot(B,C,A));
path omega=circumcircle(A,B,C), c=circumcircle(H,B,C);
pair O=circumcenter(H,B,C), Op=2*H-O, Z=bisectorpoint(O,Op), X=IP(omega,L(H,Z,0,50)), Y=IP(omega,L(H,Z,50,0)), D=2*H-A, K=extension(B,C,X,Y), E=extension(A,origin,X,Y), L=foot(H,B,C);
draw(K--C^^omega^^c^^L(A,D,0,1)); draw(Y--K^^L(A,origin,0,1.5)^^L(X,D,0.7,3.5),fuchsia+0.4);
draw(CR(H,length(H-L)),royalblue);
draw(C--K,p);
dot("$A$",A,dir(120)); dot("$B$",B,down); dot("$C$",C,down); dot("$H$",H,down); dot("$X$",X,up); dot("$Y$",Y,down);
dot("$O$",origin,down); dot("$O'$",O,down); dot("$K$",K,up); dot("$L$",L,dir(H-X));
dot("$D$",D,down); dot("$E$",E,down);
//draw(O--origin,p);
//draw(origin--4*dir(57),fuchsia);
//draw(Arc(H,3,160,200),royalblue); //draw(Arc(H,2,30,70),heavygreen); //draw(Arc(H,6,220,240),fuchsia);
// Block 2
defaultpen(fontsize(12)+0.6); size(250); pen p=fontsize(10)+gray+0.4; var phi=75.5, theta=130, r=4.8; pair A=r*dir(270-phi-57), C=r*dir(270+phi-57), B=r*dir(theta-57), H=extension(A,foot(A,B,C),B,foot(B,C,A)); path omega=circumcircle(A,B,C), c=circumcircle(H,B,C); pair O=circumcenter(H,B,C), Op=2*H-O, Z=bisectorpoint(O,Op), X=IP(omega,L(H,Z,0,50)), Y=IP(omega,L(H,Z,50,0)), D=2*H-A, K=extension(B,C,X,Y), E=extension(A,origin,X,Y), L=foot(H,B,C); draw(K--C^^omega^^c^^L(A,D,0,1)); draw(Y--K^^L(A,origin,0,1.5)^^L(X,D,0.7,3.5),fuchsia+0.4); draw(CR(H,length(H-L)),royalblue); draw(C--K,p); dot("$A$",A,dir(120)); dot("$B$",B,down); dot("$C$",C,down); dot("$H$",H,down); dot("$X$",X,up); dot("$Y$",Y,down); dot("$O$",origin,down); dot("$O'$",O,down); dot("$K$",K,up); dot("$L$",L,dir(H-X)); dot("$D$",D,down); dot("$E$",E,down); //draw(O--origin,p); //draw(origin--4*dir(57),fuchsia); //draw(Arc(H,3,160,200),royalblue); //draw(Arc(H,2,30,70),heavygreen); //draw(Arc(H,6,220,240),fuchsia); | [] |
227 | Let $\triangle ABC$ be an acute triangle with circumcircle $\omega,$ and let $H$ be the intersection of the altitudes of $\triangle ABC.$ Suppose the tangent to the circumcircle of $\triangle HBC$ at $H$ intersects $\omega$ at points $X$ and $Y$ with $HA=3,HX=2,$ and $HY=6.$ The area of $\triangle ABC$ can be written in the form $m\sqrt{n},$ where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n.$ | 2020 AIME I Problem 15 | Diagram not to scale.
We first observe that $H'$, the image of the reflection of $H$ over line $BC$, lies on circle $O$. This is because $\angle HBC = 90 - \angle C = \angle H'AC = \angle H'BC$. This is a well known lemma. The result of this observation is that circle $O'$, the circumcircle of $\triangle BHC$ is the image of circle $O$ over line $BC$, which in turn implies that $\overline{AH} = \overline{OO'}$ and thus $AHO'O$ is a parallelogram. That $AHO'O$ is a parallelogram implies that the radius $AO$ is perpendicular to $\overline{XY}$, and thus divides segment $\overline{XY}$ in two equal pieces, $\overline{XD}$ and $\overline{DY}$, of length $4$.
Using Power of a Point,
\[\overline{AH} \cdot \overline{HH'} = \overline{XH} \cdot \overline{HY} \Longrightarrow 3 \cdot \overline{HH'} = 2 \cdot 6 \Longrightarrow \overline{HH'} = 4\]
This means that $\overline{HL} = \frac12 \cdot 4 = 2$ and $\overline{AL} = 2 + 3 = 5$, where $L$ is the foot of the altitude from $A$ onto $BC$. All that remains to be found is the length of segment $\overline{BC}$.
Looking at right triangle $\triangle AHD$, we find that
\[\overline{AD} = \sqrt{\overline{AH}^2 - \overline{HD}^2} = \sqrt{3^2 - 2^2} = \sqrt{5}\]
Looking at right triangle $\triangle ODY$, we get the equation
\[\overline{OY}^2 - \overline{DY}^2 = \overline{OD}^2 = \left(\overline{AO} - \overline{AD}\right)^2\]
Plugging in known values, and letting $R$ be the radius of the circle, we find that
\[R^2 - 16 = (R - \sqrt{5})^2 = R^2 - 2\sqrt5 R + 5 \Longrightarrow R = \frac{21\sqrt5}{10}\]
Recall that $AHO'O$ is a parallelogram, so $\overline{AH} = \overline{OO'} = 3$. So, $\overline{OM} = \frac32$, where $M$ is the midpoint of $\overline{BC}$. This means that
\[\overline{BC} = 2\overline{BM} = 2\sqrt{R^2 - \left(\frac32\right)^2} = 2\sqrt{\frac{441}{20} - \frac{9}{4}} = \frac{6\sqrt{55}}{5}\]
Thus, the area of triangle $\triangle ABC$ is
\[\frac{\overline{AL} \cdot \overline{BC}}{2} = \frac{5 \cdot \frac{6\sqrt{55}}{5}}{2} = {3\sqrt{55}}\]
The answer is $3 + 55 = \boxed{058}$. | // Block 1
size(10cm);
pair A, B, C, D, H, K, O, P, L, M, X, Y;
A = (-15, 27);
B = (-24, 0);
C = (24, 0);
D = (-8.28, 18.04);
O = (0, 7);
P = (0, -7);
H = (-15, 13);
K = (-15, -13);
M = (0, 0);
L = (-15, 0);
X = (-24.9569, 5.53234);
Y = (8.39688, 30.5477);
draw(circle(O, 25));
draw(circle(P, 25));
draw(A--B--C--cycle);
draw(H -- K);
draw(A -- O -- P -- H -- cycle);
draw(X -- Y);
draw(O -- X, dashed);
draw(O -- Y, dashed);
draw(O -- B, dashed);
draw(O -- C, dashed);
label("$O$", O, ENE);
label("$A$", A, NW);
label("$B$", B, W);
label("$C$", C, E);
label("$H$", H, E);
label("$H'$", K, NE);
label("$X$", X, W);
label("$Y$", Y, NE);
label("$O'$", P, E);
label("$M$", M, NE);
label("$L$", L, NE);
label("$D$", D, NNE);
label("$2$", X -- H, NW);
label("$3$", H -- A, SW);
label("$6$", H -- Y, NW);
label("$R$", O -- Y, E);
dot(O);
dot(P);
dot(D);
dot(H);
// Block 2
size(10cm); pair A, B, C, D, H, K, O, P, L, M, X, Y; A = (-15, 27); B = (-24, 0); C = (24, 0); D = (-8.28, 18.04); O = (0, 7); P = (0, -7); H = (-15, 13); K = (-15, -13); M = (0, 0); L = (-15, 0); X = (-24.9569, 5.53234); Y = (8.39688, 30.5477); draw(circle(O, 25)); draw(circle(P, 25)); draw(A--B--C--cycle); draw(H -- K); draw(A -- O -- P -- H -- cycle); draw(X -- Y); draw(O -- X, dashed); draw(O -- Y, dashed); draw(O -- B, dashed); draw(O -- C, dashed); label("$O$", O, ENE); label("$A$", A, NW); label("$B$", B, W); label("$C$", C, E); label("$H$", H, E); label("$H'$", K, NE); label("$X$", X, W); label("$Y$", Y, NE); label("$O'$", P, E); label("$M$", M, NE); label("$L$", L, NE); label("$D$", D, NNE); label("$2$", X -- H, NW); label("$3$", H -- A, SW); label("$6$", H -- Y, NW); label("$R$", O -- Y, E); dot(O); dot(P); dot(D); dot(H); | [] |
227 | Let $\triangle ABC$ be an acute triangle with circumcircle $\omega,$ and let $H$ be the intersection of the altitudes of $\triangle ABC.$ Suppose the tangent to the circumcircle of $\triangle HBC$ at $H$ intersects $\omega$ at points $X$ and $Y$ with $HA=3,HX=2,$ and $HY=6.$ The area of $\triangle ABC$ can be written in the form $m\sqrt{n},$ where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n.$ | 2020 AIME I Problem 15 | Extend $\overline{AH}$ to intersect $\omega$ again at $P$. The Power of a Point Theorem yields $HP = \tfrac{HX \cdot HY}{HA} = 4$. Because $\angle CAP=\angle CBP$, and $\angle CAP$ and $\angle CBH$ are both complements to $\angle C$, it follows that $\angle CBP = \angle CBH$, implying that $\overline{BC}$ bisects $\overline{HP}$, so the length of the altitude from $A$ to $\overline{BC}$ is $h_a = AH + \tfrac12 HP = 5$.
Let the circumcircle of $\triangle BCH$ be $\omega'$. Because $\triangle BCH \cong \triangle BCP$, the two triangles must have the same circumradius. Because the circumcircle of $\triangle BCP$ is $\omega$, the circles $\omega$ and $\omega'$ have the same radius $R$. Denote the centers of $\omega$ and $\omega'$ by $O$ and $O'$, respectively, and let $M$ be the midpoint of $\overline{XY}$. Note that trapezoid $HMOO'$ has $\angle H = \angle M = 90^\circ$. Also $HM = XM - XH = \frac12\cdot XY - HX = 2$ and $HO' = R$. Because $\omega$ is a translation of $\omega'$ in the direction of $\overline{AH}$, it follows that $OO' = AH = 3$. Finally, the Pythagorean Theorem applied to $\triangle XMO$ yields $MO = \sqrt{R^2-16}$. Let $T$ be the projection of $O$ onto $\overline{HO'}$. Then $TO' = R-MO$, so the Pythagorean Theorem applied to $\triangle TOO'$ yields
\[R - \sqrt{R^2-16} = \sqrt{3^2 - 2^2} = \sqrt{5}.\]Solving for $R$ gives $R = \tfrac{21}{2\sqrt5}$. It follows from properties of the orthocenter $H$ that\[\cos\angle A = \dfrac{AH}{2R} = \dfrac{\sqrt5}{7},\]so\[\sin\angle A = \sqrt{1 - \cos^2\angle A} = \dfrac{2\sqrt{11}}{7}.\]Therefore by the Extended Law of Sines\[a = BC = 2R \sin\angle A = \dfrac{6\sqrt{11}}{\sqrt5},\]so
\[[\triangle ABC] = \frac12 a h_a = \frac12 \cdot \frac{6\sqrt{11}}{\sqrt{5}} \cdot 5 = 3\sqrt{55}.\]The requested sum is $3+55 = 58$. | // Block 1
unitsize(0.6 cm);
pair A, B, C, D, H, M, O, Op, P, T, X, Y, Z;
real R = 21/(2*sqrt(5));
A = (7/sqrt(5),7/2);
O = (0,0);
B = intersectionpoint((0,-3/2)--(R,-3/2),Circle(O,R));
C = intersectionpoint((0,-3/2)--(-R,-3/2),Circle(O,R));
H = A + B + C;
P = reflect(B,C)*(H);
D = (H + P)/2;
Op = reflect(B,C)*(O);
X = intersectionpoint(H--(H + scale(2)*rotate(90)*(Op - H)), Circle(O,R));
Y = intersectionpoint(H--(H + scale(2)*rotate(90)*(H - Op)), Circle(O,R));
Z = extension(X, Y, B, C);
M = (X + Y)/2;
T = H + O - M;
draw(Circle(O,R));
draw(Circle(Op,R));
draw(A--B--C--cycle);
draw(A--P);
draw(B--Z--Y);
draw(H--Op--O--M);
draw(O--T);
draw(O--X);
dot("$A$", A, NE);
dot("$B$", B, SW);
dot("$C$", C, W);
dot("$D$", D, SW);
dot("$H$", H, NE);
dot("$M$", M, NE);
dot("$O$", O, W);
dot("$O'$", Op, W);
dot("$P$", P, SE);
dot("$T$", T, N);
dot("$X$", X, E);
dot("$Y$", Y, NW);
dot("$Z$", Z, E);
label("$\omega$", R*dir(140), dir(140));
label("$\omega'$", Op + R*dir(220), dir(220));
// Block 2
unitsize(0.6 cm); pair A, B, C, D, H, M, O, Op, P, T, X, Y, Z; real R = 21/(2*sqrt(5)); A = (7/sqrt(5),7/2); O = (0,0); B = intersectionpoint((0,-3/2)--(R,-3/2),Circle(O,R)); C = intersectionpoint((0,-3/2)--(-R,-3/2),Circle(O,R)); H = A + B + C; P = reflect(B,C)*(H); D = (H + P)/2; Op = reflect(B,C)*(O); X = intersectionpoint(H--(H + scale(2)*rotate(90)*(Op - H)), Circle(O,R)); Y = intersectionpoint(H--(H + scale(2)*rotate(90)*(H - Op)), Circle(O,R)); Z = extension(X, Y, B, C); M = (X + Y)/2; T = H + O - M; draw(Circle(O,R)); draw(Circle(Op,R)); draw(A--B--C--cycle); draw(A--P); draw(B--Z--Y); draw(H--Op--O--M); draw(O--T); draw(O--X); dot("$A$", A, NE); dot("$B$", B, SW); dot("$C$", C, W); dot("$D$", D, SW); dot("$H$", H, NE); dot("$M$", M, NE); dot("$O$", O, W); dot("$O'$", Op, W); dot("$P$", P, SE); dot("$T$", T, N); dot("$X$", X, E); dot("$Y$", Y, NW); dot("$Z$", Z, E); label("$\omega$", R*dir(140), dir(140)); label("$\omega'$", Op + R*dir(220), dir(220)); | [] |
228 | Let $P$ be a point chosen uniformly at random in the interior of the unit square with vertices at $(0,0), (1,0), (1,1)$, and $(0,1)$. The probability that the slope of the line determined by $P$ and the point $\left(\frac58, \frac38 \right)$ is greater than or equal to $\frac12$ can be written as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2020 AIME II Problem 2 | The line through the fixed point $\left(\frac58,\frac38\right)$ with slope $\frac12$ has equation $y=\frac12 x + \frac1{16}$. The slope between $P$ and the fixed point exceeds $\frac12$ if $P$ falls within the shaded region in the diagram below consisting of two trapezoids with area
\[\frac{\frac1{16}+\frac38}2\cdot\frac58 + \frac{\frac58+\frac7{16}}2\cdot\frac38 = \frac{43}{128}.\]Because the entire square has area $1,$ the required probability is $\frac{43}{128}$. The requested sum is $43+128 = 171$. | // Block 1
defaultpen(fontsize(8pt));
unitsize(4cm);
pair A = (0,0);
pair B = (1,0);
pair C = (1,1);
pair D = (0,1);
pair F = (0, 1/16);
pair G = (1, 9/16);
pair H = (5/8, 0);
pair J = (5/8, 1);
pair K = IP(H--J, F--G);
pair P = (13/16, 12/16);
pair Q = extension(P,K,A,B);
pair R = extension(K,P,C,D);
draw(A--B--C--D--cycle);
label("$(0,0)$", A, SW);
label("$(1,0)$", B, SE);
label("$(1,1)$", C, E);
label("$(0,1)$", D, W);
filldraw(A--H--K--F--cycle, lightgray);
filldraw(K--G--C--J--cycle, lightgray);
dot(K);
dot("$P$", P, W);
draw(Q -- R, dashed);
label("$\frac 38$", H--K, E);
label("$\frac 58$", K--J, W);
label("$\frac 7{16}$", G--C, E);
label("$\frac 38$", C--J, N);
label("$\frac 1{16}$", A--F, dir(160));
// Block 2
defaultpen(fontsize(8pt)); unitsize(4cm); pair A = (0,0); pair B = (1,0); pair C = (1,1); pair D = (0,1); pair F = (0, 1/16); pair G = (1, 9/16); pair H = (5/8, 0); pair J = (5/8, 1); pair K = IP(H--J, F--G); pair P = (13/16, 12/16); pair Q = extension(P,K,A,B); pair R = extension(K,P,C,D); draw(A--B--C--D--cycle); label("$(0,0)$", A, SW); label("$(1,0)$", B, SE); label("$(1,1)$", C, E); label("$(0,1)$", D, W); filldraw(A--H--K--F--cycle, lightgray); filldraw(K--G--C--J--cycle, lightgray); dot(K); dot("$P$", P, W); draw(Q -- R, dashed); label("$\frac 38$", H--K, E); label("$\frac 58$", K--J, W); label("$\frac 7{16}$", G--C, E); label("$\frac 38$", C--J, N); label("$\frac 1{16}$", A--F, dir(160)); | [] |
229 | Triangles $\triangle ABC$ and $\triangle A'B'C'$ lie in the coordinate plane with vertices $A(0,0)$, $B(0,12)$, $C(16,0)$, $A'(24,18)$, $B'(36,18)$, $C'(24,2)$. A rotation of $m$ degrees clockwise around the point $(x,y)$ where $0<m<180$, will transform $\triangle ABC$ to $\triangle A'B'C'$. Find $m+x+y$. | 2020 AIME II Problem 4 | We first draw a diagram with the correct Cartesian coordinates and a center of rotation $P$. Note that $PC=PC'$ because $P$ lies on the perpendicular bisector of $CC'$ (it must be equidistant from $C$ and $C'$ by properties of a rotation).
Since $AB$ is vertical while $A'B'$ is horizontal, we have that the angle of rotation must be $90^{\circ}$, and therefore $\angle P = 90^{\circ}$. Therefore, $CPC'$ is a 45-45-90 right triangle, and $CD=DP$.
We calculate $D$ to be $(20,1)$. Since we translate $4$ right and $1$ up to get from point $C$ to point $D$, we must translate $1$ right and $4$ down to get to point $P$. This gives us $P(21,-3)$. Our answer is then $90+21-3=\boxed{108}$. ~Lopkiloinm & samrocksnature | // Block 1
/* Geogebra to Asymptote conversion by samrocksnature, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -8.0451801958033, xmax = 47.246151591238494, ymin = -10.271454747548662, ymax = 21.426040258770957; /* image dimensions */
pen zzttqq = rgb(0.6,0.2,0); pen qqwuqq = rgb(0,0.39215686274509803,0); pen qqffff = rgb(0,1,1);
draw((16,0)--(0,0)--(0,12)--cycle, linewidth(2) + zzttqq);
draw((24,2)--(24,18)--(36,18)--cycle, linewidth(2) + blue);
draw((16,0)--(21,-3)--(24,2)--cycle, linewidth(2) + qqwuqq);
draw((21.39134584768662,-2.3477569205223032)--(20.73910276820892,-1.9564110728356852)--(20.347756920522304,-2.608654152313382)--(21,-3)--cycle, linewidth(2) + qqffff);
/* draw figures */
draw((16,0)--(0,0), linewidth(2) + zzttqq);
draw((0,0)--(0,12), linewidth(2) + zzttqq);
draw((0,12)--(16,0), linewidth(2) + zzttqq);
draw((24,2)--(24,18), linewidth(2) + blue);
draw((24,18)--(36,18), linewidth(2) + blue);
draw((36,18)--(24,2), linewidth(2) + blue);
draw((16,0)--(24,2), linewidth(2));
draw((16,0)--(21,-3), linewidth(2) + qqwuqq);
draw((21,-3)--(24,2), linewidth(2) + qqwuqq);
draw((24,2)--(16,0), linewidth(2) + qqwuqq);
draw((21,-3)--(20,1), linewidth(2.8) + qqffff);
/* dots and labels */
dot((0,0),linewidth(4pt) + dotstyle);
label("$A$", (-0.6228029714727868,0.12704474547474198), NE * labelscalefactor);
dot((0,12),dotstyle);
label("$B$", (0.1301918194013232,12.354245873478124), NE * labelscalefactor);
dot((16,0),dotstyle);
label("$C$", (16.15822379657881,0.34218611429591583), NE * labelscalefactor);
dot((24,18),dotstyle);
label("$A'$", (24.154311337765787,18.342347305667463), NE * labelscalefactor);
dot((24,2),dotstyle);
label("$C'$", (23.186175178070503,1.95574638045472), NE * labelscalefactor);
dot((36,18),dotstyle);
label("$B'$", (36.13051420214449,18.342347305667463), NE * labelscalefactor);
dot((21,-3),dotstyle);
label("$P$", (21.35747354309052,-3.458644734878156), NE * labelscalefactor);
dot((20,1),linewidth(4pt) + dotstyle);
label("$D$", (20.13833911977053,1.2744653791876692), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
// Block 2
/* Geogebra to Asymptote conversion by samrocksnature, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -8.0451801958033, xmax = 47.246151591238494, ymin = -10.271454747548662, ymax = 21.426040258770957; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); pen qqwuqq = rgb(0,0.39215686274509803,0); pen qqffff = rgb(0,1,1); draw((16,0)--(0,0)--(0,12)--cycle, linewidth(2) + zzttqq); draw((24,2)--(24,18)--(36,18)--cycle, linewidth(2) + blue); draw((16,0)--(21,-3)--(24,2)--cycle, linewidth(2) + qqwuqq); draw((21.39134584768662,-2.3477569205223032)--(20.73910276820892,-1.9564110728356852)--(20.347756920522304,-2.608654152313382)--(21,-3)--cycle, linewidth(2) + qqffff); /* draw figures */ draw((16,0)--(0,0), linewidth(2) + zzttqq); draw((0,0)--(0,12), linewidth(2) + zzttqq); draw((0,12)--(16,0), linewidth(2) + zzttqq); draw((24,2)--(24,18), linewidth(2) + blue); draw((24,18)--(36,18), linewidth(2) + blue); draw((36,18)--(24,2), linewidth(2) + blue); draw((16,0)--(24,2), linewidth(2)); draw((16,0)--(21,-3), linewidth(2) + qqwuqq); draw((21,-3)--(24,2), linewidth(2) + qqwuqq); draw((24,2)--(16,0), linewidth(2) + qqwuqq); draw((21,-3)--(20,1), linewidth(2.8) + qqffff); /* dots and labels */ dot((0,0),linewidth(4pt) + dotstyle); label("$A$", (-0.6228029714727868,0.12704474547474198), NE * labelscalefactor); dot((0,12),dotstyle); label("$B$", (0.1301918194013232,12.354245873478124), NE * labelscalefactor); dot((16,0),dotstyle); label("$C$", (16.15822379657881,0.34218611429591583), NE * labelscalefactor); dot((24,18),dotstyle); label("$A'$", (24.154311337765787,18.342347305667463), NE * labelscalefactor); dot((24,2),dotstyle); label("$C'$", (23.186175178070503,1.95574638045472), NE * labelscalefactor); dot((36,18),dotstyle); label("$B'$", (36.13051420214449,18.342347305667463), NE * labelscalefactor); dot((21,-3),dotstyle); label("$P$", (21.35747354309052,-3.458644734878156), NE * labelscalefactor); dot((20,1),linewidth(4pt) + dotstyle); label("$D$", (20.13833911977053,1.2744653791876692), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ | [] |
230 | Two congruent right circular cones each with base radius $3$ and height $8$ have the axes of symmetry that intersect at right angles at a point in the interior of the cones a distance $3$ from the base of each cone. A sphere with radius $r$ lies within both cones. The maximum possible value of $r^2$ is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2020 AIME II Problem 7 | Consider the cross section of the cones and sphere by a plane that contains the two axes of symmetry of the cones as shown below. The sphere with maximum radius will be tangent to the sides of each of the cones. The center of that sphere must be on the axis of symmetry of each of the cones and thus must be at the intersection of their axes of symmetry. Let $A$ be the point in the cross section where the bases of the cones meet, and let $C$ be the center of the sphere. Let the axis of symmetry of one of the cones extend from its vertex, $B$, to the center of its base, $D$. Let the sphere be tangent to $\overline{AB}$ at $E$. The right triangles $\triangle ABD$ and $\triangle CBE$ are similar, implying that the radius of the sphere is\[CE = AD \cdot\frac{BC}{AB} = AD \cdot\frac{BD-CD}{AB} =3\cdot\frac5{\sqrt{8^2+3^2}} = \frac{15}{\sqrt{73}}=\sqrt{\frac{225}{73}}.\]The requested sum is $225+73=298$.
Not part of MAA's solution, but this: https://www.geogebra.org/calculator/xv4nm97a is a good visual of the cones in GeoGebra. | // Block 1
unitsize(0.6cm);
pair A = (0,0);
pair TriangleOneLeft = (-6,0);
pair TriangleOneDown = (-3,-8);
pair TriangleOneMid = (-3,0);
pair D = (0,-3);
pair TriangleTwoDown = (0,-6);
pair B = (-8,-3);
pair C = IP(TriangleOneMid -- TriangleOneDown, B--D);
pair EE = foot(C, A, B);
real radius = arclength(C--EE);
path circ = Circle(C, radius);
draw(A--B--TriangleTwoDown--cycle);
draw(B--D);
draw(A--TriangleOneLeft--TriangleOneDown--cycle);
draw(circ);
draw(C--EE);
draw(TriangleOneMid -- TriangleOneDown, gray);
dot("$B$", B, W);
dot("$E$", EE, NW);
dot("$A$", A, NE);
dot("$D$", D, E);
dot("$C$", C, SE);
// Block 2
unitsize(0.6cm); pair A = (0,0); pair TriangleOneLeft = (-6,0); pair TriangleOneDown = (-3,-8); pair TriangleOneMid = (-3,0); pair D = (0,-3); pair TriangleTwoDown = (0,-6); pair B = (-8,-3); pair C = IP(TriangleOneMid -- TriangleOneDown, B--D); pair EE = foot(C, A, B); real radius = arclength(C--EE); path circ = Circle(C, radius); draw(A--B--TriangleTwoDown--cycle); draw(B--D); draw(A--TriangleOneLeft--TriangleOneDown--cycle); draw(circ); draw(C--EE); draw(TriangleOneMid -- TriangleOneDown, gray); dot("$B$", B, W); dot("$E$", EE, NW); dot("$A$", A, NE); dot("$D$", D, E); dot("$C$", C, SE); | [] |
230 | Two congruent right circular cones each with base radius $3$ and height $8$ have the axes of symmetry that intersect at right angles at a point in the interior of the cones a distance $3$ from the base of each cone. A sphere with radius $r$ lies within both cones. The maximum possible value of $r^2$ is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2020 AIME II Problem 7 | Let $A$ be the origin in the above diagram. Then $B$ is $(6,0)$, $C$ is $(0,6)$, $D$ is $(3,8)$, and $E$ is $(8,3)$. Also, it is easy to see that the inscribed sphere is simply the inscribed circle of $AEF$. Then we want to find the intersection of $AD$ and $CE$ to determine the coordinates of point $F$. Note that $AD$ is line $y=\frac{8}{3}x$, and $CE$ is line $y=-\frac{3}{8}x+6$. Then, you can see that these lines are perpendicular, indicating that $AEF$ is a right triangle with right angle at $F$. Finding the intersection of the lines by solving the system, we get that $F$ is the point $(\frac{144}{73},\frac{384}{73})$ in this plane. Then, we can find the distances $EF$ and $AF$ by the distance formula, which are $\frac{55}{\sqrt{73}}$ and $\frac{48}{\sqrt{73}}$ respectively. Also, $AE=\sqrt{73}$. Then, to find the radius of this triangle's incircle, we use the formula $a=rs$ from which we get that $r=\frac{15}{\sqrt{73}}$ and $r^2=\frac{225}{73} \implies \boxed{298}$ is the answer.
~SirAppel | // Block 1
unitsize(0.6cm);
// Coordinates
pair A = (0,0);
pair B = (6,0);
pair C = (0,6);
// Calculate point C
pair D = (3,8);
pair E = (8,3);
pair F = (144/73,384/73);
// Draw triangles (cones)
draw(A--B--D--cycle);
draw(A--C--E--cycle);
draw(incircle(A,E,F));
pair EE = foot(C, A, B);
real radius = arclength(C--EE);
path circ = Circle(C, radius);
// Label points
dot("$A$", A, NW);
dot("$B$", B, SW);
dot("$C$", C, NE);
dot("$D$", D, SE);
dot("$E$", E, SE);
dot("$F$", F, NE);
// Block 2
unitsize(0.6cm); // Coordinates pair A = (0,0); pair B = (6,0); pair C = (0,6); // Calculate point C pair D = (3,8); pair E = (8,3); pair F = (144/73,384/73); // Draw triangles (cones) draw(A--B--D--cycle); draw(A--C--E--cycle); draw(incircle(A,E,F)); pair EE = foot(C, A, B); real radius = arclength(C--EE); path circ = Circle(C, radius); // Label points dot("$A$", A, NW); dot("$B$", B, SW); dot("$C$", C, NE); dot("$D$", D, SE); dot("$E$", E, SE); dot("$F$", F, NE); | [] |
230 | Two congruent right circular cones each with base radius $3$ and height $8$ have the axes of symmetry that intersect at right angles at a point in the interior of the cones a distance $3$ from the base of each cone. A sphere with radius $r$ lies within both cones. The maximum possible value of $r^2$ is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2020 AIME II Problem 7 | We graph this on graph paper, with the scale of $\sqrt{2}:1$. So, we can find $OT$ then divide by $\sqrt{2}$ to convert to our desired units, then square the result. With 5 minutes' worth of coordbashing, we finally arrive at $298$.
~samrocksnature | // Block 1
/* Geogebra to Asymptote conversion by samrocksnature, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -8.325025958411356, xmax = 8, ymin = -0.6033105644019334, ymax = 12.237120576121757; /* image dimensions */
pen zzttqq = rgb(0.6,0.2,0); pen qqwuqq = rgb(0,0.39215686274509803,0); pen cqcqcq = rgb(0.7529411764705882,0.7529411764705882,0.7529411764705882);
draw((5,11)--(0,0)--(-6,6)--cycle, linewidth(1) + zzttqq);
draw((6,6)--(-5,11)--(0,0)--cycle, linewidth(1) + qqwuqq);
draw((0.2328977836854361,5.767102216314564)--(0.4657955673708722,6)--(0.2328977836854361,6.232897783685436)--(0,6)--cycle, linewidth(1) + qqwuqq);
/* draw grid of horizontal/vertical lines */
pen gridstyle = linewidth(0.7) + cqcqcq; real gridx = 1, gridy = 1; /* grid intervals */
for(real i = ceil(xmin/gridx)*gridx; i <= floor(xmax/gridx)*gridx; i += gridx)
draw((i,ymin)--(i,ymax), gridstyle);
for(real i = ceil(ymin/gridy)*gridy; i <= floor(ymax/gridy)*gridy; i += gridy)
draw((xmin,i)--(xmax,i), gridstyle);
/* end grid */
Label laxis; laxis.p = fontsize(10);
xaxis(xmin, xmax, Ticks(laxis, Step = 1, Size = 2, NoZero),EndArrow(6), above = true);
yaxis(ymin, ymax, Ticks(laxis, Step = 1, Size = 2, NoZero),EndArrow(6), above = true); /* draws axes; NoZero hides '0' label */
/* draw figures */
draw((5,11)--(0,0), linewidth(1) + zzttqq);
draw((0,0)--(-6,6), linewidth(1) + zzttqq);
draw((-6,6)--(5,11), linewidth(1) + zzttqq);
draw((6,6)--(-5,11), linewidth(1) + qqwuqq);
draw((-5,11)--(0,0), linewidth(1) + qqwuqq);
draw((0,0)--(6,6), linewidth(1) + qqwuqq);
draw((-3,3)--(5,11), linewidth(1));
draw((-5,11)--(3,3), linewidth(1));
draw(circle((0,6), 2.482817665807104), linewidth(1));
draw((0,6)--(2.2602739726027394,4.972602739726027), linewidth(1));
/* dots and labels */
dot((0,0),linewidth(1pt) + dotstyle);
dot((3,3),dotstyle);
dot((-3,3),dotstyle);
dot((6,6),dotstyle);
dot((-6,6),dotstyle);
dot((5,11),dotstyle);
dot((-5,11),dotstyle);
dot((0,6),linewidth(1pt) + dotstyle);
label("$O$", (0.059294254264342997,6.119672124650978), NE * labelscalefactor);
dot((2.2602739726027394,4.972602739726027),linewidth(1pt) + dotstyle);
label("$T$", (2.326166015469254,5.094921876435061), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
// Block 2
/* Geogebra to Asymptote conversion by samrocksnature, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -8.325025958411356, xmax = 8, ymin = -0.6033105644019334, ymax = 12.237120576121757; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); pen qqwuqq = rgb(0,0.39215686274509803,0); pen cqcqcq = rgb(0.7529411764705882,0.7529411764705882,0.7529411764705882); draw((5,11)--(0,0)--(-6,6)--cycle, linewidth(1) + zzttqq); draw((6,6)--(-5,11)--(0,0)--cycle, linewidth(1) + qqwuqq); draw((0.2328977836854361,5.767102216314564)--(0.4657955673708722,6)--(0.2328977836854361,6.232897783685436)--(0,6)--cycle, linewidth(1) + qqwuqq); /* draw grid of horizontal/vertical lines */ pen gridstyle = linewidth(0.7) + cqcqcq; real gridx = 1, gridy = 1; /* grid intervals */ for(real i = ceil(xmin/gridx)*gridx; i <= floor(xmax/gridx)*gridx; i += gridx) draw((i,ymin)--(i,ymax), gridstyle); for(real i = ceil(ymin/gridy)*gridy; i <= floor(ymax/gridy)*gridy; i += gridy) draw((xmin,i)--(xmax,i), gridstyle); /* end grid */ Label laxis; laxis.p = fontsize(10); xaxis(xmin, xmax, Ticks(laxis, Step = 1, Size = 2, NoZero),EndArrow(6), above = true); yaxis(ymin, ymax, Ticks(laxis, Step = 1, Size = 2, NoZero),EndArrow(6), above = true); /* draws axes; NoZero hides '0' label */ /* draw figures */ draw((5,11)--(0,0), linewidth(1) + zzttqq); draw((0,0)--(-6,6), linewidth(1) + zzttqq); draw((-6,6)--(5,11), linewidth(1) + zzttqq); draw((6,6)--(-5,11), linewidth(1) + qqwuqq); draw((-5,11)--(0,0), linewidth(1) + qqwuqq); draw((0,0)--(6,6), linewidth(1) + qqwuqq); draw((-3,3)--(5,11), linewidth(1)); draw((-5,11)--(3,3), linewidth(1)); draw(circle((0,6), 2.482817665807104), linewidth(1)); draw((0,6)--(2.2602739726027394,4.972602739726027), linewidth(1)); /* dots and labels */ dot((0,0),linewidth(1pt) + dotstyle); dot((3,3),dotstyle); dot((-3,3),dotstyle); dot((6,6),dotstyle); dot((-6,6),dotstyle); dot((5,11),dotstyle); dot((-5,11),dotstyle); dot((0,6),linewidth(1pt) + dotstyle); label("$O$", (0.059294254264342997,6.119672124650978), NE * labelscalefactor); dot((2.2602739726027394,4.972602739726027),linewidth(1pt) + dotstyle); label("$T$", (2.326166015469254,5.094921876435061), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ | [] |
231 | Convex pentagon $ABCDE$ has side lengths $AB=5$, $BC=CD=DE=6$, and $EA=7$. Moreover, the pentagon has an inscribed circle (a circle tangent to each side of the pentagon). Find the area of $ABCDE$. | 2020 AIME II Problem 13 | Let $\omega$ be the inscribed circle, $I$ be its center, and $r$ be its radius. The area of $ABCDE$ is equal to its semiperimeter, $15,$ times $r$, so the problem is reduced to finding $r$. Let $a$ be the length of the tangent segment from $A$ to $\omega$, and analogously define $b$, $c$, $d$, and $e$. Then $a+b=5$, $b+c= c+d=d+e=6$, and $e+a=7$, with a total of $a+b+c+d+e=15$. Hence $a=3$, $b=d=2$, and $c=e=4$. It follows that $\angle B= \angle D$ and $\angle C= \angle E$. Let $Q$ be the point where $\omega$ is tangent to $\overline{CD}$. Then $\angle IAE = \angle IAB =\frac{1}{2}\angle A$. Now we claim that points $A, I, Q$ are collinear, which can be proved if $\angle{AIQ}=\angle{QIA}=180^{\circ}$. The sum of the internal angles in polygons $ABCQI$ and $AIQDE$ are equal, so $\angle IAE + \angle AIQ + \angle IQD + \angle D + \angle E = \angle IAB + \angle B + \angle C + \angle CQI + \angle QIA$, which implies that $\angle AIQ$ must be $180^\circ$. Therefore points $A$, $I$, and $Q$ are collinear.
Because $\overline{AQ} \perp \overline{CD}$, it follows that\[AC^2-AD^2=CQ^2-DQ^2=c^2-d^2=12.\]Another expression for $AC^2-AD^2$ can be found as follows. Note that $\tan \left(\frac{\angle B}{2}\right) = \frac{r}{2}$ and $\tan \left(\frac{\angle E}{2}\right) = \frac{r}{4}$, so
\[\cos (\angle B) =\frac{1-\tan^2 \left(\frac{\angle B}{2}\right)}{1+\tan^2 \left(\frac{\angle B}{2}\right)} = \frac{4-r^2}{4+r^2}\]and
\[\cos (\angle E) = \frac{1-\tan^2 \left(\frac{\angle E}{2}\right)}{1+\tan^2 \left(\frac{\angle E}{2}\right)}= \frac{16-r^2}{16+r^2}.\]Applying the Law of Cosines to $\triangle ABC$ and $\triangle AED$ gives
\[AC^2=AB^2+BC^2-2\cdot AB\cdot BC\cdot \cos (\angle B) = 5^2+6^2-2 \cdot 5 \cdot 6 \cdot \frac{4-r^2}{4+r^2}\]
and
\[AD^2=AE^2+DE^2-2 \cdot AE \cdot DE \cdot \cos(\angle E) = 7^2+6^2-2 \cdot 7 \cdot 6 \cdot \frac{16-r^2}{16+r^2}.\]
Hence
\[12=AC^2- AD^2= 5^2-2\cdot 5 \cdot 6\cdot \frac{4-r^2}{4+r^2} -7^2+2\cdot 7 \cdot 6 \cdot \frac{16-r^2}{16+r^2},\]
yielding
\[2\cdot 7 \cdot 6 \cdot \frac{16-r^2}{16+r^2}- 2\cdot 5 \cdot 6\cdot \frac{4-r^2}{4+r^2}= 36;\]
equivalently
\[7(16-r^2)(4+r^2)-5(4-r^2)(16+r^2) = 3(16+r^2)(4+r^2).\]
Substituting $x=r^2$ gives the quadratic equation $5x^2-84x+64=0$, with solutions $\frac{42 - 38}{5}=\frac45$, and $\frac{42 + 38}{5}= 16$. The solution $r^2=\frac45$ corresponds to a five-pointed star, which is not convex. Indeed, if $r<3$, then $\tan \left(\frac{\angle A}{2}\right)$, $\tan \left(\frac{\angle C}{2}\right)$, and $\tan \left(\frac{\angle E}{2}\right)$ are less than $1,$ implying that $\angle A$, $\angle C$, and $\angle E$ are acute, which cannot happen in a convex pentagon. Thus $r^2=16$ and $r=4$. The requested area is $15\cdot4 = \boxed{60}$. | // Block 1
defaultpen(fontsize(8pt));
unitsize(0.025cm);
pair[] vertices = {(0,0), (5,0), (8.6,4.8), (3.8,8.4), (-1.96, 6.72)};
string[] labels = {"$A$", "$B$", "$C$", "$D$", "$E$"};
pair[] dirs = {SW, SE,E, N, NW};
string[] smallLabels = {"$a$", "$b$", "$c$", "$d$", "$e$"};
pair I = (3,4);
real rad = 4;
pair Q = foot(I, vertices[2], vertices[3]);
pair[] interpoints = {};
for(int i =0; i<vertices.length; ++i){
interpoints.push(foot(I, vertices[i], vertices[(i+1)%vertices.length]));
}
for(int i = 0; i< vertices.length; ++i){
draw(vertices[i]--vertices[(i+1)%vertices.length]);
dot(labels[i],vertices[i],dirs[i]);
draw(I--vertices[i]);
}
draw(Circle(I, rad));
dot("$I$", I, dir(200));
draw(I--Q);
dot("$Q$", Q, NE);
for(int i = 0; i < vertices.length; ++i){
label(smallLabels[i], vertices[i] --interpoints[i]);
//dot(interpoints[i], blue);
label(smallLabels[i], interpoints[(i-1)%vertices.length] -- vertices[i]);
}
// Block 2
defaultpen(fontsize(8pt)); unitsize(0.025cm); pair[] vertices = {(0,0), (5,0), (8.6,4.8), (3.8,8.4), (-1.96, 6.72)}; string[] labels = {"$A$", "$B$", "$C$", "$D$", "$E$"}; pair[] dirs = {SW, SE,E, N, NW}; string[] smallLabels = {"$a$", "$b$", "$c$", "$d$", "$e$"}; pair I = (3,4); real rad = 4; pair Q = foot(I, vertices[2], vertices[3]); pair[] interpoints = {}; for(int i =0; i<vertices.length; ++i){ interpoints.push(foot(I, vertices[i], vertices[(i+1)%vertices.length])); } for(int i = 0; i< vertices.length; ++i){ draw(vertices[i]--vertices[(i+1)%vertices.length]); dot(labels[i],vertices[i],dirs[i]); draw(I--vertices[i]); } draw(Circle(I, rad)); dot("$I$", I, dir(200)); draw(I--Q); dot("$Q$", Q, NE); for(int i = 0; i < vertices.length; ++i){ label(smallLabels[i], vertices[i] --interpoints[i]); //dot(interpoints[i], blue); label(smallLabels[i], interpoints[(i-1)%vertices.length] -- vertices[i]); } | [] |
232 | Let $\triangle ABC$ be an acute scalene triangle with circumcircle $\omega$. The tangents to $\omega$ at $B$ and $C$ intersect at $T$. Let $X$ and $Y$ be the projections of $T$ onto lines $AB$ and $AC$, respectively. Suppose $BT = CT = 16$, $BC = 22$, and $TX^2 + TY^2 + XY^2 = 1143$. Find $XY^2$. | 2020 AIME II Problem 15 | Let $M$ denote the midpoint of $\overline{BC}$. The critical claim is that $M$ is the orthocenter of $\triangle AXY$, which has the circle with diameter $\overline{AT}$ as its circumcircle. To see this, note that because $\angle BXT = \angle BMT = 90^\circ$, the quadrilateral $MBXT$ is cyclic, it follows that
\[\angle MXA = \angle MXB = \angle MTB = 90^\circ - \angle TBM = 90^\circ - \angle A,\] implying that $\overline{MX} \perp \overline{AC}$. Similarly, $\overline{MY} \perp \overline{AB}$. In particular, $MXTY$ is a parallelogram.
Hence, by the Parallelogram Law,
\[TM^2 + XY^2 = 2(TX^2 + TY^2) = 2(1143-XY^2).\] But $TM^2 = TB^2 - BM^2 = 16^2 - 11^2 = 135$. Therefore \[XY^2 = \frac13(2 \cdot 1143-135) = \boxed{717}.\] | // Block 1
defaultpen(fontsize(8pt));
unitsize(0.8cm);
pair A = (0,0);
pair B = (-1.26,-4.43);
pair C = (-1.26+3.89, -4.43);
pair M = (B+C)/2;
pair O = circumcenter(A,B,C);
pair T = (0.68, -6.49);
pair X = foot(T,A,B);
pair Y = foot(T,A,C);
path omega = circumcircle(A,B,C);
real rad = circumradius(A,B,C);
filldraw(A--B--C--cycle, 0.2*royalblue+white);
label("$\omega$", O + rad*dir(45), SW);
//filldraw(T--Y--M--X--cycle, rgb(150, 247, 254));
filldraw(T--Y--M--X--cycle, 0.2*heavygreen+white);
draw(M--T);
draw(X--Y);
draw(B--T--C);
draw(A--X--Y--cycle);
draw(omega);
dot("$X$", X, W);
dot("$Y$", Y, E);
dot("$O$", O, W);
dot("$T$", T, S);
dot("$A$", A, N);
dot("$B$", B, W);
dot("$C$", C, E);
dot("$M$", M, N);
// Block 2
defaultpen(fontsize(8pt)); unitsize(0.8cm); pair A = (0,0); pair B = (-1.26,-4.43); pair C = (-1.26+3.89, -4.43); pair M = (B+C)/2; pair O = circumcenter(A,B,C); pair T = (0.68, -6.49); pair X = foot(T,A,B); pair Y = foot(T,A,C); path omega = circumcircle(A,B,C); real rad = circumradius(A,B,C); filldraw(A--B--C--cycle, 0.2*royalblue+white); label("$\omega$", O + rad*dir(45), SW); //filldraw(T--Y--M--X--cycle, rgb(150, 247, 254)); filldraw(T--Y--M--X--cycle, 0.2*heavygreen+white); draw(M--T); draw(X--Y); draw(B--T--C); draw(A--X--Y--cycle); draw(omega); dot("$X$", X, W); dot("$Y$", Y, E); dot("$O$", O, W); dot("$T$", T, S); dot("$A$", A, N); dot("$B$", B, W); dot("$C$", C, E); dot("$M$", M, N); | [] |
233 | Find the number of integers $c$ such that the equation \[\left||20|x|-x^2|-c\right|=21\]has $12$ distinct real solutions. | 2021 AIME I Problem 8 | We take cases for the outermost absolute value, then rearrange: \[\left|20|x|-x^2\right|=c\pm21.\]
Let $f(x)=\left|20|x|-x^2\right|.$ We rewrite $f(x)$ as a piecewise function without using absolute values:
\[f(x) = \begin{cases} \left|-20x-x^2\right| & \mathrm{if} \ x \le 0 \begin{cases} 20x+x^2 & \mathrm{if} \ x\le-20 \\ -20x-x^2 & \mathrm{if} \ -20<x\leq0 \end{cases} \\ \left|20x-x^2\right| & \mathrm{if} \ x > 0 \begin{cases} 20x-x^2 & \mathrm{if} \ 0<x\leq20 \\ -20x+x^2 & \mathrm{if} \ x>20 \end{cases} \end{cases}.\]
We graph $y=f(x)$ with all extremum points labeled, as shown below. The fact that $f(x)$ is an even function ($f(x)=f(-x)$ holds for all real numbers $x,$ so the graph of $y=f(x)$ is symmetric about the $y$-axis) should facilitate the process of graphing.
Since $f(x)=c\pm21$ has $12$ distinct real solutions, it is clear that each case has $6$ distinct real solutions geometrically. We shift the graphs of $y=\pm21$ up $c$ units, where $c\geq0:$
For $f(x)=c+21$ to have $6$ distinct real solutions, we need $0\leq c<79.$
For $f(x)=c-21$ to have $6$ distinct real solutions, we need $21<c<121.$
Taking the intersection of these two cases gives $21<c<79,$ from which there are $79-21-1=\boxed{057}$ such integers $c.$
~MRENTHUSIASM | // Block 1
/* Made by MRENTHUSIASM */
size(1200,300);
real xMin = -65;
real xMax = 65;
real yMin = -50;
real yMax = 125;
draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5));
draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5));
label("$x$",(xMax,0),(2,0));
label("$y$",(0,yMax),(0,2));
real f(real x) { return abs(20*abs(x)-x^2); }
real g(real x) { return 21; }
real h(real x) { return -21; }
draw(graph(f,-25,25),red,"$y=\left|20|x|-x^2\right|$");
draw(graph(g,-65,65),blue,"$y=\pm21$");
draw(graph(h,-65,65),blue);
pair A[];
A[0] = (-20,0);
A[1] = (-10,100);
A[2] = (0,0);
A[3] = (10,100);
A[4] = (20,0);
for(int i = 0; i <= 4; ++i) {
dot(A[i],red+linewidth(4.5));
}
label("$(-20,0)$",A[0],(-1.5,-1.5),red,UnFill);
label("$(-10,100)$",A[1],(-1.5,1.5),red);
label("$(0,0)$",A[2],(0,-1.5),red,UnFill);
label("$(10,100)$",A[3],(1.5,1.5),red);
label("$(20,0)$",A[4],(1.5,-1.5),red,UnFill);
add(legend(),point(E),40E,UnFill);
// Block 2
/* Made by MRENTHUSIASM */ size(1200,300); real xMin = -65; real xMax = 65; real yMin = -50; real yMax = 125; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("$x$",(xMax,0),(2,0)); label("$y$",(0,yMax),(0,2)); real f(real x) { return abs(20*abs(x)-x^2); } real g(real x) { return 21; } real h(real x) { return -21; } draw(graph(f,-25,25),red,"$y=\left|20|x|-x^2\right|$"); draw(graph(g,-65,65),blue,"$y=\pm21$"); draw(graph(h,-65,65),blue); pair A[]; A[0] = (-20,0); A[1] = (-10,100); A[2] = (0,0); A[3] = (10,100); A[4] = (20,0); for(int i = 0; i <= 4; ++i) { dot(A[i],red+linewidth(4.5)); } label("$(-20,0)$",A[0],(-1.5,-1.5),red,UnFill); label("$(-10,100)$",A[1],(-1.5,1.5),red); label("$(0,0)$",A[2],(0,-1.5),red,UnFill); label("$(10,100)$",A[3],(1.5,1.5),red); label("$(20,0)$",A[4],(1.5,-1.5),red,UnFill); add(legend(),point(E),40E,UnFill); | [] |
234 | Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$ | 2021 AIME I Problem 9 | Let $\overline{AE}, \overline{AF},$ and $\overline{AG}$ be the perpendiculars from $A$ to $\overleftrightarrow{BC}, \overleftrightarrow{CD},$ and $\overleftrightarrow{BD},$ respectively. Next, let $H$ be the intersection of $\overline{AF}$ and $\overline{BD}.$
We set $AB=x$ and $AH=y,$ as shown below.
From here, we obtain $HF=18-y$ by segment subtraction, and $BG=\sqrt{x^2-10^2}$ and $HG=\sqrt{y^2-10^2}$ by the Pythagorean Theorem.
Since $\angle ABG$ and $\angle HAG$ are both complementary to $\angle AHB,$ we have $\angle ABG = \angle HAG,$ from which $\triangle ABG \sim \triangle HAG$ by AA. It follows that $\frac{BG}{AG}=\frac{AG}{HG},$ so $BG\cdot HG=AG^2,$ or \[\sqrt{x^2-10^2}\cdot\sqrt{y^2-10^2}=10^2. \hspace{10mm} (1)\]
Since $\angle AHB = \angle FHD$ by vertical angles, we have $\triangle AHB \sim \triangle FHD$ by AA, with the ratio of similitude $\frac{AH}{FH}=\frac{BA}{DF}.$ It follows that $DF=BA\cdot\frac{FH}{AH}=x\cdot\frac{18-y}{y}.$
Since $\angle EBA = \angle ECD = \angle FDA$ by angle chasing, we have $\triangle EBA \sim \triangle FDA$ by AA, with the ratio of similitude $\frac{EA}{FA}=\frac{BA}{DA}.$ It follows that $DA=BA\cdot\frac{FA}{EA}=x\cdot\frac{18}{15}=\frac{6}{5}x.$
By the Pythagorean Theorem on right $\triangle ADF,$ we have $DF^2+AF^2=AD^2,$ or \[\left(x\cdot\frac{18-y}{y}\right)^2+18^2=\left(\frac{6}{5}x\right)^2. \hspace{7mm} (2)\]
Solving this system of equations ($(1)$ and $(2)$), we get $x=\frac{45\sqrt2}{4}$ and $y=\frac{90}{7},$ so $AB=x=\frac{45\sqrt2}{4}$ and $CD=AB+2DF=x+2\left(x\cdot\frac{18-y}{y}\right)=\frac{81\sqrt2}{4}.$ Finally, the area of $ABCD$ is \[K=\frac{AB+CD}{2}\cdot AF=\frac{567\sqrt2}{2},\] from which $\sqrt2 \cdot K=\boxed{567}.$
~MRENTHUSIASM
Remark
Instead of solving the system of equations $(1)$ and $(2),$ which can be time consuming, by noting that $\triangle ACF \sim \triangle ABG$ by AA, we could find out $\frac{AB}{AG} = \frac{AC}{AF}$, which gives $AC = \frac{9}{5}x$. We also know that $EB = \sqrt{x^2 - 15^2}$ by Pythagorean Theorem on $\triangle ABE$. From $BC = AD = \frac{6}{5}x,$ we apply the Pythagorean Theorem to $\triangle ACE$ and obtain
\[AC^2 = (EB+BC)^2 + AE^2.\]
Substituting, we get
\[\frac{81}{25}x^2 = \left(\sqrt{x^2 -225}+\frac{6}{5}x\right)^2+225 \iff x = 3\sqrt{x^2 - 15^2},\]
from which $x = \frac{45\sqrt{2}}{4}.$
~Chupdogs | // Block 1
/* Made by MRENTHUSIASM */
size(250);
pair A, B, C, D, E, F, G, H;
A = (-45sqrt(2)/8,18);
B = (45sqrt(2)/8,18);
C = (81sqrt(2)/8,0);
D = (-81sqrt(2)/8,0);
E = foot(A,C,B);
F = foot(A,C,D);
G = foot(A,B,D);
H = intersectionpoint(A--F,B--D);
markscalefactor=0.1;
draw(rightanglemark(A,E,B),red);
draw(rightanglemark(A,F,C),red);
draw(rightanglemark(A,G,D),red);
dot("$A$",A,1.5*NW,linewidth(4));
dot("$B$",B,1.5*NE,linewidth(4));
dot("$C$",C,1.5*SE,linewidth(4));
dot("$D$",D,1.5*SW,linewidth(4));
dot("$E$",E,1.5*dir(E),linewidth(4));
dot("$F$",F,1.5*S,linewidth(4));
dot("$G$",G,SE,linewidth(4));
dot("$H$",H,SE,linewidth(4));
draw(A--B--C--D--cycle^^B--D^^B--E);
draw(A--E^^A--F^^A--G,dashed);
label("$10$",midpoint(A--G),1.5*(1,0));
label("$15$",midpoint(A--E),1.5*N);
Label L = Label("$18$", align=(0,0), position=MidPoint, filltype=Fill(0,3,white));
draw(C+(5,0)--(81sqrt(2)/8,18)+(5,0), L=L, arrow=Arrows(),bar=Bars(15));
label("$x$",midpoint(A--B),N);
label("$y$",midpoint(A--H),W);
// Block 2
/* Made by MRENTHUSIASM */ size(250); pair A, B, C, D, E, F, G, H; A = (-45sqrt(2)/8,18); B = (45sqrt(2)/8,18); C = (81sqrt(2)/8,0); D = (-81sqrt(2)/8,0); E = foot(A,C,B); F = foot(A,C,D); G = foot(A,B,D); H = intersectionpoint(A--F,B--D); markscalefactor=0.1; draw(rightanglemark(A,E,B),red); draw(rightanglemark(A,F,C),red); draw(rightanglemark(A,G,D),red); dot("$A$",A,1.5*NW,linewidth(4)); dot("$B$",B,1.5*NE,linewidth(4)); dot("$C$",C,1.5*SE,linewidth(4)); dot("$D$",D,1.5*SW,linewidth(4)); dot("$E$",E,1.5*dir(E),linewidth(4)); dot("$F$",F,1.5*S,linewidth(4)); dot("$G$",G,SE,linewidth(4)); dot("$H$",H,SE,linewidth(4)); draw(A--B--C--D--cycle^^B--D^^B--E); draw(A--E^^A--F^^A--G,dashed); label("$10$",midpoint(A--G),1.5*(1,0)); label("$15$",midpoint(A--E),1.5*N); Label L = Label("$18$", align=(0,0), position=MidPoint, filltype=Fill(0,3,white)); draw(C+(5,0)--(81sqrt(2)/8,18)+(5,0), L=L, arrow=Arrows(),bar=Bars(15)); label("$x$",midpoint(A--B),N); label("$y$",midpoint(A--H),W); | [] |
234 | Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$ | 2021 AIME I Problem 9 | Let the points formed by dropping altitudes from $A$ to the lines $BC$, $CD$, and $BD$ be $E$, $F$, and $G$, respectively.
We have
\[\triangle ABE \sim \triangle ADF \implies \frac{AD}{18} = \frac{AB}{15} \implies AD = \frac{6}{5}AB\]
and
\[BD\cdot10 = 2[ABD] = AB\cdot18 \implies BD = \frac{9}{5}AB.\]
For convenience, let $AB = 5x$. By Heron's formula on $\triangle ABD$, we have sides $5x,6x,9x$ and semiperimeter $10x$, so
\[\sqrt{10x\cdot5x\cdot4x\cdot1x} = [ABD] = \frac{AB\cdot18}{2} = 45x \implies 10\sqrt{2}x^2 = 45x \implies x= \frac{45}{10\sqrt{2}},\]
so $AB = 5x = \frac{45}{2\sqrt{2}}$.
Then,
\[BE = \sqrt{AB^2 - CA^2} = \sqrt{\left(\frac{45}{2\sqrt{2}}\right)^2 - 15^2} = \sqrt{\frac{225}{8}} = \frac{15}{2\sqrt{2}}\]
and
\[\triangle ABE \sim \triangle ADF \implies DF = \frac{6}{5}BE = \frac{6}{5}\cdot\frac{15}{2\sqrt{2}} = \frac{18}{2\sqrt{2}}.\]
Finally, recalling that $ABCD$ is isosceles,
\[K = [ABCD] = \frac{18}{2}(AB + (AB + 2DF)) = 18(AB + DF) = 18\left(\frac{45}{2\sqrt{2}} + \frac{18}{2\sqrt{2}}\right) = \frac{567}{\sqrt{2}},\]
so $\sqrt{2}\cdot K = \boxed{567}$.
~emerald_block | // Block 1
size(250);
pair A, B, C, D, E, F, G, H;
A = (-45sqrt(2)/8,18);
B = (45sqrt(2)/8,18);
C = (81sqrt(2)/8,0);
D = (-81sqrt(2)/8,0);
E = foot(A,C,B);
F = foot(A,C,D);
G = foot(A,B,D);
H = intersectionpoint(A--F,B--D);
markscalefactor=0.1;
draw(rightanglemark(A,E,B),red);
draw(rightanglemark(A,F,C),red);
draw(rightanglemark(A,G,D),red);
filldraw(A--D--F--cycle,yellow,black+linewidth(1.5));
filldraw(A--B--E--cycle,yellow,black+linewidth(1.5));
dot("$A$",A,1.5*NW,linewidth(4));
dot("$B$",B,1.5*NE,linewidth(4));
dot("$C$",C,1.5*SE,linewidth(4));
dot("$D$",D,1.5*SW,linewidth(4));
dot(E,linewidth(4));
dot(F,linewidth(4));
dot(G,linewidth(4));
label("$E$",E,NE);
label("$F$",F, S);
label("$G$",G,SE);
draw(A--B--C--D--cycle^^B--D^^B--E);
draw(A--E^^A--F^^A--G,dashed);
label("$10$",midpoint(A--G),1.5*(1,0));
label("$15$",midpoint(A--E),1.5*N);
label("$5x$",midpoint(A--B),S);
label("$6x$",midpoint(A--D),1.5*(-1,0));
Label L = Label("$18$", align=(0,0), position=MidPoint, filltype=Fill(0,3,white));
draw(C+(5,0)--(81sqrt(2)/8,18)+(5,0), L=L, arrow=Arrows(),bar=Bars(15));
// Block 2
size(250); pair A, B, C, D, E, F, G, H; A = (-45sqrt(2)/8,18); B = (45sqrt(2)/8,18); C = (81sqrt(2)/8,0); D = (-81sqrt(2)/8,0); E = foot(A,C,B); F = foot(A,C,D); G = foot(A,B,D); H = intersectionpoint(A--F,B--D); markscalefactor=0.1; draw(rightanglemark(A,E,B),red); draw(rightanglemark(A,F,C),red); draw(rightanglemark(A,G,D),red); filldraw(A--D--F--cycle,yellow,black+linewidth(1.5)); filldraw(A--B--E--cycle,yellow,black+linewidth(1.5)); dot("$A$",A,1.5*NW,linewidth(4)); dot("$B$",B,1.5*NE,linewidth(4)); dot("$C$",C,1.5*SE,linewidth(4)); dot("$D$",D,1.5*SW,linewidth(4)); dot(E,linewidth(4)); dot(F,linewidth(4)); dot(G,linewidth(4)); label("$E$",E,NE); label("$F$",F, S); label("$G$",G,SE); draw(A--B--C--D--cycle^^B--D^^B--E); draw(A--E^^A--F^^A--G,dashed); label("$10$",midpoint(A--G),1.5*(1,0)); label("$15$",midpoint(A--E),1.5*N); label("$5x$",midpoint(A--B),S); label("$6x$",midpoint(A--D),1.5*(-1,0)); Label L = Label("$18$", align=(0,0), position=MidPoint, filltype=Fill(0,3,white)); draw(C+(5,0)--(81sqrt(2)/8,18)+(5,0), L=L, arrow=Arrows(),bar=Bars(15)); | [] |
235 | Let $ABCD$ be a cyclic quadrilateral with $AB=4,BC=5,CD=6,$ and $DA=7.$ Let $A_1$ and $C_1$ be the feet of the perpendiculars from $A$ and $C,$ respectively, to line $BD,$ and let $B_1$ and $D_1$ be the feet of the perpendiculars from $B$ and $D,$ respectively, to line $AC.$ The perimeter of $A_1B_1C_1D_1$ is $\frac mn,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | 2021 AIME I Problem 11 | This solution refers to the Diagram section.
By the Converse of the Inscribed Angle Theorem, if distinct points $X$ and $Y$ lie on the same side of $\overline{PQ}$ (but not on $\overline{PQ}$ itself) for which $\angle PXQ=\angle PYQ,$ then $P,Q,X,$ and $Y$ are cyclic. From the Converse of the Inscribed Angle Theorem, quadrilaterals $ABA_1B_1,BCC_1B_1,CDC_1D_1,$ and $DAA_1D_1$ are all cyclic.
Suppose $\overline{AC}$ and $\overline{BD}$ intersect at $E,$ and let $\angle AEB=\theta.$ It follows that $\angle CED=\theta$ and $\angle BEC=\angle DEA=180^\circ-\theta.$
We obtain the following diagram:
In every cyclic quadrilateral, each pair of opposite angles is supplementary. So, we have $\angle EA_1B_1=\angle EAB$ (both supplementary to $\angle B_1A_1B$) and $\angle EB_1A_1=\angle EBA$ (both supplementary to $\angle A_1B_1A$), from which $\triangle A_1B_1E \sim \triangle ABE$ by AA, with the ratio of similitude \[\frac{A_1B_1}{AB}=\underbrace{\frac{A_1E}{AE}}_{\substack{\text{right} \\ \triangle A_1AE}}=\underbrace{\frac{B_1E}{BE}}_{\substack{\text{right} \\ \triangle B_1BE}}=\cos\theta. \hspace{15mm}(1)\]
Similarly, we have $\angle EC_1D_1=\angle ECD$ (both supplementary to $\angle D_1C_1D$) and $\angle ED_1C_1=\angle EDC$ (both supplementary to $\angle C_1D_1C$), from which $\triangle C_1D_1E \sim \triangle CDE$ by AA, with the ratio of similitude \[\frac{C_1D_1}{CD}=\underbrace{\frac{C_1E}{CE}}_{\substack{\text{right} \\ \triangle C_1CE}}=\underbrace{\frac{D_1E}{DE}}_{\substack{\text{right} \\ \triangle D_1DE}}=\cos\theta. \hspace{14.75mm}(2)\]
We apply the Transitive Property to $(1)$ and $(2):$
We get $\frac{B_1E}{BE}=\frac{C_1E}{CE}=\cos\theta,$ so $\triangle B_1C_1E \sim \triangle BCE$ by SAS, with the ratio of similitude \[\frac{B_1C_1}{BC}=\frac{B_1E}{BE}=\frac{C_1E}{CE}=\cos\theta. \hspace{14.75mm}(3)\]
We get $\frac{D_1E}{DE}=\frac{A_1E}{AE}=\cos\theta,$ so $\triangle D_1A_1E \sim \triangle DAE$ by SAS, with the ratio of similitude \[\frac{D_1A_1}{DA}=\frac{D_1E}{DE}=\frac{A_1E}{AE}=\cos\theta. \hspace{14mm}(4)\]
From $(1),(2),(3),$ and $(4),$ the perimeter of $A_1B_1C_1D_1$ is
\begin{align*} A_1B_1+B_1C_1+C_1D_1+D_1A_1&=AB\cos\theta+BC\cos\theta+CD\cos\theta+DA\cos\theta \\ &=(AB+BC+CD+DA)\cos\theta \\ &=22\cos\theta. &&\hspace{5mm}(\bigstar) \end{align*}
Two solutions follow from here:
Solution 1.1 (Law of Cosines)
Note that $\cos(180^\circ-\theta)=-\cos\theta$ holds for all $\theta.$ We apply the Law of Cosines to $\triangle ABE, \triangle BCE, \triangle CDE,$ and $\triangle DAE,$ respectively:
\begin{alignat*}{12} &&&AE^2+BE^2-2\cdot AE\cdot BE\cdot\cos\angle AEB&&=AB^2&&\quad\implies\quad AE^2+BE^2-2\cdot AE\cdot BE\cdot\cos\theta&&=4^2, \hspace{15mm} &(1\star) \\ &&&BE^2+CE^2-2\cdot BE\cdot CE\cdot\cos\angle BEC&&=BC^2&&\quad\implies\quad BE^2+CE^2+2\cdot BE\cdot CE\cdot\cos\theta&&=5^2, \hspace{15mm} &(2\star) \\ &&&CE^2+DE^2-2\cdot CE\cdot DE\cdot\cos\angle CED&&=CD^2&&\quad\implies\quad CE^2+DE^2-2\cdot CE\cdot DE\cdot\cos\theta&&=6^2, \hspace{15mm} &(3\star) \\ &&&DE^2+AE^2-2\cdot DE\cdot AE\cdot\cos\angle DEA&&=DA^2&&\quad\implies\quad DE^2+AE^2+2\cdot DE\cdot AE\cdot\cos\theta&&=7^2. \hspace{15mm} &(4\star) \\ \end{alignat*}
We subtract $(1\star)+(3\star)$ from $(2\star)+(4\star):$
\begin{align*} 2\cdot AE\cdot BE\cdot\cos\theta+2\cdot BE\cdot CE\cdot\cos\theta+2\cdot CE\cdot DE\cdot\cos\theta+2\cdot DE\cdot AE\cdot\cos\theta&=22 \\ 2\cdot\cos\theta\cdot(\phantom{ }\underbrace{AE\cdot BE+BE\cdot CE+CE\cdot DE+DE\cdot AE}_{\text{Use the result from }\textbf{Remark}\text{.}}\phantom{ })&=22 \\ 2\cdot\cos\theta\cdot59&=22 \\ \cos\theta&=\frac{11}{59}. \end{align*}
Finally, substituting this result into $(\bigstar)$ gives $22\cos\theta=\frac{242}{59},$ from which the answer is $242+59=\boxed{301}.$
~MRENTHUSIASM (credit given to Math Jams's 2021 AIME I Discussion)
Solution 1.2 (Area Formulas)
Let the brackets denote areas.
We find $[ABCD]$ in two different ways:
Note that $\sin(180^\circ-\theta)=\sin\theta$ holds for all $\theta.$ By area addition, we get
\begin{align*} [ABCD]&=[ABE]+[BCE]+[CDE]+[DAE] \\ &=\frac12\cdot AE\cdot BE\cdot\sin\angle AEB+\frac12\cdot BE\cdot CE\cdot\sin\angle BEC+\frac12\cdot CE\cdot DE\cdot\sin\angle CED+\frac12\cdot DE\cdot AE\cdot\sin\angle DEA \\ &=\frac12\cdot AE\cdot BE\cdot\sin\theta+\frac12\cdot BE\cdot CE\cdot\sin\theta+\frac12\cdot CE\cdot DE\cdot\sin\theta+\frac12\cdot DE\cdot AE\cdot\sin\theta \\ &=\frac12\cdot\sin\theta\cdot(\phantom{ }\underbrace{AE\cdot BE+BE\cdot CE+CE\cdot DE+DE\cdot AE}_{\text{Use the result from }\textbf{Remark}\text{.}}\phantom{ }) \\ &=\frac12\cdot\sin\theta\cdot59. \end{align*}
By Brahmagupta's Formula, we get \[[ABCD]=\sqrt{(s-AB)(s-BC)(s-CD)(s-DA)}=2\sqrt{210},\] where $s=\frac{AB+BC+CD+DA}{2}=11$ is the semiperimeter of $ABCD.$
Equating the expressions for $[ABCD],$ we have
\[\frac12\cdot\sin\theta\cdot59=2\sqrt{210},\] so $\sin\theta=\frac{4\sqrt{210}}{59}.$ Since $0^\circ<\theta<90^\circ,$ we have $\cos\theta>0.$ It follows that \[\cos\theta=\sqrt{1-\sin^2\theta}=\frac{11}{59}.\]
Finally, substituting this result into $(\bigstar)$ gives $22\cos\theta=\frac{242}{59},$ from which the answer is $242+59=\boxed{301}.$
~MRENTHUSIASM (credit given to Leonard my dude)
Remark (Ptolemy's Theorem)
In $ABCD,$ we have
\begin{align*} AE\cdot BE+BE\cdot CE+CE\cdot DE+DE\cdot AE &= (AE+CE)(BE+DE) &&\hspace{10mm}\text{Factor by Grouping} \\ &=AC\cdot BD &&\hspace{10mm}\text{Segment Addition} \\ &=AB\cdot CD+BC\cdot DA &&\hspace{10mm}\text{Ptolemy's Theorem} \\ &=59. &&\hspace{10mm}\text{Substitution} \end{align*}
~MRENTHUSIASM | // Block 1
/* Made by MRENTHUSIASM */
size(300);
pair A, B, C, D, A1, B1, C1, D1, P, M1, M2;
A = origin;
C = (sqrt(53041)/31,0);
B = intersectionpoints(Circle(A,4),Circle(C,5))[0];
D = intersectionpoints(Circle(A,7),Circle(C,6))[1];
A1 = foot(A,B,D);
C1 = foot(C,B,D);
B1 = foot(B,A,C);
D1 = foot(D,A,C);
P = intersectionpoint(A--C,B--D);
M1 = midpoint(A--B);
M2 = midpoint(C--D);
markscalefactor=0.025;
draw(rightanglemark(A,A1,B),red);
draw(rightanglemark(B,B1,A),red);
draw(rightanglemark(C,C1,D),red);
draw(rightanglemark(D,D1,C),red);
draw(Arc(M1,A,B)^^Arc(M2,C,D),blue);
draw(A1--B1--C1--D1--cycle,green);
dot("$A$",A,1.5*W,linewidth(4));
dot("$B$",B,1.5*dir(180-aCos(11/59)),linewidth(4));
dot("$C$",C,1.5*E,linewidth(4));
dot("$D$",D,1.5*dir(-aCos(11/59)),linewidth(4));
dot("$A_1$",A1,1.5*dir(A1-A),linewidth(4));
dot("$B_1$",B1,1.5*S,linewidth(4));
dot("$C_1$",C1,1.5*dir(C1-C),linewidth(4));
dot("$D_1$",D1,1.5*N,linewidth(4));
dot("$E$",P,dir((180-aCos(11/59))/2),linewidth(4));
label("$\theta$",P,dir(180-aCos(11/59)/2),red);
draw(A--B--C--D--cycle^^A--C^^B--D^^circumcircle(A,B,C));
draw(A--A1^^B--B1^^C--C1^^D--D1,dashed);
// Block 2
/* Made by MRENTHUSIASM */ size(300); pair A, B, C, D, A1, B1, C1, D1, P, M1, M2; A = origin; C = (sqrt(53041)/31,0); B = intersectionpoints(Circle(A,4),Circle(C,5))[0]; D = intersectionpoints(Circle(A,7),Circle(C,6))[1]; A1 = foot(A,B,D); C1 = foot(C,B,D); B1 = foot(B,A,C); D1 = foot(D,A,C); P = intersectionpoint(A--C,B--D); M1 = midpoint(A--B); M2 = midpoint(C--D); markscalefactor=0.025; draw(rightanglemark(A,A1,B),red); draw(rightanglemark(B,B1,A),red); draw(rightanglemark(C,C1,D),red); draw(rightanglemark(D,D1,C),red); draw(Arc(M1,A,B)^^Arc(M2,C,D),blue); draw(A1--B1--C1--D1--cycle,green); dot("$A$",A,1.5*W,linewidth(4)); dot("$B$",B,1.5*dir(180-aCos(11/59)),linewidth(4)); dot("$C$",C,1.5*E,linewidth(4)); dot("$D$",D,1.5*dir(-aCos(11/59)),linewidth(4)); dot("$A_1$",A1,1.5*dir(A1-A),linewidth(4)); dot("$B_1$",B1,1.5*S,linewidth(4)); dot("$C_1$",C1,1.5*dir(C1-C),linewidth(4)); dot("$D_1$",D1,1.5*N,linewidth(4)); dot("$E$",P,dir((180-aCos(11/59))/2),linewidth(4)); label("$\theta$",P,dir(180-aCos(11/59)/2),red); draw(A--B--C--D--cycle^^A--C^^B--D^^circumcircle(A,B,C)); draw(A--A1^^B--B1^^C--C1^^D--D1,dashed); | [] |
236 | Circles $\omega_1$ and $\omega_2$ with radii $961$ and $625$, respectively, intersect at distinct points $A$ and $B$. A third circle $\omega$ is externally tangent to both $\omega_1$ and $\omega_2$. Suppose line $AB$ intersects $\omega$ at two points $P$ and $Q$ such that the measure of minor arc $\widehat{PQ}$ is $120^{\circ}$. Find the distance between the centers of $\omega_1$ and $\omega_2$. | 2021 AIME I Problem 13 | Let $O_i$ and $r_i$ be the center and radius of $\omega_i$, and let $O$ and $r$ be the center and radius of $\omega$.
Since $\overline{AB}$ extends to an arc with arc $120^\circ$, the distance from $O$ to $\overline{AB}$ is $r/2$. Let $X=\overline{AB}\cap \overline{O_1O_2}$. Consider $\triangle OO_1O_2$. The line $\overline{AB}$ is perpendicular to $\overline{O_1O_2}$ and passes through $X$. Let $H$ be the foot from $O$ to $\overline{O_1O_2}$; so $HX=r/2$. We have by tangency $OO_1=r+r_1$ and $OO_2=r+r_2$. Let $O_1O_2=d$.
Since $X$ is on the radical axis of $\omega_1$ and $\omega_2$, it has equal power with respect to both circles, so
\[O_1X^2 - r_1^2 = O_2X^2-r_2^2 \implies O_1X-O_2X = \frac{r_1^2-r_2^2}{d}\]since $O_1X+O_2X=d$. Now we can solve for $O_1X$ and $O_2X$, and in particular,
\begin{align*} O_1H &= O_1X - HX = \frac{d+\frac{r_1^2-r_2^2}{d}}{2} - \frac{r}{2} \\ O_2H &= O_2X + HX = \frac{d-\frac{r_1^2-r_2^2}{d}}{2} + \frac{r}{2}. \end{align*}
We want to solve for $d$. By the Pythagorean Theorem (twice):
\begin{align*} &\qquad -OH^2 = O_2H^2 - (r+r_2)^2 = O_1H^2 - (r+r_1)^2 \\ &\implies \left(d+r-\tfrac{r_1^2-r_2^2}{d}\right)^2 - 4(r+r_2)^2 = \left(d-r+\tfrac{r_1^2-r_2^2}{d}\right)^2 - 4(r+r_1)^2 \\ &\implies 2dr - 2(r_1^2-r_2^2)-8rr_2-4r_2^2 = -2dr+2(r_1^2-r_2^2)-8rr_1-4r_1^2 \\ &\implies 4dr = 8rr_2-8rr_1 \\ &\implies d=2r_2-2r_1 \end{align*}
Therefore, $d=2(r_2-r_1) = 2(961-625)=\boxed{672}$. | // Block 1
unitsize(3cm);
pointpen=black; pointfontpen=fontsize(9);
pair A=dir(110), B=dir(230), C=dir(310);
DPA(A--B--C--A);
pair H = foot(A, B, C);
draw(A--H);
pair X = 0.3*B + 0.7*C;
pair Y = A+X-H;
draw(X--1.3*Y-0.3*X);
draw(A--Y, dotted);
pair R1 = 1.3*X-0.3*Y;
pair R2 = 0.7*X+0.3*Y;
draw(R1--X);
D("O",A,dir(A));
D("O_1",B,dir(B));
D("O_2",C,dir(C));
D("H",H,dir(270));
D("X",X,dir(225));
D("A",R1,dir(180));
D("B",R2,dir(180));
draw(rightanglemark(Y,X,C,3));
// Block 2
unitsize(3cm); pointpen=black; pointfontpen=fontsize(9); pair A=dir(110), B=dir(230), C=dir(310); DPA(A--B--C--A); pair H = foot(A, B, C); draw(A--H); pair X = 0.3*B + 0.7*C; pair Y = A+X-H; draw(X--1.3*Y-0.3*X); draw(A--Y, dotted); pair R1 = 1.3*X-0.3*Y; pair R2 = 0.7*X+0.3*Y; draw(R1--X); D("O",A,dir(A)); D("O_1",B,dir(B)); D("O_2",C,dir(C)); D("H",H,dir(270)); D("X",X,dir(225)); D("A",R1,dir(180)); D("B",R2,dir(180)); draw(rightanglemark(Y,X,C,3)); | [] |
236 | Circles $\omega_1$ and $\omega_2$ with radii $961$ and $625$, respectively, intersect at distinct points $A$ and $B$. A third circle $\omega$ is externally tangent to both $\omega_1$ and $\omega_2$. Suppose line $AB$ intersects $\omega$ at two points $P$ and $Q$ such that the measure of minor arc $\widehat{PQ}$ is $120^{\circ}$. Find the distance between the centers of $\omega_1$ and $\omega_2$. | 2021 AIME I Problem 13 | Denote by $O_1$, $O_2$, and $O$ the centers of $\omega_1$, $\omega_2$, and $\omega$, respectively. Let $R_1 = 961$ and $R_2 = 625$ denote the radii of $\omega_1$ and $\omega_2$ respectively, $r$ be the radius of $\omega$, and $\ell$ the distance from $O$ to the line $AB$. We claim that\[\dfrac{\ell}{r} = \dfrac{R_2-R_1}{d},\]where $d = O_1O_2$. This solves the problem, for then the $\widehat{PQ} = 120^\circ$ condition implies $\tfrac{\ell}r = \cos 60^\circ = \tfrac{1}{2}$, and then we can solve to get $d = \boxed{672}$.
Denote by $O_1$ and $O_2$ the centers of $\omega_1$ and $\omega_2$ respectively. Set $X$ as the projection of $O$ onto $O_1O_2$, and denote by $Y$ the intersection of $AB$ with $O_1O_2$. Note that $\ell = XY$. Now recall that\[d(O_2Y-O_1Y) = O_2Y^2 - O_1Y^2 = R_2^2 - R_1^2.\]Furthermore, note that\begin{align*}d(O_2X - O_1X) &= O_2X^2 - O_1X^2= O_2O^2 - O_1O^2 \\ &= (R_2 + r)^2 - (R_1+r)^2 = (R_2^2 - R_1^2) + 2r(R_2 - R_1).\end{align*}Substituting the first equality into the second one and subtracting yields\[2r(R_2 - R_1) = d(O_2X - O_1X) - d(O_2Y - O_1Y) = 2dXY,\]which rearranges to the desired. | // Block 1
import olympiad;
size(230pt);
defaultpen(linewidth(0.8)+fontsize(10pt));
real r1 = 17, r2 = 27, d = 35, r = 18;
pair O1 = origin, O2 = (d,0);
path w1 = circle(origin,r1), w2 = circle((d,0),r2), w1p = circle(origin,r1+r), w2p = circle((d,0), r2 + r);
pair[] X = intersectionpoints(w1,w2), Y = intersectionpoints(w1p,w2p);
pair O = Y[1];
path w = circle(Y[1],r);
pair Xp = 5 * X[1] - 4 * X[0];
pair[] P = intersectionpoints(Xp--X[0],w);
label("$O_1$",origin,N);
label("$O_2$",(d,0),N);
label("$O$",Y[1],SW);
draw(origin--Y[1]--(d,0)--cycle,gray(0.6));
pair T = foot(O,O1,O2), Tp = foot(O,X[0],X[1]);
draw(Tp--O--T^^rightanglemark(O,T,O1,60)^^rightanglemark(O,Tp,X[0],60),gray(0.6));
draw(w^^w1^^w2^^P[0]--X[0]);
dot(Y[1]^^origin^^(d,0));
label("$X$",T,N,gray(0.6));
label("$Y$",foot(X[0],O1,O2),NE,gray(0.6));
label("$\ell$",(O+Tp)/2,S,gray(0.6));
// Block 2
import olympiad; size(230pt); defaultpen(linewidth(0.8)+fontsize(10pt)); real r1 = 17, r2 = 27, d = 35, r = 18; pair O1 = origin, O2 = (d,0); path w1 = circle(origin,r1), w2 = circle((d,0),r2), w1p = circle(origin,r1+r), w2p = circle((d,0), r2 + r); pair[] X = intersectionpoints(w1,w2), Y = intersectionpoints(w1p,w2p); pair O = Y[1]; path w = circle(Y[1],r); pair Xp = 5 * X[1] - 4 * X[0]; pair[] P = intersectionpoints(Xp--X[0],w); label("$O_1$",origin,N); label("$O_2$",(d,0),N); label("$O$",Y[1],SW); draw(origin--Y[1]--(d,0)--cycle,gray(0.6)); pair T = foot(O,O1,O2), Tp = foot(O,X[0],X[1]); draw(Tp--O--T^^rightanglemark(O,T,O1,60)^^rightanglemark(O,Tp,X[0],60),gray(0.6)); draw(w^^w1^^w2^^P[0]--X[0]); dot(Y[1]^^origin^^(d,0)); label("$X$",T,N,gray(0.6)); label("$Y$",foot(X[0],O1,O2),NE,gray(0.6)); label("$\ell$",(O+Tp)/2,S,gray(0.6)); | [] |
236 | Circles $\omega_1$ and $\omega_2$ with radii $961$ and $625$, respectively, intersect at distinct points $A$ and $B$. A third circle $\omega$ is externally tangent to both $\omega_1$ and $\omega_2$. Suppose line $AB$ intersects $\omega$ at two points $P$ and $Q$ such that the measure of minor arc $\widehat{PQ}$ is $120^{\circ}$. Find the distance between the centers of $\omega_1$ and $\omega_2$. | 2021 AIME I Problem 13 | Suppose we label the points as shown below.
By radical axis, the tangents to $\omega$ at $D$ and $E$ intersect on $AB$. Thus $PDQE$ is harmonic, so the tangents to $\omega$ at $P$ and $Q$ intersect at $X \in DE$. Moreover, $OX \parallel O_1O_2$ because both $OX$ and $O_1O_2$ are perpendicular to $AB$, and $OX = 2OP$ because $\angle POQ = 120^{\circ}$. Thus\[O_1O_2 = O_1Y - O_2Y = 2 \cdot 961 - 2\cdot 625 = \boxed{672}\]by similar triangles.
~mathman3880 | // Block 1
defaultpen(fontsize(12)+0.6); size(300);
pen p=fontsize(10)+royalblue+0.4;
var r=1200;
pair O1=origin, O2=(672,0), O=OP(CR(O1,961+r),CR(O2,625+r));
path c1=CR(O1,961), c2=CR(O2,625), c=CR(O,r);
pair
A=IP(CR(O1,961),CR(O2,625)), B=OP(CR(O1,961),CR(O2,625)),
P=IP(L(A,B,0,0.2),c),
Q=IP(L(A,B,0,200),c),
F=IP(CR(O,625+r),O--O1),
M=(F+O2)/2,
D=IP(CR(O,r),O--O1),
E=IP(CR(O,r),O--O2),
X=extension(E,D,O,O+O1-O2),
Y=extension(D,E,O1,O2);
draw(c1^^c2); draw(c,blue+0.6); draw(O1--O2--O--cycle,black+0.6); draw(O--X^^Y--O2,black+0.6); draw(X--Y,heavygreen+0.6);
draw((X+O)/2--O,MidArrow); draw(O2--Y-(300,0),MidArrow);
dot("$A$",A,dir(A-O2/2));
dot("$B$",B,dir(B-O2/2));
dot("$O_2$",O2,right+up); dot("$O_1$",O1,left+up);
dot("$O$",O,dir(O-O2));
dot("$D$",D,dir(170));
dot("$E$",E,dir(E-O1));
dot("$X$",X,dir(X-D));
dot("$Y$",Y,dir(Y-D));
label("$R$",O--E,right+up,p);
label("$R$",O--D,left+down,p);
label("$2R$",(X+O)/2-(150,0),down,p);
label("$961$",O1--D,2*(left+down),p);
label("$625$",O2--E,2*(right+up),p);
MA("",E,D,O1,100,fuchsia+linewidth(1));
MA("",X,D,O,100,fuchsia+linewidth(1));
MA("",Y,E,O2,100,orange+linewidth(1));
MA("",D,E,O,100,orange+linewidth(1));
// Block 2
defaultpen(fontsize(12)+0.6); size(300); pen p=fontsize(10)+royalblue+0.4; var r=1200; pair O1=origin, O2=(672,0), O=OP(CR(O1,961+r),CR(O2,625+r)); path c1=CR(O1,961), c2=CR(O2,625), c=CR(O,r); pair A=IP(CR(O1,961),CR(O2,625)), B=OP(CR(O1,961),CR(O2,625)), P=IP(L(A,B,0,0.2),c), Q=IP(L(A,B,0,200),c), F=IP(CR(O,625+r),O--O1), M=(F+O2)/2, D=IP(CR(O,r),O--O1), E=IP(CR(O,r),O--O2), X=extension(E,D,O,O+O1-O2), Y=extension(D,E,O1,O2); draw(c1^^c2); draw(c,blue+0.6); draw(O1--O2--O--cycle,black+0.6); draw(O--X^^Y--O2,black+0.6); draw(X--Y,heavygreen+0.6); draw((X+O)/2--O,MidArrow); draw(O2--Y-(300,0),MidArrow); dot("$A$",A,dir(A-O2/2)); dot("$B$",B,dir(B-O2/2)); dot("$O_2$",O2,right+up); dot("$O_1$",O1,left+up); dot("$O$",O,dir(O-O2)); dot("$D$",D,dir(170)); dot("$E$",E,dir(E-O1)); dot("$X$",X,dir(X-D)); dot("$Y$",Y,dir(Y-D)); label("$R$",O--E,right+up,p); label("$R$",O--D,left+down,p); label("$2R$",(X+O)/2-(150,0),down,p); label("$961$",O1--D,2*(left+down),p); label("$625$",O2--E,2*(right+up),p); MA("",E,D,O1,100,fuchsia+linewidth(1)); MA("",X,D,O,100,fuchsia+linewidth(1)); MA("",Y,E,O2,100,orange+linewidth(1)); MA("",D,E,O,100,orange+linewidth(1)); | [] |
237 | For positive real numbers $s$, let $\tau(s)$ denote the set of all obtuse triangles that have area $s$ and two sides with lengths $4$ and $10$. The set of all $s$ for which $\tau(s)$ is nonempty, but all triangles in $\tau(s)$ are congruent, is an interval $[a,b)$. Find $a^2+b^2$. | 2021 AIME II Problem 5 | We have the diagram below.
We proceed by taking cases on the angles that can be obtuse, and finding the ranges for $s$ that they yield .
If angle $\theta$ is obtuse, then we have that $s \in (0,20)$. This is because $s=20$ is attained at $\theta = 90^{\circ}$, and the area of the triangle is strictly decreasing as $\theta$ increases beyond $90^{\circ}$. This can be observed from
\[s=\frac{1}{2}(4)(10)\sin\theta\]by noting that $\sin\theta$ is decreasing in $\theta \in (90^{\circ},180^{\circ})$.
Then, we note that if $\alpha$ is obtuse, we have $s \in (0,4\sqrt{21})$. This is because we get $x=\sqrt{10^2-4^2}=\sqrt{84}=2\sqrt{21}$ when $\alpha=90^{\circ}$, yielding $s=4\sqrt{21}$. Then, $s$ is decreasing as $\alpha$ increases by the same argument as before.
$\angle{ACB}$ cannot be obtuse since $AC>AB$.
Now we have the intervals $s \in (0,20)$ and $s \in (0,4\sqrt{21})$ for the cases where $\theta$ and $\alpha$ are obtuse, respectively. We are looking for the $s$ that are in exactly one of these intervals, and because $4\sqrt{21}<20$, the desired range is
\[s\in [4\sqrt{21},20)\]giving \[a^2+b^2=\boxed{736}. \Box\] | // Block 1
draw((0,0)--(1,2*sqrt(3)));
draw((1,2*sqrt(3))--(10,0));
draw((10,0)--(0,0));
label("$A$",(0,0),SW);
label("$B$",(1,2*sqrt(3)),N);
label("$C$",(10,0),SE);
label("$\theta$",(0,0),NE);
label("$\alpha$",(1,2*sqrt(3)),SSE);
label("$4$",(0,0)--(1,2*sqrt(3)),WNW);
label("$10$",(0,0)--(10,0),S);
// Block 2
draw((0,0)--(1,2*sqrt(3))); draw((1,2*sqrt(3))--(10,0)); draw((10,0)--(0,0)); label("$A$",(0,0),SW); label("$B$",(1,2*sqrt(3)),N); label("$C$",(10,0),SE); label("$\theta$",(0,0),NE); label("$\alpha$",(1,2*sqrt(3)),SSE); label("$4$",(0,0)--(1,2*sqrt(3)),WNW); label("$10$",(0,0)--(10,0),S); | [] |
237 | For positive real numbers $s$, let $\tau(s)$ denote the set of all obtuse triangles that have area $s$ and two sides with lengths $4$ and $10$. The set of all $s$ for which $\tau(s)$ is nonempty, but all triangles in $\tau(s)$ are congruent, is an interval $[a,b)$. Find $a^2+b^2$. | 2021 AIME II Problem 5 | For $\triangle ABC,$ we fix $AB=10$ and $BC=4.$ Without the loss of generality, we consider $C$ on only one side of $\overline{AB}.$
As shown below, all locations for $C$ at which $\triangle ABC$ is an obtuse triangle are indicated in red, excluding the endpoints.
Note that:
The region in which $\angle B$ is obtuse is determined by construction.
The region in which $\angle C$ is obtuse is determined by the corollaries of the Inscribed Angle Theorem.
For any fixed value of $s,$ the height from $C$ is fixed. We need obtuse $\triangle ABC$ to be unique, so there can only be one possible location for $C.$ As shown below, all possible locations for $C$ are on minor arc $\widehat{C_1C_2},$ including $C_1$ but excluding $C_2.$
Let the brackets denote areas:
If $C=C_1,$ then $[ABC]$ will be minimized (attainable). By the same base and height and the Inscribed Angle Theorem, we have
\begin{align*} [ABC]&=[ABD] \\ &=\frac12\cdot BD\cdot DA \\ &=\frac12\cdot BD\cdot \sqrt{AB^2-BD^2} \\ &=\frac12\cdot 4\cdot \sqrt{10^2-4^2} \\ &=2\sqrt{84}. \end{align*}
If $C=C_2,$ then $[ABC]$ will be maximized (unattainable). For this right triangle, we have
\begin{align*} [ABC]&=\frac12\cdot AB\cdot BC \\ &=\frac12\cdot 10\cdot 4 \\ &=20. \end{align*}
Finally, the set of all such $s$ is $[a,b)=\left[2\sqrt{84},20\right),$ from which $a^2+b^2=\boxed{736}.$
~MRENTHUSIASM (credit given to Snowfan) | // Block 1
/* Made by MRENTHUSIASM */
size(300);
pair A, B, O, P, Q, C1, C2, D;
A = origin;
B = (10,0);
O = midpoint(A--B);
P = B - (4,0);
Q = B + (4,0);
C1 = intersectionpoints(D--D+(100,0),Arc(B,Q,P))[1];
C2 = B + (0,4);
D = intersectionpoint(Arc(O,B,A),Arc(B,Q,P));
draw(Arc(O,B,A)^^Arc(B,C2,D)^^A--Q);
draw(Arc(B,Q,C2)^^Arc(B,D,P),red);
dot("$A$", A, 1.5*S, linewidth(4.5));
dot("$B$", B, 1.5*S, linewidth(4.5));
dot(O, linewidth(4.5));
dot(P^^C2^^D^^Q, linewidth(0.8), UnFill);
Label L1 = Label("$10$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white));
Label L2 = Label("$4$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white));
draw(A-(0,2)--B-(0,2), L=L1, arrow=Arrows(),bar=Bars(15));
draw(B-(0,2)--Q-(0,2), L=L2, arrow=Arrows(),bar=Bars(15));
label("$\angle C$ obtuse",(midpoint(Arc(B,D,P)).x,2),2.5*W,red);
label("$\angle B$ obtuse",(midpoint(Arc(B,Q,C2)).x,2),5*E,red);
// Block 2
/* Made by MRENTHUSIASM */
size(250);
pair A, B, O, P, Q, C1, C2, D;
A = origin;
B = (10,0);
O = midpoint(A--B);
P = B - (4,0);
Q = B + (4,0);
C2 = B + (0,4);
D = intersectionpoint(Arc(O,B,A),Arc(B,Q,P));
C1 = intersectionpoint(D--D+(100,0),Arc(B,Q,C2));
draw(Arc(O,B,A)^^Arc(B,C2,D)^^A--Q);
draw(Arc(B,Q,C1)^^Arc(B,D,P),red);
draw(Arc(B,C1,C2),green);
draw((A.x,D.y)--(Q.x,D.y),dashed);
dot("$A$", A, 1.5*S, linewidth(4.5));
dot("$B$", B, 1.5*S, linewidth(4.5));
dot("$D$", D, 1.5*dir(75), linewidth(0.8), UnFill);
dot("$C_2$", C2, 1.5*N, linewidth(4.5));
dot("$C_1$", C1, 1.5*dir(C1-B), linewidth(4.5));
dot(O, linewidth(4.5));
dot(P^^C2^^Q, linewidth(0.8), UnFill);
dot(C1, green+linewidth(4.5));
Label L1 = Label("$10$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white));
Label L2 = Label("$4$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white));
draw(A-(0,2)--B-(0,2), L=L1, arrow=Arrows(),bar=Bars(15));
draw(B-(0,2)--Q-(0,2), L=L2, arrow=Arrows(),bar=Bars(15));
// Block 3
/* Made by MRENTHUSIASM */ size(300); pair A, B, O, P, Q, C1, C2, D; A = origin; B = (10,0); O = midpoint(A--B); P = B - (4,0); Q = B + (4,0); C1 = intersectionpoints(D--D+(100,0),Arc(B,Q,P))[1]; C2 = B + (0,4); D = intersectionpoint(Arc(O,B,A),Arc(B,Q,P)); draw(Arc(O,B,A)^^Arc(B,C2,D)^^A--Q); draw(Arc(B,Q,C2)^^Arc(B,D,P),red); dot("$A$", A, 1.5*S, linewidth(4.5)); dot("$B$", B, 1.5*S, linewidth(4.5)); dot(O, linewidth(4.5)); dot(P^^C2^^D^^Q, linewidth(0.8), UnFill); Label L1 = Label("$10$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); Label L2 = Label("$4$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); draw(A-(0,2)--B-(0,2), L=L1, arrow=Arrows(),bar=Bars(15)); draw(B-(0,2)--Q-(0,2), L=L2, arrow=Arrows(),bar=Bars(15)); label("$\angle C$ obtuse",(midpoint(Arc(B,D,P)).x,2),2.5*W,red); label("$\angle B$ obtuse",(midpoint(Arc(B,Q,C2)).x,2),5*E,red);
// Block 4
/* Made by MRENTHUSIASM */ size(250); pair A, B, O, P, Q, C1, C2, D; A = origin; B = (10,0); O = midpoint(A--B); P = B - (4,0); Q = B + (4,0); C2 = B + (0,4); D = intersectionpoint(Arc(O,B,A),Arc(B,Q,P)); C1 = intersectionpoint(D--D+(100,0),Arc(B,Q,C2)); draw(Arc(O,B,A)^^Arc(B,C2,D)^^A--Q); draw(Arc(B,Q,C1)^^Arc(B,D,P),red); draw(Arc(B,C1,C2),green); draw((A.x,D.y)--(Q.x,D.y),dashed); dot("$A$", A, 1.5*S, linewidth(4.5)); dot("$B$", B, 1.5*S, linewidth(4.5)); dot("$D$", D, 1.5*dir(75), linewidth(0.8), UnFill); dot("$C_2$", C2, 1.5*N, linewidth(4.5)); dot("$C_1$", C1, 1.5*dir(C1-B), linewidth(4.5)); dot(O, linewidth(4.5)); dot(P^^C2^^Q, linewidth(0.8), UnFill); dot(C1, green+linewidth(4.5)); Label L1 = Label("$10$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); Label L2 = Label("$4$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); draw(A-(0,2)--B-(0,2), L=L1, arrow=Arrows(),bar=Bars(15)); draw(B-(0,2)--Q-(0,2), L=L2, arrow=Arrows(),bar=Bars(15)); | [] |
238 | An ant makes a sequence of moves on a cube where a move consists of walking from one vertex to an adjacent vertex along an edge of the cube. Initially the ant is at a vertex of the bottom face of the cube and chooses one of the three adjacent vertices to move to as its first move. For all moves after the first move, the ant does not return to its previous vertex, but chooses to move to one of the other two adjacent vertices. All choices are selected at random so that each of the possible moves is equally likely. The probability that after exactly $8$ moves that ant is at a vertex of the top face on the cube is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n.$ | 2021 AIME II Problem 8 | For all positive integers $k,$ let
$N(k,\mathrm{BB})$ be the number of ways to make a sequence of exactly $k$ moves, where the last move is from the bottom face to the bottom face.
$N(k,\mathrm{BT})$ be the number of ways to make a sequence of exactly $k$ moves, where the last move is from the bottom face to the top face.
$N(k,\mathrm{TB})$ be the number of ways to make a sequence of exactly $k$ moves, where the last move is from the top face to the bottom face.
$N(k,\mathrm{TT})$ be the number of ways to make a sequence of exactly $k$ moves, where the last move is from the top face to the top face.
The base case occurs at $k=1,$ from which $\left(N(1,\mathrm{BB}),N(1,\mathrm{BT}),N(1,\mathrm{TB}),N(1,\mathrm{TT})\right)=(2,1,0,0).$
Suppose the ant makes exactly $k$ moves for some $k\geq2.$ We perform casework on its last move:
If its last move is from the bottom face to the bottom face, then its next move has
$1$ way to move from the bottom face to the bottom face.
$1$ way to move from the bottom face to the top face.
If its last move is from the bottom face to the top face, then its next move has $2$ ways to move from the top face to the top face.
If its last move is from the top face to the bottom face, then its next move has $2$ ways to move from the bottom face to the bottom face.
If its last move is from the top face to the top face, then its next move has
$1$ way to move from the top face to the bottom face.
$1$ way to move from the top face to the top face.
Alternatively, this recursion argument is illustrated below, where each dashed arrow indicates $1$ way, and each solid arrow indicates $2$ ways:
Therefore, we have the following relationships:
\begin{align*} N(1,\mathrm{BB})&=2, \\ N(1,\mathrm{BT})&=1, \\ N(1,\mathrm{TB})&=0, \\ N(1,\mathrm{TT})&=0, \\ N(k+1,\mathrm{BB})&=N(k,\mathrm{BB})+2\cdot N(k,\mathrm{TB}), \\ N(k+1,\mathrm{BT})&=N(k,\mathrm{BB}), \\ N(k+1,\mathrm{TB})&=N(k,\mathrm{TT}), \\ N(k+1,\mathrm{TT})&=N(k,\mathrm{TT})+2\cdot N(k,\mathrm{BT}). \end{align*}
Using these equations, we recursively fill out the table below:
\[\begin{array}{c||c|c|c|c|c|c|c|c} \hspace{7mm}&\hspace{6.5mm}&\hspace{6.5mm}&\hspace{6.75mm}&\hspace{6.75mm}&\hspace{6.75mm}&\hspace{6.75mm}&& \\ [-2.5ex] \boldsymbol{k} & \boldsymbol{1} & \boldsymbol{2} & \boldsymbol{3} & \boldsymbol{4} & \boldsymbol{5} & \boldsymbol{6} & \boldsymbol{7} & \boldsymbol{8} \\ \hline \hline &&&&&&&& \\ [-2.25ex] \boldsymbol{N(k,\mathrm{BB})} &2&2&2&6&18&38&66&118 \\ \hline &&&&&&&& \\ [-2.25ex] \boldsymbol{N(k,\mathrm{BT})} &1&2&2&2&6&18&38&66 \\ \hline &&&&&&&& \\ [-2.25ex] \boldsymbol{N(k,\mathrm{TB})} &0&0&2&6&10&14&26&62 \\ \hline &&&&&&&& \\ [-2.25ex] \boldsymbol{N(k,\mathrm{TT})} &0&2&6&10&14&26&62&138 \\ \hline \hline &&&&&&&& \\ [-2.25ex] \textbf{Total}&\boldsymbol{3}&\boldsymbol{6}&\boldsymbol{12}&\boldsymbol{24}&\boldsymbol{48}&\boldsymbol{96}&\boldsymbol{192}&\boldsymbol{384} \end{array}\]
By the Multiplication Principle, there are $3\cdot2^{k-1}$ ways to make exactly $k$ moves. So, we must get \[N(k,\mathrm{BB})+N(k,\mathrm{BT})+N(k,\mathrm{TB})+N(k,\mathrm{TT})=3\cdot2^{k-1}\] for all values of $k.$
Finally, the requested probability is \[\frac{N(8,\mathrm{BT})+N(8,\mathrm{TT})}{N(8,\mathrm{BB})+N(8,\mathrm{BT})+N(8,\mathrm{TB})+N(8,\mathrm{TT})}=\frac{66+138}{118+66+62+138}=\frac{204}{384}=\frac{17}{32},\] from which the answer is $17+32=\boxed{049}.$
~Arcticturn ~MRENTHUSIASM | // Block 1
/* Made by MRENTHUSIASM */
size(9cm);
pair A, B, C, D, E, F, G, H, X, Y;
A=(0,6);
B=(0,4);
C=(0,2);
D=(0,0);
E=(10,6);
F=(10,4);
G=(10,2);
H=(10,0);
X=(-1,8);
Y=(11,8);
label("BB", A, (-2,0));
label("BT", B, (-2,0));
label("TB", C, (-2,0));
label("TT", D, (-2,0));
label("BB", E, (2,0));
label("BT", F, (2,0));
label("TB", G, (2,0));
label("TT", H, (2,0));
label("\textbf{The \boldmath{$k$}th Move}", shift(0.3,0)*X);
label("\textbf{The \boldmath{$(k+1)$}th Move}", shift(-0.3,-0.085)*Y);
draw(A--E,0.8+black+dashed,EndArrow);
draw(A--F,0.8+black+dashed,EndArrow);
draw(B--H,0.8+black,EndArrow);
draw(C--E,0.8+black,EndArrow);
draw(D--G,0.8+black+dashed,EndArrow);
draw(D--H,0.8+black+dashed,EndArrow);
dot(A^^B^^C^^D^^E^^F^^G^^H, 5+black);
// Block 2
/* Made by MRENTHUSIASM */ size(9cm); pair A, B, C, D, E, F, G, H, X, Y; A=(0,6); B=(0,4); C=(0,2); D=(0,0); E=(10,6); F=(10,4); G=(10,2); H=(10,0); X=(-1,8); Y=(11,8); label("BB", A, (-2,0)); label("BT", B, (-2,0)); label("TB", C, (-2,0)); label("TT", D, (-2,0)); label("BB", E, (2,0)); label("BT", F, (2,0)); label("TB", G, (2,0)); label("TT", H, (2,0)); label("\textbf{The \boldmath{$k$}th Move}", shift(0.3,0)*X); label("\textbf{The \boldmath{$(k+1)$}th Move}", shift(-0.3,-0.085)*Y); draw(A--E,0.8+black+dashed,EndArrow); draw(A--F,0.8+black+dashed,EndArrow); draw(B--H,0.8+black,EndArrow); draw(C--E,0.8+black,EndArrow); draw(D--G,0.8+black+dashed,EndArrow); draw(D--H,0.8+black+dashed,EndArrow); dot(A^^B^^C^^D^^E^^F^^G^^H, 5+black); | [] |
239 | Two spheres with radii $36$ and one sphere with radius $13$ are each externally tangent to the other two spheres and to two different planes $\mathcal{P}$ and $\mathcal{Q}$. The intersection of planes $\mathcal{P}$ and $\mathcal{Q}$ is the line $\ell$. The distance from line $\ell$ to the point where the sphere with radius $13$ is tangent to plane $\mathcal{P}$ is $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$. | 2021 AIME II Problem 10 | This solution refers to the Diagram section.
As shown below, let $O_1,O_2,O_3$ be the centers of the spheres (where sphere $O_3$ has radius $13$) and $T_1,T_2,T_3$ be their respective points of tangency to plane $\mathcal{P}.$ Let $\mathcal{R}$ be the plane that is determined by $O_1,O_2,$ and $O_3.$ Suppose $A$ is the foot of the perpendicular from $O_3$ to line $\ell,$ so $\overleftrightarrow{O_3A}$ is the perpendicular bisector of $\overline{O_1O_2}.$ We wish to find $T_3A.$
Note that:
In $\triangle O_1O_2O_3,$ we get $O_1O_2=72$ and $O_1O_3=O_2O_3=49.$
Both $\triangle O_1O_2O_3$ and $\overline{O_3A}$ lie in plane $\mathcal{R}.$ Both $\triangle T_1T_2T_3$ and $\overline{T_3A}$ lie in plane $\mathcal{P}.$
By symmetry, since planes $\mathcal{P}$ and $\mathcal{Q}$ are reflections of each other about plane $\mathcal{R},$ the three planes are concurrent to line $\ell.$
Since $\overline{O_1T_1}\perp\mathcal{P}$ and $\overline{O_3T_3}\perp\mathcal{P},$ it follows that $\overline{O_1T_1}\parallel\overline{O_3T_3},$ from which $O_1,O_3,T_1,$ and $T_3$ are coplanar.
Now, we focus on cross-sections $O_1O_3T_3T_1$ and $\mathcal{R}:$
In the three-dimensional space, the intersection of a line and a plane must be exactly one of the empty set, a point, or a line.
Clearly, cross-section $O_1O_3T_3T_1$ intersects line $\ell$ at exactly one point. Furthermore, as the intersection of planes $\mathcal{R}$ and $\mathcal{P}$ is line $\ell,$ we conclude that $\overrightarrow{O_1O_3}$ and $\overrightarrow{T_1T_3}$ must intersect line $\ell$ at the same point. Let $B$ be the point of concurrency of $\overrightarrow{O_1O_3},\overrightarrow{T_1T_3},$ and line $\ell.$
In cross-section $\mathcal{R},$ let $C$ be the foot of the perpendicular from $O_1$ to line $\ell,$ and $D$ be the foot of the perpendicular from $O_3$ to $\overline{O_1C}.$
We have the following diagram:
In cross-section $O_1O_3T_3T_1,$ since $\overline{O_1T_1}\parallel\overline{O_3T_3}$ as discussed, we obtain $\triangle O_1T_1B\sim\triangle O_3T_3B$ by AA, with the ratio of similitude $\frac{O_1T_1}{O_3T_3}=\frac{36}{13}.$ Therefore, we get $\frac{O_1B}{O_3B}=\frac{49+O_3B}{O_3B}=\frac{36}{13},$ or $O_3B=\frac{637}{23}.$
In cross-section $\mathcal{R},$ note that $O_1O_3=49$ and $DO_3=\frac{O_1O_2}{2}=36.$ Applying the Pythagorean Theorem to right $\triangle O_1DO_3,$ we have $O_1D=\sqrt{1105}.$ Moreover, since $\ell\perp\overline{O_1C}$ and $\overline{DO_3}\perp\overline{O_1C},$ we obtain $\ell\parallel\overline{DO_3}$ so that $\triangle O_1CB\sim\triangle O_1DO_3$ by AA, with the ratio of similitude $\frac{O_1B}{O_1O_3}=\frac{49+\frac{637}{23}}{49}.$ Therefore, we get $\frac{O_1C}{O_1D}=\frac{\sqrt{1105}+DC}{\sqrt{1105}}=\frac{49+\frac{637}{23}}{49},$ or $DC=\frac{13\sqrt{1105}}{23}.$
Finally, note that $\overline{O_3T_3}\perp\overline{T_3A}$ and $O_3T_3=13.$ Since quadrilateral $DCAO_3$ is a rectangle, we have $O_3A=DC=\frac{13\sqrt{1105}}{23}.$ Applying the Pythagorean Theorem to right $\triangle O_3T_3A$ gives $T_3A=\frac{312}{23},$ from which the answer is $312+23=\boxed{335}.$
~MRENTHUSIASM | // Block 1
/* Made by MRENTHUSIASM */
size(300);
import graph3;
import solids;
currentprojection=orthographic((1,1/2,0));
triple O1, O2, O3, T1, T2, T3, A, L1, L2;
O1 = (0,-36,0);
O2 = (0,36,0);
O3 = (0,0,-sqrt(1105));
T1 = (864*sqrt(1105)/1105,-36,-828*sqrt(1105)/1105);
T2 = (864*sqrt(1105)/1105,36,-828*sqrt(1105)/1105);
T3 = (24*sqrt(1105)/85,0,-108*sqrt(1105)/85);
A = (0,0,-36*sqrt(1105)/23);
L1 = shift(0,-80,0)*A;
L2 = shift(0,80,0)*A;
draw(surface(L1--L2--(-T2.x,L2.y,T2.z)--(-T1.x,L1.y,T1.z)--cycle),pink);
draw(surface(L1--L2--(L2.x,L2.y,40)--(L1.x,L1.y,40)--cycle),gray);
draw(shift(O1)*rotate(90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White);
draw(shift(O2)*rotate(90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White);
draw(shift(O3)*rotate(90,O1,O2)*scale3(13)*unithemisphere,red,light=White);
draw(surface(L1--L2--(T2.x,L2.y,T2.z)--(T1.x,L1.y,T1.z)--cycle),palegreen);
draw(surface(L1--L2--(-T2.x,L2.y,L2.z-abs(T2.z))--(-T1.x,L1.y,L2.z-abs(T1.z))--cycle),palegreen);
draw(surface(L1--L2--(L2.x,L2.y,L2.z-abs(T1.z))--(L1.x,L1.y,L1.z-abs(T2.z))--cycle),gray);
draw(surface(L1--L2--(T2.x,L2.y,L2.z-abs(T1.z))--(T1.x,L1.y,L1.z-abs(T2.z))--cycle),pink);
draw(O1--O2--O3--cycle^^O3--A,dashed);
draw(T1--T2--T3--cycle^^T3--A,heavygreen);
draw(O1--T1^^O2--T2^^O3--T3,mediumblue+dashed);
draw(L1--L2,L=Label("$\ell$",position=EndPoint,align=3*E),red);
label("$\mathcal{P}$",midpoint(L1--(T1.x,L1.y,T1.z)),(0,-3,0),heavygreen);
label("$\mathcal{Q}$",midpoint(L1--(T1.x,L1.y,L1.z-abs(T2.z))),(0,-3,0),heavymagenta);
label("$\mathcal{R}$",O1,(0,-24,0));
dot("$O_1$",O1,(0,-1,1),linewidth(4.5));
dot("$O_2$",O2,(0,1,1),linewidth(4.5));
dot("$O_3$",O3,(0,-1.5,0),linewidth(4.5));
dot("$T_1$",T1,(0,-1,-1),heavygreen+linewidth(4.5));
dot("$T_2$",T2,(0,1,-1),heavygreen+linewidth(4.5));
dot("$T_3$",T3,(0,-1,-1),heavygreen+linewidth(4.5));
dot("$A$",A,(0,0,-2),red+linewidth(4.5));
// Block 2
size(300);
import graph3;
import solids;
currentprojection=orthographic((1,1/2,0));
triple O1, O3, T1, T3, A, B, C, D;
O1 = (0,-36,0);
O3 = (0,0,-sqrt(1105));
T1 = (864*sqrt(1105)/1105,-36,-828*sqrt(1105)/1105);
T3 = (24*sqrt(1105)/85,0,-108*sqrt(1105)/85);
A = (0,0,-36*sqrt(1105)/23);
B = intersectionpoint(O1--O1+100*(O3-O1),T1--T1+100*(T3-T1));
C = (0,-36,-36*sqrt(1105)/23);
D = (0,-36,-sqrt(1105));
draw(C--O1--O3--A^^D--O3--B,dashed);
draw(T1--T3--A^^T3--B,heavygreen);
draw(O1--T1^^O3--T3,mediumblue+dashed);
draw(shift(0,-80,0)*A--shift(0,80,0)*A,L=Label("$\ell$",position=EndPoint,align=3*E),red);
dot("$O_1$",O1,(0,-1,1),linewidth(4.5));
dot("$O_3$",O3,(0,1,1),linewidth(4.5));
dot("$T_1$",T1,(0,-1,-1),heavygreen+linewidth(4.5));
dot("$T_3$",T3,(0,-1,-1),heavygreen+linewidth(4.5));
dot("$A$",A,(0,0,-2),red+linewidth(4.5));
dot("$B$",B,(0,0,-2),red+linewidth(4.5));
dot("$C$",C,(0,0,-2),red+linewidth(4.5));
dot("$D$",D,(0,-2,0),linewidth(4.5));
// Block 3
/* Made by MRENTHUSIASM */ size(300); import graph3; import solids; currentprojection=orthographic((1,1/2,0)); triple O1, O2, O3, T1, T2, T3, A, L1, L2; O1 = (0,-36,0); O2 = (0,36,0); O3 = (0,0,-sqrt(1105)); T1 = (864*sqrt(1105)/1105,-36,-828*sqrt(1105)/1105); T2 = (864*sqrt(1105)/1105,36,-828*sqrt(1105)/1105); T3 = (24*sqrt(1105)/85,0,-108*sqrt(1105)/85); A = (0,0,-36*sqrt(1105)/23); L1 = shift(0,-80,0)*A; L2 = shift(0,80,0)*A; draw(surface(L1--L2--(-T2.x,L2.y,T2.z)--(-T1.x,L1.y,T1.z)--cycle),pink); draw(surface(L1--L2--(L2.x,L2.y,40)--(L1.x,L1.y,40)--cycle),gray); draw(shift(O1)*rotate(90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White); draw(shift(O2)*rotate(90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White); draw(shift(O3)*rotate(90,O1,O2)*scale3(13)*unithemisphere,red,light=White); draw(surface(L1--L2--(T2.x,L2.y,T2.z)--(T1.x,L1.y,T1.z)--cycle),palegreen); draw(surface(L1--L2--(-T2.x,L2.y,L2.z-abs(T2.z))--(-T1.x,L1.y,L2.z-abs(T1.z))--cycle),palegreen); draw(surface(L1--L2--(L2.x,L2.y,L2.z-abs(T1.z))--(L1.x,L1.y,L1.z-abs(T2.z))--cycle),gray); draw(surface(L1--L2--(T2.x,L2.y,L2.z-abs(T1.z))--(T1.x,L1.y,L1.z-abs(T2.z))--cycle),pink); draw(O1--O2--O3--cycle^^O3--A,dashed); draw(T1--T2--T3--cycle^^T3--A,heavygreen); draw(O1--T1^^O2--T2^^O3--T3,mediumblue+dashed); draw(L1--L2,L=Label("$\ell$",position=EndPoint,align=3*E),red); label("$\mathcal{P}$",midpoint(L1--(T1.x,L1.y,T1.z)),(0,-3,0),heavygreen); label("$\mathcal{Q}$",midpoint(L1--(T1.x,L1.y,L1.z-abs(T2.z))),(0,-3,0),heavymagenta); label("$\mathcal{R}$",O1,(0,-24,0)); dot("$O_1$",O1,(0,-1,1),linewidth(4.5)); dot("$O_2$",O2,(0,1,1),linewidth(4.5)); dot("$O_3$",O3,(0,-1.5,0),linewidth(4.5)); dot("$T_1$",T1,(0,-1,-1),heavygreen+linewidth(4.5)); dot("$T_2$",T2,(0,1,-1),heavygreen+linewidth(4.5)); dot("$T_3$",T3,(0,-1,-1),heavygreen+linewidth(4.5)); dot("$A$",A,(0,0,-2),red+linewidth(4.5));
// Block 4
size(300); import graph3; import solids; currentprojection=orthographic((1,1/2,0)); triple O1, O3, T1, T3, A, B, C, D; O1 = (0,-36,0); O3 = (0,0,-sqrt(1105)); T1 = (864*sqrt(1105)/1105,-36,-828*sqrt(1105)/1105); T3 = (24*sqrt(1105)/85,0,-108*sqrt(1105)/85); A = (0,0,-36*sqrt(1105)/23); B = intersectionpoint(O1--O1+100*(O3-O1),T1--T1+100*(T3-T1)); C = (0,-36,-36*sqrt(1105)/23); D = (0,-36,-sqrt(1105)); draw(C--O1--O3--A^^D--O3--B,dashed); draw(T1--T3--A^^T3--B,heavygreen); draw(O1--T1^^O3--T3,mediumblue+dashed); draw(shift(0,-80,0)*A--shift(0,80,0)*A,L=Label("$\ell$",position=EndPoint,align=3*E),red); dot("$O_1$",O1,(0,-1,1),linewidth(4.5)); dot("$O_3$",O3,(0,1,1),linewidth(4.5)); dot("$T_1$",T1,(0,-1,-1),heavygreen+linewidth(4.5)); dot("$T_3$",T3,(0,-1,-1),heavygreen+linewidth(4.5)); dot("$A$",A,(0,0,-2),red+linewidth(4.5)); dot("$B$",B,(0,0,-2),red+linewidth(4.5)); dot("$C$",C,(0,0,-2),red+linewidth(4.5)); dot("$D$",D,(0,-2,0),linewidth(4.5)); | [] |
239 | Two spheres with radii $36$ and one sphere with radius $13$ are each externally tangent to the other two spheres and to two different planes $\mathcal{P}$ and $\mathcal{Q}$. The intersection of planes $\mathcal{P}$ and $\mathcal{Q}$ is the line $\ell$. The distance from line $\ell$ to the point where the sphere with radius $13$ is tangent to plane $\mathcal{P}$ is $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$. | 2021 AIME II Problem 10 | This solution refers to the Diagram section.
The isosceles triangle of centers $O_1 O_2 O$ ($O$ is the center of sphere of radii $13$) has sides $O_1 O = O_2 O = 36 + 13 = 49,$ and $O_1 O_2 = 36 + 36 = 72.$
Let $N$ be the midpoint $O_1 O_2$.
The isosceles triangle of points of tangency $T_1 T_2 T$ has sides $T_1 T = T_2 T = 2 \sqrt{13 \cdot 36} = 12 \sqrt{13}$ and $T_1 T_2 = 72.$
Let $M$ be the midpoint $T_1 T_2.$
The height $TM$ is $\sqrt {12^2 \cdot 13 - 36^2} = 12 \sqrt {13-9} = 24.$
The tangents of the half-angle between the planes is $\frac {TO}{AT} = \frac {MN - TO}{TM},$ so $\frac {13}{AT} = \frac {36 - 13}{24},$ \[AT = \frac{24\cdot 13}{23} = \frac {312}{23} \implies 312 + 23 = \boxed{335}.\]
vladimir.shelomovskii@gmail.com, vvsss | // Block 1
/* Made by MRENTHUSIASM */
size(300);
import graph3;
import solids;
currentprojection=orthographic((10,-3,-40));
triple O1, O2, O3, T1, T2, T3, A, L1, L2, M;
O1 = (0,-36,0);
O2 = (0,36,0);
O3 = (0,0,-sqrt(1105));
T1 = (864*sqrt(1105)/1105,-36,-828*sqrt(1105)/1105);
T2 = (864*sqrt(1105)/1105,36,-828*sqrt(1105)/1105);
T3 = (24*sqrt(1105)/85,0,-108*sqrt(1105)/85);
A = (0,0,-36*sqrt(1105)/23);
L1 = shift(0,-80,0)*A;
L2 = shift(0,80,0)*A;
M = midpoint(T1--T2);
draw(shift(O1)*rotate(-90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White);
draw(shift(O2)*rotate(-90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White);
draw(shift(O3)*rotate(-90,O1,O2)*scale3(13)*unithemisphere,red,light=White);
draw(surface(L1--L2--(L2.x,L2.y,40)--(L1.x,L1.y,40)--cycle),gray);
draw(shift(O1)*rotate(90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White);
draw(shift(O2)*rotate(90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White);
draw(shift(O3)*rotate(90,O1,O2)*scale3(13)*unithemisphere,red,light=White);
draw(surface(T2--T1--T3--A--cycle),cyan);
draw(surface(L1--L2--(L2.x,L2.y,L2.z-abs(T1.z))--(L1.x,L1.y,L1.z-abs(T2.z))--cycle),gray);
draw(T1--T2--T3--cycle^^M--A--T2,blue);
dot("$O_1$",O1,(0,-1,1),linewidth(4.5));
dot("$O_2$",O2,(0,1,1),linewidth(4.5));
dot("$O$",O3,(0.5,-1,0),linewidth(4.5));
dot("$T_1$",T1,(0,-1,-1),blue+linewidth(4.5));
dot("$T_2$",T2,(0,1,-1),blue+linewidth(4.5));
dot("$T$",T3,(1,1,2),blue+linewidth(4.5));
dot("$M$",M,(0,0,5),blue+linewidth(4.5));
dot("$A$",A,(-0.5,-1.5,0),red+linewidth(4.5));
// Block 2
/* Made by MRENTHUSIASM */ size(300); import graph3; import solids; currentprojection=orthographic((10,-3,-40)); triple O1, O2, O3, T1, T2, T3, A, L1, L2, M; O1 = (0,-36,0); O2 = (0,36,0); O3 = (0,0,-sqrt(1105)); T1 = (864*sqrt(1105)/1105,-36,-828*sqrt(1105)/1105); T2 = (864*sqrt(1105)/1105,36,-828*sqrt(1105)/1105); T3 = (24*sqrt(1105)/85,0,-108*sqrt(1105)/85); A = (0,0,-36*sqrt(1105)/23); L1 = shift(0,-80,0)*A; L2 = shift(0,80,0)*A; M = midpoint(T1--T2); draw(shift(O1)*rotate(-90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White); draw(shift(O2)*rotate(-90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White); draw(shift(O3)*rotate(-90,O1,O2)*scale3(13)*unithemisphere,red,light=White); draw(surface(L1--L2--(L2.x,L2.y,40)--(L1.x,L1.y,40)--cycle),gray); draw(shift(O1)*rotate(90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White); draw(shift(O2)*rotate(90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White); draw(shift(O3)*rotate(90,O1,O2)*scale3(13)*unithemisphere,red,light=White); draw(surface(T2--T1--T3--A--cycle),cyan); draw(surface(L1--L2--(L2.x,L2.y,L2.z-abs(T1.z))--(L1.x,L1.y,L1.z-abs(T2.z))--cycle),gray); draw(T1--T2--T3--cycle^^M--A--T2,blue); dot("$O_1$",O1,(0,-1,1),linewidth(4.5)); dot("$O_2$",O2,(0,1,1),linewidth(4.5)); dot("$O$",O3,(0.5,-1,0),linewidth(4.5)); dot("$T_1$",T1,(0,-1,-1),blue+linewidth(4.5)); dot("$T_2$",T2,(0,1,-1),blue+linewidth(4.5)); dot("$T$",T3,(1,1,2),blue+linewidth(4.5)); dot("$M$",M,(0,0,5),blue+linewidth(4.5)); dot("$A$",A,(-0.5,-1.5,0),red+linewidth(4.5)); | [] |
240 | A teacher was leading a class of four perfectly logical students. The teacher chose a set $S$ of four integers and gave a different number in $S$ to each student. Then the teacher announced to the class that the numbers in $S$ were four consecutive two-digit positive integers, that some number in $S$ was divisible by $6$, and a different number in $S$ was divisible by $7$. The teacher then asked if any of the students could deduce what $S$ is, but in unison, all of the students replied no.
However, upon hearing that all four students replied no, each student was able to determine the elements of $S$. Find the sum of all possible values of the greatest element of $S$. | 2021 AIME II Problem 11 | Note that $\operatorname{lcm}(6,7)=42.$ It is clear that $42\not\in S$ and $84\not\in S,$ otherwise the three other elements in $S$ are divisible by neither $6$ nor $7.$
In the table below, the multiples of $6$ are colored in yellow, and the multiples of $7$ are colored in green. By the least common multiple, we obtain cycles: If $n$ is a possible maximum value of $S,$ then $n+42$ must be another possible maximum value of $S,$ and vice versa. By observations, we circle all possible maximum values of $S.$
From the second row of the table above, we perform casework on the possible maximum value of $S:$
\[\begin{array}{c||c|c|l} & & & \\ [-2.5ex] \textbf{Max Value} & \boldsymbol{S} & \textbf{Valid?} & \hspace{16.25mm}\textbf{Reasoning/Conclusion} \\ [0.5ex] \hline & & & \\ [-2ex] 49 & \{46,47,48,49\} & & \text{The student who gets } 46 \text{ will reply yes.} \\ 50 & \{47,48,49,50\} & \checkmark & \text{Another possibility is } S=\{89,90,91,92\}. \\ 51 & \{48,49,50,51\} & & \text{The student who gets } 51 \text{ will reply yes.} \\ 56 & \{53,54,55,56\} & & \text{The student who gets } 53 \text{ will reply yes.} \\ 57 & \{54,55,56,57\} & & \text{The student who gets } 57 \text{ will reply yes.} \\ 63 & \{60,61,62,63\} & & \text{The students who get } 60,61,62 \text{ will reply yes.} \\ 66 & \{63,64,65,66\} & & \text{The students who get } 64,65,66 \text{ will reply yes.} \\ 72 & \{69,70,71,72\} & & \text{The student who gets } 69 \text{ will reply yes.} \\ 73 & \{70,71,72,73\} & & \text{The student who gets } 73 \text{ will reply yes.} \\ 78 & \{75,76,77,78\} & & \text{The student who gets } 75 \text{ will reply yes.} \\ 79 & \{76,77,78,79\} & \checkmark & \text{Another possibility is } S=\{34,35,36,37\}. \\ 80 & \{77,78,79,80\} & & \text{The student who gets } 80 \text{ will reply yes.} \end{array}\]
Finally, all possibilities for $S$ are $\{34,35,36,37\}, \{47,48,49,50\}, \{76,77,78,79\},$ and $\{89,90,91,92\},$ from which the answer is $37+50+79+92=\boxed{258}.$
Remarks
Alternatively, we can reconstruct the second table in this solution as follows, where Y and N denote the replies of "yes" and "no", respectively. Notice that this table has some kind of symmetry!
As a confirmation, we can verify that each student will be able to deduce what $S$ is upon hearing the four replies of "no" in unison. For example, if $S=\{47,48,49,50\},$ then all students will know that no one gets $46$ or $51,$ otherwise that student will reply yes (as discussed). Therefore, all students will conclude that $S$ has only one possibility.
~MRENTHUSIASM | // Block 1
/* Made by MRENTHUSIASM */
size(20cm);
fill((5,0)--(6,0)--(6,2)--(5,2)--cycle,yellow);
fill((11,0)--(12,0)--(12,3)--(11,3)--cycle,yellow);
fill((17,1)--(18,1)--(18,3)--(17,3)--cycle,yellow);
fill((23,1)--(24,1)--(24,3)--(23,3)--cycle,yellow);
fill((29,1)--(30,1)--(30,3)--(29,3)--cycle,yellow);
fill((35,1)--(36,1)--(36,3)--(35,3)--cycle,yellow);
fill((6,0)--(7,0)--(7,2)--(6,2)--cycle,green);
fill((13,0)--(14,0)--(14,3)--(13,3)--cycle,green);
fill((20,1)--(21,1)--(21,3)--(20,3)--cycle,green);
fill((27,1)--(28,1)--(28,3)--(27,3)--cycle,green);
fill((34,1)--(35,1)--(35,3)--(34,3)--cycle,green);
fill((42,3)--(41,3)--(41,2)--cycle,yellow);
fill((42,2)--(41,2)--(41,1)--cycle,yellow);
fill((42,3)--(42,2)--(41,2)--cycle,green);
fill((42,2)--(42,1)--(41,1)--cycle,green);
for (real i=9.5; i<=41.5; ++i) {
label("$"+string(i+0.5)+"$",(i,2.5),fontsize(9pt));
}
for (real i=0.5; i<=41.5; ++i) {
label("$"+string(i+42.5)+"$",(i,1.5),fontsize(9pt));
}
for (real i=0.5; i<=14.5; ++i) {
label("$"+string(i+84.5)+"$",(i,0.5),fontsize(9pt));
}
draw(circle((6.5,1.5),0.45)); draw(circle((6.5,0.5),0.45));
draw(circle((7.5,1.5),0.45)); draw(circle((7.5,0.5),0.45));
draw(circle((8.5,1.5),0.45)); draw(circle((8.5,0.5),0.45));
draw(circle((13.5,2.5),0.45)); draw(circle((13.5,1.5),0.45)); draw(circle((13.5,0.5),0.45));
draw(circle((14.5,2.5),0.45)); draw(circle((14.5,1.5),0.45)); draw(circle((14.5,0.5),0.45));
draw(circle((20.5,2.5),0.45)); draw(circle((20.5,1.5),0.45));
draw(circle((23.5,2.5),0.45)); draw(circle((23.5,1.5),0.45));
draw(circle((29.5,2.5),0.45)); draw(circle((29.5,1.5),0.45));
draw(circle((30.5,2.5),0.45)); draw(circle((30.5,1.5),0.45));
draw(circle((35.5,2.5),0.45)); draw(circle((35.5,1.5),0.45));
draw(circle((36.5,2.5),0.45)); draw(circle((36.5,1.5),0.45));
draw(circle((37.5,2.5),0.45)); draw(circle((37.5,1.5),0.45));
draw((9,3)--(42,3));
draw((0,2)--(42,2));
draw((0,1)--(42,1));
draw((0,0)--(15,0));
for (real i=0; i<9; ++i)
{
draw((i,2)--(i,0));
}
for (real i=9; i<16; ++i)
{
draw((i,3)--(i,0));
}
for (real i=16; i<=42; ++i)
{
draw((i,3)--(i,1));
}
// Block 2
/* Made by MRENTHUSIASM */
size(20cm);
for (real j=5.5; j<41.5; j+=6)
{
fill((j+0.5,4)--(j-0.5,4)--(j-0.5,3)--(j+0.5,3)--cycle,yellow);
}
for (real j=6.5; j<41.5; j+=7)
{
fill((j+0.5,4)--(j-0.5,4)--(j-0.5,3)--(j+0.5,3)--cycle,green);
}
fill((4,1)--(8,1)--(8,2)--(4,2)--cycle,mediumgray);
fill((33,1)--(37,1)--(37,2)--(33,2)--cycle,mediumgray);
fill((42,4)--(41,4)--(41,3)--cycle,yellow);
fill((42,4)--(42,3)--(41,3)--cycle,green);
for (real i=0.5; i<=41.5; ++i) {
label("$"+string(i+42.5)+"$",(i,3.5),fontsize(9pt));
}
draw(circle((6.5,3.5),0.45));
draw(circle((7.5,3.5),0.45));
draw(circle((8.5,3.5),0.45));
draw(circle((13.5,3.5),0.45));
draw(circle((14.5,3.5),0.45));
draw(circle((20.5,3.5),0.45));
draw(circle((23.5,3.5),0.45));
draw(circle((29.5,3.5),0.45));
draw(circle((30.5,3.5),0.45));
draw(circle((35.5,3.5),0.45));
draw(circle((36.5,3.5),0.45));
draw(circle((37.5,3.5),0.45));
label("Y",(3.5,2.5),blue); label("N",(4.5,2.5),blue);
label("N",(5.5,2.5),blue); label("N",(6.5,2.5),blue);
label("N",(4.5,1.5),blue); label("N",(5.5,1.5),blue);
label("N",(6.5,1.5),blue); label("N",(7.5,1.5),blue);
label("N",(5.5,0.5),blue); label("N",(6.5,0.5),blue);
label("N",(7.5,0.5),blue); label("Y",(8.5,0.5),blue);
label("Y",(10.5,2.5),blue); label("N",(11.5,2.5),blue);
label("N",(12.5,2.5),blue); label("N",(13.5,2.5),blue);
label("N",(11.5,1.5),blue); label("N",(12.5,1.5),blue);
label("N",(13.5,1.5),blue); label("Y",(14.5,1.5),blue);
label("Y",(17.5,2.5),blue); label("Y",(18.5,2.5),blue);
label("Y",(19.5,2.5),blue); label("N",(20.5,2.5),blue);
label("N",(20.5,1.5),blue); label("Y",(21.5,1.5),blue);
label("Y",(22.5,1.5),blue); label("Y",(23.5,1.5),blue);
label("Y",(26.5,2.5),blue); label("N",(27.5,2.5),blue);
label("N",(28.5,2.5),blue); label("N",(29.5,2.5),blue);
label("N",(27.5,1.5),blue); label("N",(28.5,1.5),blue);
label("N",(29.5,1.5),blue); label("Y",(30.5,1.5),blue);
label("Y",(32.5,2.5),blue); label("N",(33.5,2.5),blue);
label("N",(34.5,2.5),blue); label("N",(35.5,2.5),blue);
label("N",(33.5,1.5),blue); label("N",(34.5,1.5),blue);
label("N",(35.5,1.5),blue); label("N",(36.5,1.5),blue);
label("N",(34.5,0.5),blue); label("N",(35.5,0.5),blue);
label("N",(36.5,0.5),blue); label("Y",(37.5,0.5),blue);
for (real i=0; i<=42; ++i)
{
draw((i,4)--(i,3));
}
draw((0,4)--(42,4));
draw((0,3)--(42,3));
// Block 3
/* Made by MRENTHUSIASM */ size(20cm); fill((5,0)--(6,0)--(6,2)--(5,2)--cycle,yellow); fill((11,0)--(12,0)--(12,3)--(11,3)--cycle,yellow); fill((17,1)--(18,1)--(18,3)--(17,3)--cycle,yellow); fill((23,1)--(24,1)--(24,3)--(23,3)--cycle,yellow); fill((29,1)--(30,1)--(30,3)--(29,3)--cycle,yellow); fill((35,1)--(36,1)--(36,3)--(35,3)--cycle,yellow); fill((6,0)--(7,0)--(7,2)--(6,2)--cycle,green); fill((13,0)--(14,0)--(14,3)--(13,3)--cycle,green); fill((20,1)--(21,1)--(21,3)--(20,3)--cycle,green); fill((27,1)--(28,1)--(28,3)--(27,3)--cycle,green); fill((34,1)--(35,1)--(35,3)--(34,3)--cycle,green); fill((42,3)--(41,3)--(41,2)--cycle,yellow); fill((42,2)--(41,2)--(41,1)--cycle,yellow); fill((42,3)--(42,2)--(41,2)--cycle,green); fill((42,2)--(42,1)--(41,1)--cycle,green); for (real i=9.5; i<=41.5; ++i) { label("$"+string(i+0.5)+"$",(i,2.5),fontsize(9pt)); } for (real i=0.5; i<=41.5; ++i) { label("$"+string(i+42.5)+"$",(i,1.5),fontsize(9pt)); } for (real i=0.5; i<=14.5; ++i) { label("$"+string(i+84.5)+"$",(i,0.5),fontsize(9pt)); } draw(circle((6.5,1.5),0.45)); draw(circle((6.5,0.5),0.45)); draw(circle((7.5,1.5),0.45)); draw(circle((7.5,0.5),0.45)); draw(circle((8.5,1.5),0.45)); draw(circle((8.5,0.5),0.45)); draw(circle((13.5,2.5),0.45)); draw(circle((13.5,1.5),0.45)); draw(circle((13.5,0.5),0.45)); draw(circle((14.5,2.5),0.45)); draw(circle((14.5,1.5),0.45)); draw(circle((14.5,0.5),0.45)); draw(circle((20.5,2.5),0.45)); draw(circle((20.5,1.5),0.45)); draw(circle((23.5,2.5),0.45)); draw(circle((23.5,1.5),0.45)); draw(circle((29.5,2.5),0.45)); draw(circle((29.5,1.5),0.45)); draw(circle((30.5,2.5),0.45)); draw(circle((30.5,1.5),0.45)); draw(circle((35.5,2.5),0.45)); draw(circle((35.5,1.5),0.45)); draw(circle((36.5,2.5),0.45)); draw(circle((36.5,1.5),0.45)); draw(circle((37.5,2.5),0.45)); draw(circle((37.5,1.5),0.45)); draw((9,3)--(42,3)); draw((0,2)--(42,2)); draw((0,1)--(42,1)); draw((0,0)--(15,0)); for (real i=0; i<9; ++i) { draw((i,2)--(i,0)); } for (real i=9; i<16; ++i) { draw((i,3)--(i,0)); } for (real i=16; i<=42; ++i) { draw((i,3)--(i,1)); }
// Block 4
/* Made by MRENTHUSIASM */ size(20cm); for (real j=5.5; j<41.5; j+=6) { fill((j+0.5,4)--(j-0.5,4)--(j-0.5,3)--(j+0.5,3)--cycle,yellow); } for (real j=6.5; j<41.5; j+=7) { fill((j+0.5,4)--(j-0.5,4)--(j-0.5,3)--(j+0.5,3)--cycle,green); } fill((4,1)--(8,1)--(8,2)--(4,2)--cycle,mediumgray); fill((33,1)--(37,1)--(37,2)--(33,2)--cycle,mediumgray); fill((42,4)--(41,4)--(41,3)--cycle,yellow); fill((42,4)--(42,3)--(41,3)--cycle,green); for (real i=0.5; i<=41.5; ++i) { label("$"+string(i+42.5)+"$",(i,3.5),fontsize(9pt)); } draw(circle((6.5,3.5),0.45)); draw(circle((7.5,3.5),0.45)); draw(circle((8.5,3.5),0.45)); draw(circle((13.5,3.5),0.45)); draw(circle((14.5,3.5),0.45)); draw(circle((20.5,3.5),0.45)); draw(circle((23.5,3.5),0.45)); draw(circle((29.5,3.5),0.45)); draw(circle((30.5,3.5),0.45)); draw(circle((35.5,3.5),0.45)); draw(circle((36.5,3.5),0.45)); draw(circle((37.5,3.5),0.45)); label("Y",(3.5,2.5),blue); label("N",(4.5,2.5),blue); label("N",(5.5,2.5),blue); label("N",(6.5,2.5),blue); label("N",(4.5,1.5),blue); label("N",(5.5,1.5),blue); label("N",(6.5,1.5),blue); label("N",(7.5,1.5),blue); label("N",(5.5,0.5),blue); label("N",(6.5,0.5),blue); label("N",(7.5,0.5),blue); label("Y",(8.5,0.5),blue); label("Y",(10.5,2.5),blue); label("N",(11.5,2.5),blue); label("N",(12.5,2.5),blue); label("N",(13.5,2.5),blue); label("N",(11.5,1.5),blue); label("N",(12.5,1.5),blue); label("N",(13.5,1.5),blue); label("Y",(14.5,1.5),blue); label("Y",(17.5,2.5),blue); label("Y",(18.5,2.5),blue); label("Y",(19.5,2.5),blue); label("N",(20.5,2.5),blue); label("N",(20.5,1.5),blue); label("Y",(21.5,1.5),blue); label("Y",(22.5,1.5),blue); label("Y",(23.5,1.5),blue); label("Y",(26.5,2.5),blue); label("N",(27.5,2.5),blue); label("N",(28.5,2.5),blue); label("N",(29.5,2.5),blue); label("N",(27.5,1.5),blue); label("N",(28.5,1.5),blue); label("N",(29.5,1.5),blue); label("Y",(30.5,1.5),blue); label("Y",(32.5,2.5),blue); label("N",(33.5,2.5),blue); label("N",(34.5,2.5),blue); label("N",(35.5,2.5),blue); label("N",(33.5,1.5),blue); label("N",(34.5,1.5),blue); label("N",(35.5,1.5),blue); label("N",(36.5,1.5),blue); label("N",(34.5,0.5),blue); label("N",(35.5,0.5),blue); label("N",(36.5,0.5),blue); label("Y",(37.5,0.5),blue); for (real i=0; i<=42; ++i) { draw((i,4)--(i,3)); } draw((0,4)--(42,4)); draw((0,3)--(42,3)); | [] |
241 | A convex quadrilateral has area $30$ and side lengths $5, 6, 9,$ and $7,$ in that order. Denote by $\theta$ the measure of the acute angle formed by the diagonals of the quadrilateral. Then $\tan \theta$ can be written in the form $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$. | 2021 AIME II Problem 12 | Since we are asked to find $\tan \theta$, we can find $\sin \theta$ and $\cos \theta$ separately and use their values to get $\tan \theta$. We can start by drawing a diagram. Let the vertices of the quadrilateral be $A$, $B$, $C$, and $D$. Let $AB = 5$, $BC = 6$, $CD = 9$, and $DA = 7$. Let $AX = a$, $BX = b$, $CX = c$, and $DX = d$. We know that $\theta$ is the acute angle formed between the intersection of the diagonals $AC$ and $BD$.
We are given that the area of quadrilateral $ABCD$ is $30$. We can express this area using the areas of triangles $AXB$, $BXC$, $CXD$, and $DXA$. Since we want to find $\sin \theta$ and $\cos \theta$, we can represent these areas using $\sin \theta$ as follows:
\begin{align*} 30 &=[ABCD] \\ &=[AXB] + [BXC] + [CXD] + [DXA] \\ &=\frac{1}{2} ab \sin (\angle AXB) + \frac{1}{2} bc \sin (\angle BXC) + \frac{1}{2} cd \sin (\angle CXD) + \frac{1}{2} da \sin (\angle AXD) \\ &=\frac{1}{2} ab \sin (180^\circ - \theta) + \frac{1}{2} bc \sin (\theta) + \frac{1}{2} cd \sin (180^\circ - \theta) + \frac{1}{2} da \sin (\theta). \end{align*}
We know that $\sin (180^\circ - \theta) = \sin \theta$. Therefore it follows that:
\begin{align*} 30 &=\frac{1}{2} ab \sin (180^\circ - \theta) + \frac{1}{2} bc \sin (\theta) + \frac{1}{2} cd \sin (180^\circ - \theta) + \frac{1}{2} da \sin (\theta) \\ &=\frac{1}{2} ab \sin (\theta) + \frac{1}{2} bc \sin (\theta) + \frac{1}{2} cd \sin (\theta) + \frac{1}{2} da \sin (\theta) \\ &=\frac{1}{2}\sin\theta (ab + bc + cd + da). \end{align*}
From here we see that $\sin \theta = \frac{60}{ab + bc + cd + da}$. Now we need to find $\cos \theta$. Using the Law of Cosines on each of the four smaller triangles, we get following equations:
\begin{align*} 5^2 &= a^2 + b^2 - 2ab\cos(180^\circ-\theta), \\ 6^2 &= b^2 + c^2 - 2bc\cos \theta, \\ 9^2 &= c^2 + d^2 - 2cd\cos(180^\circ-\theta), \\ 7^2 &= d^2 + a^2 - 2da\cos \theta. \end{align*}
We know that $\cos (180^\circ - \theta) = -\cos \theta$ for all $\theta$. We can substitute this value into our equations to get:
\begin{align*} 5^2 &= a^2 + b^2 + 2ab\cos \theta, &&(1) \\ 6^2 &= b^2 + c^2 - 2bc\cos \theta, &&(2) \\ 9^2 &= c^2 + d^2 + 2cd\cos \theta, &&(3) \\ 7^2 &= d^2 + a^2 - 2da\cos \theta. &&(4) \end{align*}
If we subtract $(2)+(4)$ from $(1)+(3)$, the squared terms cancel, leaving us with:
\begin{align*} 5^2 + 9^2 - 6^2 - 7^2 &= 2ab \cos \theta + 2bc \cos \theta + 2cd \cos \theta + 2da \cos \theta \\ 21 &= 2\cos \theta (ab + bc + cd + da). \end{align*}
From here we see that $\cos \theta = \frac{21/2}{ab + bc + cd + da}$.
Since we have figured out $\sin \theta$ and $\cos \theta$, we can calculate $\tan \theta$: \[\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{60}{ab + bc + cd + da}}{\frac{21/2}{ab + bc + cd + da}} = \frac{60}{21/2} = \frac{120}{21} = \frac{40}{7}.\]
Therefore our answer is $40 + 7 = \boxed{047}$.
~ Steven Chen (www.professorchenedu.com)
~ my_aops_lessons | // Block 1
unitsize(4cm);
pair A,B,C,D,X;
A = (0,0);
B = (1,0);
C = (1.25,-1);
D = (-0.75,-0.75);
draw(A--B--C--D--cycle,black+1bp);
X = intersectionpoint(A--C,B--D);
draw(A--C);
draw(B--D);
label("$A$",A,NW);
label("$B$",B,NE);
label("$C$",C,SE);
label("$D$",D,SW);
dot(X);
label("$X$",X,S);
label("$5$",(A+B)/2,N);
label("$6$",(B+C)/2,E);
label("$9$",(C+D)/2,S);
label("$7$",(D+A)/2,W);
label("$\theta$",X,2.5E);
label("$a$",(A+X)/2,NE);
label("$b$",(B+X)/2,NW);
label("$c$",(C+X)/2,SW);
label("$d$",(D+X)/2,SE);
// Block 2
unitsize(4cm); pair A,B,C,D,X; A = (0,0); B = (1,0); C = (1.25,-1); D = (-0.75,-0.75); draw(A--B--C--D--cycle,black+1bp); X = intersectionpoint(A--C,B--D); draw(A--C); draw(B--D); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); dot(X); label("$X$",X,S); label("$5$",(A+B)/2,N); label("$6$",(B+C)/2,E); label("$9$",(C+D)/2,S); label("$7$",(D+A)/2,W); label("$\theta$",X,2.5E); label("$a$",(A+X)/2,NE); label("$b$",(B+X)/2,NW); label("$c$",(C+X)/2,SW); label("$d$",(D+X)/2,SE); | [] |
241 | A convex quadrilateral has area $30$ and side lengths $5, 6, 9,$ and $7,$ in that order. Denote by $\theta$ the measure of the acute angle formed by the diagonals of the quadrilateral. Then $\tan \theta$ can be written in the form $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$. | 2021 AIME II Problem 12 | In convex quadrilateral $ABCD,$ let $AB=5,BC=6,CD=9,$ and $DA=7.$ Let $A'$ and $C'$ be the feet of the perpendiculars from $A$ and $C,$ respectively, to $\overline{BD}.$ We obtain the following diagram:
Let $BC'=p,C'E=q,EA'=r,A'D=s,AA'=h_1,$ and $CC'=h_2.$ We apply the Pythagorean Theorem to right triangles $\triangle ABA',\triangle BCC',\triangle CDC',$ and $\triangle DAA',$ respectively:
\[\begin{array}{ccccccccccccccccc} (p+q+r)^2&+&h_1^2&=&5^2, &&&&&&&&&&&&\hspace{36mm}(1) \\ [1ex] p^2&+&h_2^2&=&6^2, &&&&&&&&&&&&\hspace{36mm}(2) \\ [1ex] (q+r+s)^2&+&h_2^2&=&9^2, &&&&&&&&&&&&\hspace{36mm}(3) \\ [1ex] s^2&+&h_1^2&=&7^2. &&&&&&&&&&&&\hspace{36mm}(4) \end{array}\]
Let the brackets denote areas. We get
\begin{align*} [ABD]+[CBD]&=[ABCD] \\ \frac12(p+q+r+s)h_1+\frac12(p+q+r+s)h_2&=30 \\ \frac12(p+q+r+s)(h_1+h_2)&=30 \\ (p+q+r+s)(h_1+h_2)&=60. \hspace{49.25mm}(5) \end{align*}
We subtract $(2)+(4)$ from $(1)+(3):$
\begin{align*} (p+q+r)^2+(q+r+s)^2-p^2-s^2&=21 \\ \left[(p+q+r)^2-s^2\right]+\left[(q+r+s)^2-p^2\right]&=21 \\ (p+q+r+s)(p+q+r-s)+(p+q+r+s)(-p+q+r+s)&=21 \\ (p+q+r+s)(2q+2r)&=21 \\ 2(p+q+r+s)(q+r)&=21 \\ (p+q+r+s)(q+r)&=\frac{21}{2}. \hspace{9.5mm}(6) \end{align*}
From right triangles $\triangle AEA'$ and $\triangle CEC',$ we have $\tan\theta=\frac{h_1}{r}=\frac{h_2}{q}.$ It follows that
\begin{alignat*}{8} \tan\theta&=\frac{h_1}{r}\qquad&\implies\qquad h_1&=r\tan\theta, \hspace{64mm}&(1\star)\\ \tan\theta&=\frac{h_2}{q}\qquad&\implies\qquad h_2&=q\tan\theta. &(2\star) \end{alignat*}
Finally, we divide $(5)$ by $(6):$
\begin{align*} \frac{h_1+h_2}{q+r}&=\frac{40}{7} \\ \frac{r\tan\theta+q\tan\theta}{q+r}&=\frac{40}{7} \hspace{15mm} &&\text{by }(1\star)\text{ and }(2\star)\\ \frac{(r+q)\tan\theta}{q+r}&=\frac{40}{7} \\ \tan\theta&=\frac{40}{7}, \end{align*}
from which the answer is $40+7=\boxed{047}.$
~MRENTHUSIASM | // Block 1
/* Made by MRENTHUSIASM */
size(500);
pair A, B, C, D, P, A1, C1;
B = origin;
D = (3*sqrt(32498*(29400*sqrt(47)+312523))/32498,0);
A = intersectionpoints(Circle(B,5),Circle(D,7))[0];
C = intersectionpoints(Circle(B,6),Circle(D,9))[1];
P = intersectionpoint(A--C,B--D);
A1 = foot(A,B,D);
C1 = foot(C,B,D);
markscalefactor=3/160;
draw(rightanglemark(A,A1,D),red);
draw(rightanglemark(C,C1,B),red);
dot("$A$",A,1.5*dir(aCos(7/sqrt(1649))));
dot("$B$",B,1.5*W);
dot("$C$",C,1.5*dir(180+aCos(7/sqrt(1649))));
dot("$D$",D,1.5*E);
dot("$E$",P,dir(180-(180-aCos(7/sqrt(1649)))/2));
dot("$A'$",A1,dir(-75));
dot("$C'$",C1,N);
label("$\theta$",P,dir(180+aCos(7/sqrt(1649))/2),red);
draw(A--A1^^C--C1,dashed);
draw(A--B--C--D--cycle^^A--C^^B--D);
// Block 2
/* Made by MRENTHUSIASM */ size(500); pair A, B, C, D, P, A1, C1; B = origin; D = (3*sqrt(32498*(29400*sqrt(47)+312523))/32498,0); A = intersectionpoints(Circle(B,5),Circle(D,7))[0]; C = intersectionpoints(Circle(B,6),Circle(D,9))[1]; P = intersectionpoint(A--C,B--D); A1 = foot(A,B,D); C1 = foot(C,B,D); markscalefactor=3/160; draw(rightanglemark(A,A1,D),red); draw(rightanglemark(C,C1,B),red); dot("$A$",A,1.5*dir(aCos(7/sqrt(1649)))); dot("$B$",B,1.5*W); dot("$C$",C,1.5*dir(180+aCos(7/sqrt(1649)))); dot("$D$",D,1.5*E); dot("$E$",P,dir(180-(180-aCos(7/sqrt(1649)))/2)); dot("$A'$",A1,dir(-75)); dot("$C'$",C1,N); label("$\theta$",P,dir(180+aCos(7/sqrt(1649))/2),red); draw(A--A1^^C--C1,dashed); draw(A--B--C--D--cycle^^A--C^^B--D); | [] |
241 | A convex quadrilateral has area $30$ and side lengths $5, 6, 9,$ and $7,$ in that order. Denote by $\theta$ the measure of the acute angle formed by the diagonals of the quadrilateral. Then $\tan \theta$ can be written in the form $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$. | 2021 AIME II Problem 12 | Bretschneider's Formula
Given quadrilateral $ABCD$, let, $a, b, c, d$, be the sides, $s$ the semiperimeter, and $u, v$, the diagonals. Then the area, $K$, is given by \[K = \tfrac 14 \sqrt {4u^2v^2-(b^2+d^2-a^2-c^2)^2}\]
Solution
By Bretschneider's Formula, \[30=\tfrac{1}{4}\sqrt{4u^2v^2-(b^2+d^2-a^2-c^2)^2}=\tfrac{1}{4}\sqrt{4u^2v^2-441}.\] Thus, $uv=\tfrac{3\sqrt{1649}}{2}$. Also, \[[ABCD]=\tfrac 12 \cdot uv\sin{\theta};\] solving for $\sin{\theta}$ yields $\sin{\theta}=\tfrac{40}{\sqrt{1649}}$. Since $\theta$ is acute, $\cos{\theta}$ is positive, from which $\cos{\theta}=\tfrac{7}{\sqrt{1649}}$. Solving for $\tan{\theta}$ yields \[\tan{\theta}=\frac{\sin{\theta}}{\cos{\theta}}=\frac{40}{7},\] for a final answer of $\boxed{047}$.
~ Leo.Euler | size(200); import olympiad; defaultpen(linewidth(0.8)+fontsize(10)); pair A,B,C,D; A=origin; B=(1.25,0); D=dir(65); C=D+0.85*dir(90)*(A-D); draw(A--B--C--D--cycle); draw(A--C^^B--D, gray+0.5); dot("$A$", A, SW); dot("$B$", B, SE); dot("$C$", C, NE); dot("$D$",D,NW); label("$a$", A--B, S); label("$b$", B--C, E); label("$c$", D--C, N); label("$d$",D--A,W); label("$u$",D--B,2*dir(170)); label("$v$",A--C,S); | [] |
242 | Let $\Delta ABC$ be an acute triangle with circumcenter $O$ and centroid $G$. Let $X$ be the intersection of the line tangent to the circumcircle of $\Delta ABC$ at $A$ and the line perpendicular to $GO$ at $G$. Let $Y$ be the intersection of lines $XG$ and $BC$. Given that the measures of $\angle ABC, \angle BCA,$ and $\angle XOY$ are in the ratio $13 : 2 : 17,$ the degree measure of $\angle BAC$ can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2021 AIME II Problem 14 | $\angle OAX = \angle OGX = 90^\circ \implies$ quadrilateral $XAGO$ is cyclic $\implies$
$\angle GXO = \angle GAO,$ as they share the same intersept $\overset{\Large\frown} {GO}.$
$\angle OGY = \angle OMY = 90^\circ \implies$ quadrilateral $OGYM$ is cyclic $\implies$
$\angle GYO = \angle OMG,$ as they share the same intercept $\overset{\Large\frown} {GO}.$
In triangles $\triangle XOY$ and $\triangle AOM,$ two pairs of angles are equal, which means that the third angles $\angle XOY = \angle AOM$ are also equal.
$\angle ABC : \angle BCA : \angle AOM = 13 : 2 : 17,$ so $\angle AOM = \angle ABC + 2 \angle BCA.$
According to the Claim, $\triangle ABC$ is isosceles,
\[\angle ABC : \angle BCA : \angle BAC = 13 : 2 : 13.\]
\[\angle BAC = \frac{13} {13 + 2 + 13} \cdot 180^\circ = \frac {585^\circ}{7} \implies 585 + 7 = \boxed{592}.\]
[Image: images/aime/2021_AIME_II_Problem_14_0.png]
Claim
Let $\triangle ABC$ be an acute triangle with circumcenter $O.$
Let $M$ be the midpoint of $BC$ so $MO\perp BC.$
If $\angle AOM = 2\angle ACB + \angle ABC,$ then $AC = BC.$
We define $\angle AOM$ as the sum of $\angle AOB + \angle BOM,$ this angle can be greater than $180^\circ.$
Proof
$\angle BAC = \angle BOM$ as they share the same intercept $\overset{\Large\frown} {BC}$ (an inscribed angle and half of central angle).
$\angle AOB = 2\angle ACB$ as they share the same intercept $\overset{\Large\frown} {AB}.$
\[\angle AOM = \angle AOB + \angle BOM = 2 \angle ACB + \angle CAB.\]
If $\angle AOM = 2 \angle ACB + \angle ABC,$ then $\angle ABC = \angle CAB, AC = BC.$
vladimir.shelomovskii@gmail.com, vvsss | // Block 1
/* Made by MRENTHUSIASM */
size(375);
pair A, B, C, O, G, X, Y;
A = origin;
B = (1,0);
C = extension(A,A+10*dir(585/7),B,B+10*dir(180-585/7));
O = circumcenter(A,B,C);
G = centroid(A,B,C);
Y = intersectionpoint(G--G+(100,0),B--C);
X = intersectionpoint(G--G-(100,0),A--scale(100)*rotate(90)*dir(O-A));
pair O1=circumcenter(O,G,A);
real r1=length(O1-O);
markscalefactor=3/160;
filldraw(O--X--Y--cycle, rgb(255,255,0));
draw(rightanglemark(O,G,X),red);
draw(A--O--B,fuchsia+0.4);
draw(Arc(O1,r1,-40,50),royalblue+0.5);
draw(circumcircle(O,G,Y), heavygreen+0.5);
dot("$A$",A,1.5*dir(180+585/7),linewidth(4));
dot("$B$",B,1.5*dir(-585/7),linewidth(4));
dot("$C$",C,1.5N,linewidth(4));
dot("$O$",O,1.5N,linewidth(4));
dot("$G$",G,1.5S,linewidth(4));
dot("$Y$",Y,1.5E,linewidth(4));
dot("$X$",X,1.5W,linewidth(4));
draw(A--B--C--cycle^^X--O--Y--cycle^^A--X^^O--G^^circumcircle(A,B,C));
// Block 2
/* Made by MRENTHUSIASM */ size(375); pair A, B, C, O, G, X, Y; A = origin; B = (1,0); C = extension(A,A+10*dir(585/7),B,B+10*dir(180-585/7)); O = circumcenter(A,B,C); G = centroid(A,B,C); Y = intersectionpoint(G--G+(100,0),B--C); X = intersectionpoint(G--G-(100,0),A--scale(100)*rotate(90)*dir(O-A)); pair O1=circumcenter(O,G,A); real r1=length(O1-O); markscalefactor=3/160; filldraw(O--X--Y--cycle, rgb(255,255,0)); draw(rightanglemark(O,G,X),red); draw(A--O--B,fuchsia+0.4); draw(Arc(O1,r1,-40,50),royalblue+0.5); draw(circumcircle(O,G,Y), heavygreen+0.5); dot("$A$",A,1.5*dir(180+585/7),linewidth(4)); dot("$B$",B,1.5*dir(-585/7),linewidth(4)); dot("$C$",C,1.5N,linewidth(4)); dot("$O$",O,1.5N,linewidth(4)); dot("$G$",G,1.5S,linewidth(4)); dot("$Y$",Y,1.5E,linewidth(4)); dot("$X$",X,1.5W,linewidth(4)); draw(A--B--C--cycle^^X--O--Y--cycle^^A--X^^O--G^^circumcircle(A,B,C)); | ["https://artofproblemsolving.com/wiki/images/thumb/6/6c/AIME-II-2021-14.png/230px-AIME-II-2021-14.png"] |
243 | In isosceles trapezoid $ABCD$, parallel bases $\overline{AB}$ and $\overline{CD}$ have lengths $500$ and $650$, respectively, and $AD=BC=333$. The angle bisectors of $\angle{A}$ and $\angle{D}$ meet at $P$, and the angle bisectors of $\angle{B}$ and $\angle{C}$ meet at $Q$. Find $PQ$. | 2022 AIME I Problem 3 | We have the following diagram:
Let $X$ and $W$ be the points where $AP$ and $BQ$ extend to meet $CD$, and $YZ$ be the height of $\triangle AZB$. As proven in Solution 2, triangles $APD$ and $DPW$ are congruent right triangles. Therefore, $AD = DW = 333$. We can apply this logic to triangles $BCQ$ and $XCQ$ as well, giving us $BC = CX = 333$. Since $CD = 650$, $XW = DW + CX - CD = 16$.
Additionally, we can see that $\triangle XZW$ is similar to $\triangle PQZ$ and $\triangle AZB$. We know that $\frac{XW}{AB} = \frac{16}{500}$. So, we can say that the height of the triangle $AZB$ is $500u$ while the height of the triangle $XZW$ is $16u$. After that, we can figure out the distance from $Y$ to $PQ: \frac{500+16}{2} = 258u$ and the height of triangle $PZQ: 500-258 = 242u$.
Finally, since the ratio between the height of $PZQ$ to the height of $AZB$ is $242:500$ and $AB$ is $500$, $PQ = \boxed{242}.$
~Cytronical | // Block 1
/* Made by MRENTHUSIASM , modified by Cytronical */
size(300);
pair A, B, C, D, A1, B1, C1, D1, P, Q, X, Y, Z, W;
A = (-250,6*sqrt(731));
B = (250,6*sqrt(731));
C = (325,-6*sqrt(731));
D = (-325,-6*sqrt(731));
A1 = bisectorpoint(B,A,D);
B1 = bisectorpoint(A,B,C);
C1 = bisectorpoint(B,C,D);
D1 = bisectorpoint(A,D,C);
P = intersectionpoint(A--300*(A1-A)+A,D--300*(D1-D)+D);
Q = intersectionpoint(B--300*(B1-B)+B,C--300*(C1-C)+C);
X = intersectionpoint(B--5*(Q-B)+B,C--D);
Y = (0,6*sqrt(731));
Z = intersectionpoint(A--4*(P-A)+A,B--4*(Q-B)+B);
W = intersectionpoint(A--5*(P-A)+A,C--D);
draw(anglemark(P,A,B,1000),red);
draw(anglemark(D,A,P,1000),red);
draw(anglemark(A,B,Q,1000),red);
draw(anglemark(Q,B,C,1000),red);
draw(anglemark(P,D,A,1000),red);
draw(anglemark(C,D,P,1000),red);
draw(anglemark(Q,C,D,1000),red);
draw(anglemark(B,C,Q,1000),red);
add(pathticks(anglemark(P,A,B,1000), n = 1, r = 0.15, s = 750, red));
add(pathticks(anglemark(D,A,P,1000), n = 1, r = 0.15, s = 750, red));
add(pathticks(anglemark(A,B,Q,1000), n = 1, r = 0.15, s = 750, red));
add(pathticks(anglemark(Q,B,C,1000), n = 1, r = 0.15, s = 750, red));
add(pathticks(anglemark(P,D,A,1000), n = 2, r = 0.12, spacing = 150, s = 750, red));
add(pathticks(anglemark(C,D,P,1000), n = 2, r = 0.12, spacing = 150, s = 750, red));
add(pathticks(anglemark(Q,C,D,1000), n = 2, r = 0.12, spacing = 150, s = 750, red));
add(pathticks(anglemark(B,C,Q,1000), n = 2, r = 0.12, spacing = 150, s = 750, red));
dot("$A$",A,1.5*dir(A),linewidth(4));
dot("$B$",B,1.5*dir(B),linewidth(4));
dot("$C$",C,1.5*dir(C),linewidth(4));
dot("$D$",D,1.5*dir(D),linewidth(4));
dot("$P$",P,1.5*(-1,0),linewidth(4));
dot("$Q$",Q,1.5*E,linewidth(4));
dot("$X$",X,1.5*dir(-105),linewidth(4));
dot("$Y$",Y,1.5*N,linewidth(4));
dot("$Z$",Z,4.5*dir(75),linewidth(4));
dot("$W$",W,1.5*dir(-75),linewidth(4));
draw(A--B--C--D--cycle^^A--P--D^^B--Q--C^^P--Q);
draw(P--W^^Q--X^^Y--Z,dashed);
// Block 2
/* Made by MRENTHUSIASM , modified by Cytronical */ size(300); pair A, B, C, D, A1, B1, C1, D1, P, Q, X, Y, Z, W; A = (-250,6*sqrt(731)); B = (250,6*sqrt(731)); C = (325,-6*sqrt(731)); D = (-325,-6*sqrt(731)); A1 = bisectorpoint(B,A,D); B1 = bisectorpoint(A,B,C); C1 = bisectorpoint(B,C,D); D1 = bisectorpoint(A,D,C); P = intersectionpoint(A--300*(A1-A)+A,D--300*(D1-D)+D); Q = intersectionpoint(B--300*(B1-B)+B,C--300*(C1-C)+C); X = intersectionpoint(B--5*(Q-B)+B,C--D); Y = (0,6*sqrt(731)); Z = intersectionpoint(A--4*(P-A)+A,B--4*(Q-B)+B); W = intersectionpoint(A--5*(P-A)+A,C--D); draw(anglemark(P,A,B,1000),red); draw(anglemark(D,A,P,1000),red); draw(anglemark(A,B,Q,1000),red); draw(anglemark(Q,B,C,1000),red); draw(anglemark(P,D,A,1000),red); draw(anglemark(C,D,P,1000),red); draw(anglemark(Q,C,D,1000),red); draw(anglemark(B,C,Q,1000),red); add(pathticks(anglemark(P,A,B,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(D,A,P,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(A,B,Q,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(Q,B,C,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(P,D,A,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(C,D,P,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(Q,C,D,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(B,C,Q,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$D$",D,1.5*dir(D),linewidth(4)); dot("$P$",P,1.5*(-1,0),linewidth(4)); dot("$Q$",Q,1.5*E,linewidth(4)); dot("$X$",X,1.5*dir(-105),linewidth(4)); dot("$Y$",Y,1.5*N,linewidth(4)); dot("$Z$",Z,4.5*dir(75),linewidth(4)); dot("$W$",W,1.5*dir(-75),linewidth(4)); draw(A--B--C--D--cycle^^A--P--D^^B--Q--C^^P--Q); draw(P--W^^Q--X^^Y--Z,dashed); | [] |
243 | In isosceles trapezoid $ABCD$, parallel bases $\overline{AB}$ and $\overline{CD}$ have lengths $500$ and $650$, respectively, and $AD=BC=333$. The angle bisectors of $\angle{A}$ and $\angle{D}$ meet at $P$, and the angle bisectors of $\angle{B}$ and $\angle{C}$ meet at $Q$. Find $PQ$. | 2022 AIME I Problem 3 | Extend line $PQ$ to meet $AD$ at $P'$ and $BC$ at $Q'$. The diagram looks like this:
Because the trapezoid is isosceles, by symmetry $PQ$ is parallel to $AB$ and $CD$. Therefore, $\angle PAB \cong \angle APP'$ by interior angles and $\angle PAB \cong \angle PAD$ by the problem statement. Thus, $\triangle P'AP$ is isosceles with $P'P = P'A$. By symmetry, $P'DP$ is also isosceles, and thus $P'A = \frac{AD}{2}$. Similarly, the same thing is happening on the right side of the trapezoid, and thus $P'Q'$ is the midline of the trapezoid. Then, $PQ = P'Q' - (P'P + Q'Q)$.
Since $P'P = P'A = \frac{AD}{2}, Q'Q = Q'B = \frac{BC}{2}$ and $AD = BC = 333$, we have $P'P + Q'Q = \frac{333}{2} + \frac{333}{2} = 333$. The length of the midline of a trapezoid is the average of their bases, so $P'Q' = \frac{500+650}{2} = 575$. Finally, $PQ = 575 - 333 = \boxed{242}$.
~KingRavi | // Block 1
/* Made by MRENTHUSIASM */
size(300);
pair A, B, C, D, A1, B1, C1, D1, P, Q, P1, Q1;
A = (-250,6*sqrt(731));
B = (250,6*sqrt(731));
C = (325,-6*sqrt(731));
D = (-325,-6*sqrt(731));
A1 = bisectorpoint(B,A,D);
B1 = bisectorpoint(A,B,C);
C1 = bisectorpoint(B,C,D);
D1 = bisectorpoint(A,D,C);
P = intersectionpoint(A--300*(A1-A)+A,D--300*(D1-D)+D);
Q = intersectionpoint(B--300*(B1-B)+B,C--300*(C1-C)+C);
P1 = intersectionpoint(A--D,P--(-300)*(Q-P)+P);
Q1 = intersectionpoint(B--C,Q--300*(Q-P)+Q);
draw(anglemark(P,A,B,1000),red);
draw(anglemark(D,A,P,1000),red);
draw(anglemark(A,B,Q,1000),red);
draw(anglemark(Q,B,C,1000),red);
draw(anglemark(P,D,A,1000),red);
draw(anglemark(C,D,P,1000),red);
draw(anglemark(Q,C,D,1000),red);
draw(anglemark(B,C,Q,1000),red);
add(pathticks(anglemark(P,A,B,1000), n = 1, r = 0.15, s = 750, red));
add(pathticks(anglemark(D,A,P,1000), n = 1, r = 0.15, s = 750, red));
add(pathticks(anglemark(A,B,Q,1000), n = 1, r = 0.15, s = 750, red));
add(pathticks(anglemark(Q,B,C,1000), n = 1, r = 0.15, s = 750, red));
add(pathticks(anglemark(P,D,A,1000), n = 2, r = 0.12, spacing = 150, s = 750, red));
add(pathticks(anglemark(C,D,P,1000), n = 2, r = 0.12, spacing = 150, s = 750, red));
add(pathticks(anglemark(Q,C,D,1000), n = 2, r = 0.12, spacing = 150, s = 750, red));
add(pathticks(anglemark(B,C,Q,1000), n = 2, r = 0.12, spacing = 150, s = 750, red));
dot("$A$",A,1.5*dir(A),linewidth(4));
dot("$B$",B,1.5*dir(B),linewidth(4));
dot("$C$",C,1.5*dir(C),linewidth(4));
dot("$D$",D,1.5*dir(D),linewidth(4));
dot("$P$",P,1.5*NE,linewidth(4));
dot("$Q$",Q,1.5*NW,linewidth(4));
dot("$P'$",P1,1.5*W,linewidth(4));
dot("$Q'$",Q1,1.5*E,linewidth(4));
draw(A--B--C--D--cycle^^A--P--D^^B--Q--C^^P--Q);
draw(P--P1^^Q--Q1,dashed);
// Block 2
/* Made by MRENTHUSIASM */ size(300); pair A, B, C, D, A1, B1, C1, D1, P, Q, P1, Q1; A = (-250,6*sqrt(731)); B = (250,6*sqrt(731)); C = (325,-6*sqrt(731)); D = (-325,-6*sqrt(731)); A1 = bisectorpoint(B,A,D); B1 = bisectorpoint(A,B,C); C1 = bisectorpoint(B,C,D); D1 = bisectorpoint(A,D,C); P = intersectionpoint(A--300*(A1-A)+A,D--300*(D1-D)+D); Q = intersectionpoint(B--300*(B1-B)+B,C--300*(C1-C)+C); P1 = intersectionpoint(A--D,P--(-300)*(Q-P)+P); Q1 = intersectionpoint(B--C,Q--300*(Q-P)+Q); draw(anglemark(P,A,B,1000),red); draw(anglemark(D,A,P,1000),red); draw(anglemark(A,B,Q,1000),red); draw(anglemark(Q,B,C,1000),red); draw(anglemark(P,D,A,1000),red); draw(anglemark(C,D,P,1000),red); draw(anglemark(Q,C,D,1000),red); draw(anglemark(B,C,Q,1000),red); add(pathticks(anglemark(P,A,B,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(D,A,P,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(A,B,Q,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(Q,B,C,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(P,D,A,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(C,D,P,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(Q,C,D,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(B,C,Q,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$D$",D,1.5*dir(D),linewidth(4)); dot("$P$",P,1.5*NE,linewidth(4)); dot("$Q$",Q,1.5*NW,linewidth(4)); dot("$P'$",P1,1.5*W,linewidth(4)); dot("$Q'$",Q1,1.5*E,linewidth(4)); draw(A--B--C--D--cycle^^A--P--D^^B--Q--C^^P--Q); draw(P--P1^^Q--Q1,dashed); | [] |
243 | In isosceles trapezoid $ABCD$, parallel bases $\overline{AB}$ and $\overline{CD}$ have lengths $500$ and $650$, respectively, and $AD=BC=333$. The angle bisectors of $\angle{A}$ and $\angle{D}$ meet at $P$, and the angle bisectors of $\angle{B}$ and $\angle{C}$ meet at $Q$. Find $PQ$. | 2022 AIME I Problem 3 | We have the following diagram:
Extend lines $AP$ and $BQ$ to meet line $DC$ at points $W$ and $X$, respectively, and extend lines $DP$ and $CQ$ to meet $AB$ at points $Z$ and $Y$, respectively.
Claim: quadrilaterals $AZWD$ and $BYXC$ are rhombuses.
Proof: Since $\angle DAB + \angle ADC = 180^{\circ}$, $\angle ADP + \angle PAD = 90^{\circ}$. Therefore, triangles $APD$, $APZ$, $DPW$ and $PZW$ are all right triangles. By SAA congruence, the first three triangles are congruent; by SAS congruence, $\triangle PZW$ is congruent to the other three. Therefore, $AD = DW = WZ = AZ$, so $AZWD$ is a rhombus. By symmetry, $BYXC$ is also a rhombus.
Extend line $PQ$ to meet $\overline{AD}$ and $\overline{BC}$ at $R$ and $S$, respectively. Because of rhombus properties, $RP = QS = \frac{333}{2}$. Also, by rhombus properties, $R$ and $S$ are the midpoints of segments $AD$ and $BC$, respectively; therefore, by trapezoid properties, $RS = \frac{AB + CD}{2} = 575$. Finally, $PQ = RS - RP - QS = \boxed{242}$.
~ihatemath123 | // Block 1
/* Made by MRENTHUSIASM */
size(300);
pair A, B, C, D, A1, B1, C1, D1, P, Q, X, Y, Z, W;
A = (-250,6*sqrt(731));
B = (250,6*sqrt(731));
C = (325,-6*sqrt(731));
D = (-325,-6*sqrt(731));
A1 = bisectorpoint(B,A,D);
B1 = bisectorpoint(A,B,C);
C1 = bisectorpoint(B,C,D);
D1 = bisectorpoint(A,D,C);
P = intersectionpoint(A--300*(A1-A)+A,D--300*(D1-D)+D);
Q = intersectionpoint(B--300*(B1-B)+B,C--300*(C1-C)+C);
X = intersectionpoint(B--5*(Q-B)+B,C--D);
Y = intersectionpoint(C--5*(Q-C)+C,A--B);
Z = intersectionpoint(D--5*(P-D)+D,A--B);
W = intersectionpoint(A--5*(P-A)+A,C--D);
draw(anglemark(P,A,B,1000),red);
draw(anglemark(D,A,P,1000),red);
draw(anglemark(A,B,Q,1000),red);
draw(anglemark(Q,B,C,1000),red);
draw(anglemark(P,D,A,1000),red);
draw(anglemark(C,D,P,1000),red);
draw(anglemark(Q,C,D,1000),red);
draw(anglemark(B,C,Q,1000),red);
add(pathticks(anglemark(P,A,B,1000), n = 1, r = 0.15, s = 750, red));
add(pathticks(anglemark(D,A,P,1000), n = 1, r = 0.15, s = 750, red));
add(pathticks(anglemark(A,B,Q,1000), n = 1, r = 0.15, s = 750, red));
add(pathticks(anglemark(Q,B,C,1000), n = 1, r = 0.15, s = 750, red));
add(pathticks(anglemark(P,D,A,1000), n = 2, r = 0.12, spacing = 150, s = 750, red));
add(pathticks(anglemark(C,D,P,1000), n = 2, r = 0.12, spacing = 150, s = 750, red));
add(pathticks(anglemark(Q,C,D,1000), n = 2, r = 0.12, spacing = 150, s = 750, red));
add(pathticks(anglemark(B,C,Q,1000), n = 2, r = 0.12, spacing = 150, s = 750, red));
dot("$A$",A,1.5*dir(A),linewidth(4));
dot("$B$",B,1.5*dir(B),linewidth(4));
dot("$C$",C,1.5*dir(C),linewidth(4));
dot("$D$",D,1.5*dir(D),linewidth(4));
dot("$P$",P,1.5*(-1,0),linewidth(4));
dot("$Q$",Q,1.5*E,linewidth(4));
dot("$X$",X,1.5*dir(-105),linewidth(4));
dot("$Y$",Y,1.5*N,linewidth(4));
dot("$Z$",Z,1.5*N,linewidth(4));
dot("$W$",W,1.5*dir(-75),linewidth(4));
draw(A--B--C--D--cycle^^A--P--D^^B--Q--C^^P--Q);
draw(P--Z^^P--W^^Q--X^^Q--Y,dashed);
// Block 2
/* Made by MRENTHUSIASM */ size(300); pair A, B, C, D, A1, B1, C1, D1, P, Q, X, Y, Z, W; A = (-250,6*sqrt(731)); B = (250,6*sqrt(731)); C = (325,-6*sqrt(731)); D = (-325,-6*sqrt(731)); A1 = bisectorpoint(B,A,D); B1 = bisectorpoint(A,B,C); C1 = bisectorpoint(B,C,D); D1 = bisectorpoint(A,D,C); P = intersectionpoint(A--300*(A1-A)+A,D--300*(D1-D)+D); Q = intersectionpoint(B--300*(B1-B)+B,C--300*(C1-C)+C); X = intersectionpoint(B--5*(Q-B)+B,C--D); Y = intersectionpoint(C--5*(Q-C)+C,A--B); Z = intersectionpoint(D--5*(P-D)+D,A--B); W = intersectionpoint(A--5*(P-A)+A,C--D); draw(anglemark(P,A,B,1000),red); draw(anglemark(D,A,P,1000),red); draw(anglemark(A,B,Q,1000),red); draw(anglemark(Q,B,C,1000),red); draw(anglemark(P,D,A,1000),red); draw(anglemark(C,D,P,1000),red); draw(anglemark(Q,C,D,1000),red); draw(anglemark(B,C,Q,1000),red); add(pathticks(anglemark(P,A,B,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(D,A,P,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(A,B,Q,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(Q,B,C,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(P,D,A,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(C,D,P,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(Q,C,D,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(B,C,Q,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$D$",D,1.5*dir(D),linewidth(4)); dot("$P$",P,1.5*(-1,0),linewidth(4)); dot("$Q$",Q,1.5*E,linewidth(4)); dot("$X$",X,1.5*dir(-105),linewidth(4)); dot("$Y$",Y,1.5*N,linewidth(4)); dot("$Z$",Z,1.5*N,linewidth(4)); dot("$W$",W,1.5*dir(-75),linewidth(4)); draw(A--B--C--D--cycle^^A--P--D^^B--Q--C^^P--Q); draw(P--Z^^P--W^^Q--X^^Q--Y,dashed); | [] |
243 | In isosceles trapezoid $ABCD$, parallel bases $\overline{AB}$ and $\overline{CD}$ have lengths $500$ and $650$, respectively, and $AD=BC=333$. The angle bisectors of $\angle{A}$ and $\angle{D}$ meet at $P$, and the angle bisectors of $\angle{B}$ and $\angle{C}$ meet at $Q$. Find $PQ$. | 2022 AIME I Problem 3 | Let $X$ and $Y$ be the feet of the altitudes from $P$ and $Q$, respectively, to $AB$, and let $Z$ and $W$ be the feet of the altitudes from $P$ and $Q$, respectively, to $CD$. Side $AB$ is parallel to side $CD$, so $XYWZ$ is a rectangle with width $PQ$. Furthermore, because $CD - AB = 650-500 = 150$ and trapezoid $ABCD$ is isosceles, $WC - YB = ZD - XA = 75$.
Also because $ABCD$ is isosceles, $\angle ABC + \angle BCD$ is half the total sum of angles in $ABCD$, or $180^{\circ}$. Since $BQ$ and $CQ$ bisect $\angle ABC$ and $\angle BCD$, respectively, we have $\angle QBC + \angle QCB = 90^{\circ}$, so $\angle BQC = 90^{\circ}$.
Letting $BQ = 333k$, applying Pythagoras to $\triangle BQC$ yields $QC = 333\sqrt{1-k^2}$. We then proceed using similar triangles: $\angle BYQ = \angle BQC = 90^{\circ}$ and $\angle YBQ = \angle QBC$, so by AA similarity $YB = 333k^2$. Likewise, $\angle CWQ = \angle BQC = 90^{\circ}$ and $\angle WCQ = \angle QCB$, so by AA similarity $WC = 333(1 - k^2)$. Thus $WC + YB = 333$.
Adding our two equations for $WC$ and $YB$ gives $2WC = 75 + 333 = 408$. Therefore, the answer is $PQ = ZW = CD - 2WC = 650 - 408 = \boxed{242}$.
~Orange_Quail_9 | // Block 1
/* Made by MRENTHUSIASM */
size(300);
pair A, B, C, D, A1, B1, C1, D1, P, Q, X, Y, Z, W;
A = (-250,6*sqrt(731));
B = (250,6*sqrt(731));
C = (325,-6*sqrt(731));
D = (-325,-6*sqrt(731));
A1 = bisectorpoint(B,A,D);
B1 = bisectorpoint(A,B,C);
C1 = bisectorpoint(B,C,D);
D1 = bisectorpoint(A,D,C);
P = intersectionpoint(A--300*(A1-A)+A,D--300*(D1-D)+D);
Q = intersectionpoint(B--300*(B1-B)+B,C--300*(C1-C)+C);
draw(anglemark(P,A,B,1000),red);
draw(anglemark(D,A,P,1000),red);
draw(anglemark(A,B,Q,1000),red);
draw(anglemark(Q,B,C,1000),red);
draw(anglemark(P,D,A,1000),red);
draw(anglemark(C,D,P,1000),red);
draw(anglemark(Q,C,D,1000),red);
draw(anglemark(B,C,Q,1000),red);
add(pathticks(anglemark(P,A,B,1000), n = 1, r = 0.15, s = 750, red));
add(pathticks(anglemark(D,A,P,1000), n = 1, r = 0.15, s = 750, red));
add(pathticks(anglemark(A,B,Q,1000), n = 1, r = 0.15, s = 750, red));
add(pathticks(anglemark(Q,B,C,1000), n = 1, r = 0.15, s = 750, red));
add(pathticks(anglemark(P,D,A,1000), n = 2, r = 0.12, spacing = 150, s = 750, red));
add(pathticks(anglemark(C,D,P,1000), n = 2, r = 0.12, spacing = 150, s = 750, red));
add(pathticks(anglemark(Q,C,D,1000), n = 2, r = 0.12, spacing = 150, s = 750, red));
add(pathticks(anglemark(B,C,Q,1000), n = 2, r = 0.12, spacing = 150, s = 750, red));
dot("$A$",A,1.5*dir(A),linewidth(4));
dot("$B$",B,1.5*dir(B),linewidth(4));
dot("$C$",C,1.5*dir(C),linewidth(4));
dot("$D$",D,1.5*dir(D),linewidth(4));
dot("$P$",P,1.5*NE,linewidth(4));
dot("$Q$",Q,1.5*NW,linewidth(4));
draw(A--B--C--D--cycle^^A--P--D^^B--Q--C^^P--Q);
X = (-121,6*sqrt(731));
Y = (121,6*sqrt(731));
Z = (-121,-6*sqrt(731));
W = (121,-6*sqrt(731));
draw(X--Z^^Y--W,dashed);
draw(rightanglemark(A,X,Z,500),red);
draw(rightanglemark(B,Y,W,500),red);
draw(rightanglemark(C,W,Y,500),red);
draw(rightanglemark(D,Z,X,500),red);
dot("$X$",X,1.5*N,linewidth(4));
dot("$Y$",Y,1.5*N,linewidth(4));
dot("$Z$",Z,1.5*S,linewidth(4));
dot("$W$",W,1.5*S,linewidth(4));
// Block 2
/* Made by MRENTHUSIASM */ size(300); pair A, B, C, D, A1, B1, C1, D1, P, Q, X, Y, Z, W; A = (-250,6*sqrt(731)); B = (250,6*sqrt(731)); C = (325,-6*sqrt(731)); D = (-325,-6*sqrt(731)); A1 = bisectorpoint(B,A,D); B1 = bisectorpoint(A,B,C); C1 = bisectorpoint(B,C,D); D1 = bisectorpoint(A,D,C); P = intersectionpoint(A--300*(A1-A)+A,D--300*(D1-D)+D); Q = intersectionpoint(B--300*(B1-B)+B,C--300*(C1-C)+C); draw(anglemark(P,A,B,1000),red); draw(anglemark(D,A,P,1000),red); draw(anglemark(A,B,Q,1000),red); draw(anglemark(Q,B,C,1000),red); draw(anglemark(P,D,A,1000),red); draw(anglemark(C,D,P,1000),red); draw(anglemark(Q,C,D,1000),red); draw(anglemark(B,C,Q,1000),red); add(pathticks(anglemark(P,A,B,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(D,A,P,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(A,B,Q,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(Q,B,C,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(P,D,A,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(C,D,P,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(Q,C,D,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(B,C,Q,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$D$",D,1.5*dir(D),linewidth(4)); dot("$P$",P,1.5*NE,linewidth(4)); dot("$Q$",Q,1.5*NW,linewidth(4)); draw(A--B--C--D--cycle^^A--P--D^^B--Q--C^^P--Q); X = (-121,6*sqrt(731)); Y = (121,6*sqrt(731)); Z = (-121,-6*sqrt(731)); W = (121,-6*sqrt(731)); draw(X--Z^^Y--W,dashed); draw(rightanglemark(A,X,Z,500),red); draw(rightanglemark(B,Y,W,500),red); draw(rightanglemark(C,W,Y,500),red); draw(rightanglemark(D,Z,X,500),red); dot("$X$",X,1.5*N,linewidth(4)); dot("$Y$",Y,1.5*N,linewidth(4)); dot("$Z$",Z,1.5*S,linewidth(4)); dot("$W$",W,1.5*S,linewidth(4)); | [] |
243 | In isosceles trapezoid $ABCD$, parallel bases $\overline{AB}$ and $\overline{CD}$ have lengths $500$ and $650$, respectively, and $AD=BC=333$. The angle bisectors of $\angle{A}$ and $\angle{D}$ meet at $P$, and the angle bisectors of $\angle{B}$ and $\angle{C}$ meet at $Q$. Find $PQ$. | 2022 AIME I Problem 3 | Extend line $PQ$ to meet $AD$ at $P'$ and $BC$ at $Q'$. The diagram looks like this:
Since $\angle A + \angle D=\angle B + \angle C = 180^{\circ}$, it follows that $\angle P'AP+\angle P'DP = \angle Q'BQ + \angle Q'CQ = 90^{\circ}$. Thus, $\angle APD = \angle BQC = 90^{\circ}$, implying that $\triangle APD$ and $\triangle BQC$ are right triangles. Since $P'P$ and $Q'Q$ are medians, $P'P+Q'Q=\frac{333\times2}{2}=333$. Since $P'Q'=\frac{500+650}{2}=575$, we have $PQ+P'P+Q'Q=575$, or $PQ=575-333=\boxed{242}$.
~sigma | // Block 1
/* Made by MRENTHUSIASM */
size(300);
pair A, B, C, D, A1, B1, C1, D1, P, Q, P1, Q1;
A = (-250,6*sqrt(731));
B = (250,6*sqrt(731));
C = (325,-6*sqrt(731));
D = (-325,-6*sqrt(731));
A1 = bisectorpoint(B,A,D);
B1 = bisectorpoint(A,B,C);
C1 = bisectorpoint(B,C,D);
D1 = bisectorpoint(A,D,C);
P = intersectionpoint(A--300*(A1-A)+A,D--300*(D1-D)+D);
Q = intersectionpoint(B--300*(B1-B)+B,C--300*(C1-C)+C);
P1 = intersectionpoint(A--D,P--(-300)*(Q-P)+P);
Q1 = intersectionpoint(B--C,Q--300*(Q-P)+Q);
draw(anglemark(P,A,B,1000),red);
draw(anglemark(D,A,P,1000),red);
draw(anglemark(A,B,Q,1000),red);
draw(anglemark(Q,B,C,1000),red);
draw(anglemark(P,D,A,1000),red);
draw(anglemark(C,D,P,1000),red);
draw(anglemark(Q,C,D,1000),red);
draw(anglemark(B,C,Q,1000),red);
add(pathticks(anglemark(P,A,B,1000), n = 1, r = 0.15, s = 750, red));
add(pathticks(anglemark(D,A,P,1000), n = 1, r = 0.15, s = 750, red));
add(pathticks(anglemark(A,B,Q,1000), n = 1, r = 0.15, s = 750, red));
add(pathticks(anglemark(Q,B,C,1000), n = 1, r = 0.15, s = 750, red));
add(pathticks(anglemark(P,D,A,1000), n = 2, r = 0.12, spacing = 150, s = 750, red));
add(pathticks(anglemark(C,D,P,1000), n = 2, r = 0.12, spacing = 150, s = 750, red));
add(pathticks(anglemark(Q,C,D,1000), n = 2, r = 0.12, spacing = 150, s = 750, red));
add(pathticks(anglemark(B,C,Q,1000), n = 2, r = 0.12, spacing = 150, s = 750, red));
dot("$A$",A,1.5*dir(A),linewidth(4));
dot("$B$",B,1.5*dir(B),linewidth(4));
dot("$C$",C,1.5*dir(C),linewidth(4));
dot("$D$",D,1.5*dir(D),linewidth(4));
dot("$P$",P,1.5*NE,linewidth(4));
dot("$Q$",Q,1.5*NW,linewidth(4));
dot("$P'$",P1,1.5*W,linewidth(4));
dot("$Q'$",Q1,1.5*E,linewidth(4));
draw(A--B--C--D--cycle^^A--P--D^^B--Q--C^^P--Q);
draw(P--P1^^Q--Q1,dashed);
// Block 2
/* Made by MRENTHUSIASM */ size(300); pair A, B, C, D, A1, B1, C1, D1, P, Q, P1, Q1; A = (-250,6*sqrt(731)); B = (250,6*sqrt(731)); C = (325,-6*sqrt(731)); D = (-325,-6*sqrt(731)); A1 = bisectorpoint(B,A,D); B1 = bisectorpoint(A,B,C); C1 = bisectorpoint(B,C,D); D1 = bisectorpoint(A,D,C); P = intersectionpoint(A--300*(A1-A)+A,D--300*(D1-D)+D); Q = intersectionpoint(B--300*(B1-B)+B,C--300*(C1-C)+C); P1 = intersectionpoint(A--D,P--(-300)*(Q-P)+P); Q1 = intersectionpoint(B--C,Q--300*(Q-P)+Q); draw(anglemark(P,A,B,1000),red); draw(anglemark(D,A,P,1000),red); draw(anglemark(A,B,Q,1000),red); draw(anglemark(Q,B,C,1000),red); draw(anglemark(P,D,A,1000),red); draw(anglemark(C,D,P,1000),red); draw(anglemark(Q,C,D,1000),red); draw(anglemark(B,C,Q,1000),red); add(pathticks(anglemark(P,A,B,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(D,A,P,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(A,B,Q,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(Q,B,C,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(P,D,A,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(C,D,P,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(Q,C,D,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(B,C,Q,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$D$",D,1.5*dir(D),linewidth(4)); dot("$P$",P,1.5*NE,linewidth(4)); dot("$Q$",Q,1.5*NW,linewidth(4)); dot("$P'$",P1,1.5*W,linewidth(4)); dot("$Q'$",Q1,1.5*E,linewidth(4)); draw(A--B--C--D--cycle^^A--P--D^^B--Q--C^^P--Q); draw(P--P1^^Q--Q1,dashed); | [] |
243 | In isosceles trapezoid $ABCD$, parallel bases $\overline{AB}$ and $\overline{CD}$ have lengths $500$ and $650$, respectively, and $AD=BC=333$. The angle bisectors of $\angle{A}$ and $\angle{D}$ meet at $P$, and the angle bisectors of $\angle{B}$ and $\angle{C}$ meet at $Q$. Find $PQ$. | 2022 AIME I Problem 3 | As in solution 4, $\angle APD = 90^{\circ}$. Set $k = AX$ and $x = DP$.
We know that $DZ = AX + \frac{DC-AB}{2}$, so $DZ = k + \frac{650-500}{2} = k + 75$.
$\triangle DPZ \sim \triangle APD$ by AA, so we have $\frac{PD}{AD} = \frac{ZD}{PD}$, resulting in
\[\frac{x}{333} = \frac{k+75}{x} \text{ (1)}\]
$\triangle APX \sim \triangle ADP$ by AA, so we have $\frac{AP}{AD} = \frac{AX}{AP}$, resulting in
\[\frac{\sqrt{333^2-x^2}}{333} = \frac{k}{\sqrt{333^2-k^2}} \text{ (2)}\]
From $\text{(1)}$, we have $x^2 = 333k + 333(75) = 333k + 24975$. From $\text{(2)}$, we have $333^2 - x^2 = 333k$, or $x^2 = 333^2 - 333k$. Thus, $333k + 24975 = 333^2 - 333k$. Solving for $k$ yields $k = 129$.
By symmetry, $YB = AX = 129$. Thus, $PQ = XY = AB - 2AX = 500 - 2(129) = \boxed{242}$.
~ adam_zheng | // Block 1
/* Made by MRENTHUSIASM */
size(300);
pair A, B, C, D, A1, B1, C1, D1, P, Q, X, Y, Z, W;
A = (-250,6*sqrt(731));
B = (250,6*sqrt(731));
C = (325,-6*sqrt(731));
D = (-325,-6*sqrt(731));
A1 = bisectorpoint(B,A,D);
B1 = bisectorpoint(A,B,C);
C1 = bisectorpoint(B,C,D);
D1 = bisectorpoint(A,D,C);
P = intersectionpoint(A--300*(A1-A)+A,D--300*(D1-D)+D);
Q = intersectionpoint(B--300*(B1-B)+B,C--300*(C1-C)+C);
draw(anglemark(P,A,B,1000),red);
draw(anglemark(D,A,P,1000),red);
draw(anglemark(A,B,Q,1000),red);
draw(anglemark(Q,B,C,1000),red);
draw(anglemark(P,D,A,1000),red);
draw(anglemark(C,D,P,1000),red);
draw(anglemark(Q,C,D,1000),red);
draw(anglemark(B,C,Q,1000),red);
add(pathticks(anglemark(P,A,B,1000), n = 1, r = 0.15, s = 750, red));
add(pathticks(anglemark(D,A,P,1000), n = 1, r = 0.15, s = 750, red));
add(pathticks(anglemark(A,B,Q,1000), n = 1, r = 0.15, s = 750, red));
add(pathticks(anglemark(Q,B,C,1000), n = 1, r = 0.15, s = 750, red));
add(pathticks(anglemark(P,D,A,1000), n = 2, r = 0.12, spacing = 150, s = 750, red));
add(pathticks(anglemark(C,D,P,1000), n = 2, r = 0.12, spacing = 150, s = 750, red));
add(pathticks(anglemark(Q,C,D,1000), n = 2, r = 0.12, spacing = 150, s = 750, red));
add(pathticks(anglemark(B,C,Q,1000), n = 2, r = 0.12, spacing = 150, s = 750, red));
dot("$A$",A,1.5*dir(A),linewidth(4));
dot("$B$",B,1.5*dir(B),linewidth(4));
dot("$C$",C,1.5*dir(C),linewidth(4));
dot("$D$",D,1.5*dir(D),linewidth(4));
dot("$P$",P,1.5*NE,linewidth(4));
dot("$Q$",Q,1.5*NW,linewidth(4));
draw(A--B--C--D--cycle^^A--P--D^^B--Q--C^^P--Q);
X = (-121,6*sqrt(731));
Y = (121,6*sqrt(731));
Z = (-121,-6*sqrt(731));
W = (121,-6*sqrt(731));
draw(X--Z^^Y--W,dashed);
draw(rightanglemark(A,X,Z,500),red);
draw(rightanglemark(B,Y,W,500),red);
draw(rightanglemark(C,W,Y,500),red);
draw(rightanglemark(D,Z,X,500),red);
dot("$X$",X,1.5*N,linewidth(4));
dot("$Y$",Y,1.5*N,linewidth(4));
dot("$Z$",Z,1.5*S,linewidth(4));
dot("$W$",W,1.5*S,linewidth(4));
// Block 2
/* Made by MRENTHUSIASM */ size(300); pair A, B, C, D, A1, B1, C1, D1, P, Q, X, Y, Z, W; A = (-250,6*sqrt(731)); B = (250,6*sqrt(731)); C = (325,-6*sqrt(731)); D = (-325,-6*sqrt(731)); A1 = bisectorpoint(B,A,D); B1 = bisectorpoint(A,B,C); C1 = bisectorpoint(B,C,D); D1 = bisectorpoint(A,D,C); P = intersectionpoint(A--300*(A1-A)+A,D--300*(D1-D)+D); Q = intersectionpoint(B--300*(B1-B)+B,C--300*(C1-C)+C); draw(anglemark(P,A,B,1000),red); draw(anglemark(D,A,P,1000),red); draw(anglemark(A,B,Q,1000),red); draw(anglemark(Q,B,C,1000),red); draw(anglemark(P,D,A,1000),red); draw(anglemark(C,D,P,1000),red); draw(anglemark(Q,C,D,1000),red); draw(anglemark(B,C,Q,1000),red); add(pathticks(anglemark(P,A,B,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(D,A,P,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(A,B,Q,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(Q,B,C,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(P,D,A,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(C,D,P,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(Q,C,D,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(B,C,Q,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$D$",D,1.5*dir(D),linewidth(4)); dot("$P$",P,1.5*NE,linewidth(4)); dot("$Q$",Q,1.5*NW,linewidth(4)); draw(A--B--C--D--cycle^^A--P--D^^B--Q--C^^P--Q); X = (-121,6*sqrt(731)); Y = (121,6*sqrt(731)); Z = (-121,-6*sqrt(731)); W = (121,-6*sqrt(731)); draw(X--Z^^Y--W,dashed); draw(rightanglemark(A,X,Z,500),red); draw(rightanglemark(B,Y,W,500),red); draw(rightanglemark(C,W,Y,500),red); draw(rightanglemark(D,Z,X,500),red); dot("$X$",X,1.5*N,linewidth(4)); dot("$Y$",Y,1.5*N,linewidth(4)); dot("$Z$",Z,1.5*S,linewidth(4)); dot("$W$",W,1.5*S,linewidth(4)); | [] |
244 | A straight river that is $264$ meters wide flows from west to east at a rate of $14$ meters per minute. Melanie and Sherry sit on the south bank of the river with Melanie a distance of $D$ meters downstream from Sherry. Relative to the water, Melanie swims at $80$ meters per minute, and Sherry swims at $60$ meters per minute. At the same time, Melanie and Sherry begin swimming in straight lines to a point on the north bank of the river that is equidistant from their starting positions. The two women arrive at this point simultaneously. Find $D$. | 2022 AIME I Problem 5 | Define $m$ as the number of minutes they swim for.
Let their meeting point be $A$. Melanie is swimming against the current, so she must aim upstream from point $A$, to compensate for this; in particular, since she is swimming for $m$ minutes, the current will push her $14m$ meters downstream in that time, so she must aim for a point $B$ that is $14m$ meters upstream from point $A$. Similarly, Sherry is swimming downstream for $m$ minutes, so she must also aim at point $B$ to compensate for the flow of the current.
If Melanie and Sherry were to both aim at point $B$ in a currentless river with the same dimensions, they would still both meet at that point simultaneously. Since there is no current in this scenario, the distances that Melanie and Sherry travel, respectively, are $80m$ and $60m$ meters. We can draw out this new scenario, with the dimensions that we have:
(While it is indeed true that the triangle above with side lengths $60m$, $80m$ and $D$ is a right triangle, we do not know this yet, so we cannot assume this based on the diagram.)
By the Pythagorean Theorem, we have
\begin{align*} 264^{2} + \left( \frac{D}{2} - 14m \right) ^{2} &= 3600m^{2} \\ 264^{2} + \left( \frac{D}{2} + 14m \right) ^{2} &= 6400m^{2}. \end{align*}
Subtracting the first equation from the second gives us $28Dm = 2800m^{2}$, so $D = 100m$. Substituting this into our first equation, we have that
\begin{align*}264^{2} + 36^{2} m^{2} &= 60^{2}m^{2} \\ 264^{2} &= 96 \cdot 24 \cdot m^{2} \\ 11^{2} &= 4 \cdot m^{2} \\ m &= \frac{11}{2}. \end{align*}
So $D = 100m = \boxed{550}$.
~ihatemath123 | // Block 1
unitsize(0.02cm);
draw((0,0)--(0,264)--(550,264)--(550,0)--cycle);
pair B = (198,264);
dot(B^^(0,0)^^(550,0),linewidth(5));
draw((0,0)--B,dashed);
draw((550,0)--B,dashed);
label("$60m$", (0,0)--B, E);
label("$80m$", (550,0)--B, W);
label("$264$", (0,0)--(0,264), W);
label("$\frac{D}{2} - 14m$", (0,264)--B, N);
label("$\frac{D}{2} + 14m$", B--(550,264), N);
label("$D$", (0,0)--(550,0), S);
label("$B$", B, N);
label("Downstream", (350,325), E);
label("Upstream", (200,325), W);
draw((225,325)--(325,325), Arrows);
// Block 2
unitsize(0.02cm); draw((0,0)--(0,264)--(550,264)--(550,0)--cycle); pair B = (198,264); dot(B^^(0,0)^^(550,0),linewidth(5)); draw((0,0)--B,dashed); draw((550,0)--B,dashed); label("$60m$", (0,0)--B, E); label("$80m$", (550,0)--B, W); label("$264$", (0,0)--(0,264), W); label("$\frac{D}{2} - 14m$", (0,264)--B, N); label("$\frac{D}{2} + 14m$", B--(550,264), N); label("$D$", (0,0)--(550,0), S); label("$B$", B, N); label("Downstream", (350,325), E); label("Upstream", (200,325), W); draw((225,325)--(325,325), Arrows); | [] |
244 | A straight river that is $264$ meters wide flows from west to east at a rate of $14$ meters per minute. Melanie and Sherry sit on the south bank of the river with Melanie a distance of $D$ meters downstream from Sherry. Relative to the water, Melanie swims at $80$ meters per minute, and Sherry swims at $60$ meters per minute. At the same time, Melanie and Sherry begin swimming in straight lines to a point on the north bank of the river that is equidistant from their starting positions. The two women arrive at this point simultaneously. Find $D$. | 2022 AIME I Problem 5 | We have the following diagram:
Since Melanie and Sherry swim for the same distance and the same amount of time, they swim at the same net speed.
Let $x$ and $y$ be some positive numbers. We have the following table:
\[\begin{array}{c||c|c|c} & \textbf{Net Velocity Vector (m/min)} & \textbf{Natural Velocity Vector (m/min)} & \textbf{Natural Speed (m/min)} \\ \hline \hline &&& \\ [-2.25ex] \textbf{Melanie} & \langle -x,y\rangle & \langle -x-14,y\rangle & 80 \\ \hline &&& \\ [-2.25ex] \textbf{Sherry} & \langle x,y\rangle & \langle x-14,y\rangle & 60 \end{array}\]
Recall that $|\text{velocity}|=\text{speed},$ so
\begin{align*} (-x-14)^2 + y^2 &= 80^2, &&(1) \\ (x-14)^2 + y^2 &= 60^2. &&(2) \end{align*}
We subtract $(2)$ from $(1)$ to get $56x=2800,$ from which $x=50.$ Substituting this into either equation, we have $y=48.$
It follows that Melanie and Sherry both swim for $264\div y=5.5$ minutes. Therefore, the answer is \[D=2x\cdot5.5=\boxed{550}.\]
~MRENTHUSIASM | // Block 1
/* Made by MRENTHUSIASM */
size(350);
pair A, B, C;
A = (0,264);
B = (-275,0);
C = (275,0);
draw((-300,0)--(300,0)^^(-300,264)--(300,264)^^A--B^^A--C,linewidth(2));
dot("Finish",A,1.75*N,linewidth(5));
dot("Sherry",B,1.75*S,linewidth(5));
dot("Melanie",C,1.75*S,linewidth(5));
Label L1 = Label("$D$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white));
Label L2 = Label("$264$", align=(0,0), position=MidPoint, filltype=Fill(0,3,white));
Label L3 = Label("Current $(14)$", position=EndPoint, filltype=Fill(3,0,white));
Label L4 = Label("$y$", align=(-1,0), position=Relative(0.4));
Label L5 = Label("$x$", align=(0,1), position=Relative(0.4));
Label L6 = Label("$y$", align=(1,0), position=Relative(0.4));
Label L7 = Label("$x$", align=(0,1), position=Relative(0.4));
draw(B-(0,75)--C-(0,75), L=L1, arrow=Arrows(),bar=Bars(15));
draw((-350,0)--(-350,264), L=L2, arrow=Arrows(),bar=Bars(15));
draw((-300,-120)--(300,-120), L=L3, arrow=EndArrow());
draw(B--B+(0,48), L=L4, arrow=EndArrow());
draw(B+(0,48)--B+(50,48), L=L5, arrow=EndArrow());
draw(C--C+(0,48), L=L6, arrow=EndArrow());
draw(C+(0,48)--C+(-50,48), L=L7, arrow=EndArrow());
// Block 2
/* Made by MRENTHUSIASM */ size(350); pair A, B, C; A = (0,264); B = (-275,0); C = (275,0); draw((-300,0)--(300,0)^^(-300,264)--(300,264)^^A--B^^A--C,linewidth(2)); dot("Finish",A,1.75*N,linewidth(5)); dot("Sherry",B,1.75*S,linewidth(5)); dot("Melanie",C,1.75*S,linewidth(5)); Label L1 = Label("$D$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); Label L2 = Label("$264$", align=(0,0), position=MidPoint, filltype=Fill(0,3,white)); Label L3 = Label("Current $(14)$", position=EndPoint, filltype=Fill(3,0,white)); Label L4 = Label("$y$", align=(-1,0), position=Relative(0.4)); Label L5 = Label("$x$", align=(0,1), position=Relative(0.4)); Label L6 = Label("$y$", align=(1,0), position=Relative(0.4)); Label L7 = Label("$x$", align=(0,1), position=Relative(0.4)); draw(B-(0,75)--C-(0,75), L=L1, arrow=Arrows(),bar=Bars(15)); draw((-350,0)--(-350,264), L=L2, arrow=Arrows(),bar=Bars(15)); draw((-300,-120)--(300,-120), L=L3, arrow=EndArrow()); draw(B--B+(0,48), L=L4, arrow=EndArrow()); draw(B+(0,48)--B+(50,48), L=L5, arrow=EndArrow()); draw(C--C+(0,48), L=L6, arrow=EndArrow()); draw(C+(0,48)--C+(-50,48), L=L7, arrow=EndArrow()); | [] |
244 | A straight river that is $264$ meters wide flows from west to east at a rate of $14$ meters per minute. Melanie and Sherry sit on the south bank of the river with Melanie a distance of $D$ meters downstream from Sherry. Relative to the water, Melanie swims at $80$ meters per minute, and Sherry swims at $60$ meters per minute. At the same time, Melanie and Sherry begin swimming in straight lines to a point on the north bank of the river that is equidistant from their starting positions. The two women arrive at this point simultaneously. Find $D$. | 2022 AIME I Problem 5 | We can break down movement into two components: the $x$-component and the $y$-component. Suppose that Melanie travels a distance of $a$ in the $x$-direction and a distance of $c$ in the $y$-direction in one minute when there is no current. Similarly, suppose that Sherry travels a distance of $a$ in the $x$-direction but a distance of $b$ in the $y$-direction in one minute when there is no current. The current only affects the $x$-components because it goes in the $x$-direction.
Now, note that $a^2 + b^2 = 60^2$ because Sherry travels 60 meters in a minute. Thus, $a^2 + c^2 = 80^2$ because Melanie travels 80 meters in a minute. Also, the distance they travel with the current must be the same in one minute because they reach the point equidistant from them at the same time. That means $b + 14 = c - 14$ or $b = c - 28$. So now we can plug that into the two equations to get:
\begin{align*} a^2 + c^2 &= 80^2, \\ a^2 + (c-28)^2 &= 60^2. \end{align*}
We can solve the system of equations to get $a = 48$ and $c = 64$. From this, we can figure out that it must've taken them $5.5$ minutes to get to the other side, because $264/48 = 5.5$. This means that there are $5.5$ lengths of $48$ in each person's travel. Also, $D$ must be equal to $11(b+14) = 11(c-14)$ because there are $(5.5)2 = 11$ lengths of $b-14$ between them, $5.5$ on each person's side. Since $c = 64$, we have $c-14 = 50$, so the answer is \[D=11\cdot50=\boxed{550}.\]
~Curious_crow | // Block 1
/* Ruthlessly plagiarized from MRENTHUSIASM by Curious_crow */
size(350);
pair A, B, C;
A = (0,264);
B = (-275,0);
C = (275,0);
draw((-300,0)--(300,0)^^(-300,264)--(300,264)^^A--B^^A--C,linewidth(2));
dot("Finish",A,1.75*N,linewidth(5));
dot("Sherry",B,1.75*S,linewidth(5));
dot("Melanie",C,1.75*S,linewidth(5));
Label L1 = Label("$D$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white));
Label L2 = Label("$264$", align=(0,0), position=MidPoint, filltype=Fill(0,3,white));
Label L3 = Label("Current $(14)$", position=EndPoint, filltype=Fill(3,0,white));
Label L4 = Label("$a$", align=(-1,0), position=Relative(0.4));
Label L5 = Label("$b+14$", align=(0,1), position=Relative(0.4));
Label L6 = Label("$a$", align=(1,0), position=Relative(0.4));
Label L7 = Label("$c-14$", align=(0,1), position=Relative(0.4));
draw(B-(0,75)--C-(0,75), L=L1, arrow=Arrows(),bar=Bars(15));
draw((-350,0)--(-350,264), L=L2, arrow=Arrows(),bar=Bars(15));
draw((-300,-120)--(300,-120), L=L3, arrow=EndArrow());
draw(B--B+(0,48), L=L4, arrow=EndArrow());
draw(B+(0,48)--B+(50,48), L=L5, arrow=EndArrow());
draw(C--C+(0,48), L=L6, arrow=EndArrow());
draw(C+(0,48)--C+(-50,48), L=L7, arrow=EndArrow());
// Block 2
/* Ruthlessly plagiarized from MRENTHUSIASM by Curious_crow */ size(350); pair A, B, C; A = (0,264); B = (-275,0); C = (275,0); draw((-300,0)--(300,0)^^(-300,264)--(300,264)^^A--B^^A--C,linewidth(2)); dot("Finish",A,1.75*N,linewidth(5)); dot("Sherry",B,1.75*S,linewidth(5)); dot("Melanie",C,1.75*S,linewidth(5)); Label L1 = Label("$D$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); Label L2 = Label("$264$", align=(0,0), position=MidPoint, filltype=Fill(0,3,white)); Label L3 = Label("Current $(14)$", position=EndPoint, filltype=Fill(3,0,white)); Label L4 = Label("$a$", align=(-1,0), position=Relative(0.4)); Label L5 = Label("$b+14$", align=(0,1), position=Relative(0.4)); Label L6 = Label("$a$", align=(1,0), position=Relative(0.4)); Label L7 = Label("$c-14$", align=(0,1), position=Relative(0.4)); draw(B-(0,75)--C-(0,75), L=L1, arrow=Arrows(),bar=Bars(15)); draw((-350,0)--(-350,264), L=L2, arrow=Arrows(),bar=Bars(15)); draw((-300,-120)--(300,-120), L=L3, arrow=EndArrow()); draw(B--B+(0,48), L=L4, arrow=EndArrow()); draw(B+(0,48)--B+(50,48), L=L5, arrow=EndArrow()); draw(C--C+(0,48), L=L6, arrow=EndArrow()); draw(C+(0,48)--C+(-50,48), L=L7, arrow=EndArrow()); | [] |
245 | Equilateral triangle $\triangle ABC$ is inscribed in circle $\omega$ with radius $18.$ Circle $\omega_A$ is tangent to sides $\overline{AB}$ and $\overline{AC}$ and is internally tangent to $\omega.$ Circles $\omega_B$ and $\omega_C$ are defined analogously. Circles $\omega_A,$ $\omega_B,$ and $\omega_C$ meet in six points---two points for each pair of circles. The three intersection points closest to the vertices of $\triangle ABC$ are the vertices of a large equilateral triangle in the interior of $\triangle ABC,$ and the other three intersection points are the vertices of a smaller equilateral triangle in the interior of $\triangle ABC.$ The side length of the smaller equilateral triangle can be written as $\sqrt{a} - \sqrt{b},$ where $a$ and $b$ are positive integers. Find $a+b.$ | 2022 AIME I Problem 8 | We can extend $AB$ and $AC$ to $B'$ and $C'$ respectively such that circle $\omega_A$ is the incircle of $\triangle AB'C'$.
Since the diameter of the circle is the height of this triangle, the height of this triangle is $36$. We can use inradius or equilateral triangle properties to get the inradius of this triangle is $12$ (The incenter is also a centroid in an equilateral triangle, and the distance from a side to the centroid is a third of the height). Therefore, the radius of each of the smaller circles is $12$.
Let $O=\omega$ be the center of the largest circle. We will set up a coordinate system with $O$ as the origin. The center of $\omega_A$ will be at $(0,-6)$ because it is directly beneath $O$ and is the length of the larger radius minus the smaller radius, or $18-12 = 6$. By rotating this point $120^{\circ}$ around $O$, we get the center of $\omega_B$. This means that the magnitude of vector $\overrightarrow{O\omega_B}$ is $6$ and is at a $30$ degree angle from the horizontal. Therefore, the coordinates of this point are $(3\sqrt{3},3)$ and by symmetry the coordinates of the center of $\omega_C$ is $(-3\sqrt{3},3)$.
The upper left and right circles intersect at two points, the lower of which is $X$. The equations of these two circles are:
\begin{align*} (x+3\sqrt3)^2 + (y-3)^2 &= 12^2, \\ (x-3\sqrt3)^2 + (y-3)^2 &= 12^2. \end{align*}
We solve this system by subtracting to get $x = 0$. Plugging back in to the first equation, we have $(3\sqrt{3})^2 + (y-3)^2 = 144 \implies (y-3)^2 = 117 \implies y-3 = \pm \sqrt{117} \implies y = 3 \pm \sqrt{117}$. Since we know $X$ is the lower solution, we take the negative value to get $X = (0,3-\sqrt{117})$.
We can solve the problem two ways from here. We can find $Y$ by rotation and use the distance formula to find the length, or we can be somewhat more clever. We notice that it is easier to find $OX$ as they lie on the same vertical, $\angle XOY$ is $120$ degrees so we can make use of $30-60-90$ triangles, and $OX = OY$ because $O$ is the center of triangle $XYZ$. We can draw the diagram as such:
Note that $OX = OY = \sqrt{117} - 3$. It follows that
\begin{align*} XY &= 2 \cdot \frac{OX\cdot\sqrt{3}}{2} \\ &= OX \cdot \sqrt{3} \\ &= (\sqrt{117}-3) \cdot \sqrt{3} \\ &= \sqrt{351}-\sqrt{27}. \end{align*}
Finally, the answer is $351+27 = \boxed{378}$.
~KingRavi | // Block 1
/* Made by MRENTHUSIASM */
size(300);
pair A, B, C, B1, C1, W, WA, WB, WC, X, Y, Z;
A = 18*dir(90);
B = 18*dir(210);
C = 18*dir(330);
B1 = A+24*sqrt(3)*dir(B-A);
C1 = A+24*sqrt(3)*dir(C-A);
W = (0,0);
WA = 6*dir(270);
WB = 6*dir(30);
WC = 6*dir(150);
X = (sqrt(117)-3)*dir(270);
Y = (sqrt(117)-3)*dir(30);
Z = (sqrt(117)-3)*dir(150);
filldraw(X--Y--Z--cycle,green,dashed);
draw(Circle(WA,12)^^Circle(WB,12)^^Circle(WC,12),blue);
draw(Circle(W,18)^^A--B--C--cycle);
draw(B--B1--C1--C,dashed);
dot("$A$",A,1.5*dir(A),linewidth(4));
dot("$B$",B,1.5*(-1,0),linewidth(4));
dot("$C$",C,1.5*(1,0),linewidth(4));
dot("$B'$",B1,1.5*dir(B1),linewidth(4));
dot("$C'$",C1,1.5*dir(C1),linewidth(4));
dot("$O$",W,1.5*dir(90),linewidth(4));
dot("$X$",X,1.5*dir(X),linewidth(4));
dot("$Y$",Y,1.5*dir(Y),linewidth(4));
dot("$Z$",Z,1.5*dir(Z),linewidth(4));
// Block 2
/* Made by MRENTHUSIASM */
size(300);
pair A, B, C, B1, C1, W, WA, WB, WC, X, Y, Z;
A = 18*dir(90);
B = 18*dir(210);
C = 18*dir(330);
B1 = A+24*sqrt(3)*dir(B-A);
C1 = A+24*sqrt(3)*dir(C-A);
W = (0,0);
WA = 6*dir(270);
WB = 6*dir(30);
WC = 6*dir(150);
X = (sqrt(117)-3)*dir(270);
Y = (sqrt(117)-3)*dir(30);
Z = (sqrt(117)-3)*dir(150);
filldraw(X--Y--Z--cycle,green,dashed);
draw(Circle(WA,12)^^Circle(WB,12)^^Circle(WC,12),blue);
draw(Circle(W,18)^^A--B--C--cycle);
draw(B--B1--C1--C^^W--X^^W--Y^^W--midpoint(X--Y),dashed);
dot("$A$",A,1.5*dir(A),linewidth(4));
dot("$B$",B,1.5*(-1,0),linewidth(4));
dot("$C$",C,1.5*(1,0),linewidth(4));
dot("$B'$",B1,1.5*dir(B1),linewidth(4));
dot("$C'$",C1,1.5*dir(C1),linewidth(4));
dot("$O$",W,1.5*dir(90),linewidth(4));
dot("$X$",X,1.5*dir(X),linewidth(4));
dot("$Y$",Y,1.5*dir(Y),linewidth(4));
dot("$Z$",Z,1.5*dir(Z),linewidth(4));
// Block 3
/* Made by MRENTHUSIASM */ size(300); pair A, B, C, B1, C1, W, WA, WB, WC, X, Y, Z; A = 18*dir(90); B = 18*dir(210); C = 18*dir(330); B1 = A+24*sqrt(3)*dir(B-A); C1 = A+24*sqrt(3)*dir(C-A); W = (0,0); WA = 6*dir(270); WB = 6*dir(30); WC = 6*dir(150); X = (sqrt(117)-3)*dir(270); Y = (sqrt(117)-3)*dir(30); Z = (sqrt(117)-3)*dir(150); filldraw(X--Y--Z--cycle,green,dashed); draw(Circle(WA,12)^^Circle(WB,12)^^Circle(WC,12),blue); draw(Circle(W,18)^^A--B--C--cycle); draw(B--B1--C1--C,dashed); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*(-1,0),linewidth(4)); dot("$C$",C,1.5*(1,0),linewidth(4)); dot("$B'$",B1,1.5*dir(B1),linewidth(4)); dot("$C'$",C1,1.5*dir(C1),linewidth(4)); dot("$O$",W,1.5*dir(90),linewidth(4)); dot("$X$",X,1.5*dir(X),linewidth(4)); dot("$Y$",Y,1.5*dir(Y),linewidth(4)); dot("$Z$",Z,1.5*dir(Z),linewidth(4));
// Block 4
/* Made by MRENTHUSIASM */ size(300); pair A, B, C, B1, C1, W, WA, WB, WC, X, Y, Z; A = 18*dir(90); B = 18*dir(210); C = 18*dir(330); B1 = A+24*sqrt(3)*dir(B-A); C1 = A+24*sqrt(3)*dir(C-A); W = (0,0); WA = 6*dir(270); WB = 6*dir(30); WC = 6*dir(150); X = (sqrt(117)-3)*dir(270); Y = (sqrt(117)-3)*dir(30); Z = (sqrt(117)-3)*dir(150); filldraw(X--Y--Z--cycle,green,dashed); draw(Circle(WA,12)^^Circle(WB,12)^^Circle(WC,12),blue); draw(Circle(W,18)^^A--B--C--cycle); draw(B--B1--C1--C^^W--X^^W--Y^^W--midpoint(X--Y),dashed); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*(-1,0),linewidth(4)); dot("$C$",C,1.5*(1,0),linewidth(4)); dot("$B'$",B1,1.5*dir(B1),linewidth(4)); dot("$C'$",C1,1.5*dir(C1),linewidth(4)); dot("$O$",W,1.5*dir(90),linewidth(4)); dot("$X$",X,1.5*dir(X),linewidth(4)); dot("$Y$",Y,1.5*dir(Y),linewidth(4)); dot("$Z$",Z,1.5*dir(Z),linewidth(4)); | [] |
245 | Equilateral triangle $\triangle ABC$ is inscribed in circle $\omega$ with radius $18.$ Circle $\omega_A$ is tangent to sides $\overline{AB}$ and $\overline{AC}$ and is internally tangent to $\omega.$ Circles $\omega_B$ and $\omega_C$ are defined analogously. Circles $\omega_A,$ $\omega_B,$ and $\omega_C$ meet in six points---two points for each pair of circles. The three intersection points closest to the vertices of $\triangle ABC$ are the vertices of a large equilateral triangle in the interior of $\triangle ABC,$ and the other three intersection points are the vertices of a smaller equilateral triangle in the interior of $\triangle ABC.$ The side length of the smaller equilateral triangle can be written as $\sqrt{a} - \sqrt{b},$ where $a$ and $b$ are positive integers. Find $a+b.$ | 2022 AIME I Problem 8 | For equilateral triangle with side length $l$, height $h$, and circumradius $r$, there are relationships: $h = \frac{\sqrt{3}}{2} l$, $r = \frac{2}{3} h = \frac{\sqrt{3}}{3} l$, and $l = \sqrt{3}r$.
There is a lot of symmetry in the figure. The radius of the big circle $\odot \omega$ is $R = 18$, let the radius of the small circles $\odot \omega_A$, $\odot \omega_B$, $\odot \omega_C$ be $r$.
We are going to solve this problem in $3$ steps:
$\textbf{Step 1:}$
We have $\triangle A \omega_A D$ is a $30-60-90$ triangle, and $A \omega_A = 2 \cdot \omega_A D$, $A \omega_A = 2R-r$ ($\odot \omega$ and $\odot \omega_A$ are tangent), and $\omega_A D = r$. So, we get $2R-r = 2r$ and $r = \frac{2}{3} \cdot R = 12$.
Since $\odot \omega$ and $\odot \omega_A$ are tangent, we get $\omega \omega_A = R - r = \frac{1}{3} \cdot R = 6$.
Note that $\triangle \omega_A \omega_B \omega_C$ is an equilateral triangle, and $\omega$ is its center, so $\omega_B \omega_C = \sqrt{3} \cdot \omega \omega_A = 6 \sqrt{3}$.
$\textbf{Step 2:}$
Note that $\triangle \omega_C E X$ is an isosceles triangle, so \[EX = 2 \sqrt{(\omega_C E)^2 - \left(\frac{\omega_B \omega_C}{2}\right)^2} = 2 \sqrt{r^2 - \left(\frac{\omega_B \omega_C}{2}\right)^2} = 2 \sqrt{12^2 - (3 \sqrt{3})^2} = 2 \sqrt{117}.\]
$\textbf{Step 3:}$
In $\odot \omega_C$, Power of a Point gives $\omega X \cdot \omega E = r^2 - (\omega_C \omega)^2$ and $\omega E = EX - \omega X = 2\sqrt{117} - \omega X$.
It follows that $\omega X \cdot (2\sqrt{117} - \omega X) = 12^2 - 6^2$. We solve this quadratic equation: $\omega X = \sqrt{117} - 3$.
Since $\omega X$ is the circumradius of equilateral $\triangle XYZ$, we have $XY = \sqrt{3} \cdot \omega X = \sqrt{3} \cdot (\sqrt{117} - 3) = \sqrt{351}-\sqrt{27}$.
Therefore, the answer is $351+27 = \boxed{378}$.
~isabelchen | // Block 1
/* Made by MRENTHUSIASM */
/* Modified by isabelchen */
size(250);
pair A, B, C, W, WA, WB, WC, X, Y, Z, D, E;
A = 18*dir(90);
B = 18*dir(210);
C = 18*dir(330);
W = (0,0);
WA = 6*dir(270);
WB = 6*dir(30);
WC = 6*dir(150);
X = (sqrt(117)-3)*dir(270);
Y = (sqrt(117)-3)*dir(30);
Z = (sqrt(117)-3)*dir(150);
D = intersectionpoint(Circle(WA,12),A--C);
E = intersectionpoints(Circle(WB,12),Circle(WC,12))[0];
filldraw(X--Y--Z--cycle,green,dashed);
draw(Circle(WA,12)^^Circle(WB,12)^^Circle(WC,12),blue);
draw(Circle(W,18)^^A--B--C--cycle);
dot("$A$",A,1.5*dir(A),linewidth(4));
dot("$B$",B,1.5*dir(B),linewidth(4));
dot("$C$",C,1.5*dir(C),linewidth(4));
dot("$\omega$",W,1.5*dir(270),linewidth(4));
dot("$\omega_A$",WA,1.5*dir(-WA),linewidth(4));
dot("$\omega_B$",WB,1.5*dir(-WB),linewidth(4));
dot("$\omega_C$",WC,1.5*dir(-WC),linewidth(4));
dot("$X$",X,1.5*dir(X),linewidth(4));
dot("$Y$",Y,1.5*dir(Y),linewidth(4));
dot("$Z$",Z,1.5*dir(Z),linewidth(4));
dot("$E$",E,1.5*dir(E),linewidth(4));
dot("$D$",D,1.5*dir(D),linewidth(4));
draw(WC--WB^^WC--X^^WC--E^^WA--D^^A--X);
// Block 2
/* Made by MRENTHUSIASM */ /* Modified by isabelchen */ size(250); pair A, B, C, W, WA, WB, WC, X, Y, Z, D, E; A = 18*dir(90); B = 18*dir(210); C = 18*dir(330); W = (0,0); WA = 6*dir(270); WB = 6*dir(30); WC = 6*dir(150); X = (sqrt(117)-3)*dir(270); Y = (sqrt(117)-3)*dir(30); Z = (sqrt(117)-3)*dir(150); D = intersectionpoint(Circle(WA,12),A--C); E = intersectionpoints(Circle(WB,12),Circle(WC,12))[0]; filldraw(X--Y--Z--cycle,green,dashed); draw(Circle(WA,12)^^Circle(WB,12)^^Circle(WC,12),blue); draw(Circle(W,18)^^A--B--C--cycle); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$\omega$",W,1.5*dir(270),linewidth(4)); dot("$\omega_A$",WA,1.5*dir(-WA),linewidth(4)); dot("$\omega_B$",WB,1.5*dir(-WB),linewidth(4)); dot("$\omega_C$",WC,1.5*dir(-WC),linewidth(4)); dot("$X$",X,1.5*dir(X),linewidth(4)); dot("$Y$",Y,1.5*dir(Y),linewidth(4)); dot("$Z$",Z,1.5*dir(Z),linewidth(4)); dot("$E$",E,1.5*dir(E),linewidth(4)); dot("$D$",D,1.5*dir(D),linewidth(4)); draw(WC--WB^^WC--X^^WC--E^^WA--D^^A--X); | [] |
246 | Three spheres with radii $11$, $13$, and $19$ are mutually externally tangent. A plane intersects the spheres in three congruent circles centered at $A$, $B$, and $C$, respectively, and the centers of the spheres all lie on the same side of this plane. Suppose that $AB^2 = 560$. Find $AC^2$. | 2022 AIME I Problem 10 | This solution refers to the Diagram section.
We let $\ell$ be the plane that passes through the spheres and $O_A$ and $O_B$ be the centers of the spheres with radii $11$ and $13$. We take a cross-section that contains $A$ and $B$, which contains these two spheres but not the third, as shown below:
Because the plane cuts out congruent circles, they have the same radius and from the given information, $AB = \sqrt{560}$. Since $ABO_BO_A$ is a trapezoid, we can drop an altitude from $O_A$ to $BO_B$ to create a rectangle and triangle to use Pythagorean theorem. We know that the length of the altitude is $\sqrt{560}$ and let the distance from $O_B$ to $D$ be $x$. Then we have $x^2 = 576-560 \implies x = 4$.
We have $AO_A = BD$ because of the rectangle, so $\sqrt{11^2-r^2} = \sqrt{13^2-r^2}-4$.
Squaring, we have $121-r^2 = 169-r^2 + 16 - 8 \cdot \sqrt{169-r^2}$.
Subtracting, we get $8 \cdot \sqrt{169-r^2} = 64 \implies \sqrt{169-r^2} = 8 \implies 169-r^2 = 64 \implies r^2 = 105$.
We also notice that since we had $\sqrt{169-r^2} = 8$ means that $BO_B = 8$ and since we know that $x = 4$, $AO_A = 4$.
We take a cross-section that contains $A$ and $C$, which contains these two spheres but not the third, as shown below:
We have $CO_C = \sqrt{19^2-r^2} = \sqrt{361 - 105} = \sqrt{256} = 16$. Since $AO_A = 4$, we have $EO_C = 16-4 = 12$. Using Pythagorean theorem, $O_AE = \sqrt{30^2 - 12^2} = \sqrt{900-144} = \sqrt{756}$. Therefore, $O_AE^2 = AC^2 = \boxed{756}$.
~KingRavi | // Block 1
size(400);
pair A, B, OA, OB;
B = (0,0);
A = (-23.6643191,0);
OB = (0,8);
OA = (-23.6643191,4);
draw(circle(OB,13));
draw(circle(OA,11));
draw((-48,0)--(24,0));
label("$\ell$",(-42,0),S);
label("$A$",A,S);
label("$B$",B,S);
label("$O_A$",OA,N);
label("$O_B$",OB,N);
draw(A--OA);
draw(B--OB);
draw(OA--OB);
draw(OA--(0,4));
draw(OA--(-33.9112699,0));
draw(OB--(10.2469508,0));
label("$24$",midpoint(OA--OB),N);
label("$\sqrt{560}$",midpoint(A--B),S);
label("$11$",midpoint(OA--(-33.9112699,0)),NW);
label("$13$",midpoint(OB--(10.2469508,0)),NE);
label("$r$",midpoint(midpoint(A--B)--A),S);
label("$r$",midpoint(midpoint(A--B)--B),S);
label("$r$",midpoint(A--(-33.9112699,0)),S);
label("$r$",midpoint(B--(10.2469508,0)),S);
label("$x$",midpoint(midpoint(B--OB)--OB),W);
label("$D$",midpoint(B--OB),E);
// Block 2
size(400);
pair A, C, OA, OC, M;
C = (0,0);
A = (-27.4954541697,0);
OC = (0,16);
OA = (-27.4954541697,4);
M = midpoint(A--C);
draw(circle(OC,19));
draw(circle(OA,11));
draw((-48,0)--(24,0));
label("$\ell$",(-42,0),S);
label("$A$",A,S);
label("$C$",C,S);
label("$O_A$",OA,N);
label("$O_C$",OC,N);
draw(A--OA);
draw(C--OC);
draw(OA--OC);
draw(OA--(0,4));
draw(OA--(-37.8877590151,0));
draw(OC--(10.2469508,0));
label("$30$",midpoint(OA--OC),NW);
label("$11$",midpoint(OA--(-37.8877590151,0)),NW);
label("$19$",midpoint(OC--(10.2469508,0)),NE);
label("$r$",midpoint(midpoint(M--A)--A),S);
label("$r$",midpoint(midpoint(M--C)--C),S);
label("$r$",midpoint(A--(-37.8877590151,0)),S);
label("$r$",midpoint(C--(10.2469508,0)),S);
label("$E$",(0,4),E);
// Block 3
size(400); pair A, B, OA, OB; B = (0,0); A = (-23.6643191,0); OB = (0,8); OA = (-23.6643191,4); draw(circle(OB,13)); draw(circle(OA,11)); draw((-48,0)--(24,0)); label("$\ell$",(-42,0),S); label("$A$",A,S); label("$B$",B,S); label("$O_A$",OA,N); label("$O_B$",OB,N); draw(A--OA); draw(B--OB); draw(OA--OB); draw(OA--(0,4)); draw(OA--(-33.9112699,0)); draw(OB--(10.2469508,0)); label("$24$",midpoint(OA--OB),N); label("$\sqrt{560}$",midpoint(A--B),S); label("$11$",midpoint(OA--(-33.9112699,0)),NW); label("$13$",midpoint(OB--(10.2469508,0)),NE); label("$r$",midpoint(midpoint(A--B)--A),S); label("$r$",midpoint(midpoint(A--B)--B),S); label("$r$",midpoint(A--(-33.9112699,0)),S); label("$r$",midpoint(B--(10.2469508,0)),S); label("$x$",midpoint(midpoint(B--OB)--OB),W); label("$D$",midpoint(B--OB),E);
// Block 4
size(400); pair A, C, OA, OC, M; C = (0,0); A = (-27.4954541697,0); OC = (0,16); OA = (-27.4954541697,4); M = midpoint(A--C); draw(circle(OC,19)); draw(circle(OA,11)); draw((-48,0)--(24,0)); label("$\ell$",(-42,0),S); label("$A$",A,S); label("$C$",C,S); label("$O_A$",OA,N); label("$O_C$",OC,N); draw(A--OA); draw(C--OC); draw(OA--OC); draw(OA--(0,4)); draw(OA--(-37.8877590151,0)); draw(OC--(10.2469508,0)); label("$30$",midpoint(OA--OC),NW); label("$11$",midpoint(OA--(-37.8877590151,0)),NW); label("$19$",midpoint(OC--(10.2469508,0)),NE); label("$r$",midpoint(midpoint(M--A)--A),S); label("$r$",midpoint(midpoint(M--C)--C),S); label("$r$",midpoint(A--(-37.8877590151,0)),S); label("$r$",midpoint(C--(10.2469508,0)),S); label("$E$",(0,4),E); | [] |
247 | Let $ABCD$ be a parallelogram with $\angle BAD < 90^\circ.$ A circle tangent to sides $\overline{DA},$ $\overline{AB},$ and $\overline{BC}$ intersects diagonal $\overline{AC}$ at points $P$ and $Q$ with $AP < AQ,$ as shown. Suppose that $AP=3,$ $PQ=9,$ and $QC=16.$ Then the area of $ABCD$ can be expressed in the form $m\sqrt{n},$ where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n.$ | 2022 AIME I Problem 11 | Let's redraw the diagram, but extend some helpful lines.
We obviously see that we must use power of a point since they've given us lengths in a circle and there are intersection points. Let $T_1, T_2, T_3$ be our tangents from the circle to the parallelogram. By the secant power of a point, the power of $A = 3 \cdot (3+9) = 36$. Then $AT_2 = AT_3 = \sqrt{36} = 6$. Similarly, the power of $C = 16 \cdot (16+9) = 400$ and $CT_1 = \sqrt{400} = 20$. We let $BT_3 = BT_1 = x$ and label the diagram accordingly.
Notice that because $BC = AD, 20+x = 6+DT_2 \implies DT_2 = 14+x$. Let $O$ be the center of the circle. Since $OT_1$ and $OT_2$ intersect $BC$ and $AD$, respectively, at right angles, we have $T_2T_1CD$ is a right-angled trapezoid and more importantly, the diameter of the circle is the height of the triangle. Therefore, we can drop an altitude from $D$ to $BC$ and $C$ to $AD$, and both are equal to $2r$. Since $T_1E = T_2D$, $20 - CE = 14+x \implies CE = 6-x$. Since $CE = DF, DF = 6-x$ and $AF = 6+14+x+6-x = 26$. We can now use Pythagorean theorem on $\triangle ACF$; we have $26^2 + (2r)^2 = (3+9+16)^2 \implies 4r^2 = 784-676 \implies 4r^2 = 108 \implies 2r = 6\sqrt{3}$ and $r^2 = 27$.
We know that $CD = 6+x$ because $ABCD$ is a parallelogram. Using Pythagorean theorem on $\triangle CDF$, $(6+x)^2 = (6-x)^2 + 108 \implies (6+x)^2-(6-x)^2 = 108 \implies 12 \cdot 2x = 108 \implies 2x = 9 \implies x = \frac{9}{2}$. Therefore, base $BC = 20 + \frac{9}{2} = \frac{49}{2}$. Thus the area of the parallelogram is the base times the height, which is $\frac{49}{2} \cdot 6\sqrt{3} = 147\sqrt{3}$ and the answer is $\boxed{150}$
~KingRavi | // Block 1
size(10cm);
pair A,B,C,D,EE,F,P,Q,O;
A=(0,0);
EE = (24,15);
F = (30,0);
O = (10.5,7.5);
label("$A$", A, SW);
B=(6,15);
label("$B$", B, NW);
C=(30,15);
label("$C$", C, NE);
D=(24,0);
label("$D$", D, SE);
P=(5.2,2.6);
label("$P$", (5.8,2.6), N);
Q=(18.3,9.1);
label("$Q$", (18.1,9.7), W);
draw(A--B--C--D--cycle);
draw(C--A);
draw(Circle((10.95,7.45), 7.45));
dot(A^^B^^C^^D^^P^^Q);
dot(O);
label("$O$",O,W);
draw((10.5,15)--(10.5,0));
draw(D--(24,15),dashed);
draw(C--(30,0),dashed);
draw(D--(30,0));
dot(EE);
dot(F);
label("$3$", midpoint(A--P), S);
label("$9$", midpoint(P--Q), S);
label("$16$", midpoint(Q--C), S);
label("$x$", (5.5,13.75), W);
label("$20$", (20.25,15), N);
label("$6$", (5.25,0), S);
label("$6$", (1.5,3.75), W);
label("$x$", (8.25,15),N);
label("$14+x$", (17.25,0), S);
label("$6-x$", (27,15), N);
label("$6+x$", (27,7.5), S);
label("$6\sqrt{3}$", (30,7.5), E);
label("$T_1$", (10.5,15), N);
label("$T_2$", (10.5,0), S);
label("$T_3$", (4.5,11.25), W);
label("$E$", EE, N);
label("$F$", F, S);
// Block 2
size(10cm); pair A,B,C,D,EE,F,P,Q,O; A=(0,0); EE = (24,15); F = (30,0); O = (10.5,7.5); label("$A$", A, SW); B=(6,15); label("$B$", B, NW); C=(30,15); label("$C$", C, NE); D=(24,0); label("$D$", D, SE); P=(5.2,2.6); label("$P$", (5.8,2.6), N); Q=(18.3,9.1); label("$Q$", (18.1,9.7), W); draw(A--B--C--D--cycle); draw(C--A); draw(Circle((10.95,7.45), 7.45)); dot(A^^B^^C^^D^^P^^Q); dot(O); label("$O$",O,W); draw((10.5,15)--(10.5,0)); draw(D--(24,15),dashed); draw(C--(30,0),dashed); draw(D--(30,0)); dot(EE); dot(F); label("$3$", midpoint(A--P), S); label("$9$", midpoint(P--Q), S); label("$16$", midpoint(Q--C), S); label("$x$", (5.5,13.75), W); label("$20$", (20.25,15), N); label("$6$", (5.25,0), S); label("$6$", (1.5,3.75), W); label("$x$", (8.25,15),N); label("$14+x$", (17.25,0), S); label("$6-x$", (27,15), N); label("$6+x$", (27,7.5), S); label("$6\sqrt{3}$", (30,7.5), E); label("$T_1$", (10.5,15), N); label("$T_2$", (10.5,0), S); label("$T_3$", (4.5,11.25), W); label("$E$", EE, N); label("$F$", F, S); | [] |
248 | A circle with radius $6$ is externally tangent to a circle with radius $24$. Find the area of the triangular region bounded by the three common tangent lines of these two circles. | 2022 AIME II Problem 7 | $r_1 = O_1A = 24$, $r_2 = O_2B = 6$, $AG = BO_2 = r_2 = 6$, $O_1G = r_1 - r_2 = 24 - 6 = 18$, $O_1O_2 = r_1 + r_2 = 30$
$\triangle O_2BD \sim \triangle O_1GO_2$, $\frac{O_2D}{O_1O_2} = \frac{BO_2}{GO_1}$, $\frac{O_2D}{30} = \frac{6}{18}$, $O_2D = 10$
$CD = O_2D + r_2 = 10 + 6 = 16$,
$EF = 2EC = EA + EB = AB = GO_2 = \sqrt{(O_1O_2)^2-(O_1G)^2} = \sqrt{30^2-18^2} = 24$
$DEF = \frac12 \cdot EF \cdot CD = \frac12 \cdot 24 \cdot 16 = \boxed{\textbf{192}}$
~isabelchen
Alternative Finish
Note that $\triangle{O_1 O_2 G} \sim \triangle{O_1 D A}$ by AA similarity. Setting up the ratio $\frac{18}{24}=\frac{24}{AB+8}$ and then substituting $AB$ with our known values will lead us to the same solution.
mathboy282 | // Block 1
//Created by isabelchen
size(12cm, 12cm);
draw(circle((0,0),24));
draw(circle((30,0),6));
draw((72/5, 96/5) -- (40,0));
draw((72/5, -96/5) -- (40,0));
draw((24, 12) -- (24, -12));
draw((0, 0) -- (40, 0));
draw((72/5, 96/5) -- (0,0));
draw((168/5, 24/5) -- (30,0));
draw((54/5, 72/5) -- (30,0));
dot((72/5, 96/5));
label("$A$",(72/5, 96/5),NE);
dot((168/5, 24/5));
label("$B$",(168/5, 24/5),NE);
dot((24,0));
label("$C$",(24,0),NW);
dot((40, 0));
label("$D$",(40, 0),NE);
dot((24, 12));
label("$E$",(24, 12),NE);
dot((24, -12));
label("$F$",(24, -12),SE);
dot((54/5, 72/5));
label("$G$",(54/5, 72/5),NW);
dot((0, 0));
label("$O_1$",(0, 0),S);
dot((30, 0));
label("$O_2$",(30, 0),S);
// Block 2
//Created by isabelchen size(12cm, 12cm); draw(circle((0,0),24)); draw(circle((30,0),6)); draw((72/5, 96/5) -- (40,0)); draw((72/5, -96/5) -- (40,0)); draw((24, 12) -- (24, -12)); draw((0, 0) -- (40, 0)); draw((72/5, 96/5) -- (0,0)); draw((168/5, 24/5) -- (30,0)); draw((54/5, 72/5) -- (30,0)); dot((72/5, 96/5)); label("$A$",(72/5, 96/5),NE); dot((168/5, 24/5)); label("$B$",(168/5, 24/5),NE); dot((24,0)); label("$C$",(24,0),NW); dot((40, 0)); label("$D$",(40, 0),NE); dot((24, 12)); label("$E$",(24, 12),NE); dot((24, -12)); label("$F$",(24, -12),SE); dot((54/5, 72/5)); label("$G$",(54/5, 72/5),NW); dot((0, 0)); label("$O_1$",(0, 0),S); dot((30, 0)); label("$O_2$",(30, 0),S); | [] |
248 | A circle with radius $6$ is externally tangent to a circle with radius $24$. Find the area of the triangular region bounded by the three common tangent lines of these two circles. | 2022 AIME II Problem 7 | First, we want to find $O_2D$. We know that $\angle O_1AD = \angle O_2BD = 90^{\circ}$, so by AA similarity, $\triangle O_1AD \sim \triangle O_2BD$. We want to find the length of $x$, and using the similar triangles, we write an equation: $\frac{30 + x}{4} = x$. Solving, we get $x=10$. Therefore, $CD = 10 + 6 = 16$. Next, we find that using AA similarity, $\triangle O_2BD \sim \triangle HO_2D \sim \triangle ECD$ and they are 3-4-5 triangles. We can quickly compute $EF = 2EC = 2 \cdot \left( \frac{3}{4} \cdot 16 \right) = 2 \cdot 12 = 24$. Therefore, the area is $\frac{1}{2} \cdot 16 \cdot 24 = \boxed{192}$.
~afly | // Block 1
//Created by isabelchen and edited by afly
size(12cm, 12cm);
draw(circle((0,0),24));
draw(circle((30,0),6));
draw((72/5, 96/5) -- (40,0));
draw((72/5, -96/5) -- (40,0));
draw((24, 12) -- (24, -12));
draw((0, 0) -- (40, 0));
draw((72/5, 96/5) -- (0,0));
draw((168/5, 24/5) -- (30,0));
draw((54/5, 72/5) -- (30,0));
draw((30,0)--(30,15/2));
dot((72/5, 96/5));
label("$A$",(72/5, 96/5),NE);
dot((168/5, 24/5));
label("$B$",(168/5, 24/5),NE);
dot((24,0));
label("$C$",(24,0),NW);
dot((40, 0));
label("$D$",(40, 0),NE);
dot((24, 12));
label("$E$",(24, 12),NE);
dot((24, -12));
label("$F$",(24, -12),SE);
dot((54/5, 72/5));
label("$G$",(54/5, 72/5),NW);
dot((0, 0));
label("$O_1$",(0, 0),S);
dot((30, 0));
label("$O_2$",(30, 0),S);
dot((30,15/2));
label("$H$",(30,15/2),N);
label("$x$",(30,0)--(40,0),N);
// Block 2
//Created by isabelchen and edited by afly size(12cm, 12cm); draw(circle((0,0),24)); draw(circle((30,0),6)); draw((72/5, 96/5) -- (40,0)); draw((72/5, -96/5) -- (40,0)); draw((24, 12) -- (24, -12)); draw((0, 0) -- (40, 0)); draw((72/5, 96/5) -- (0,0)); draw((168/5, 24/5) -- (30,0)); draw((54/5, 72/5) -- (30,0)); draw((30,0)--(30,15/2)); dot((72/5, 96/5)); label("$A$",(72/5, 96/5),NE); dot((168/5, 24/5)); label("$B$",(168/5, 24/5),NE); dot((24,0)); label("$C$",(24,0),NW); dot((40, 0)); label("$D$",(40, 0),NE); dot((24, 12)); label("$E$",(24, 12),NE); dot((24, -12)); label("$F$",(24, -12),SE); dot((54/5, 72/5)); label("$G$",(54/5, 72/5),NW); dot((0, 0)); label("$O_1$",(0, 0),S); dot((30, 0)); label("$O_2$",(30, 0),S); dot((30,15/2)); label("$H$",(30,15/2),N); label("$x$",(30,0)--(40,0),N); | [] |
249 | Let $ABCD$ be a convex quadrilateral with $AB=2, AD=7,$ and $CD=3$ such that the bisectors of acute angles $\angle{DAB}$ and $\angle{ADC}$ intersect at the midpoint of $\overline{BC}.$ Find the square of the area of $ABCD.$ | 2022 AIME II Problem 11 | According to the problem, we have $AB=AB'=2$, $DC=DC'=3$, $MB=MB'$, $MC=MC'$, and $B'C'=7-2-3=2$
Because $M$ is the midpoint of $BC$, we have $BM=MC$, so: \[MB=MB'=MC'=MC.\]
Then, we can see that $\bigtriangleup{MB'C'}$ is an isosceles triangle with $MB'=MC'$
Therefore, we could start our angle chasing: $\angle{MB'C'}=\angle{MC'B'}=180^\circ-\angle{MC'D}=180^\circ-\angle{MCD}$.
This is when we found that points $M$, $C$, $D$, and $B'$ are on a circle. Thus, $\angle{BMB'}=\angle{CDC'} \Rightarrow \angle{B'MA}=\angle{C'DM}$. This is the time we found that $\bigtriangleup{AB'M} \sim \bigtriangleup{MC'D}$.
Thus, $\frac{AB'}{B'M}=\frac{MC'}{C'D} \Longrightarrow (B'M)^2=AB' \cdot C'D = 6$
Point $H$ is the midpoint of $B'C'$, and $MH \perp AD$. $B'H=HC'=1 \Longrightarrow MH=\sqrt{B'M^2-B'H^2}=\sqrt{6-1}=\sqrt{5}$.
The area of this quadrilateral is the sum of areas of triangles: \[S_{\bigtriangleup{ABM}}+S_{\bigtriangleup{AB'M}}+S_{\bigtriangleup{CDM}}+S_{\bigtriangleup{C'DM}}+S_{\bigtriangleup{B'C'M}}\]
\[=S_{\bigtriangleup{AB'M}}\cdot 2 + S_{\bigtriangleup{B'C'M}} + S_{\bigtriangleup{C'DM}}\cdot 2\]
\[=2 \cdot \frac{1}{2} \cdot AB' \cdot MH + \frac{1}{2} \cdot B'C' \cdot MH + 2 \cdot \frac{1}{2} \cdot C'D \cdot MH\]
\[=2\sqrt{5}+\sqrt{5}+3\sqrt{5}=6\sqrt{5}\]
Finally, the square of the area is $(6\sqrt{5})^2=\boxed{180}$ | // Block 1
defaultpen(fontsize(12)+0.6); size(300);
pair A,B,C,D,M,H; real xb=71, xd=121;
A=origin; D=(7,0); B=2*dir(xb); C=3*dir(xd)+D; M=(B+C)/2; H=foot(M,A,D); path c=CR(D,3); pair A1=bisectorpoint(D,A,B), D1=bisectorpoint(C,D,A), Bp=IP(CR(A,2),A--H), Cp=IP(CR(D,3),D--H);
draw(B--A--D--C--B); draw(A--M--D^^M--H^^Bp--M--Cp, gray+0.4); draw(rightanglemark(A,H,M,5));
dot("$A$",A,SW); dot("$D$",D,SE); dot("$B$",B,NW); dot("$C$",C,NE); dot("$M$",M,up); dot("$H$",H,down); dot("$B'$",Bp,down); dot("$C'$",Cp,down);
// Block 2
defaultpen(fontsize(12)+0.6); size(300); pair A,B,C,D,M,H; real xb=71, xd=121; A=origin; D=(7,0); B=2*dir(xb); C=3*dir(xd)+D; M=(B+C)/2; H=foot(M,A,D); path c=CR(D,3); pair A1=bisectorpoint(D,A,B), D1=bisectorpoint(C,D,A), Bp=IP(CR(A,2),A--H), Cp=IP(CR(D,3),D--H); draw(B--A--D--C--B); draw(A--M--D^^M--H^^Bp--M--Cp, gray+0.4); draw(rightanglemark(A,H,M,5)); dot("$A$",A,SW); dot("$D$",D,SE); dot("$B$",B,NW); dot("$C$",C,NE); dot("$M$",M,up); dot("$H$",H,down); dot("$B'$",Bp,down); dot("$C'$",Cp,down); | [] |
250 | Two externally tangent circles $\omega_1$ and $\omega_2$ have centers $O_1$ and $O_2$, respectively. A third circle $\Omega$ passing through $O_1$ and $O_2$ intersects $\omega_1$ at $B$ and $C$ and $\omega_2$ at $A$ and $D$, as shown. Suppose that $AB = 2$, $O_1O_2 = 15$, $CD = 16$, and $ABO_1CDO_2$ is a convex hexagon. Find the area of this hexagon. | 2022 AIME II Problem 15 | First observe that $AO_2 = O_2D$ and $BO_1 = O_1C$. Let points $A'$ and $B'$ be the reflections of $A$ and $B$, respectively, about the perpendicular bisector of $\overline{O_1O_2}$. Then quadrilaterals $ABO_1O_2$ and $B'A'O_2O_1$ are congruent, so hexagons $ABO_1CDO_2$ and $A'B'O_1CDO_2$ have the same area. Furthermore, triangles $DO_2A'$ and $B'O_1C$ are congruent, so $A'D = B'C$ and quadrilateral $A'B'CD$ is an isosceles trapezoid.
Next, remark that $B'O_1 = DO_2$, so quadrilateral $B'O_1DO_2$ is also an isosceles trapezoid; in turn, $B'D = O_1O_2 = 15$, and similarly $A'C = 15$. Thus, Ptolemy's theorem on $A'B'CD$ yields $A'D\cdot B'C + 2\cdot 16 = 15^2$, whence $A'D = B'C = \sqrt{193}$. Let $\alpha = \angle A'B'D$. The Law of Cosines on triangle $A'B'D$ yields
\[\cos\alpha = \frac{15^2 + 2^2 - (\sqrt{193})^2}{2\cdot 2\cdot 15} = \frac{36}{60} = \frac 35,\] and hence $\sin\alpha = \tfrac 45$. Thus the distance between bases $A’B’$ and $CD$ is $12$ (in fact, $\triangle A'B'D$ is a $9-12-15$ triangle with a $7-12-\sqrt{193}$ triangle removed), which implies the area of $A'B'CD$ is $\tfrac12\cdot 12\cdot(2+16) = 108$.
Now let $O_1C = O_2A' = r_1$ and $O_2D = O_1B' = r_2$; the tangency of circles $\omega_1$ and $\omega_2$ implies $r_1 + r_2 = 15$. Furthermore, angles $A'O_2D$ and $A'B'D$ are opposite angles in cyclic quadrilateral $B'A'O_2D$, which implies the measure of angle $A'O_2D$ is $180^\circ - \alpha$. Therefore, the Law of Cosines applied to triangle $\triangle A'O_2D$ yields
\begin{align*} 193 &= r_1^2 + r_2^2 - 2r_1r_2(-\tfrac 35) = (r_1^2 + 2r_1r_2 + r_2^2) - \tfrac45r_1r_2\\ &= (r_1+r_2)^2 - \tfrac45 r_1r_2 = 225 - \tfrac45r_1r_2. \end{align*}
Thus $r_1r_2 = 40$, and so the area of triangle $A'O_2D$ is $\tfrac12r_1r_2\sin\alpha = 16$.
Thus, the area of hexagon $ABO_{1}CDO_{2}$ is $108 + 2\cdot 16 = \boxed{140}$.
~djmathman
Additional Graph for Better Understanding (Rearranging of the Pizza Slices)
This is how the reflection mentioned above is actually done. The reflection is actually a reorganization of the red and blue triangles, creating a symmetric figure with a isosceles trapezoid without changing the area. Basically you rearrange them so that each side contains a red triangle above a blue one. Then you calculate the area of the trapezoid and the two congruent triangles beside.
[Image: images/aime/2022_AIME_II_Problem_15_0.png]
~cassphe | // Block 1
import olympiad;
size(180);
defaultpen(linewidth(0.7));
pair Ap = dir(105), Bp = dir(75), O1 = dir(25), C = dir(320), D = dir(220), O2 = dir(175);
draw(unitcircle^^Ap--Bp--O1--C--D--O2--cycle);
label("$A'$",Ap,dir(origin--Ap));
label("$B'$",Bp,dir(origin--Bp));
label("$O_1$",O1,dir(origin--O1));
label("$C$",C,dir(origin--C));
label("$D$",D,dir(origin--D));
label("$O_2$",O2,dir(origin--O2));
draw(O2--O1,linetype("4 4"));
draw(Ap--D^^Bp--C,linetype("2 2"));
// Block 2
import olympiad; size(180); defaultpen(linewidth(0.7)); pair Ap = dir(105), Bp = dir(75), O1 = dir(25), C = dir(320), D = dir(220), O2 = dir(175); draw(unitcircle^^Ap--Bp--O1--C--D--O2--cycle); label("$A'$",Ap,dir(origin--Ap)); label("$B'$",Bp,dir(origin--Bp)); label("$O_1$",O1,dir(origin--O1)); label("$C$",C,dir(origin--C)); label("$D$",D,dir(origin--D)); label("$O_2$",O2,dir(origin--O2)); draw(O2--O1,linetype("4 4")); draw(Ap--D^^Bp--C,linetype("2 2")); | ["https://artofproblemsolving.com/wiki/images/thumb/9/98/Actual_Graph_for_Solution_1.PNG/500px-Actual_Graph_for_Solution_1.PNG"] |
250 | Two externally tangent circles $\omega_1$ and $\omega_2$ have centers $O_1$ and $O_2$, respectively. A third circle $\Omega$ passing through $O_1$ and $O_2$ intersects $\omega_1$ at $B$ and $C$ and $\omega_2$ at $A$ and $D$, as shown. Suppose that $AB = 2$, $O_1O_2 = 15$, $CD = 16$, and $ABO_1CDO_2$ is a convex hexagon. Find the area of this hexagon. | 2022 AIME II Problem 15 | Start by noting that the radius of $\omega_1$ and $\omega_2$ sum up to 15.
Let the radius of $\omega_1$ be $x$, then the radius of $\omega_2$ will be $15-x$. We can find the area of the hexagon as the sum of the area of the quadrilateral $[O_1BAO_2]$ and $[O_1O_2DC]$, which is given by the Brahmagupta's Formula as: $\sqrt{(16-2)(16-15)(16-x)(1+x)} + \sqrt{(23-15)(23-16)(23-x)(8+x)}$, we only need $x$, and the condition is sufficient already. Note that the total number of diagonals and sides of the hexagon is $\frac{6 \times 5}{2} = 15$, 2 pairs of them $O_1B, O_1C$, and $O_2A, O_2D$ can all be represented by the same variable $x$, and 3 of them ($AB, O_1O_2, CD$) are given. Therefore, the total number of variables is $15-4+1-3=9$. And we have $\frac{6 \times 5}{2} = 15$ equations by doing ptolemy on the quadrilaterals inside the hexagon, we can list some of them out in hope of bashing for a solution:
\[2\times 15 + x(15-x) = BO_2\times AO_1\]
\[15\times 16 + x(15-x) = DO_1\times CO_2\]
\[AC\times x + 2x = BC\times AO_1\]
\[AC\times (15-x) + 16(15-x) = AD\times CO_2\]
\[BD\times x + 16x = BC\times DO_1\]
\[BD\times (15-x) + 2(15-x) = AD\times BO_2\]
\[2CO_2 + BC\times (15-x) = AC\times BO_2\]
\[2DO_1 + AD\times x = BD\times AO_1\]
\[BO_2\times x + CO_2\times x = 15BC\]
\[AD\times x + 16AO_1 = AC\times DO_1\]
\[BC\times (15-x) + 16BO_2 = BD\times CO_2\]
\[AO_1\times (15-x) + DO_1\times (15-x) = 15AD\]
\[AO_1\times CO_2 + x(15-x) = 15AC\]
\[BC\times AD + 2\times 16 = AC\times BD\]
\[BO_2\times DO_1 + x(15-x) = 15BD\]
15 equations, 9 variables, hopefully one can solve this.
The question is especially disgusting in the sense that there is not a lot of geometric properties or hypothesis that one can conduct. The constructions seemed hard even to USAJMO level and lacked motivation. The normal trig took way too much time and this solution took even more. The question is a straightforward and brutal one when using ptolemy, all the conditions given and inherent are used and it guarantees a solution.
~IDKHowtoaddsolution | // Block 1
import geometry;
size(10cm);
point O1=(0,0),O2=(15,0),B=9*dir(30);
circle w1=circle(O1,9),w2=circle(O2,6),o=circle(O1,O2,B);
point A=intersectionpoints(o,w2)[1],D=intersectionpoints(o,w2)[0],C=intersectionpoints(o,w1)[0];
filldraw(A--B--O1--C--D--O2--cycle,0.2*fuchsia+white,black);
draw(w1);
draw(w2);
draw(O1--O2,dashed);
draw(o);
dot(O1);
dot(O2);
dot(A);
dot(D);
dot(C);
dot(B);
label("$\omega_1$",8*dir(110),SW);
label("$\omega_2$",5*dir(70)+(15,0),SE);
label("$O_1$",O1,W);
label("$O_2$",O2,E);
label("$B$",B,N+1/2*E);
label("$A$",A,N+1/2*W);
label("$C$",C,S+1/4*W);
label("$D$",D,S+1/4*E);
label("$15$",midpoint(O1--O2),N);
label("$16$",midpoint(C--D),N);
label("$2$",midpoint(A--B),S);
label("$x$", midpoint(O1--B),N);
label("$x$", midpoint(O1--C),E);
label("$15-x$", midpoint(O2--A),W);
label("$15-x$", midpoint(O2--D),E);
label("$\Omega$",o.C+(o.r-1)*dir(270));
// Block 2
import geometry; size(10cm); point O1=(0,0),O2=(15,0),B=9*dir(30); circle w1=circle(O1,9),w2=circle(O2,6),o=circle(O1,O2,B); point A=intersectionpoints(o,w2)[1],D=intersectionpoints(o,w2)[0],C=intersectionpoints(o,w1)[0]; filldraw(A--B--O1--C--D--O2--cycle,0.2*fuchsia+white,black); draw(w1); draw(w2); draw(O1--O2,dashed); draw(o); dot(O1); dot(O2); dot(A); dot(D); dot(C); dot(B); label("$\omega_1$",8*dir(110),SW); label("$\omega_2$",5*dir(70)+(15,0),SE); label("$O_1$",O1,W); label("$O_2$",O2,E); label("$B$",B,N+1/2*E); label("$A$",A,N+1/2*W); label("$C$",C,S+1/4*W); label("$D$",D,S+1/4*E); label("$15$",midpoint(O1--O2),N); label("$16$",midpoint(C--D),N); label("$2$",midpoint(A--B),S); label("$x$", midpoint(O1--B),N); label("$x$", midpoint(O1--C),E); label("$15-x$", midpoint(O2--A),W); label("$15-x$", midpoint(O2--D),E); label("$\Omega$",o.C+(o.r-1)*dir(270)); | [] |
251 | Let $P$ be a point on the circle circumscribing square $ABCD$ that satisfies $PA \cdot PC = 56$ and $PB \cdot PD = 90.$ Find the area of $ABCD.$ | 2023 AIME I Problem 5 | By the Inscribed Angle Theorem, we conclude that $\triangle PAC$ and $\triangle PBD$ are right triangles.
Let the brackets denote areas. We are given that
\begin{alignat*}{8} 2[PAC] &= PA \cdot PC &&= 56, \\ 2[PBD] &= PB \cdot PD &&= 90. \end{alignat*}
Let $O$ be the center of the circle, $X$ be the foot of the perpendicular from $P$ to $\overline{AC},$ and $Y$ be the foot of the perpendicular from $P$ to $\overline{BD},$ as shown below:
Let $d$ be the diameter of $\odot O.$ It follows that
\begin{alignat*}{8} 2[PAC] &= d\cdot PX &&= 56, \\ 2[PBD] &= d\cdot PY &&= 90. \end{alignat*}
Moreover, note that $OXPY$ is a rectangle. By the Pythagorean Theorem, we have \[PX^2+PY^2=PO^2.\]
We rewrite this equation in terms of $d:$ \[\left(\frac{56}{d}\right)^2+\left(\frac{90}{d}\right)^2=\left(\frac d2\right)^2,\] from which $d^2=212.$ Therefore, we get \[[ABCD] = \frac{d^2}{2} = \boxed{106}.\]
~MRENTHUSIASM | // Block 1
/* Made by MRENTHUSIASM */
size(200);
pair A, B, C, D, O, P, X, Y;
A = (-sqrt(106)/2,sqrt(106)/2);
B = (-sqrt(106)/2,-sqrt(106)/2);
C = (sqrt(106)/2,-sqrt(106)/2);
D = (sqrt(106)/2,sqrt(106)/2);
O = origin;
path p;
p = Circle(O,sqrt(212)/2);
draw(p);
P = intersectionpoints(Circle(A,4),p)[1];
X = foot(P,A,C);
Y = foot(P,B,D);
draw(A--B--C--D--cycle);
draw(P--A--C--cycle,red);
draw(P--B--D--cycle,blue);
draw(P--X,red+dashed);
draw(P--Y,blue+dashed);
markscalefactor=0.075;
draw(rightanglemark(A,P,C),red);
draw(rightanglemark(P,X,C),red);
draw(rightanglemark(B,P,D),blue);
draw(rightanglemark(P,Y,D),blue);
dot("$A$", A, 1.5*NW, linewidth(4));
dot("$B$", B, 1.5*SW, linewidth(4));
dot("$C$", C, 1.5*SE, linewidth(4));
dot("$D$", D, 1.5*NE, linewidth(4));
dot("$P$", P, 1.5*dir(P), linewidth(4));
dot("$X$", X, 1.5*dir(20), linewidth(4));
dot("$Y$", Y, 1.5*dir(Y-P), linewidth(4));
dot("$O$", O, 1.5*E, linewidth(4));
// Block 2
/* Made by MRENTHUSIASM */ size(200); pair A, B, C, D, O, P, X, Y; A = (-sqrt(106)/2,sqrt(106)/2); B = (-sqrt(106)/2,-sqrt(106)/2); C = (sqrt(106)/2,-sqrt(106)/2); D = (sqrt(106)/2,sqrt(106)/2); O = origin; path p; p = Circle(O,sqrt(212)/2); draw(p); P = intersectionpoints(Circle(A,4),p)[1]; X = foot(P,A,C); Y = foot(P,B,D); draw(A--B--C--D--cycle); draw(P--A--C--cycle,red); draw(P--B--D--cycle,blue); draw(P--X,red+dashed); draw(P--Y,blue+dashed); markscalefactor=0.075; draw(rightanglemark(A,P,C),red); draw(rightanglemark(P,X,C),red); draw(rightanglemark(B,P,D),blue); draw(rightanglemark(P,Y,D),blue); dot("$A$", A, 1.5*NW, linewidth(4)); dot("$B$", B, 1.5*SW, linewidth(4)); dot("$C$", C, 1.5*SE, linewidth(4)); dot("$D$", D, 1.5*NE, linewidth(4)); dot("$P$", P, 1.5*dir(P), linewidth(4)); dot("$X$", X, 1.5*dir(20), linewidth(4)); dot("$Y$", Y, 1.5*dir(Y-P), linewidth(4)); dot("$O$", O, 1.5*E, linewidth(4)); | [] |
251 | Let $P$ be a point on the circle circumscribing square $ABCD$ that satisfies $PA \cdot PC = 56$ and $PB \cdot PD = 90.$ Find the area of $ABCD.$ | 2023 AIME I Problem 5 | Let the center of the circle be $O$, and the radius of the circle be $r$. Since $ABCD$ is a rhombus with diagonals $2r$ and $2r$, its area is $\dfrac{1}{2}(2r)(2r) = 2r^2$. Since $AC$ and $BD$ are diameters of the circle, $\triangle APC$ and $\triangle BPD$ are right triangles. Let $X$ and $Y$ be the foot of the altitudes to $AC$ and $BD$, respectively. We have
\[[\triangle APC] = \frac{1}{2}(PA)(PC) = \frac{1}{2}(PX)(AC),\]
so $PX = \dfrac{(PA)(PC)}{AC} = \dfrac{28}{r}$. Similarly,
\[[\triangle BPD] = \frac{1}{2}(PB)(PD) = \frac{1}{2}(PY)(PB),\]
so $PY = \dfrac{(PB)(PD)}{BD} = \dfrac{45}{r}$. Since $\triangle APX \sim \triangle PCX,$
\[\frac{AX}{PX} = \frac{PX}{CX}\]
\[\frac{AO - XO}{PX} = \frac{PX}{OC + XO}.\]
But $PXOY$ is a rectangle, so $PY = XO$, and our equation becomes
\[\frac{r - PY}{PX} = \frac{PX}{r + PY}.\]
Cross multiplying and rearranging gives us $r^2 = PX^2 + PY^2 = \left(\dfrac{28}{r}\right)^2 + \left(\dfrac{45}{r}\right)^2$, which rearranges to $r^4 = 2809$. Therefore $[ABCD] = 2r^2 = \boxed{106}$.
~Cantalon | // Block 1
/* Made by MRENTHUSIASM */
size(200);
pair A, B, C, D, O, P, X, Y;
A = (-sqrt(106)/2,sqrt(106)/2);
B = (-sqrt(106)/2,-sqrt(106)/2);
C = (sqrt(106)/2,-sqrt(106)/2);
D = (sqrt(106)/2,sqrt(106)/2);
O = origin;
path p;
p = Circle(O,sqrt(212)/2);
draw(p);
P = intersectionpoints(Circle(A,4),p)[1];
X = foot(P,A,C);
Y = foot(P,B,D);
draw(A--B--C--D--cycle);
draw(P--A--C--cycle,red);
draw(P--B--D--cycle,blue);
draw(P--X,red+dashed);
draw(P--Y,blue+dashed);
markscalefactor=0.075;
draw(rightanglemark(A,P,C),red);
draw(rightanglemark(P,X,C),red);
draw(rightanglemark(B,P,D),blue);
draw(rightanglemark(P,Y,D),blue);
dot("$A$", A, 1.5*NW, linewidth(4));
dot("$B$", B, 1.5*SW, linewidth(4));
dot("$C$", C, 1.5*SE, linewidth(4));
dot("$D$", D, 1.5*NE, linewidth(4));
dot("$P$", P, 1.5*dir(P), linewidth(4));
dot("$X$", X, 1.5*dir(20), linewidth(4));
dot("$Y$", Y, 1.5*dir(Y-P), linewidth(4));
dot("$O$", O, 1.5*E, linewidth(4));
// Block 2
/* Made by MRENTHUSIASM */ size(200); pair A, B, C, D, O, P, X, Y; A = (-sqrt(106)/2,sqrt(106)/2); B = (-sqrt(106)/2,-sqrt(106)/2); C = (sqrt(106)/2,-sqrt(106)/2); D = (sqrt(106)/2,sqrt(106)/2); O = origin; path p; p = Circle(O,sqrt(212)/2); draw(p); P = intersectionpoints(Circle(A,4),p)[1]; X = foot(P,A,C); Y = foot(P,B,D); draw(A--B--C--D--cycle); draw(P--A--C--cycle,red); draw(P--B--D--cycle,blue); draw(P--X,red+dashed); draw(P--Y,blue+dashed); markscalefactor=0.075; draw(rightanglemark(A,P,C),red); draw(rightanglemark(P,X,C),red); draw(rightanglemark(B,P,D),blue); draw(rightanglemark(P,Y,D),blue); dot("$A$", A, 1.5*NW, linewidth(4)); dot("$B$", B, 1.5*SW, linewidth(4)); dot("$C$", C, 1.5*SE, linewidth(4)); dot("$D$", D, 1.5*NE, linewidth(4)); dot("$P$", P, 1.5*dir(P), linewidth(4)); dot("$X$", X, 1.5*dir(20), linewidth(4)); dot("$Y$", Y, 1.5*dir(Y-P), linewidth(4)); dot("$O$", O, 1.5*E, linewidth(4)); | [] |
251 | Let $P$ be a point on the circle circumscribing square $ABCD$ that satisfies $PA \cdot PC = 56$ and $PB \cdot PD = 90.$ Find the area of $ABCD.$ | 2023 AIME I Problem 5 | First draw a diagram.
Let's say that the radius is $r$. Then the area of the $ABCD$ is $(\sqrt2r)^2 = 2r^2$
Using the formula for the length of a chord subtended by an angle, we get
\[PA = 2r\sin\left(\dfrac{\theta}2\right)\]
\[PC = 2r\sin\left(\dfrac{180-\theta}2\right) = 2r\sin\left(90 - \dfrac{\theta}2\right) = 2r\cos\left(\dfrac{\theta}2\right)\]
Multiplying and simplifying these 2 equations gives
\[PA \cdot PC = 4r^2 \sin \left(\dfrac{\theta}2 \right) \cos \left(\dfrac{\theta}2 \right) = 2r^2 \sin\left(\theta \right) = 56\]
Similarly $PB = 2r\sin\left(\dfrac{90 +\theta}2\right)$ and $PD =2r\sin\left(\dfrac{90 -\theta}2\right)$. Again, multiplying gives
\[PB \cdot PD = 4r^2 \sin\left(\dfrac{90 +\theta}2\right) \sin\left(\dfrac{90 -\theta}2\right) = 4r^2 \sin\left(90 -\dfrac{90 -\theta}2\right) \sin\left(\dfrac{90 -\theta}2\right)\]
\[=4r^2 \sin\left(\dfrac{90 -\theta}2\right) \cos\left(\dfrac{90 -\theta}2\right) = 2r^2 \sin\left(90 - \theta \right) = 2r^2 \cos\left(\theta \right) = 90\]
Dividing $2r^2 \sin \left(\theta \right)$ by $2r^2 \cos \left( \theta \right)$ gives $\tan \left(\theta \right) = \dfrac{28}{45}$, so $\theta = \tan^{-1} \left(\dfrac{28}{45} \right)$.
Pluging this back into one of the equations, gives
\[2r^2 = \dfrac{90}{\cos\left(\tan^{-1}\left(\dfrac{28}{45}\right)\right)}\]
If we imagine a $28$-$45$-$53$ right triangle, we see that if $28$ is opposite and $45$ is adjacent, $\cos\left(\theta\right) = \dfrac{\text{adj}}{\text{hyp}} = \dfrac{45}{53}$. Now we see that
\[2r^2 = \dfrac{90}{\frac{45}{53}} = \boxed{106}.\]
~Voldemort101 | // Block 1
pair A, B, C, D, O, P;
A = (0,sqrt(106));
B = (0,0);
C = (sqrt(106),0);
D = (sqrt(106),sqrt(106));
O = (sqrt(106)/2, sqrt(106)/2);
P = intersectionpoint(circle(A, sqrt(212)*sin(atan(28/45)/2)), circle(O, sqrt(212)/2));
draw(A--B--C--D--cycle);
draw(circle(O, sqrt(212)/2));
label("$A$", A, NW);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, NE);
label("$P$", P, NW);
label("$O$", O, 1.5*S);
label("$\theta$", O, dir(120)*5);
draw(P--A--C--cycle, red);
draw(P--B--D--cycle, blue);
draw(P--O);
draw(anglemark(P,O,A,30));
dot(P);
dot(O);
// Block 2
pair A, B, C, D, O, P; A = (0,sqrt(106)); B = (0,0); C = (sqrt(106),0); D = (sqrt(106),sqrt(106)); O = (sqrt(106)/2, sqrt(106)/2); P = intersectionpoint(circle(A, sqrt(212)*sin(atan(28/45)/2)), circle(O, sqrt(212)/2)); draw(A--B--C--D--cycle); draw(circle(O, sqrt(212)/2)); label("$A$", A, NW); label("$B$", B, SW); label("$C$", C, SE); label("$D$", D, NE); label("$P$", P, NW); label("$O$", O, 1.5*S); label("$\theta$", O, dir(120)*5); draw(P--A--C--cycle, red); draw(P--B--D--cycle, blue); draw(P--O); draw(anglemark(P,O,A,30)); dot(P); dot(O); | [] |
251 | Let $P$ be a point on the circle circumscribing square $ABCD$ that satisfies $PA \cdot PC = 56$ and $PB \cdot PD = 90.$ Find the area of $ABCD.$ | 2023 AIME I Problem 5 | Let $P=(a,b)$ on the upper quarter of the circle, and let $k$ be the side length of the square. Hence, we want to find $k^2$. Let the center of the circle be $(0,0)$.
The two equations would thus become:
\[\left(\left(a+\dfrac{k}2\right)^2+\left(b-\dfrac{k}2\right)^2\right)\left(\left(a-\dfrac{k}2\right)^2+\left(b+\dfrac{k}2\right)^2\right)=56^2\]
\[\left(\left(a-\dfrac{k}2\right)^2+\left(b-\dfrac{k}2\right)^2\right)\left(\left(a+\dfrac{k}2\right)^2+\left(b+\dfrac{k}2\right)^2\right)=90^2\]
Now, let $m=\left(a+\dfrac{k}2\right)^2$, $n=\left(a-\dfrac{k}2\right)^2$, $o=\left(b+\dfrac{k}2\right)^2$, and $p=\left(b-\dfrac{k}2\right)^2$. Our equations now change to $(m+p)(n+o)=56^2=mn+op+mo+pn$ and $(n+p)(m+o)=90^2=mn+op+no+pm$. Subtracting the first from the second, we have $pm+no-mo-pn=p(m-n)-o(m-n)=(m-n)(p-o)=34\cdot146$. Substituting back in and expanding, we have $2ak\cdot-2bk=34\cdot146$, so $abk^2=-17\cdot73$. We now have one of our terms we need ($k^2$). Therefore, we only need to find $ab$ to find $k^2$.
We now write the equation of the circle, which point $P$ satisfies: \[a^2+b^2=\left(\dfrac{k\sqrt{2}}{2}\right)^2=\dfrac{k^2}2\]
We can expand the second equation, yielding \[\left(a^2+b^2+\dfrac{k^2}2+(ak+bk)\right)\left(a^2+b^2+\dfrac{k^2}2-(ak+bk)\right)=(k^2+k(a+b))(k^2-k(a+b))=8100.\]
Now, with difference of squares, we get $k^4-k^2\cdot(a+b)^2=k^2(k^2-(a+b)^2)=8100$. We can add $2abk^2=-17\cdot73\cdot2=-2482$ to this equation, which we can factor into $k^2(k^2-(a+b)^2+2ab)=k^2(k^2-(a^2+b^2))=8100-2482$. We realize that $a^2+b^2$ is the same as the equation of the circle, so we plug its equation in: $k^2\left(k^2-\dfrac{k^2}2\right)=5618$. We can combine like terms to get $k^2\cdot\dfrac{k^2}2=5618$, so $(k^2)^2=11236$.
Since the answer is an integer, we know $11236$ is a perfect square. Since it is even, it is divisible by $4$, so we can factor $11236=2^2\cdot2809$. With some testing with approximations and last-digit methods, we can find that $53^2=2809$. Therefore, taking the square root, we find that $k^2$, the area of square $ABCD$, is $2\cdot53=\boxed{106}$.
~wuwang2002 | // Block 1
pair A,B,C,D,P;
A=(-3,3);
B=(3,3);
C=(3,-3);
D=(-3,-3);
draw(A--B--C--D--cycle);
label(A,"$A$",NW);
label(B,"$B$",NE);
label(C,"$C$",SE);
label(D,"$D$",SW);
draw(circle((0,0),4.24264068712));
P=(-1,4.12310562562);
label(P,"$P$", NW);
pen k=red+dashed;
draw(P--A,k);
draw(P--B,k);
draw(P--C,k);
draw(P--D,k);
dot(P);
// Block 2
pair A,B,C,D,P; A=(-3,3); B=(3,3); C=(3,-3); D=(-3,-3); draw(A--B--C--D--cycle); label(A,"$A$",NW); label(B,"$B$",NE); label(C,"$C$",SE); label(D,"$D$",SW); draw(circle((0,0),4.24264068712)); P=(-1,4.12310562562); label(P,"$P$", NW); pen k=red+dashed; draw(P--A,k); draw(P--B,k); draw(P--C,k); draw(P--D,k); dot(P); | [] |
252 | Rhombus $ABCD$ has $\angle BAD < 90^\circ.$ There is a point $P$ on the incircle of the rhombus such that the distances from $P$ to the lines $DA,AB,$ and $BC$ are $9,$ $5,$ and $16,$ respectively. Find the perimeter of $ABCD.$ | 2023 AIME I Problem 8 | This solution refers to the Diagram section.
Let $O$ be the incenter of $ABCD$ for which $\odot O$ is tangent to $\overline{DA},\overline{AB},$ and $\overline{BC}$ at $X,Y,$ and $Z,$ respectively. Moreover, suppose that $R,S,$ and $T$ are the feet of the perpendiculars from $P$ to $\overleftrightarrow{DA},\overleftrightarrow{AB},$ and $\overleftrightarrow{BC},$ respectively, such that $\overline{RT}$ intersects $\odot O$ at $P$ and $Q.$
We obtain the following diagram:
Note that $\angle RXZ = \angle TZX = 90^\circ$ by the properties of tangents, so $RTZX$ is a rectangle. It follows that the diameter of $\odot O$ is $XZ = RT = 25.$
Let $x=PQ$ and $y=RX=TZ.$
We apply the Power of a Point Theorem to $R$ and $T:$
\begin{align*} y^2 &= 9(9+x), \\ y^2 &= 16(16-x). \end{align*}
We solve this system of equations to get $x=7$ and $y=12.$ Alternatively, we can find these results by the symmetry on rectangle $RTZX$ and semicircle $\widehat{XPZ}.$
We extend $\overline{SP}$ beyond $P$ to intersect $\odot O$ and $\overleftrightarrow{CD}$ at $E$ and $F,$ respectively, where $E\neq P.$ So, we have $EF=SP=5$ and $PE=25-SP-EF=15.$ On the other hand, we have $PX=15$ by the Pythagorean Theorem on right $\triangle PRX.$ Together, we conclude that $E=X.$ Therefore, points $S,P,$ and $X$ must be collinear.
Let $G$ be the foot of the perpendicular from $D$ to $\overline{AB}.$ Note that $\overline{DG}\parallel\overline{XP},$ as shown below:
As $\angle PRX = \angle AGD = 90^\circ$ and $\angle PXR = \angle ADG,$ we conclude that $\triangle PRX \sim \triangle AGD$ by the AA Similarity. The ratio of similitude is \[\frac{PX}{AD} = \frac{RX}{GD}.\] We get $\frac{15}{AD} = \frac{12}{25},$ from which $AD = \frac{125}{4}.$
Finally, the perimeter of $ABCD$ is $4AD = \boxed{125}.$
~MRENTHUSIASM (inspired by awesomeming327. and WestSuburb) | // Block 1
/* Made by MRENTHUSIASM; inspired by Math Jams. */
size(300);
pair A, B, C, D, O, P, R, S, T, X, Y, Z, Q;
A = origin;
B = (125/4,0);
C = B + 125/4 * dir((3,4));
D = A + 125/4 * dir((3,4));
O = (25,25/2);
P = (15,5);
R = foot(P,A,D);
S = foot(P,A,B);
T = foot(P,B,C);
X = (15,20);
Y = (25,0);
Z = (35,5);
Q = intersectionpoints(Circle(O,25/2),R--T)[1];
fill(R--T--Z--X--cycle,cyan);
markscalefactor=0.15;
draw(rightanglemark(P,R,D)^^rightanglemark(P,S,B)^^rightanglemark(P,T,C),red);
draw(Circle(O,25/2)^^A--B--C--D--cycle^^B--T);
draw(P--R^^P--S^^P--T,red+dashed);
draw(O--X^^O--Y^^O--Z);
dot("$A$",A,1.5*dir(225),linewidth(4.5));
dot("$B$",B,1.5*dir(-45),linewidth(4.5));
dot("$C$",C,1.5*dir(45),linewidth(4.5));
dot("$D$",D,1.5*dir(135),linewidth(4.5));
dot("$P$",P,1.5*dir(60),linewidth(4.5));
dot("$R$",R,1.5*dir(135),linewidth(4.5));
dot("$S$",S,1.5*dir(-90),linewidth(4.5));
dot("$T$",T,1.5*dir(-45),linewidth(4.5));
dot("$O$",O,1.5*dir(45),linewidth(4.5));
dot("$X$",X,1.5*dir(135),linewidth(4.5));
dot("$Y$",Y,1.5*dir(-90),linewidth(4.5));
dot("$Z$",Z,1.5*dir(-45),linewidth(4.5));
dot("$Q$",Q,1.5*dir(60),linewidth(4.5));
label("$9$",midpoint(P--R),dir(A-D),red);
label("$5$",midpoint(P--S),dir(180),red);
label("$16$",midpoint(P--T),dir(A-D),red);
// Block 2
/* Made by MRENTHUSIASM; inspired by Math Jams. */
size(300);
pair A, B, C, D, O, P, R, S, T, X, Y, Z, Q, G;
A = origin;
B = (125/4,0);
C = B + 125/4 * dir((3,4));
D = A + 125/4 * dir((3,4));
O = (25,25/2);
P = (15,5);
R = foot(P,A,D);
S = foot(P,A,B);
T = foot(P,B,C);
X = (15,20);
Y = (25,0);
Z = (35,5);
Q = intersectionpoints(Circle(O,25/2),R--T)[1];
G = foot(D,A,B);
fill(D--A--G--cycle,green);
fill(P--R--X--cycle,yellow);
markscalefactor=0.15;
draw(rightanglemark(P,R,D)^^rightanglemark(D,G,A),red);
draw(Circle(O,25/2)^^A--B--C--D--cycle^^X--P^^D--G);
draw(P--R,red+dashed);
dot("$A$",A,1.5*dir(225),linewidth(4.5));
dot("$B$",B,1.5*dir(-45),linewidth(4.5));
dot("$C$",C,1.5*dir(45),linewidth(4.5));
dot("$D$",D,1.5*dir(135),linewidth(4.5));
dot("$P$",P,1.5*dir(60),linewidth(4.5));
dot("$R$",R,1.5*dir(135),linewidth(4.5));
dot("$O$",O,1.5*dir(45),linewidth(4.5));
dot("$X$",X,1.5*dir(135),linewidth(4.5));
dot("$G$",G,1.5*dir(-90),linewidth(4.5));
draw(P--X,MidArrow(0.3cm,Fill(red)));
draw(G--D,MidArrow(0.3cm,Fill(red)));
label("$9$",midpoint(P--R),dir(A-D),red);
label("$12$",midpoint(R--X),dir(135),red);
label("$15$",midpoint(X--P),dir(0),red);
label("$25$",midpoint(G--D),dir(0),red);
// Block 3
/* Made by MRENTHUSIASM; inspired by Math Jams. */ size(300); pair A, B, C, D, O, P, R, S, T, X, Y, Z, Q; A = origin; B = (125/4,0); C = B + 125/4 * dir((3,4)); D = A + 125/4 * dir((3,4)); O = (25,25/2); P = (15,5); R = foot(P,A,D); S = foot(P,A,B); T = foot(P,B,C); X = (15,20); Y = (25,0); Z = (35,5); Q = intersectionpoints(Circle(O,25/2),R--T)[1]; fill(R--T--Z--X--cycle,cyan); markscalefactor=0.15; draw(rightanglemark(P,R,D)^^rightanglemark(P,S,B)^^rightanglemark(P,T,C),red); draw(Circle(O,25/2)^^A--B--C--D--cycle^^B--T); draw(P--R^^P--S^^P--T,red+dashed); draw(O--X^^O--Y^^O--Z); dot("$A$",A,1.5*dir(225),linewidth(4.5)); dot("$B$",B,1.5*dir(-45),linewidth(4.5)); dot("$C$",C,1.5*dir(45),linewidth(4.5)); dot("$D$",D,1.5*dir(135),linewidth(4.5)); dot("$P$",P,1.5*dir(60),linewidth(4.5)); dot("$R$",R,1.5*dir(135),linewidth(4.5)); dot("$S$",S,1.5*dir(-90),linewidth(4.5)); dot("$T$",T,1.5*dir(-45),linewidth(4.5)); dot("$O$",O,1.5*dir(45),linewidth(4.5)); dot("$X$",X,1.5*dir(135),linewidth(4.5)); dot("$Y$",Y,1.5*dir(-90),linewidth(4.5)); dot("$Z$",Z,1.5*dir(-45),linewidth(4.5)); dot("$Q$",Q,1.5*dir(60),linewidth(4.5)); label("$9$",midpoint(P--R),dir(A-D),red); label("$5$",midpoint(P--S),dir(180),red); label("$16$",midpoint(P--T),dir(A-D),red);
// Block 4
/* Made by MRENTHUSIASM; inspired by Math Jams. */ size(300); pair A, B, C, D, O, P, R, S, T, X, Y, Z, Q, G; A = origin; B = (125/4,0); C = B + 125/4 * dir((3,4)); D = A + 125/4 * dir((3,4)); O = (25,25/2); P = (15,5); R = foot(P,A,D); S = foot(P,A,B); T = foot(P,B,C); X = (15,20); Y = (25,0); Z = (35,5); Q = intersectionpoints(Circle(O,25/2),R--T)[1]; G = foot(D,A,B); fill(D--A--G--cycle,green); fill(P--R--X--cycle,yellow); markscalefactor=0.15; draw(rightanglemark(P,R,D)^^rightanglemark(D,G,A),red); draw(Circle(O,25/2)^^A--B--C--D--cycle^^X--P^^D--G); draw(P--R,red+dashed); dot("$A$",A,1.5*dir(225),linewidth(4.5)); dot("$B$",B,1.5*dir(-45),linewidth(4.5)); dot("$C$",C,1.5*dir(45),linewidth(4.5)); dot("$D$",D,1.5*dir(135),linewidth(4.5)); dot("$P$",P,1.5*dir(60),linewidth(4.5)); dot("$R$",R,1.5*dir(135),linewidth(4.5)); dot("$O$",O,1.5*dir(45),linewidth(4.5)); dot("$X$",X,1.5*dir(135),linewidth(4.5)); dot("$G$",G,1.5*dir(-90),linewidth(4.5)); draw(P--X,MidArrow(0.3cm,Fill(red))); draw(G--D,MidArrow(0.3cm,Fill(red))); label("$9$",midpoint(P--R),dir(A-D),red); label("$12$",midpoint(R--X),dir(135),red); label("$15$",midpoint(X--P),dir(0),red); label("$25$",midpoint(G--D),dir(0),red); | [] |
252 | Rhombus $ABCD$ has $\angle BAD < 90^\circ.$ There is a point $P$ on the incircle of the rhombus such that the distances from $P$ to the lines $DA,AB,$ and $BC$ are $9,$ $5,$ and $16,$ respectively. Find the perimeter of $ABCD.$ | 2023 AIME I Problem 8 | This solution refers to the Diagram section.
Define points $O,R,S,$ and $T$ as Solution 1 does. Moreover, let
$H$ be the foot of the perpendicular from $P$ to $\overleftrightarrow{CD},$
$M$ be the foot of the perpendicular from $O$ to $\overleftrightarrow{HS},$ and
$N$ be the foot of the perpendicular from $O$ to $\overleftrightarrow{RT}.$
We obtain the following diagram:
Note that the diameter of $\odot O$ is $HS=RT=25,$ so $OP=\frac{25}{2}.$ It follows that:
In right $\triangle OMP,$ we have $MP=\frac{HS}{2}-PS=\frac{15}{2}$ by symmetry, from which $OM=10$ by the Pythagorean Theorem.
In right $\triangle ONP,$ we have $NP=\frac{RT}{2}-RP=\frac{7}{2}$ by symmetry, from which $ON=12$ by the Pythagorean Theorem.
Since $\overline{MO}\parallel\overline{AB}$ and $\overline{ON}\parallel\overline{DA},$ we conclude that $\angle A = \angle MON.$ We apply the Sine of a Sum Formula:
\begin{align*} \sin\angle A &= \sin\angle MON \\ &= \sin(\angle MOP + \angle PON) \\ &= \sin\angle MOP \cos\angle PON + \cos\angle MOP \sin\angle PON \\ &= \frac{3}{5}\cdot\frac{24}{25} + \frac{4}{5}\cdot\frac{7}{25} \\ &= \frac{4}{5}. \end{align*}
Note that \[\sin\angle A = \frac{HS}{DA},\] from which $\frac{4}{5} = \frac{25}{DA}.$ We solve this equation to get $DA=\frac{125}{4}.$
Finally, the perimeter of $ABCD$ is $4DA = \boxed{125}.$
~MRENTHUSIASM (credit given to TheAMCHub) | // Block 1
/* Made by MRENTHUSIASM; inspired by Math Jams. */
size(300);
pair A, B, C, D, O, P, R, S, T, H, M, N;
A = origin;
B = (125/4,0);
C = B + 125/4 * dir((3,4));
D = A + 125/4 * dir((3,4));
O = (25,25/2);
P = (15,5);
R = foot(P,A,D);
S = foot(P,A,B);
T = foot(P,B,C);
H = foot(S,C,D);
M = foot(O,S,H);
N = foot(O,R,T);
fill(O--M--P--cycle,yellow);
fill(O--N--P--cycle,green);
markscalefactor=0.15;
draw(rightanglemark(P,R,D)^^rightanglemark(P,S,B)^^rightanglemark(P,T,C)^^rightanglemark(O,M,P)^^rightanglemark(O,N,P)^^rightanglemark(S,H,D),red);
draw(Circle(O,25/2)^^A--B--C--D--cycle^^B--T^^D--H^^O--M^^O--N^^O--P);
draw(P--R^^P--S^^P--T^^P--H,red+dashed);
dot("$A$",A,1.5*dir(225),linewidth(4.5));
dot("$B$",B,1.5*dir(-45),linewidth(4.5));
dot("$C$",C,1.5*dir(45),linewidth(4.5));
dot("$D$",D,1.5*dir(90),linewidth(4.5));
dot("$P$",P,1.5*dir(60),linewidth(4.5));
dot("$R$",R,1.5*dir(135),linewidth(4.5));
dot("$S$",S,1.5*dir(-90),linewidth(4.5));
dot("$T$",T,1.5*dir(-45),linewidth(4.5));
dot("$O$",O,1.5*dir(45),linewidth(4.5));
dot("$H$",H,1.5*dir(90),linewidth(4.5));
dot("$M$",M,1.5*dir(180),linewidth(4.5));
dot("$N$",N,1.5*dir(15),linewidth(4.5));
label("$9$",midpoint(P--R),dir(A-D),red);
label("$5$",midpoint(P--S),dir(180),red);
label("$16$",midpoint(P--T),dir(A-D),red);
// Block 2
/* Made by MRENTHUSIASM; inspired by Math Jams. */ size(300); pair A, B, C, D, O, P, R, S, T, H, M, N; A = origin; B = (125/4,0); C = B + 125/4 * dir((3,4)); D = A + 125/4 * dir((3,4)); O = (25,25/2); P = (15,5); R = foot(P,A,D); S = foot(P,A,B); T = foot(P,B,C); H = foot(S,C,D); M = foot(O,S,H); N = foot(O,R,T); fill(O--M--P--cycle,yellow); fill(O--N--P--cycle,green); markscalefactor=0.15; draw(rightanglemark(P,R,D)^^rightanglemark(P,S,B)^^rightanglemark(P,T,C)^^rightanglemark(O,M,P)^^rightanglemark(O,N,P)^^rightanglemark(S,H,D),red); draw(Circle(O,25/2)^^A--B--C--D--cycle^^B--T^^D--H^^O--M^^O--N^^O--P); draw(P--R^^P--S^^P--T^^P--H,red+dashed); dot("$A$",A,1.5*dir(225),linewidth(4.5)); dot("$B$",B,1.5*dir(-45),linewidth(4.5)); dot("$C$",C,1.5*dir(45),linewidth(4.5)); dot("$D$",D,1.5*dir(90),linewidth(4.5)); dot("$P$",P,1.5*dir(60),linewidth(4.5)); dot("$R$",R,1.5*dir(135),linewidth(4.5)); dot("$S$",S,1.5*dir(-90),linewidth(4.5)); dot("$T$",T,1.5*dir(-45),linewidth(4.5)); dot("$O$",O,1.5*dir(45),linewidth(4.5)); dot("$H$",H,1.5*dir(90),linewidth(4.5)); dot("$M$",M,1.5*dir(180),linewidth(4.5)); dot("$N$",N,1.5*dir(15),linewidth(4.5)); label("$9$",midpoint(P--R),dir(A-D),red); label("$5$",midpoint(P--S),dir(180),red); label("$16$",midpoint(P--T),dir(A-D),red); | [] |
253 | Let $\triangle ABC$ be an equilateral triangle with side length $55.$ Points $D,$ $E,$ and $F$ lie on $\overline{BC},$ $\overline{CA},$ and $\overline{AB},$ respectively, with $BD = 7,$ $CE=30,$ and $AF=40.$ Point $P$ inside $\triangle ABC$ has the property that \[\angle AEP = \angle BFP = \angle CDP.\] Find $\tan^2(\angle AEP).$ | 2023 AIME I Problem 12 | By Miquel's theorem, $P=(AEF)\cap(BFD)\cap(CDE)$ (intersection of circles)$\text{*}$. The law of cosines can be used to compute $DE=42$, $EF=35$, and $FD=13$. Toss the points on the coordinate plane; let $B=(-7, 0)$, $D=(0, 0)$, and $C=(48, 0)$, where we will find $\tan^{2}\left(\measuredangle CDP\right)$ with $P=(BFD)\cap(CDE)$.
By the extended law of sines, the radius of circle $(BFD)$ is $\frac{13}{2\sin 60^{\circ}}=\frac{13}{3}\sqrt{3}$. Its center lies on the line $x=-\frac{7}{2}$, and the origin is a point on it, so $y=\frac{23}{6}\sqrt{3}$.
The radius of circle $(CDE)$ is $\frac{42}{2\sin 60^{\circ}}=14\sqrt{3}$. The origin is also a point on it, and its center is on the line $x=24$, so $y=2\sqrt{3}$.
The equations of the two circles are \begin{align*}(BFD)&:\left(x+\tfrac{7}{2}\right)^{2}+\left(y-\tfrac{23}{6}\sqrt{3}\right)^{2}=\tfrac{169}{3} \\ (CDE)&:\left(x-24\right)^{2}+\left(y-2\sqrt{3}\right)^{2}=588\end{align*} These equations simplify to \begin{align*}(BFD)&:x^{2}+7x+y^{2}-\tfrac{23}{3}\sqrt{3}y=0 \\ (CDE)&: x^{2}-48x+y^{2}-4\sqrt{3}y=0\end{align*} Subtracting these two equations gives that both their points of intersection, $D$ and $P$, lie on the line $55x-\tfrac{11}{3}\sqrt{3}y=0$. Hence, $\tan^{2}\left(\measuredangle AEP\right)=\tan^{2}\left(\measuredangle CDP\right)=\left(\frac{55}{\tfrac{11}{3}\sqrt{3}}\right)^{2}=3\left(\tfrac{55}{11}\right)^{2}=\boxed{075}$. To scale, the configuration looks like the figure below:
$\text{*}$
Basic angle chasing gives $\angle BDP=180-\angle CDP.$ Because $\angle BFP+\angle BDP=180+\angle BFP-\angle CDP=180,$ which means that $BFPD$ is cyclic, and that $P$ passes through the circumcircle of triangle $BFD.$ Similar reasoning leads us to the fact that $P$ also passes through the circumcircles of triangles $AEF$ and $CDE,$ which means that $P=(AEF)\cap(BFD)\cap(CDE).$ Continue as above.
Note: Since $D$ and $P$ are the two intersections of $(BFD)$ and $(CDE),$ we know that $DP$ is perpendicular to the line between the centers of $(BFD)$ and $(CDE).$ Thus, after getting the coordinates of the centers of $(BFD)$ and $(CDE),$ we can then immediately find the slope of line $DP,$ and then get the answer from there. | // Block 1
/* Made by MRENTHUSIASM */
size(400);
pair A, B, C, D, E, F, P;
A = 55*sqrt(3)/3 * dir(90);
B = 55*sqrt(3)/3 * dir(210);
C = 55*sqrt(3)/3 * dir(330);
D = B + 7*dir(0);
E = A + 25*dir(C-A);
F = A + 40*dir(B-A);
P = intersectionpoints(Circle(D,54*sqrt(19)/19),Circle(F,5*sqrt(19)/19))[0];
filldraw(D--E--F--cycle,yellow);
draw(A--B--C--cycle);
draw(circumcircle(A,E,F)^^circumcircle(B,D,F)^^circumcircle(C,D,E),blue);
dot("$A$",A,1.5*dir(A),linewidth(4));
dot("$B$",B,1.5*dir(B),linewidth(4));
dot("$C$",C,1.5*dir(C),linewidth(4));
dot("$D$",D,1.5*S,linewidth(4));
dot("$E$",E,1.5*dir(30),linewidth(4));
dot("$F$",F,1.5*dir(150),linewidth(4));
dot("$P$",P,1.5*dir(-30),linewidth(4));
label("$7$",midpoint(B--D),1.5*S,red);
label("$30$",midpoint(C--E),1.5*dir(30),red);
label("$40$",midpoint(A--F),1.5*dir(150),red);
// Block 2
/* Made by MRENTHUSIASM */ size(400); pair A, B, C, D, E, F, P; A = 55*sqrt(3)/3 * dir(90); B = 55*sqrt(3)/3 * dir(210); C = 55*sqrt(3)/3 * dir(330); D = B + 7*dir(0); E = A + 25*dir(C-A); F = A + 40*dir(B-A); P = intersectionpoints(Circle(D,54*sqrt(19)/19),Circle(F,5*sqrt(19)/19))[0]; filldraw(D--E--F--cycle,yellow); draw(A--B--C--cycle); draw(circumcircle(A,E,F)^^circumcircle(B,D,F)^^circumcircle(C,D,E),blue); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$D$",D,1.5*S,linewidth(4)); dot("$E$",E,1.5*dir(30),linewidth(4)); dot("$F$",F,1.5*dir(150),linewidth(4)); dot("$P$",P,1.5*dir(-30),linewidth(4)); label("$7$",midpoint(B--D),1.5*S,red); label("$30$",midpoint(C--E),1.5*dir(30),red); label("$40$",midpoint(A--F),1.5*dir(150),red); | [] |
253 | Let $\triangle ABC$ be an equilateral triangle with side length $55.$ Points $D,$ $E,$ and $F$ lie on $\overline{BC},$ $\overline{CA},$ and $\overline{AB},$ respectively, with $BD = 7,$ $CE=30,$ and $AF=40.$ Point $P$ inside $\triangle ABC$ has the property that \[\angle AEP = \angle BFP = \angle CDP.\] Find $\tan^2(\angle AEP).$ | 2023 AIME I Problem 12 | By the law of cosines,
\[FE=\sqrt{AF^2+AE^2-2AF\cdot AE\cos\angle FAE}=35.\]
Similarly we get $FD=13$ and $DE=42$. $\angle AEP=\angle BFP=\angle CDP\overset\triangle=\theta$ implies that $AFPE$, $BDPF$, and $CDPE$ are three cyclic quadrilaterals, as shown below:
Using the law of sines in each,
\[\frac{AP}{35}=\frac{AP}{FE}=\frac{CP}{42}=\frac{CP}{ED}=\frac{BP}{13}=\frac{BP}{DF}=\frac{\sin\theta}{\sin\frac\pi3}.\]
So we can set $AP=35k$, $BP=13k$, and $CP=42k$. Let $PD=d$, $PE=e$, and $PF=f$. Applying Ptolemy theorem in the cyclic quadrilaterals,
\[\begin{cases}AP\cdot FE=AF\cdot PE+AE\cdot PF,\\CP\cdot ED=CE\cdot PD+CD\cdot PE,\\BP\cdot DF=BD\cdot PF+BF\cdot PD.\end{cases} \implies \begin{cases} 1225k=40e+25f,\\1764k=30d+48e,\\169k=15d+7f, \end{cases}\]
We can solve out $d=\frac{54k}5$, $e=30k$, $f=k$. By the law of cosines in $\triangle PEF$, $FE=\sqrt{900k^2+k^2-60k\cdot\left(\frac{-1}2\right)}=\sqrt{931}k$. The law of sines yield $\frac{\sin\angle AEP}{\sin\angle FAE}=\frac{AP}{FE}=\frac{35k}{\sqrt{931}k}=\frac{35}{\sqrt{931}}$.
Lastly, $\sin\angle AEP=\frac{5\sqrt{57}}{38}$, then $\tan\angle AEP=5\sqrt3$. The answer is
\[\left(5\sqrt3\right)^2=\boxed{075}.\] | // Block 1
/* Made by MRENTHUSIASM */
size(400);
pair A, B, C, D, E, F, P;
A = 55*sqrt(3)/3 * dir(90);
B = 55*sqrt(3)/3 * dir(210);
C = 55*sqrt(3)/3 * dir(330);
D = B + 7*dir(0);
E = A + 25*dir(C-A);
F = A + 40*dir(B-A);
P = intersectionpoints(Circle(D,54*sqrt(19)/19),Circle(F,5*sqrt(19)/19))[0];
draw(anglemark(A,E,P,20),red);
draw(anglemark(B,F,P,20),red);
draw(anglemark(C,D,P,20),red);
add(pathticks(anglemark(A,E,P,20), n = 1, r = 0.2, s = 12, red));
add(pathticks(anglemark(B,F,P,20), n = 1, r = 0.2, s = 12, red));
add(pathticks(anglemark(C,D,P,20), n = 1, r = 0.2, s = 12, red));
draw(A--B--C--cycle^^P--E^^P--F^^P--D);
draw(P--A^^P--B^^P--C,dashed);
draw(circumcircle(A,E,F)^^circumcircle(B,D,F)^^circumcircle(C,D,E),blue);
dot("$A$",A,1.5*dir(A),linewidth(4));
dot("$B$",B,1.5*dir(B),linewidth(4));
dot("$C$",C,1.5*dir(C),linewidth(4));
dot("$D$",D,1.5*S,linewidth(4));
dot("$E$",E,1.5*dir(30),linewidth(4));
dot("$F$",F,1.5*dir(150),linewidth(4));
dot("$P$",P,1.5*dir(-30),linewidth(4));
label("$7$",midpoint(B--D),1.5*S,red);
label("$30$",midpoint(C--E),1.5*dir(30),red);
label("$40$",midpoint(A--F),1.5*dir(150),red);
// Block 2
/* Made by MRENTHUSIASM */ size(400); pair A, B, C, D, E, F, P; A = 55*sqrt(3)/3 * dir(90); B = 55*sqrt(3)/3 * dir(210); C = 55*sqrt(3)/3 * dir(330); D = B + 7*dir(0); E = A + 25*dir(C-A); F = A + 40*dir(B-A); P = intersectionpoints(Circle(D,54*sqrt(19)/19),Circle(F,5*sqrt(19)/19))[0]; draw(anglemark(A,E,P,20),red); draw(anglemark(B,F,P,20),red); draw(anglemark(C,D,P,20),red); add(pathticks(anglemark(A,E,P,20), n = 1, r = 0.2, s = 12, red)); add(pathticks(anglemark(B,F,P,20), n = 1, r = 0.2, s = 12, red)); add(pathticks(anglemark(C,D,P,20), n = 1, r = 0.2, s = 12, red)); draw(A--B--C--cycle^^P--E^^P--F^^P--D); draw(P--A^^P--B^^P--C,dashed); draw(circumcircle(A,E,F)^^circumcircle(B,D,F)^^circumcircle(C,D,E),blue); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$D$",D,1.5*S,linewidth(4)); dot("$E$",E,1.5*dir(30),linewidth(4)); dot("$F$",F,1.5*dir(150),linewidth(4)); dot("$P$",P,1.5*dir(-30),linewidth(4)); label("$7$",midpoint(B--D),1.5*S,red); label("$30$",midpoint(C--E),1.5*dir(30),red); label("$40$",midpoint(A--F),1.5*dir(150),red); | [] |
254 | Each face of two noncongruent parallelepipeds is a rhombus whose diagonals have lengths $\sqrt{21}$ and $\sqrt{31}$.
The ratio of the volume of the larger of the two polyhedra to the volume of the smaller is $\frac{m}{n}$, where $m$ and $n$
are relatively prime positive integers. Find $m + n$. A parallelepiped is a solid with six parallelogram faces
such as the one shown below. | 2023 AIME I Problem 13 | Since the two parallelepipeds have the same base, all we need to do is to find their respective heights.
As illustrated in the above diagram, drop a perpendicular from $D$ onto the base at $P$. Denote the center of the base by $O$. By symmetry, $P$ must be on $AO$. Now we need to find $DP$.
Apply Pythagorean theorem to $\triangle DPA$ we have
\[DP^2 = AD^2 - AP^2.\]
Apply Pythagorean theorem to $\triangle DPC$ and then $\triangle CPO$ we have
\[DP^2 = DC^2 - CP^2 = DC^2 - (CO^2 + OP^2) = DC^2 - (CO^2 + (AO-AP)^2) = DC^2 - CO^2 - (AO-AP)^2.\]
Combining the above two, we have
\[AD^2 - AP^2 = DC^2 - CO^2 - (AO-AP)^2.\]
Since $AD=\sqrt{13}$, $DC=\sqrt{21}$, $CO=\frac{\sqrt{21}}{2}$, $AO=\frac{\sqrt{31}}{2}$, plug them into the above equation and solve for the only unknown variable $AP$, we get $AP=\frac{5}{\sqrt{31}}.$
Thus the height
\[DP = \sqrt{AD^2 - AP^2} = \sqrt{13 - \frac{25}{31}} = \sqrt{\frac{378}{31}}.\]
For the other parallelepiped, using the same approach and drop a perpendicular from $E'$ onto the base at $P'$. Similarly applying Pythagorean theorem to $\triangle E'P'C'$, $\triangle E'P'A'$ and $\triangle A'P'O'$ we have
\[C'E'^2 - C'P'^2 = A'E'^2 - A'O'^2 - (C'P'-C'O')^2.\]
Plugging known values into the above equation and solve for the only unknown variable $C'P'$, we get $C'P'=\frac{5}{\sqrt{21}}.$
Thus the height
\[E'P' = \sqrt{C'E'^2 - C'P'^2} = \sqrt{13 - \frac{25}{21}} = \sqrt{\frac{248}{21}}.\]
The ratio between the two is therefore
\[\frac{DP}{E'P'} = \frac{\sqrt{\frac{378}{31}}}{\sqrt{\frac{248}{21}}} = \sqrt{\frac{2\cdot3^3\cdot7}{31}\cdot\frac{3\cdot7}{2^3\cdot31}} = \frac{3^2\cdot7}{2\cdot31} = \frac{63}{62}\], giving $\boxed{125}$.
~sgdzw | // Block 1
unitsize(2cm);
pair a = (0, 0), b = (1, 0), c = 0.8*dir(40), d = dir(70), p = 0.33*dir(20), o = (b+c)/2;
label("A",a,S);
label("B",b,S);
label("C",c,S);
label("D",d,N);
label("P",p,S);
label("O",o,E);
draw(a--b--(b+c));
draw(a--c--(b+c), dotted);
draw(shift(d)*(a--b--(b+c)--c--cycle));
draw(a--d);
draw(b--(b+d));
draw(c--(c+d), dotted);
draw((b+c)--(b+c+d));
draw(d--p, dotted);
draw(c--b, dotted);
draw(a--(b+c), dotted);
draw(p--c, dotted);
draw(d--c, dotted);
// Block 2
unitsize(2cm);
pair a = (0, 0), b = (1, 0), c = 0.8*dir(40), d = 0.7*dir(80), e = c+d, p = 0.9*dir(10), o = (b+c)/2;
label("A'",a,S);
label("B'",b,S);
label("C'",c,W);
label("E'",e,N);
label("P'",p,S);
label("O'",o,W);
draw(a--b--(b+c));
draw(a--c--(b+c), dotted);
draw(shift(d)*(a--b--(b+c)--c--cycle));
draw(a--d);
draw(b--(b+d));
draw(c--(c+d), dotted);
draw((b+c)--(b+c+d));
draw(e--p, dotted);
draw(c--b, dotted);
draw(a--(b+c), dotted);
draw(p--a, dotted);
draw(e--a, dotted);
// Block 3
unitsize(2cm); pair a = (0, 0), b = (1, 0), c = 0.8*dir(40), d = dir(70), p = 0.33*dir(20), o = (b+c)/2; label("A",a,S); label("B",b,S); label("C",c,S); label("D",d,N); label("P",p,S); label("O",o,E); draw(a--b--(b+c)); draw(a--c--(b+c), dotted); draw(shift(d)*(a--b--(b+c)--c--cycle)); draw(a--d); draw(b--(b+d)); draw(c--(c+d), dotted); draw((b+c)--(b+c+d)); draw(d--p, dotted); draw(c--b, dotted); draw(a--(b+c), dotted); draw(p--c, dotted); draw(d--c, dotted);
// Block 4
unitsize(2cm); pair a = (0, 0), b = (1, 0), c = 0.8*dir(40), d = 0.7*dir(80), e = c+d, p = 0.9*dir(10), o = (b+c)/2; label("A'",a,S); label("B'",b,S); label("C'",c,W); label("E'",e,N); label("P'",p,S); label("O'",o,W); draw(a--b--(b+c)); draw(a--c--(b+c), dotted); draw(shift(d)*(a--b--(b+c)--c--cycle)); draw(a--d); draw(b--(b+d)); draw(c--(c+d), dotted); draw((b+c)--(b+c+d)); draw(e--p, dotted); draw(c--b, dotted); draw(a--(b+c), dotted); draw(p--a, dotted); draw(e--a, dotted); | [] |
254 | Each face of two noncongruent parallelepipeds is a rhombus whose diagonals have lengths $\sqrt{21}$ and $\sqrt{31}$.
The ratio of the volume of the larger of the two polyhedra to the volume of the smaller is $\frac{m}{n}$, where $m$ and $n$
are relatively prime positive integers. Find $m + n$. A parallelepiped is a solid with six parallelogram faces
such as the one shown below. | 2023 AIME I Problem 13 | Let $\angle CAB$ be acute and let $X$ be the foot of the altitude from $C$ to $AB$. Notice that this also implies that $X$ is the foot of the altitude from $D$ to $AB$. Now $\sqrt{13} \cdot CX = AB \cdot CX = \frac{1}{2} \sqrt{21} \cdot \sqrt{31}$ so $CX = \frac{\sqrt{21} \cdot \sqrt{31}}{2 \sqrt{13}}$ and $DX$ is the same. $CD$ must either be $\sqrt{21}$ or $\sqrt{31}$ because it is a diagonal. If $CD = \sqrt{31}$, applying the Law of Cosines on $\angle DXC$, $\cos \angle DXC = -\frac{5}{21}$ so $\sin \angle DXC = \sqrt{\frac{416}{441}}$. If $CD = \sqrt{21}$, $\cos \angle DXC = \frac{5}{31}$, so $\sin \angle DXC = \sqrt{\frac{936}{961}}$. The ratios of the two parallelepipeds is equal to the ratios of the heights (since the bases are the same) which is equal to the ratio of the sines. Therefore it is \[\frac{\sqrt{\frac{936}{961}}}{\sqrt{\frac{416}{441}}} = \sqrt{\frac{936}{416}} \cdot \frac{21}{31} = \frac{3}{2} \cdot \frac{21}{31} = \frac{63}{62}\] so the answer is $63 + 62 = \boxed{125}$. | // Block 1
unitsize(2cm);
pair a = (0, 0), b = (1, 0), c = 0.8*dir(40), d = dir(70), p = 0.33*dir(20), o = (b+c)/2, x = (0.2, 0);
label("A",a,S);
label("B",b,S);
label("C",c,S);
label("D",d,N);
label("X",x,S);
draw(a--b--(b+c));
draw(a--c--(b+c), dotted);
draw(shift(d)*(a--b--(b+c)--c--cycle));
draw(a--d);
draw(b--(b+d));
draw(d--x);
draw(x--c, dotted);
draw(c--(c+d), dotted);
draw((b+c)--(b+c+d));
// Block 2
unitsize(2cm); pair a = (0, 0), b = (1, 0), c = 0.8*dir(40), d = dir(70), p = 0.33*dir(20), o = (b+c)/2, x = (0.2, 0); label("A",a,S); label("B",b,S); label("C",c,S); label("D",d,N); label("X",x,S); draw(a--b--(b+c)); draw(a--c--(b+c), dotted); draw(shift(d)*(a--b--(b+c)--c--cycle)); draw(a--d); draw(b--(b+d)); draw(d--x); draw(x--c, dotted); draw(c--(c+d), dotted); draw((b+c)--(b+c+d)); | [] |
255 | The following analog clock has two hands that can move independently of each other.
Initially, both hands point to the number $12$. The clock performs a sequence of hand movements so that on each movement, one of the two hands moves clockwise to the next number on the clock face while the other hand does not move.
Let $N$ be the number of sequences of $144$ hand movements such that during the sequence, every possible positioning of the hands appears exactly once, and at the end of the $144$ movements, the hands have returned to their initial position. Find the remainder when $N$ is divided by $1000$. | 2023 AIME I Problem 14 | This problem is, in essence, the following: A $12\times12$ coordinate grid is placed on a (flat) torus; how many loops are there that pass through each point while only moving up or right? In other words, Felix the frog starts his journey at $(0,0)$ in the coordinate plane. Every second, he jumps either to the right or up, until he reaches an $x$- or $y$-coordinate of $12$. At this point, if he tries to jump to a coordinate outside the square from $(0,0)$ to $(11,11)$, he "wraps around" and ends up at an $x$- or $y$- coordinate of $0$. How many ways are there for Felix to jump on every grid point in this square, so that he ends at $(0,0)$? This is consistent with the construction of the flat torus as $\mathbb Z^2/12\mathbb Z^2$ (2-dimensional modular arithmetic. $(\mathbb{Z}_{12})^2$)
Moving on, define a $\textit{path}$ from point $A$ to point $B$ to be a sequence of "up"s and "right"s that takes Felix from $A$ to $B$. The $\textit{distance}$ from $A$ to $B$ is the length of the shortest path from $A$ to $B$. At the crux of this problem is the following consideration: The points $A_i=(i,11-i), i\in{0,...,11}$ are pairwise equidistant, each pair having distance of $12$ in both directions.
A valid complete path then joins two $A_i$'s, say $A_i$ and $A_j$. In fact, a link between some $A_i$ and $A_j$ fully determines the rest of the cycle, as the path from $A_{i+1}$ must "hug" the path from $A_i$, to ensure that there are no gaps. We therefore see that if $A_0$ leads to $A_k$, then $A_i$ leads to $A_{i+k}$. Only the values of $k$ relatively prime to $12$ result in solutions, though, because otherwise $A_0$ would only lead to $\{A_i:\exists n\in \mathbb Z:i\equiv kn\quad\text{mod 12}\}$. The number of paths from $A_0$ to $A_k$ is ${12\choose k}$, and so the answer is
\[{12\choose1}+{12\choose5}+{12\choose7}+{12\choose11}=1\boxed{608}.\]
Notes:
- One can prove that the path from $A_{i+1}$ must "hug" the path from $A_i$ by using techniques similar to those in Solution 2.
- One can count the paths as follows: To get from $A_0$ to $A_i$, Felix takes $k$ rights and $12-k$ ups, which can be done in ${12\choose k}$ ways. | // Block 1
size(7cm);
for (int x=0; x<12; ++x){
for (int y=0; y<12; ++y){
fill(circle((x,y),0.05));}}
for (int i=0; i<12; ++i){
fill(circle((i,11-i),0.1),red);}
pen p=green+dashed;
path u=(3,8)--(4,8)--(4,9)--(4,10)--(4,11)--(5,11)--(5,11.5);
path v=(5,-0.5)--(5,0)--(5,1)--(6,1)--(6,2)--(6,3)--(6,4)--(7,4);
draw(u,p);
draw(v,p);
pen p=blue+dashed;
path u=(4,7)--(5,7)--(5,8)--(5,9)--(5,10)--(6,10)--(6,11)--(6,11.5);
path v=(6,-0.5)--(6,0)--(7,0)--(7,1)--(7,2)--(7,3)--(8,3);
draw(u,p);
draw(v,p);
// Block 2
size(7cm); for (int x=0; x<12; ++x){ for (int y=0; y<12; ++y){ fill(circle((x,y),0.05));}} for (int i=0; i<12; ++i){ fill(circle((i,11-i),0.1),red);} pen p=green+dashed; path u=(3,8)--(4,8)--(4,9)--(4,10)--(4,11)--(5,11)--(5,11.5); path v=(5,-0.5)--(5,0)--(5,1)--(6,1)--(6,2)--(6,3)--(6,4)--(7,4); draw(u,p); draw(v,p); pen p=blue+dashed; path u=(4,7)--(5,7)--(5,8)--(5,9)--(5,10)--(6,10)--(6,11)--(6,11.5); path v=(6,-0.5)--(6,0)--(7,0)--(7,1)--(7,2)--(7,3)--(8,3); draw(u,p); draw(v,p); | [] |
256 | Let $\triangle ABC$ be an isosceles triangle with $\angle A = 90^\circ.$ There exists a point $P$ inside $\triangle ABC$ such that $\angle PAB = \angle PBC = \angle PCA$ and $AP = 10.$ Find the area of $\triangle ABC.$ | 2023 AIME II Problem 3 | This solution refers to the Diagram section.
Let $\angle PAB = \angle PBC = \angle PCA = \theta,$ from which $\angle PAC = 90^\circ-\theta,$ and $\angle APC = 90^\circ.$
Moreover, we have $\angle PBA = \angle PCB = 45^\circ-\theta,$ as shown below:
Note that $\triangle PAB \sim \triangle PBC$ by AA Similarity. The ratio of similitude is $\frac{PA}{PB} = \frac{PB}{PC} = \frac{AB}{BC},$ so $\frac{10}{PB} = \frac{1}{\sqrt2}$ and thus $PB=10\sqrt2.$ Similarly, we can figure out that $PC=20$.
Finally, $AC=\sqrt{10^2+20^2}=10\sqrt{5}$, so the area of $\triangle ABC$ is \[\frac12\cdot AB\cdot AC = \frac12\cdot (10\sqrt{5})^2 = \boxed{250}.\]
~s214425
~MRENTHUSIASM | // Block 1
/* Made by MRENTHUSIASM */
size(250);
pair A, B, C, P;
A = origin;
B = (0,10*sqrt(5));
C = (10*sqrt(5),0);
P = intersectionpoints(Circle(A,10),Circle(C,20))[0];
dot("$A$",A,1.5*SW,linewidth(4));
dot("$B$",B,1.5*NW,linewidth(4));
dot("$C$",C,1.5*SE,linewidth(4));
dot("$P$",P,1.5*NE,linewidth(4));
markscalefactor=0.125;
draw(rightanglemark(B,A,C,10),red);
draw(rightanglemark(A,P,C,10),red);
draw(anglemark(P,A,B,25),red);
draw(anglemark(P,B,C,25),red);
draw(anglemark(P,C,A,25),red);
draw(anglemark(A,B,P,25),green);
draw(anglemark(B,C,P,25),green);
draw(anglemark(C,A,P,25),green);
add(pathticks(anglemark(P,A,B,25), n = 1, r = 0.1, s = 10, red));
add(pathticks(anglemark(P,B,C,25), n = 1, r = 0.1, s = 10, red));
add(pathticks(anglemark(P,C,A,25), n = 1, r = 0.1, s = 10, red));
draw(A--B--C--cycle^^P--A^^P--B^^P--C);
label("$10$",midpoint(A--P),dir(-30),blue);
label("$\theta$",A,9.5*dir(76),red);
label("$\theta$",C,9.5*dir(168),red);
label("$\theta$",B,9*dir(305),red);
label("$45^\circ-\theta$",B,6*dir(235),green);
label("$45^\circ-\theta$",C,6*dir(85),green);
label("$90^\circ-\theta$",A,2*dir(-40),green);
// Block 2
/* Made by MRENTHUSIASM */ size(250); pair A, B, C, P; A = origin; B = (0,10*sqrt(5)); C = (10*sqrt(5),0); P = intersectionpoints(Circle(A,10),Circle(C,20))[0]; dot("$A$",A,1.5*SW,linewidth(4)); dot("$B$",B,1.5*NW,linewidth(4)); dot("$C$",C,1.5*SE,linewidth(4)); dot("$P$",P,1.5*NE,linewidth(4)); markscalefactor=0.125; draw(rightanglemark(B,A,C,10),red); draw(rightanglemark(A,P,C,10),red); draw(anglemark(P,A,B,25),red); draw(anglemark(P,B,C,25),red); draw(anglemark(P,C,A,25),red); draw(anglemark(A,B,P,25),green); draw(anglemark(B,C,P,25),green); draw(anglemark(C,A,P,25),green); add(pathticks(anglemark(P,A,B,25), n = 1, r = 0.1, s = 10, red)); add(pathticks(anglemark(P,B,C,25), n = 1, r = 0.1, s = 10, red)); add(pathticks(anglemark(P,C,A,25), n = 1, r = 0.1, s = 10, red)); draw(A--B--C--cycle^^P--A^^P--B^^P--C); label("$10$",midpoint(A--P),dir(-30),blue); label("$\theta$",A,9.5*dir(76),red); label("$\theta$",C,9.5*dir(168),red); label("$\theta$",B,9*dir(305),red); label("$45^\circ-\theta$",B,6*dir(235),green); label("$45^\circ-\theta$",C,6*dir(85),green); label("$90^\circ-\theta$",A,2*dir(-40),green); | [] |
257 | A cube-shaped container has vertices $A,$ $B,$ $C,$ and $D,$ where $\overline{AB}$ and $\overline{CD}$ are parallel edges of the cube, and $\overline{AC}$ and $\overline{BD}$ are diagonals of faces of the cube, as shown. Vertex $A$ of the cube is set on a horizontal plane $\mathcal{P}$ so that the plane of the rectangle $ABDC$ is perpendicular to $\mathcal{P},$ vertex $B$ is $2$ meters above $\mathcal{P},$ vertex $C$ is $8$ meters above $\mathcal{P},$ and vertex $D$ is $10$ meters above $\mathcal{P}.$ The cube contains water whose surface is parallel to $\mathcal{P}$ at a height of $7$ meters above $\mathcal{P}.$ The volume of water is $\frac{m}{n}$ cubic meters, where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | 2023 AIME II Problem 14 | Let's first view the cube from a direction perpendicular to $ABDC$, as illustrated above. Let $x$ be the cube's side length. Since $\triangle CHA \sim \triangle AGB$, we have
\[\frac{CA}{CH} = \frac{AB}{AG}.\]
We know $AB = x$, $AG = \sqrt{x^2-2^2}$, $AC = \sqrt{2}x$, $CH = 8$. Plug them into the above equation, we get
\[\frac{\sqrt{2}x}{8} = \frac{x}{\sqrt{x^2-2^2}}.\]
Solving this we get the cube's side length $x = 6$, and $AC = 6\sqrt{2}.$
Let $PQ$ be the water's surface, both $P$ and $Q$ are $7$ meters from $\mathcal P$. Notice that $C$ is $8$ meters from $\mathcal P$, this means
\[CP = \frac{1}{8}CA = \frac{3\sqrt{2}}{4}.\]
Similarly,
\[DQ = \frac{3}{8}CA = \frac{9\sqrt{2}}{4}.\]
Now, we realize that the 3D space inside the cube without water is a frustum, with $P$ on its smaller base and $Q$ on its larger base. To find its volume, all we need is to find the areas of both bases and the height, which is $x = 6$. To find the smaller base, let's move our viewpoint onto the plane $ABDC$ and view the cube from a direction parallel to $ABDC$, as shown above. The area of the smaller base is simply
\[S_1 = CP^2 = \Bigl(\frac{3\sqrt{2}}{4}\Bigr)^2 = \frac{9}{8}.\]
Similarly, the area of the larger base is
\[S_2 = DQ^2 = \Bigl(\frac{9\sqrt{2}}{4}\Bigr)^2 = \frac{81}{8}.\]
Finally, applying the formula for a frustum's volume,
\[V = \frac{1}{3} \cdot x \cdot (S_1 + \sqrt{S_1S_2} + S_2) = \frac{1}{3} \cdot 6 \cdot \Bigl(\frac{9}{8} + \sqrt{\frac{9}{8}\cdot\frac{81}{8}} + \frac{81}{8}\Bigl) = \frac{117}{4}.\]
The water's volume is thus
\[6^3 - \frac{117}{4} = \frac{747}{4},\]
giving $\boxed{751}$. | // Block 1
defaultpen(linewidth(0.6));
pair A = (0, 0), B = (5.656,2), C = (-2.828, 8), D = B+C, P = 0.875*C, Q = B+0.625*C, H = (-2.828, 0), G = (5.656, 0);
pair P1 = (-5, 0), P2 = (10, 0);
draw(A--B--D--C--A);
filldraw(A--B--Q--P--cycle,rgb(0.35,0.7,0.9));
draw(C--H, dotted);
draw(B--G, dotted);
draw(P1--P2);
dot("$A$",A,S);
dot("$B$",B,E);
dot("$C$",C,N);
dot("$D$",D,N);
dot("$H$",H,S);
dot("$G$",G,S);
dot("$P$",P,SE);
dot("$Q$",Q,E);
label("$\mathcal P$",(11, 0));
// Block 2
defaultpen(linewidth(0.6));
pair A = (0, 0), C = (0, 2*6), X = (6, 6), Y = (-6, 6), P = (0, 1.75*6), I = (-0.25*6, 1.75*6), J = (0.25*6, 1.75*6);
pair P1 = (-8, 0), P2 = (8, 0);
draw(A--X--C--Y--A);
filldraw(A--X--J--I--Y--cycle,rgb(0.35,0.7,0.9));
draw(P1--P2, dotted);
dot("$A$",A,S);
dot("$C$",C,N);
dot("$P$",P,S);
label("$\mathcal P$",(10, 0));
// Block 3
defaultpen(linewidth(0.6)); pair A = (0, 0), B = (5.656,2), C = (-2.828, 8), D = B+C, P = 0.875*C, Q = B+0.625*C, H = (-2.828, 0), G = (5.656, 0); pair P1 = (-5, 0), P2 = (10, 0); draw(A--B--D--C--A); filldraw(A--B--Q--P--cycle,rgb(0.35,0.7,0.9)); draw(C--H, dotted); draw(B--G, dotted); draw(P1--P2); dot("$A$",A,S); dot("$B$",B,E); dot("$C$",C,N); dot("$D$",D,N); dot("$H$",H,S); dot("$G$",G,S); dot("$P$",P,SE); dot("$Q$",Q,E); label("$\mathcal P$",(11, 0));
// Block 4
defaultpen(linewidth(0.6)); pair A = (0, 0), C = (0, 2*6), X = (6, 6), Y = (-6, 6), P = (0, 1.75*6), I = (-0.25*6, 1.75*6), J = (0.25*6, 1.75*6); pair P1 = (-8, 0), P2 = (8, 0); draw(A--X--C--Y--A); filldraw(A--X--J--I--Y--cycle,rgb(0.35,0.7,0.9)); draw(P1--P2, dotted); dot("$A$",A,S); dot("$C$",C,N); dot("$P$",P,S); label("$\mathcal P$",(10, 0)); | [] |
258 | Rectangles $ABCD$ and $EFGH$ are drawn such that $D,E,C,F$ are collinear. Also, $A,D,H,G$ all lie on a circle. If $BC=16$,$AB=107$,$FG=17$, and $EF=184$, what is the length of $CE$? | 2024 AIME I Problem 5 | We use simple geometry to solve this problem.
We are given that $A$, $D$, $H$, and $G$ are concyclic; call the circle that they all pass through circle $\omega$ with center $O$. We know that, given any chord on a circle, the perpendicular bisector to the chord passes through the center; thus, given two chords, taking the intersection of their perpendicular bisectors gives the center. We therefore consider chords $HG$ and $AD$ and take the midpoints of $HG$ and $AD$ to be $P$ and $Q$, respectively.
We could draw the circumcircle, but actually it does not matter for our solution; all that matters is that $OA=OH=r$, where $r$ is the circumradius.
By the Pythagorean Theorem, $OQ^2+QA^2=OA^2$. Also, $OP^2+PH^2=OH^2$. We know that $OQ=DE+HP$, and $HP=\dfrac{184}2=92$; $QA=\dfrac{16}2=8$; $OP=DQ+HE=8+17=25$; and finally, $PH=92$. Let $DE=x$. We now know that $OA^2=(x+92)^2+8^2$ and $OH^2=25^2+92^2$. Recall that $OA=OH$; thus, $OA^2=OH^2$. We solve for $x$:
\begin{align*}
(x+92)^2+8^2&=25^2+92^2 \\
(x+92)^2&=625+(100-8)^2-8^2 \\
&=625+10000-1600+64-64 \\
&=9025 \\
x+92&=95 \\
x&=3. \\
\end{align*}
The question asks for $CE$, which is $CD-x=107-3=\boxed{104}$.
~Technodoggo | // Block 1
import graph;
unitsize(0.1cm);
pair A = (0,0);pair B = (107,0);pair C = (107,16);pair D = (0,16);pair E = (3,16);pair F = (187,16);pair G = (187,33);pair H = (3,33);
label("$A$", A, SW);label("$B$", B, SE);label("$C$", C, N);label("$D$", D, NW);label("$E$", E, S);label("$F$", F, SE);label("$G$", G, NE);label("$H$", H, NW);
draw(E--D--A--B--C--E--H--G--F--C);
/*Diagram by Technodoggo*/
// Block 2
import graph;
unitsize(0.1cm);
pair A = (0,0);pair B = (107,0);pair C = (107,16);pair D = (0,16);pair E = (3,16);pair F = (187,16);pair G = (187,33);pair H = (3,33);
label("$A$", A, SW);label("$B$", B, SE);label("$C$", C, N);label("$D$", D, NW);label("$E$", E, S);label("$F$", F, SE);label("$G$", G, NE);label("$H$", H, NW);
draw(E--D--A--B--C--E--H--G--F--C);
pair P = (95, 33);pair Q = (0, 8);
dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);dot(G);dot(H);dot(P);dot(Q);
label("$P$", P, N);label("$Q$", Q, W);
draw(Q--(107,8));draw(P--(95,0));
pair O = (95,8);
dot(O);label("$O$", O, NW);
/*Diagram by Technodoggo*/
// Block 3
import graph; unitsize(0.1cm); pair A = (0,0);pair B = (107,0);pair C = (107,16);pair D = (0,16);pair E = (3,16);pair F = (187,16);pair G = (187,33);pair H = (3,33); label("$A$", A, SW);label("$B$", B, SE);label("$C$", C, N);label("$D$", D, NW);label("$E$", E, S);label("$F$", F, SE);label("$G$", G, NE);label("$H$", H, NW); draw(E--D--A--B--C--E--H--G--F--C); /*Diagram by Technodoggo*/
// Block 4
import graph; unitsize(0.1cm); pair A = (0,0);pair B = (107,0);pair C = (107,16);pair D = (0,16);pair E = (3,16);pair F = (187,16);pair G = (187,33);pair H = (3,33); label("$A$", A, SW);label("$B$", B, SE);label("$C$", C, N);label("$D$", D, NW);label("$E$", E, S);label("$F$", F, SE);label("$G$", G, NE);label("$H$", H, NW); draw(E--D--A--B--C--E--H--G--F--C); pair P = (95, 33);pair Q = (0, 8); dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);dot(G);dot(H);dot(P);dot(Q); label("$P$", P, N);label("$Q$", Q, W); draw(Q--(107,8));draw(P--(95,0)); pair O = (95,8); dot(O);label("$O$", O, NW); /*Diagram by Technodoggo*/ | [] |
258 | Rectangles $ABCD$ and $EFGH$ are drawn such that $D,E,C,F$ are collinear. Also, $A,D,H,G$ all lie on a circle. If $BC=16$,$AB=107$,$FG=17$, and $EF=184$, what is the length of $CE$? | 2024 AIME I Problem 5 | Suppose $DE=x$. Extend $AD$ and $GH$ until they meet at $P$. From the Power of a Point Theorem, we have $(PH)(PG)=(PD)(PA)$. Substituting in these values, we get $(x)(x+184)=(17)(33)=561$. We can use guess and check to find that $x=3$, so $EC=\boxed{104}$.
~alexanderruan
~diagram by Technodoggo
rabbit47 - quadratic actually factors as (x+187)(x-3)=0, from which x=3 | // Block 1
import graph;
unitsize(0.1cm);
pair A = (0,0);pair B = (107,0);pair C = (107,16);pair D = (0,16);pair E = (3,16);pair F = (187,16);pair G = (187,33);pair H = (3,33);pair P = (0,33);
label("$A$", A, SW);label("$B$", B, SE);label("$C$", C, N);label("$D$", D, W);label("$E$", E, S);label("$F$", F, SE);label("$G$", G, NE);label("$H$", H, N);label("$P$", P, NW);
draw(E--D--A--B--C--E--H--G--F--C);
draw(D--P--H, dashed);
/*graph originally by Technodoggo, revised by alexanderruan*/
// Block 2
import graph; unitsize(0.1cm); pair A = (0,0);pair B = (107,0);pair C = (107,16);pair D = (0,16);pair E = (3,16);pair F = (187,16);pair G = (187,33);pair H = (3,33);pair P = (0,33); label("$A$", A, SW);label("$B$", B, SE);label("$C$", C, N);label("$D$", D, W);label("$E$", E, S);label("$F$", F, SE);label("$G$", G, NE);label("$H$", H, N);label("$P$", P, NW); draw(E--D--A--B--C--E--H--G--F--C); draw(D--P--H, dashed); /*graph originally by Technodoggo, revised by alexanderruan*/ | [] |
259 | Find the largest possible real part of\[(75+117i)z+\frac{96+144i}{z}\]where $z$ is a complex number with $|z|=4$. Here $i = \sqrt{-1}$. | 2024 AIME I Problem 7 | Suppose $z = 4e^{i\theta}$
\begin{align}
& \phantom{\text{ = }}\mathrm{Re}\left((75 + 117i)z + \dfrac{96 + 144i}{z}\right)\\
& = \mathrm{Re}\left((300 + 468i)e^{i\theta} + (24 + 36i)e^{-i\theta}\right)\\
& = \mathrm{Re}\left((300 + 468i)e^{i\theta} + (24 - 36i)e^{i\theta}\right)\\
& = \mathrm{Re}\left((324 + 432i)e^{i\theta}\right)\\
& \leq \sqrt{324^2 + 432^2}\\
& = \boxed{540}
\end{align}
Remark 13.1
$\overline{z_1z_2} = \overline{z_1} \cdot \overline{z_2}$,
$24 + 36i, 24 - 36i$ and $e^{-i\theta}, e^{i\theta}$ are conjugates,
The real parts of conjugates are equal,
$\mathrm{Re}(z) \leq |z|$.
Remark 13.2
The idea comes from the geometric meaning of this problem. The only difficulty is that the rotation directions are opposite. To make the rotations same, using conjugate is a good idea without changing the real part. Finally we can get a rotating parallelogram as shown in the following graph. Just find the length of the diagonal.
~reda_mandymath | unitsize(25); draw((-1, 0) -- (5, 0), Arrows(TeXHead)); draw((0, -3) -- (0, 5), Arrows(TeXHead)); draw((0, 0) -- (3, 3), EndArrow); draw((0, 0) -- (1, 2), EndArrow); draw((0, 0) -- (1, -2), dashed, EndArrow); draw((3, 3) -- (4, 1) -- (1, -2), dashed); draw((0, 0) -- (4, 1), red, EndArrow); draw(arc((0, 0), (3, 3), rotate(-20) * (3, 3), CW), blue, EndArrow(TeXHead)); draw(arc((0, 0), (1, 2), rotate(20) * (1, 2), CCW), blue, EndArrow(TeXHead)); draw(arc((0, 0), (1, -2), rotate(-20) * (1, -2), CW), blue, EndArrow(TeXHead)); draw(arc((0, 0), (4, 1), rotate(-20) * (4, 1), CW), blue, EndArrow(TeXHead)); label(scale(0.8) * "Diagram not on scale", (7, 3)); label(scale(0.9) * "$z_1$", (3, 3), N); label(scale(0.9) * "$z_2$", (1, 2), N); label(scale(0.9) * "$\overline{z_2}$", (1, -2), S); label(scale(0.9) * "$z_1 + \overline{z_2}$", (4, 1), E); | [] |
260 | Each vertex of a regular octagon is independently colored either red or blue with equal probability. The probability that the octagon can then be rotated so that all of the blue vertices end up at positions where there had been red vertices is $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2024 AIME I Problem 11 | Notice that the question's condition mandates all blues to go to reds, but reds do not necessarily have to go to blue. Let us do casework on how many blues there are.
If there are no blues whatsoever, there is only one case. This case is valid, as all of the (zero) blues have gone to reds. (One could also view it as: the location of all the blues now were not previously red.) Thus, we have $1$.
If there is a single blue somewhere, there are $8$ cases - where can the blue be? Each of these is valid.
If there are two blues, again, every case is valid, and there are $\dbinom82=28$ cases.
If there are three blues, every case is again valid; there are $\dbinom83=56$ such cases.
The case with four blues is trickier. Let us look at all possible subcases.
If all four are adjacent (as in the diagram below), it is obvious: we can simply reverse the diagram (rotate it by $4$ units) to achieve the problem's condition. There are $8$ possible ways to have $4$ adjacent blues, so this subcase contributes $8$.
If three are adjacent and one is one away (as shown in the diagram below), we can not rotate the diagram to satisfy the question. This subcase does not work.
If three are adjacent and one is two away, obviously it is not possible as there is nowhere for the three adjacent blues to go.
If there are two adjacent pairs that are $1$ apart, it is not possible since we do not have anywhere to put the two pairs.
If there are two adjacent pairs that are $2$ apart, all of these cases are possible as we can rotate the diagram by $2$ vertices to work. There are $4$ of these cases.
If there is one adjacent pair and there are two separate ones each a distance of $1$ from the other, this case does not work.
If we have one adjacent pair and two separate ones that are $2$ away from each other, we can flip the diagram by $4$ vertices. There are $8$ of these cases.
Finally, if the red and blues alternate, we can simply shift the diagram by a single vertex to satisfy the question. Thus, all of these cases work, and we have $2$ subcases.
There can not be more than $4$ blues, so we are done.
Our total is $1+8+28+56+8+4+8+2=115$. There are $2^8=256$ possible colorings, so we have $\dfrac{115}{256}$ and our answer is $115+256=\boxed{371}$.
~Technodoggo | // Block 1
import graph;
unitsize(1cm);
filldraw(circle((0.5,1.207),2.5),green,black);
void oct11(int[] pts) {
pair[] vertices = {(0,0),(1,0),(1.707,0.707),(1.707,1.707),(1,2.414),(0,2.414),(-0.707,1.707),(-0.707,0.707)};
draw((0,0)--(1,0)--(1.707,0.707)--(1.707,1.707)--(1,2.414)--(0,2.414)--(-0.707,1.707)--(-0.707,0.707)--cycle);
for (int i = 0; i < 8; i+=1) {
if (pts[i] == 0) {
dot(vertices[i], blue+5);
}
if (pts[i] == 1) {
dot(vertices[i], red+5);
}
}
};
int[] sus = {0,0,0,0,1,1,1,1};
oct11(sus);
// Block 2
import graph;
unitsize(1cm);
filldraw(circle((0.5,1.207),2.5),orange,black);
void oct11(int[] pts) {
pair[] vertices = {(0,0),(1,0),(1.707,0.707),(1.707,1.707),(1,2.414),(0,2.414),(-0.707,1.707),(-0.707,0.707)};
draw((0,0)--(1,0)--(1.707,0.707)--(1.707,1.707)--(1,2.414)--(0,2.414)--(-0.707,1.707)--(-0.707,0.707)--cycle);
for (int i = 0; i < 8; i+=1) {
if (pts[i] == 0) {
dot(vertices[i], blue+5);
}
if (pts[i] == 1) {
dot(vertices[i], red+5);
}
}
};
int[] sus = {0,0,0,1,0,1,1,1};
oct11(sus);
// Block 3
import graph;
unitsize(1cm);
filldraw(circle((0.5,1.207),2.5),orange,black);
void oct11(int[] pts) {
pair[] vertices = {(0,0),(1,0),(1.707,0.707),(1.707,1.707),(1,2.414),(0,2.414),(-0.707,1.707),(-0.707,0.707)};
draw((0,0)--(1,0)--(1.707,0.707)--(1.707,1.707)--(1,2.414)--(0,2.414)--(-0.707,1.707)--(-0.707,0.707)--cycle);
for (int i = 0; i < 8; i+=1) {
if (pts[i] == 0) {
dot(vertices[i], blue+5);
}
if (pts[i] == 1) {
dot(vertices[i], red+5);
}
}
};
int[] sus = {0,0,0,1,1,0,1,1};
oct11(sus);
// Block 4
import graph;
unitsize(1cm);
filldraw(circle((0.5,1.207),2.5),orange,black);
void oct11(int[] pts) {
pair[] vertices = {(0,0),(1,0),(1.707,0.707),(1.707,1.707),(1,2.414),(0,2.414),(-0.707,1.707),(-0.707,0.707)};
draw((0,0)--(1,0)--(1.707,0.707)--(1.707,1.707)--(1,2.414)--(0,2.414)--(-0.707,1.707)--(-0.707,0.707)--cycle);
for (int i = 0; i < 8; i+=1) {
if (pts[i] == 0) {
dot(vertices[i], blue+5);
}
if (pts[i] == 1) {
dot(vertices[i], red+5);
}
}
};
int[] sus = {0,0,1,0,0,1,1,1};
oct11(sus);
// Block 5
import graph;
unitsize(1cm);
filldraw(circle((0.5,1.207),2.5),green,black);
void oct11(int[] pts) {
pair[] vertices = {(0,0),(1,0),(1.707,0.707),(1.707,1.707),(1,2.414),(0,2.414),(-0.707,1.707),(-0.707,0.707)};
draw((0,0)--(1,0)--(1.707,0.707)--(1.707,1.707)--(1,2.414)--(0,2.414)--(-0.707,1.707)--(-0.707,0.707)--cycle);
for (int i = 0; i < 8; i+=1) {
if (pts[i] == 0) {
dot(vertices[i], blue+5);
}
if (pts[i] == 1) {
dot(vertices[i], red+5);
}
}
};
int[] sus = {0,0,1,1,0,0,1,1};
oct11(sus);
// Block 6
import graph;
unitsize(1cm);
filldraw(circle((0.5,1.207),2.5),orange,black);
void oct11(int[] pts) {
pair[] vertices = {(0,0),(1,0),(1.707,0.707),(1.707,1.707),(1,2.414),(0,2.414),(-0.707,1.707),(-0.707,0.707)};
draw((0,0)--(1,0)--(1.707,0.707)--(1.707,1.707)--(1,2.414)--(0,2.414)--(-0.707,1.707)--(-0.707,0.707)--cycle);
for (int i = 0; i < 8; i+=1) {
if (pts[i] == 0) {
dot(vertices[i], blue+5);
}
if (pts[i] == 1) {
dot(vertices[i], red+5);
}
}
};
int[] sus = {0,0,1,0,1,0,1,1};
oct11(sus);
// Block 7
import graph;
unitsize(1cm);
filldraw(circle((0.5,1.207),2.5),green,black);
void oct11(int[] pts) {
pair[] vertices = {(0,0),(1,0),(1.707,0.707),(1.707,1.707),(1,2.414),(0,2.414),(-0.707,1.707),(-0.707,0.707)};
draw((0,0)--(1,0)--(1.707,0.707)--(1.707,1.707)--(1,2.414)--(0,2.414)--(-0.707,1.707)--(-0.707,0.707)--cycle);
for (int i = 0; i < 8; i+=1) {
if (pts[i] == 0) {
dot(vertices[i], blue+5);
}
if (pts[i] == 1) {
dot(vertices[i], red+5);
}
}
};
int[] sus = {0,0,1,0,1,1,0,1};
oct11(sus);
// Block 8
import graph;
unitsize(1cm);
filldraw(circle((0.5,1.207),2.5),green,black);
void oct11(int[] pts) {
pair[] vertices = {(0,0),(1,0),(1.707,0.707),(1.707,1.707),(1,2.414),(0,2.414),(-0.707,1.707),(-0.707,0.707)};
draw((0,0)--(1,0)--(1.707,0.707)--(1.707,1.707)--(1,2.414)--(0,2.414)--(-0.707,1.707)--(-0.707,0.707)--cycle);
for (int i = 0; i < 8; i+=1) {
if (pts[i] == 0) {
dot(vertices[i], blue+5);
}
if (pts[i] == 1) {
dot(vertices[i], red+5);
}
}
};
int[] sus = {0,1,0,1,0,1,0,1};
oct11(sus);
// Block 9
import graph; unitsize(1cm); filldraw(circle((0.5,1.207),2.5),green,black); void oct11(int[] pts) { pair[] vertices = {(0,0),(1,0),(1.707,0.707),(1.707,1.707),(1,2.414),(0,2.414),(-0.707,1.707),(-0.707,0.707)}; draw((0,0)--(1,0)--(1.707,0.707)--(1.707,1.707)--(1,2.414)--(0,2.414)--(-0.707,1.707)--(-0.707,0.707)--cycle); for (int i = 0; i < 8; i+=1) { if (pts[i] == 0) { dot(vertices[i], blue+5); } if (pts[i] == 1) { dot(vertices[i], red+5); } } }; int[] sus = {0,0,0,0,1,1,1,1}; oct11(sus);
// Block 10
import graph; unitsize(1cm); filldraw(circle((0.5,1.207),2.5),orange,black); void oct11(int[] pts) { pair[] vertices = {(0,0),(1,0),(1.707,0.707),(1.707,1.707),(1,2.414),(0,2.414),(-0.707,1.707),(-0.707,0.707)}; draw((0,0)--(1,0)--(1.707,0.707)--(1.707,1.707)--(1,2.414)--(0,2.414)--(-0.707,1.707)--(-0.707,0.707)--cycle); for (int i = 0; i < 8; i+=1) { if (pts[i] == 0) { dot(vertices[i], blue+5); } if (pts[i] == 1) { dot(vertices[i], red+5); } } }; int[] sus = {0,0,0,1,0,1,1,1}; oct11(sus);
// Block 11
import graph; unitsize(1cm); filldraw(circle((0.5,1.207),2.5),orange,black); void oct11(int[] pts) { pair[] vertices = {(0,0),(1,0),(1.707,0.707),(1.707,1.707),(1,2.414),(0,2.414),(-0.707,1.707),(-0.707,0.707)}; draw((0,0)--(1,0)--(1.707,0.707)--(1.707,1.707)--(1,2.414)--(0,2.414)--(-0.707,1.707)--(-0.707,0.707)--cycle); for (int i = 0; i < 8; i+=1) { if (pts[i] == 0) { dot(vertices[i], blue+5); } if (pts[i] == 1) { dot(vertices[i], red+5); } } }; int[] sus = {0,0,0,1,1,0,1,1}; oct11(sus);
// Block 12
import graph; unitsize(1cm); filldraw(circle((0.5,1.207),2.5),orange,black); void oct11(int[] pts) { pair[] vertices = {(0,0),(1,0),(1.707,0.707),(1.707,1.707),(1,2.414),(0,2.414),(-0.707,1.707),(-0.707,0.707)}; draw((0,0)--(1,0)--(1.707,0.707)--(1.707,1.707)--(1,2.414)--(0,2.414)--(-0.707,1.707)--(-0.707,0.707)--cycle); for (int i = 0; i < 8; i+=1) { if (pts[i] == 0) { dot(vertices[i], blue+5); } if (pts[i] == 1) { dot(vertices[i], red+5); } } }; int[] sus = {0,0,1,0,0,1,1,1}; oct11(sus);
// Block 13
import graph; unitsize(1cm); filldraw(circle((0.5,1.207),2.5),green,black); void oct11(int[] pts) { pair[] vertices = {(0,0),(1,0),(1.707,0.707),(1.707,1.707),(1,2.414),(0,2.414),(-0.707,1.707),(-0.707,0.707)}; draw((0,0)--(1,0)--(1.707,0.707)--(1.707,1.707)--(1,2.414)--(0,2.414)--(-0.707,1.707)--(-0.707,0.707)--cycle); for (int i = 0; i < 8; i+=1) { if (pts[i] == 0) { dot(vertices[i], blue+5); } if (pts[i] == 1) { dot(vertices[i], red+5); } } }; int[] sus = {0,0,1,1,0,0,1,1}; oct11(sus);
// Block 14
import graph; unitsize(1cm); filldraw(circle((0.5,1.207),2.5),orange,black); void oct11(int[] pts) { pair[] vertices = {(0,0),(1,0),(1.707,0.707),(1.707,1.707),(1,2.414),(0,2.414),(-0.707,1.707),(-0.707,0.707)}; draw((0,0)--(1,0)--(1.707,0.707)--(1.707,1.707)--(1,2.414)--(0,2.414)--(-0.707,1.707)--(-0.707,0.707)--cycle); for (int i = 0; i < 8; i+=1) { if (pts[i] == 0) { dot(vertices[i], blue+5); } if (pts[i] == 1) { dot(vertices[i], red+5); } } }; int[] sus = {0,0,1,0,1,0,1,1}; oct11(sus);
// Block 15
import graph; unitsize(1cm); filldraw(circle((0.5,1.207),2.5),green,black); void oct11(int[] pts) { pair[] vertices = {(0,0),(1,0),(1.707,0.707),(1.707,1.707),(1,2.414),(0,2.414),(-0.707,1.707),(-0.707,0.707)}; draw((0,0)--(1,0)--(1.707,0.707)--(1.707,1.707)--(1,2.414)--(0,2.414)--(-0.707,1.707)--(-0.707,0.707)--cycle); for (int i = 0; i < 8; i+=1) { if (pts[i] == 0) { dot(vertices[i], blue+5); } if (pts[i] == 1) { dot(vertices[i], red+5); } } }; int[] sus = {0,0,1,0,1,1,0,1}; oct11(sus);
// Block 16
import graph; unitsize(1cm); filldraw(circle((0.5,1.207),2.5),green,black); void oct11(int[] pts) { pair[] vertices = {(0,0),(1,0),(1.707,0.707),(1.707,1.707),(1,2.414),(0,2.414),(-0.707,1.707),(-0.707,0.707)}; draw((0,0)--(1,0)--(1.707,0.707)--(1.707,1.707)--(1,2.414)--(0,2.414)--(-0.707,1.707)--(-0.707,0.707)--cycle); for (int i = 0; i < 8; i+=1) { if (pts[i] == 0) { dot(vertices[i], blue+5); } if (pts[i] == 1) { dot(vertices[i], red+5); } } }; int[] sus = {0,1,0,1,0,1,0,1}; oct11(sus); | [] |
261 | Let $ABCD$ be a tetrahedron such that $AB=CD= \sqrt{41}$, $AC=BD= \sqrt{80}$, and $BC=AD= \sqrt{89}$. There exists a point $I$ inside the tetrahedron such that the distances from $I$ to each of the faces of the tetrahedron are all equal. This distance can be written in the form $\frac{m \sqrt n}{p}$, where $m$, $n$, and $p$ are positive integers, $m$ and $p$ are relatively prime, and $n$ is not divisible by the square of any prime. Find $m+n+p$. | 2024 AIME I Problem 14 | Inscribe tetrahedron $ABCD$ in an rectangular prism as shown above.
By the Pythagorean theorem, we note
\[OA^2 + OB^2 = AB^2 = 41,\]
\[OA^2 + OC^2 = AC^2 = 80, \text{and}\]
\[OB^2 + OC^2 = BC^2 = 89.\]
Solving yields $OA = 4, OB = 5,$ and $OC = 8.$
Since each face of the tetrahedron is congruent, we know the point we seek is the center of the circumsphere of $ABCD.$ We know all rectangular prisms can be inscribed in a circumsphere, therefore the circumsphere of the rectangular prism is also the circumsphere of $ABCD.$
We know that the distance from all $4$ faces must be the same, so we only need to find the distance from the center to plane $ABC$.
Let $O = (0,0,0), A = (4,0,0), B = (0,5,0),$ and $C = (0,0,8).$ We obtain that the plane of $ABC$ can be marked as $\frac{x}{4} + \frac{y}{5} + \frac{z}{8} = 1,$ or $10x + 8y + 5z - 40 = 0,$ and the center of the prism is $(2,\frac{5}{2},4).$
Using the Point-to-Plane distance formula, our distance is
\[d = \frac{|10\cdot 2 + 8\cdot \frac{5}{2} + 5\cdot 4 - 40|}{\sqrt{10^2 + 8^2 + 5^2}} = \frac{20}{\sqrt{189}} = \frac{20\sqrt{21}}{63}.\]
Our answer is $20 + 21 + 63 = \boxed{104}.$
- spectraldragon8 | // Block 1
import three;
currentprojection = orthographic(1,1,1);
triple O = (0,0,0);
triple A = (0,2,0);
triple B = (0,0,1);
triple C = (3,0,0);
triple D = (3,2,1);
triple E = (3,2,0);
triple F = (0,2,1);
triple G = (3,0,1);
draw(A--B--C--cycle, red);
draw(A--B--D--cycle, red);
draw(A--C--D--cycle, red);
draw(B--C--D--cycle, red);
draw(E--A--O--C--cycle);
draw(D--F--B--G--cycle);
draw(O--B);
draw(A--F);
draw(E--D);
draw(C--G);
label("$O$", O, SW);
label("$A$", A, NW);
label("$B$", B, W);
label("$C$", C, S);
label("$D$", D, NE);
label("$E$", E, SE);
label("$F$", F, NW);
label("$G$", G, NE);
// Block 2
import three; currentprojection = orthographic(1,1,1); triple O = (0,0,0); triple A = (0,2,0); triple B = (0,0,1); triple C = (3,0,0); triple D = (3,2,1); triple E = (3,2,0); triple F = (0,2,1); triple G = (3,0,1); draw(A--B--C--cycle, red); draw(A--B--D--cycle, red); draw(A--C--D--cycle, red); draw(B--C--D--cycle, red); draw(E--A--O--C--cycle); draw(D--F--B--G--cycle); draw(O--B); draw(A--F); draw(E--D); draw(C--G); label("$O$", O, SW); label("$A$", A, NW); label("$B$", B, W); label("$C$", C, S); label("$D$", D, NE); label("$E$", E, SE); label("$F$", F, NW); label("$G$", G, NE); | [] |
262 | Torus $T$ is the surface produced by revolving a circle with radius $3$ around an axis in the plane of the circle that is a distance $6$ from the center of the circle (so like a donut). Let $S$ be a sphere with a radius $11$. When $T$ rests on the inside of $S$, it is internally tangent to $S$ along a circle with radius $r_i$, and when $T$ rests on the outside of $S$, it is externally tangent to $S$ along a circle with radius $r_o$. The difference $r_i-r_o$ can be written as $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2024 AIME II Problem 8 | First, let's consider a section $\mathcal{P}$ of the solids, along the axis.
By some 3D-Geomerty thinking, we can simply know that the axis crosses the sphere center. So, that is saying, the $\mathcal{P}$ we took crosses one of the equator of the sphere.
Here I drew two graphs, the first one is the case when $T$ is internally tangent to $S$,
and the second one is when $T$ is externally tangent to $S$.
For both graphs, point $O$ is the center of sphere $S$, and points $A$ and $B$ are the intersections of the sphere and the axis. Point $E$ (ignoring the subscripts) is one of the circle centers of the intersection of torus $T$ with section $\mathcal{P}$. Point $G$ (again, ignoring the subscripts) is one of the tangents between the torus $T$ and sphere $S$ on section $\mathcal{P}$. $EF\bot CD$, $HG\bot CD$.
And then, we can start our calculation.
In both cases, we know $\Delta OEF\sim \Delta OGH\Longrightarrow \frac{EF}{OE} =\frac{GH}{OG}$.
Hence, in the case of internal tangent, $\frac{E_iF_i}{OE_i} =\frac{G_iH_i}{OG_i}\Longrightarrow \frac{6}{11-3} =\frac{r_i}{11}\Longrightarrow r_i=\frac{33}{4}$.
In the case of external tangent, $\frac{E_oF_o}{OE_o} =\frac{G_oH_o}{OG_o}\Longrightarrow \frac{6}{11+3} =\frac{r_o}{11}\Longrightarrow r_o=\frac{33}{7}$.
Thereby, $r_i-r_o=\frac{33}{4}-\frac{33}{7}=\frac{99}{28}$. And there goes the answer, $99+28=\boxed{\mathbf{127} }$
~Prof_Joker | // Block 1
unitsize(0.35cm);
pair O = (0, 0);
real r1 = 11;
real r2 = 3;
draw(circle(O, r1));
pair A = O + (0, -r1);
pair B = O + (0, r1);
draw(A--B);
pair C = O + (0, -1.25*r1);
pair D = O + (0, 1.25*r1);
draw(C--D, dashed);
dot(O);
pair E = (2 * r2, -sqrt((r1 - r2) * (r1 - r2) - 4 * r2 * r2));
pair F = (0, -sqrt((r1 - r2) * (r1 - r2) - 4 * r2 * r2));
pair G = (-r2 * O + r1 * E) / (r1 - r2);
pair H = (-r2 * O + r1 * F) / (r1 - r2);
draw(circle(E, r2));
draw(circle((-2 * r2, -sqrt((r1 - r2) * (r1 - r2) - 4 * r2 * r2)), r2));
draw(O--G, dashed);
draw(F--E, dashed);
draw(G--H, dashed);
label("$O$", O, SW);
label("$A$", A, SW);
label("$B$", B, NW);
label("$C$", C, NW);
label("$D$", D, SW);
label("$E_i$", E, NE);
label("$F_i$", F, W);
label("$G_i$", G, SE);
label("$H_i$", H, W);
label("$r_i$", 0.5 * H + 0.5 * G, NE);
label("$3$", 0.5 * E + 0.5 * G, NE);
label("$11$", 0.5 * O + 0.5 * G, NE);
// Block 2
unitsize(0.35cm);
pair O = (0, 0);
real r1 = 11;
real r2 = 3;
draw(circle(O, r1));
pair A = O + (0, -r1);
pair B = O + (0, r1);
draw(A--B);
pair C = O + (0, -1.25*(r1 + r2));
pair D = O + (0, 1.25*r1);
draw(C--D, dashed);
dot(O);
pair E = (2 * r2, -sqrt((r1 + r2) * (r1 + r2) - 4 * r2 * r2));
pair F = (0, -sqrt((r1 + r2) * (r1 + r2) - 4 * r2 * r2));
pair G = (r2 * O + r1 * E) / (r1 + r2);
pair H = (r2 * O + r1 * F) / (r1 + r2);
draw(circle(E, r2));
draw(circle((-2 * r2, -sqrt((r1 + r2) * (r1 + r2) - 4 * r2 * r2)), r2));
draw(O--E, dashed);
draw(F--E, dashed);
draw(G--H, dashed);
label("$O$", O, SW);
label("$A$", A, SW);
label("$B$", B, NW);
label("$C$", C, NW);
label("$D$", D, SW);
label("$E_o$", E, NE);
label("$F_o$", F, SW);
label("$G_o$", G, S);
label("$H_o$", H, W);
label("$r_o$", 0.5 * H + 0.5 * G, NE);
label("$3$", 0.5 * E + 0.5 * G, NE);
label("$11$", 0.5 * O + 0.5 * G, NE);
// Block 3
unitsize(0.35cm); pair O = (0, 0); real r1 = 11; real r2 = 3; draw(circle(O, r1)); pair A = O + (0, -r1); pair B = O + (0, r1); draw(A--B); pair C = O + (0, -1.25*r1); pair D = O + (0, 1.25*r1); draw(C--D, dashed); dot(O); pair E = (2 * r2, -sqrt((r1 - r2) * (r1 - r2) - 4 * r2 * r2)); pair F = (0, -sqrt((r1 - r2) * (r1 - r2) - 4 * r2 * r2)); pair G = (-r2 * O + r1 * E) / (r1 - r2); pair H = (-r2 * O + r1 * F) / (r1 - r2); draw(circle(E, r2)); draw(circle((-2 * r2, -sqrt((r1 - r2) * (r1 - r2) - 4 * r2 * r2)), r2)); draw(O--G, dashed); draw(F--E, dashed); draw(G--H, dashed); label("$O$", O, SW); label("$A$", A, SW); label("$B$", B, NW); label("$C$", C, NW); label("$D$", D, SW); label("$E_i$", E, NE); label("$F_i$", F, W); label("$G_i$", G, SE); label("$H_i$", H, W); label("$r_i$", 0.5 * H + 0.5 * G, NE); label("$3$", 0.5 * E + 0.5 * G, NE); label("$11$", 0.5 * O + 0.5 * G, NE);
// Block 4
unitsize(0.35cm); pair O = (0, 0); real r1 = 11; real r2 = 3; draw(circle(O, r1)); pair A = O + (0, -r1); pair B = O + (0, r1); draw(A--B); pair C = O + (0, -1.25*(r1 + r2)); pair D = O + (0, 1.25*r1); draw(C--D, dashed); dot(O); pair E = (2 * r2, -sqrt((r1 + r2) * (r1 + r2) - 4 * r2 * r2)); pair F = (0, -sqrt((r1 + r2) * (r1 + r2) - 4 * r2 * r2)); pair G = (r2 * O + r1 * E) / (r1 + r2); pair H = (r2 * O + r1 * F) / (r1 + r2); draw(circle(E, r2)); draw(circle((-2 * r2, -sqrt((r1 + r2) * (r1 + r2) - 4 * r2 * r2)), r2)); draw(O--E, dashed); draw(F--E, dashed); draw(G--H, dashed); label("$O$", O, SW); label("$A$", A, SW); label("$B$", B, NW); label("$C$", C, NW); label("$D$", D, SW); label("$E_o$", E, NE); label("$F_o$", F, SW); label("$G_o$", G, S); label("$H_o$", H, W); label("$r_o$", 0.5 * H + 0.5 * G, NE); label("$3$", 0.5 * E + 0.5 * G, NE); label("$11$", 0.5 * O + 0.5 * G, NE); | [] |
263 | Let $\triangle$$ABC$ have incenter $I$, circumcenter $O$, inradius $6$, and circumradius $13$. Suppose that $\overline{IA} \perp \overline{OI}$. Find $AB \cdot AC$. | 2024 AIME II Problem 10 | Start off by (of course) drawing a diagram! Let $I$ and $O$ be the incenter and circumcenters of triangle $ABC$, respectively. Furthermore, extend $AI$ to meet $BC$ at $L$ and the circumcircle of triangle $ABC$ at $D$.
We'll tackle the initial steps of the problem in two different manners, both leading us to the same final calculations. | // Block 1
size(300);
import olympiad;
real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6;
pair B=(0,0),C=(c,0), D = (c/2-0.01, -2.26);
pair A = (c/3,8.65*c/10);
draw(circumcircle(A,B,C));
pair I=incenter(A,B,C);
pair O=circumcenter(A,B,C);
pair L=extension(A,I,C,B);
dot(I^^O^^A^^B^^C^^D^^L);
draw(A--L);
draw(A--D);
path midangle(pair d,pair e,pair f) {return e--e+((f-e)/length(f-e)+(d-e)/length(d-e))/2;}
draw(C--B--D--cycle);
draw(A--C--B);
draw(A--B);
draw(B--I--C^^A--I);
draw(incircle(A,B,C));
label("$B$",B,SW);
label("$C$",C,SE);
label("$A$",A,N);
label("$D$",D,S);
label("$I$",I,NW);
label("$L$",L,SW);
label("$O$",O,E);
label("$\alpha$",B,5*dir(midangle(A,B,I)),fontsize(8));
label("$\alpha$",B,5*dir(midangle(I,B,C)),fontsize(8));
label("$\beta$",C,12*dir(midangle(B,C,I)),fontsize(8));
label("$\beta$",C,12*dir(midangle(I,C,A)),fontsize(8));
label("$\gamma$",A,5*dir(midangle(B,A,I)),fontsize(8));
label("$\gamma$",A,5*dir(midangle(I,A,C)),fontsize(8));
draw(I--O);
draw(A--O);
draw(rightanglemark(A,I,O));
// Block 2
size(300); import olympiad; real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6; pair B=(0,0),C=(c,0), D = (c/2-0.01, -2.26); pair A = (c/3,8.65*c/10); draw(circumcircle(A,B,C)); pair I=incenter(A,B,C); pair O=circumcenter(A,B,C); pair L=extension(A,I,C,B); dot(I^^O^^A^^B^^C^^D^^L); draw(A--L); draw(A--D); path midangle(pair d,pair e,pair f) {return e--e+((f-e)/length(f-e)+(d-e)/length(d-e))/2;} draw(C--B--D--cycle); draw(A--C--B); draw(A--B); draw(B--I--C^^A--I); draw(incircle(A,B,C)); label("$B$",B,SW); label("$C$",C,SE); label("$A$",A,N); label("$D$",D,S); label("$I$",I,NW); label("$L$",L,SW); label("$O$",O,E); label("$\alpha$",B,5*dir(midangle(A,B,I)),fontsize(8)); label("$\alpha$",B,5*dir(midangle(I,B,C)),fontsize(8)); label("$\beta$",C,12*dir(midangle(B,C,I)),fontsize(8)); label("$\beta$",C,12*dir(midangle(I,C,A)),fontsize(8)); label("$\gamma$",A,5*dir(midangle(B,A,I)),fontsize(8)); label("$\gamma$",A,5*dir(midangle(I,A,C)),fontsize(8)); draw(I--O); draw(A--O); draw(rightanglemark(A,I,O)); | [] |
263 | Let $\triangle$$ABC$ have incenter $I$, circumcenter $O$, inradius $6$, and circumradius $13$. Suppose that $\overline{IA} \perp \overline{OI}$. Find $AB \cdot AC$. | 2024 AIME II Problem 10 | Now we have some sort of expression for $AB \cdot AC$ in terms of $ID$ and $AL$. Let's try to find $AL$ first.
Drop an altitude from $D$ to $BC$, $I$ to $AC$, and $I$ to $BC$:
Since $\angle DBE \cong \angle IAF$ and $\angle BED \cong \angle IFA$, $\triangle BDE \sim \triangle AIF$.
Furthermore, we know $BD = ID$ and $AI = ID$, so $BD = AI$. Since we have two right similar triangles and the corresponding sides are equal, these two triangles are actually congruent: this implies that $DE = IF = 6$ since $IF$ is the inradius.
Now notice that $\triangle IGL \sim \triangle DEL$ because of equal vertical angles and right angles. Furthermore, $IG$ is the inradius so it's length is $6$, which equals the length of $DE$. Therefore these two triangles are congruent, so $IL = DL$.
Since $IL+DL = ID$, $ID = 2 \cdot IL$. Furthermore, $AL = AI + IL = ID + IL = 3 \cdot IL$.
We can now plug back into our initial equations for $AB \cdot AC$:
From $1.1$, $AB \cdot AC = 2 \cdot AL \cdot ID = 2 \cdot 3 \cdot IL \cdot 2 \cdot IL$
\[\implies AB \cdot AC = 3 \cdot (2 \cdot IL) \cdot (2 \cdot IL) = 3 \cdot ID^2\]
Alternatively, from $1.2$, $AB \cdot AC = ID^2 \cdot \frac{AL}{DL}$
\[\implies AB \cdot AC = ID^2 \cdot \frac{3 \cdot IL}{IL} = 3 \cdot ID^2\]
Now all we need to do is find $ID$.
The problem now becomes very simple if one knows Euler's Formula for the distance between the incenter and the circumcenter of a triangle. This formula states that $OI^2 = R(R-2r)$, where $R$ is the circumradius and $r$ is the inradius. We will prove this formula first, but if you already know the proof, skip this part.
Theorem: in any triangle, let $d$ be the distance from the circumcenter to the incenter of the triangle. Then $d^2 = R \cdot (R-2r)$, where $R$ is the circumradius of the triangle and $r$ is the inradius of the triangle.
Proof:
Construct the following diagram:
Let $OI = d$, $OH = R$, $IF = r$. By the Power of a Point, $IH \cdot IJ = AI \cdot ID$.
$IH = R+d$ and $IJ = R-d$, so \[(R+d) \cdot (R-d) = AI \cdot ID = AI \cdot CD\]
Now consider $\triangle ACD$. Since all three points lie on the circumcircle of $\triangle ABC$, the two triangles have the same circumcircle. Thus we can apply law of sines and we get $\frac{CD}{\sin(\angle DAC)} = 2R$. This implies
\[(R+d)\cdot (R-d) = AI \cdot 2R \cdot \sin(\angle DAC)\]
Also, $\sin(\angle DAC)) = \sin(\angle IAF))$, and $\triangle IAF$ is right. Therefore \[\sin(\angle IAF) = \frac{IF}{AI} = \frac{r}{AI}\]
Plugging in, we have
\[(R+d)\cdot (R-d) = AI \cdot 2R \cdot \frac{r}{AI} = 2R \cdot r\]
Thus \[R^2-d^2 = 2R \cdot r \implies d^2 = R \cdot (R-2r)\]
Now we can finish up our solution. We know that $AB \cdot AC = 3 \cdot ID^2$. Since $ID = AI$, $AB \cdot AC = 3 \cdot AI^2$. Since $\triangle AOI$ is right, we can apply the pythagorean theorem: $AI^2 = AO^2-OI^2 = 13^2-OI^2$.
Plugging in from Euler's formula, $OI^2 = 13 \cdot (13 - 2 \cdot 6) = 13$.
Thus $AI^2 = 169-13 = 156$.
Finally $AB \cdot AC = 3 \cdot AI^2 = 3 \cdot 156 = \textbf{468}$.
~KingRavi | // Block 1
size(300);
import olympiad;
real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6;
pair B=(0,0),C=(c,0), D = (c/2-0.01, -2.26), E = (c/2-0.01,0);
pair A = (c/3,8.65*c/10);
pair F = (2*c/3-0.14, 4-0.29);
pair G = (c/2-0.68,0);
draw(circumcircle(A,B,C));
pair I=incenter(A,B,C);
pair O=circumcenter(A,B,C);
pair L=extension(A,I,C,B);
dot(I^^O^^A^^B^^C^^D^^L^^E^^F^^G);
draw(A--L);
draw(A--D);
draw(D--E);
draw(I--F);
draw(I--G);
path midangle(pair d,pair e,pair f) {return e--e+((f-e)/length(f-e)+(d-e)/length(d-e))/2;}
draw(C--B--D--cycle);
draw(A--C--B);
draw(A--B);
draw(B--I--C^^A--I);
draw(incircle(A,B,C));
label("$B$",B,SW);
label("$C$",C,SE);
label("$A$",A,N);
label("$D$",D,S);
label("$I$",I,NW);
label("$L$",L,SW);
label("$O$",O,E);
label("$E$",E,N);
label("$F$",F,NE);
label("$G$",G,SW);
label("$\alpha$",B,5*dir(midangle(A,B,I)),fontsize(8));
label("$\alpha$",B,5*dir(midangle(I,B,C)),fontsize(8));
label("$\beta$",C,12*dir(midangle(B,C,I)),fontsize(8));
label("$\beta$",C,12*dir(midangle(I,C,A)),fontsize(8));
label("$\gamma$",A,5*dir(midangle(B,A,I)),fontsize(8));
label("$\gamma$",A,5*dir(midangle(I,A,C)),fontsize(8));
draw(I--O);
draw(A--O);
draw(rightanglemark(A,I,O));
draw(rightanglemark(B,E,D));
draw(rightanglemark(I,F,A));
draw(rightanglemark(I,G,L));
// Block 2
size(300);
import olympiad;
real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6;
pair B=(0,0),C=(c,0), D = (c/2-0.01, -2.26), E = (c/2-0.01,0);
pair A = (c/3,8.65*c/10);
pair F = (2*c/3-0.14, 4-0.29);
pair G = (c/2-0.68,0);
draw(circumcircle(A,B,C));
pair I=incenter(A,B,C);
pair O=circumcenter(A,B,C);
pair L=extension(A,I,C,B);
dot(I^^O^^A^^B^^C^^D^^L^^F);
draw(A--L);
draw(A--D);
draw(I--F);
path midangle(pair d,pair e,pair f) {return e--e+((f-e)/length(f-e)+(d-e)/length(d-e))/2;}
draw(C--B--D--cycle);
draw(A--C--B);
draw(A--B);
draw(A--I);
draw(incircle(A,B,C));
label("$B$",B,SW);
label("$C$",C,SE);
label("$A$",A,N);
label("$D$",D,S);
label("$I$",I,NW);
label("$L$",L,SW);
label("$O$",O,S);
label("$F$",F,NE);
label("$\gamma$",A,5*dir(midangle(B,A,I)),fontsize(8));
label("$\gamma$",A,5*dir(midangle(I,A,C)),fontsize(8));
pair H = (10*c/8-1.46,2*c/3-1.85), J = (-0.55,1.4);
dot(H^^J);
label("$H$", H, E);
label("$J$", J, W);
draw(I--O);
draw(I--H);
draw(I--J);
draw(rightanglemark(I,F,A));
// Block 3
size(300); import olympiad; real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6; pair B=(0,0),C=(c,0), D = (c/2-0.01, -2.26), E = (c/2-0.01,0); pair A = (c/3,8.65*c/10); pair F = (2*c/3-0.14, 4-0.29); pair G = (c/2-0.68,0); draw(circumcircle(A,B,C)); pair I=incenter(A,B,C); pair O=circumcenter(A,B,C); pair L=extension(A,I,C,B); dot(I^^O^^A^^B^^C^^D^^L^^E^^F^^G); draw(A--L); draw(A--D); draw(D--E); draw(I--F); draw(I--G); path midangle(pair d,pair e,pair f) {return e--e+((f-e)/length(f-e)+(d-e)/length(d-e))/2;} draw(C--B--D--cycle); draw(A--C--B); draw(A--B); draw(B--I--C^^A--I); draw(incircle(A,B,C)); label("$B$",B,SW); label("$C$",C,SE); label("$A$",A,N); label("$D$",D,S); label("$I$",I,NW); label("$L$",L,SW); label("$O$",O,E); label("$E$",E,N); label("$F$",F,NE); label("$G$",G,SW); label("$\alpha$",B,5*dir(midangle(A,B,I)),fontsize(8)); label("$\alpha$",B,5*dir(midangle(I,B,C)),fontsize(8)); label("$\beta$",C,12*dir(midangle(B,C,I)),fontsize(8)); label("$\beta$",C,12*dir(midangle(I,C,A)),fontsize(8)); label("$\gamma$",A,5*dir(midangle(B,A,I)),fontsize(8)); label("$\gamma$",A,5*dir(midangle(I,A,C)),fontsize(8)); draw(I--O); draw(A--O); draw(rightanglemark(A,I,O)); draw(rightanglemark(B,E,D)); draw(rightanglemark(I,F,A)); draw(rightanglemark(I,G,L));
// Block 4
size(300); import olympiad; real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6; pair B=(0,0),C=(c,0), D = (c/2-0.01, -2.26), E = (c/2-0.01,0); pair A = (c/3,8.65*c/10); pair F = (2*c/3-0.14, 4-0.29); pair G = (c/2-0.68,0); draw(circumcircle(A,B,C)); pair I=incenter(A,B,C); pair O=circumcenter(A,B,C); pair L=extension(A,I,C,B); dot(I^^O^^A^^B^^C^^D^^L^^F); draw(A--L); draw(A--D); draw(I--F); path midangle(pair d,pair e,pair f) {return e--e+((f-e)/length(f-e)+(d-e)/length(d-e))/2;} draw(C--B--D--cycle); draw(A--C--B); draw(A--B); draw(A--I); draw(incircle(A,B,C)); label("$B$",B,SW); label("$C$",C,SE); label("$A$",A,N); label("$D$",D,S); label("$I$",I,NW); label("$L$",L,SW); label("$O$",O,S); label("$F$",F,NE); label("$\gamma$",A,5*dir(midangle(B,A,I)),fontsize(8)); label("$\gamma$",A,5*dir(midangle(I,A,C)),fontsize(8)); pair H = (10*c/8-1.46,2*c/3-1.85), J = (-0.55,1.4); dot(H^^J); label("$H$", H, E); label("$J$", J, W); draw(I--O); draw(I--H); draw(I--J); draw(rightanglemark(I,F,A)); | [] |
264 | Let \(O=(0,0)\), \(A=\left(\tfrac{1}{2},0\right)\), and \(B=\left(0,\tfrac{\sqrt{3}}{2}\right)\) be points in the coordinate plane. Let \(\mathcal{F}\) be the family of segments \(\overline{PQ}\) of unit length lying in the first quadrant with \(P\) on the \(x\)-axis and \(Q\) on the \(y\)-axis. There is a unique point \(C\) on \(\overline{AB}\), distinct from \(A\) and \(B\), that does not belong to any segment from \(\mathcal{F}\) other than \(\overline{AB}\). Then \(OC^2=\tfrac{p}{q}\), where \(p\) and \(q\) are relatively prime positive integers. Find \(p+q\). | 2024 AIME II Problem 12 | Let's move a little bit from $A$ to $A_1$, then $B$ must move to $B_1$ to keep $A_1B_1 = 1$. $AB$ intersects with $A_1B_1$ at $C$. Pick points $A_2$ and $B_2$ on $CA_1$ and $CB$ such that $CA_2 = CA$, $CB_2 = CB_1$, we have $A_1A_2 = BB_2$. Since $AA_1$ is very small, $\angle CA_1A \approx 60^\circ$, $\angle CBB_1 \approx 30^\circ$, so $AA_2\approx \sqrt{3}A_1A_2$, $B_1B_2 \approx \frac{1}{\sqrt{3}}BB_2$, by similarity, $\frac{CA}{CB} \approx \frac{CA}{CB_2} = \frac{AA_2}{B_1B_2} = \frac{\sqrt{3}A_1A_2}{\frac{1}{\sqrt{3}}BB_2} = 3$. So the coordinates of $C$ is $\left(\frac{1}{8}, \frac{3\sqrt{3}}{8}\right)$.
so $OC^2 = \frac{1}{64} + \frac{27}{64} = \frac{7}{16}$, the answer is $\boxed{023}$. | // Block 1
pair O=(0,0);
pair X=(1,0);
pair Y=(0,1);
pair A=(0.5,0); pair B=(0,sin(pi/3));
pair A1=(0.6,0); pair B1=(0,0.8);
pair A2=(0.575,0.04); pair B2=(0.03,0.816);
dot(O);
dot(X);
dot(Y); dot(A); dot(B);
dot(A1); dot(B1);
dot(A2); dot(B2);
draw(X--O--Y);
draw(A--B);
draw(A1--B1);
draw(A--A2);
draw(B1--B2);
label("$B$", B, W);
label("$A$", A, S);
label("$B_1$", B1, SW);
label("$A_1$", A1, S);
label("$B_2$", B2, E);
label("$A_2$", A2, NE);
label("$O$", O, SW);
pair C=(0.18,0.56);
label("$C$", C, E);
dot(C);
// Block 2
pair O=(0,0); pair X=(1,0); pair Y=(0,1); pair A=(0.5,0); pair B=(0,sin(pi/3)); pair A1=(0.6,0); pair B1=(0,0.8); pair A2=(0.575,0.04); pair B2=(0.03,0.816); dot(O); dot(X); dot(Y); dot(A); dot(B); dot(A1); dot(B1); dot(A2); dot(B2); draw(X--O--Y); draw(A--B); draw(A1--B1); draw(A--A2); draw(B1--B2); label("$B$", B, W); label("$A$", A, S); label("$B_1$", B1, SW); label("$A_1$", A1, S); label("$B_2$", B2, E); label("$A_2$", A2, NE); label("$O$", O, SW); pair C=(0.18,0.56); label("$C$", C, E); dot(C); | [] |
265 | Find the number of rectangles that can be formed inside a fixed regular dodecagon ($12$-gon) where each side of the rectangle lies on either a side or a diagonal of the dodecagon. The diagram below shows three of those rectangles. | 2024 AIME II Problem 15 | By Furaken
There are two kinds of such rectangles: those whose sides are parallel to some edges of the regular 12-gon (Case 1), and those whose sides are not (Case 2).
For Case 1, WLOG assume that the rectangle's sides are horizontal and vertical (don't forget to multiply by 3 at the end of Case 1). Then the rectangle's sides coincide with these segments as shown in the diagram.
We use inclusion-exclusion for this. There are 30 valid rectangles contained in $A_1A_5A_7A_{11}$, as well as 30 in $A_2A_4A_8A_{10}$. However, the 9 rectangles contained in $B_1B_2B_3B_4$ have been counted twice, so we subtract 9 and we have 51 rectangles in the diagram. Multiplying by 3, we get 153 rectangles for Case 1.
For Case 2, we have this diagram. To be honest, you can count the rectangles here in whatever way you like.
There are 36 rectangles contained within $A_2A_5A_8A_{11}$, and 18 that use points outside $A_2A_5A_8A_{11}$. So we get a total of $3(36+18)=162$ rectangles for Case 2.
Adding the two cases together, we get the answer $\boxed{315}$. | // Block 1
real r = pi/6;
pair A1 = (cos(r),sin(r));
pair A2 = (cos(2r),sin(2r));
pair A3 = (cos(3r),sin(3r));
pair A4 = (cos(4r),sin(4r));
pair A5 = (cos(5r),sin(5r));
pair A6 = (cos(6r),sin(6r));
pair A7 = (cos(7r),sin(7r));
pair A8 = (cos(8r),sin(8r));
pair A9 = (cos(9r),sin(9r));
pair A10 = (cos(10r),sin(10r));
pair A11 = (cos(11r),sin(11r));
pair A12 = (cos(12r),sin(12r));
dot(A1);
dot(A2);
dot(A3);
dot(A4);
dot(A5);
dot(A6);
dot(A7);
dot(A8);
dot(A9);
dot(A10);
dot(A11);
dot(A12);
pair B1 = (0.5,0.5);
pair B2 = (-0.5,0.5);
pair B3 = (-0.5,-0.5);
pair B4 = (0.5,-0.5);
dot(B1);
dot(B2);
dot(B3);
dot(B4);
draw(A1--A5--A7--A11--cycle);
draw(A2--A4--A8--A10--cycle);
draw(A3--A9);
draw(A6--A12);
label("$A_1$", A1, NE);
label("$A_2$", A2, NE);
label("$A_3$", A3, N);
label("$A_4$", A4, NW);
label("$A_5$", A5, NW);
label("$A_6$", A6, W);
label("$A_7$", A7, SW);
label("$A_8$", A8, SW);
label("$A_9$", A9, S);
label("$A_{10}$", A10, SE);
label("$A_{11}$", A11, SE);
label("$A_{12}$", A12, E);
label("$B_1$", B1, SW);
label("$B_2$", B2, SE);
label("$B_3$", B3, NE);
label("$B_4$", B4, NW);
// Block 2
real r = pi/6;
pair A1 = (cos(r),sin(r));
pair A2 = (cos(2r),sin(2r));
pair A3 = (cos(3r),sin(3r));
pair A4 = (cos(4r),sin(4r));
pair A5 = (cos(5r),sin(5r));
pair A6 = (cos(6r),sin(6r));
pair A7 = (cos(7r),sin(7r));
pair A8 = (cos(8r),sin(8r));
pair A9 = (cos(9r),sin(9r));
pair A10 = (cos(10r),sin(10r));
pair A11 = (cos(11r),sin(11r));
pair A12 = (cos(12r),sin(12r));
dot(A1);
dot(A2);
dot(A3);
dot(A4);
dot(A5);
dot(A6);
dot(A7);
dot(A8);
dot(A9);
dot(A10);
dot(A11);
dot(A12);
draw(A1--A6--A7--A12--cycle);
draw(A3--A4--A9--A10--cycle);
draw(A2--A5--A8--A11--cycle);
label("$A_1$", A1, NE);
label("$A_2$", A2, NE);
label("$A_3$", A3, N);
label("$A_4$", A4, NW);
label("$A_5$", A5, NW);
label("$A_6$", A6, W);
label("$A_7$", A7, SW);
label("$A_8$", A8, SW);
label("$A_9$", A9, S);
label("$A_{10}$", A10, SE);
label("$A_{11}$", A11, SE);
label("$A_{12}$", A12, E);
// Block 3
real r = pi/6; pair A1 = (cos(r),sin(r)); pair A2 = (cos(2r),sin(2r)); pair A3 = (cos(3r),sin(3r)); pair A4 = (cos(4r),sin(4r)); pair A5 = (cos(5r),sin(5r)); pair A6 = (cos(6r),sin(6r)); pair A7 = (cos(7r),sin(7r)); pair A8 = (cos(8r),sin(8r)); pair A9 = (cos(9r),sin(9r)); pair A10 = (cos(10r),sin(10r)); pair A11 = (cos(11r),sin(11r)); pair A12 = (cos(12r),sin(12r)); dot(A1); dot(A2); dot(A3); dot(A4); dot(A5); dot(A6); dot(A7); dot(A8); dot(A9); dot(A10); dot(A11); dot(A12); pair B1 = (0.5,0.5); pair B2 = (-0.5,0.5); pair B3 = (-0.5,-0.5); pair B4 = (0.5,-0.5); dot(B1); dot(B2); dot(B3); dot(B4); draw(A1--A5--A7--A11--cycle); draw(A2--A4--A8--A10--cycle); draw(A3--A9); draw(A6--A12); label("$A_1$", A1, NE); label("$A_2$", A2, NE); label("$A_3$", A3, N); label("$A_4$", A4, NW); label("$A_5$", A5, NW); label("$A_6$", A6, W); label("$A_7$", A7, SW); label("$A_8$", A8, SW); label("$A_9$", A9, S); label("$A_{10}$", A10, SE); label("$A_{11}$", A11, SE); label("$A_{12}$", A12, E); label("$B_1$", B1, SW); label("$B_2$", B2, SE); label("$B_3$", B3, NE); label("$B_4$", B4, NW);
// Block 4
real r = pi/6; pair A1 = (cos(r),sin(r)); pair A2 = (cos(2r),sin(2r)); pair A3 = (cos(3r),sin(3r)); pair A4 = (cos(4r),sin(4r)); pair A5 = (cos(5r),sin(5r)); pair A6 = (cos(6r),sin(6r)); pair A7 = (cos(7r),sin(7r)); pair A8 = (cos(8r),sin(8r)); pair A9 = (cos(9r),sin(9r)); pair A10 = (cos(10r),sin(10r)); pair A11 = (cos(11r),sin(11r)); pair A12 = (cos(12r),sin(12r)); dot(A1); dot(A2); dot(A3); dot(A4); dot(A5); dot(A6); dot(A7); dot(A8); dot(A9); dot(A10); dot(A11); dot(A12); draw(A1--A6--A7--A12--cycle); draw(A3--A4--A9--A10--cycle); draw(A2--A5--A8--A11--cycle); label("$A_1$", A1, NE); label("$A_2$", A2, NE); label("$A_3$", A3, N); label("$A_4$", A4, NW); label("$A_5$", A5, NW); label("$A_6$", A6, W); label("$A_7$", A7, SW); label("$A_8$", A8, SW); label("$A_9$", A9, S); label("$A_{10}$", A10, SE); label("$A_{11}$", A11, SE); label("$A_{12}$", A12, E); | [] |
266 | An isosceles trapezoid has an inscribed circle tangent to each of its four sides. The radius of the circle is $3$, and the area of the trapezoid is $72$. Let the parallel sides of the trapezoid have lengths $r$ and $s$, with $r \neq s$. Find $r^2+s^2$ | 2025 AIME I Problem 6 | Draw angle bisectors from the bottom left vertex to the center of the circle. Call the angle formed $\theta$. Drawing a line from the center of the circle to the midway point of the bottom base of the trapezoid makes a right angle, and the other angle has to be $90^{\circ} - \theta$. Then draw a line segment from the center of the circle to the top left vertex, then you have a right triangle. The smaller angle of this triangle is $180^{\circ} - (180^{\circ} - \theta) = \theta$. This means $\frac{r}{2} = \frac{3}{\tan \theta} \implies r = \frac{6}{\tan \theta}$. This also means $\frac{s}{2} = 3 \tan \theta \implies s = 6 \tan \theta$. Note that $r^2 + s^2 = (r + s)^2 - 2rs.$ $rs = \frac{6}{\tan \theta} \cdot 6 \tan \theta = 36 \implies 2rs = 72$. The area of the trapezoid is $72 = 6 \cdot \frac{r + s}{2} \implies r + s = 24$. $(r + s)^2 - 2rs = 576 - 72 = \boxed{504}$.
-alwaysgonnagiveyouup | // Block 1
size(15cm);
draw(circle((0,0), 3));
draw((-0.5 * (12 + 6sqrt(3)), -3) -- (0.5 * (12 + 6sqrt(3)), -3) -- (0.5 * (12 - 6sqrt(3)), 3) -- (-0.5 * (12 - 6sqrt(3)), 3) -- cycle);
draw((0, 0) -- (-0.5 * (12 + 6*sqrt(3)),-3) -- cycle);
draw((0, 0) -- (0.5 * (12 + 6*sqrt(3)),-3) -- cycle);
draw((0, 0) -- (-0.5 * (12 - 6*sqrt(3)),3) -- cycle);
draw((0, 0) -- (0.5 * (12 - 6*sqrt(3)),3) -- cycle);
draw((0, 0) -- (0,3) -- cycle);
draw((0, 0) -- (0,-3) -- cycle);
draw((-0.5, -3) -- (-0.5,-2.5) -- (0.5, -2.5) -- (0.5, -3) -- cycle);
draw((-0.25, 3) -- (-0.25, 2.75) -- (0.25, 2.75) -- (0.25, 3) -- cycle);
label("$\theta$", (-0.5 * (12 + 6*sqrt(3)) + 3, -3), NE);
label("$\theta$", (0.5 * (12 + 6*sqrt(3)) - 3, -3), NW);
label("$\theta$", (-0.5 * (12 + 6*sqrt(3)) + 3, -2), NE);
label("$\theta$", (0.5 * (12 + 6*sqrt(3)) - 3, -2), NW);
label("$90^{\circ} - \theta$", (0, -0.5), SW);
label("$90^{\circ} - \theta$", (0, -0.5), SE);
label("$90^{\circ}$", (0 - 0.1, 0), NW);
label("$90^{\circ}$", (0 + 0.1, 0), NE);
label("$\frac{r}{2}$", (-0.25 * (12 + 6*sqrt(3)), -3), S);
label("$\frac{r}{2}$", (0.25 * (12 + 6*sqrt(3)), -3), S);
label("$\theta$", (-0.1, 1.75), E);
label("$\theta$", (0.1, 1.75), W);
label("$\frac{s}{2}$", (-0.25 * (12 - 6*sqrt(3)), 3), N);
label("$\frac{s}{2}$", (0.25 * (12 - 6*sqrt(3)), 3), N);
// Block 2
size(15cm); draw(circle((0,0), 3)); draw((-0.5 * (12 + 6sqrt(3)), -3) -- (0.5 * (12 + 6sqrt(3)), -3) -- (0.5 * (12 - 6sqrt(3)), 3) -- (-0.5 * (12 - 6sqrt(3)), 3) -- cycle); draw((0, 0) -- (-0.5 * (12 + 6*sqrt(3)),-3) -- cycle); draw((0, 0) -- (0.5 * (12 + 6*sqrt(3)),-3) -- cycle); draw((0, 0) -- (-0.5 * (12 - 6*sqrt(3)),3) -- cycle); draw((0, 0) -- (0.5 * (12 - 6*sqrt(3)),3) -- cycle); draw((0, 0) -- (0,3) -- cycle); draw((0, 0) -- (0,-3) -- cycle); draw((-0.5, -3) -- (-0.5,-2.5) -- (0.5, -2.5) -- (0.5, -3) -- cycle); draw((-0.25, 3) -- (-0.25, 2.75) -- (0.25, 2.75) -- (0.25, 3) -- cycle); label("$\theta$", (-0.5 * (12 + 6*sqrt(3)) + 3, -3), NE); label("$\theta$", (0.5 * (12 + 6*sqrt(3)) - 3, -3), NW); label("$\theta$", (-0.5 * (12 + 6*sqrt(3)) + 3, -2), NE); label("$\theta$", (0.5 * (12 + 6*sqrt(3)) - 3, -2), NW); label("$90^{\circ} - \theta$", (0, -0.5), SW); label("$90^{\circ} - \theta$", (0, -0.5), SE); label("$90^{\circ}$", (0 - 0.1, 0), NW); label("$90^{\circ}$", (0 + 0.1, 0), NE); label("$\frac{r}{2}$", (-0.25 * (12 + 6*sqrt(3)), -3), S); label("$\frac{r}{2}$", (0.25 * (12 + 6*sqrt(3)), -3), S); label("$\theta$", (-0.1, 1.75), E); label("$\theta$", (0.1, 1.75), W); label("$\frac{s}{2}$", (-0.25 * (12 - 6*sqrt(3)), 3), N); label("$\frac{s}{2}$", (0.25 * (12 - 6*sqrt(3)), 3), N); | [] |
267 | Let $k$ be a real number such that the system
\begin{align*}
&|25 + 20i - z| = 5 \\
&|z - 4 - k| = |z - 3i - k|
\end{align*}
has exactly one complex solution $z$. The sum of all possible values of $k$ can be written as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$. Here $i = \sqrt{-1}$. | 2025 AIME I Problem 8 | The complex number $z$ must satisfy the following conditions on the complex plane:
$1.$ The magnitude between $z$ and $(25,20)$ is $5.$ This can be represented by drawing a circle with center $(25,20)$ and radius $5.$
$2.$ It is equidistant from the points $(4+k,0)$ and $(k,3).$ Hence it must lie on the perpendicular bisector of the line connecting these points.
For $z$ to have one solution, the perpendicular bisector of the segment connecting the two points must be tangent to the circle.
This bisector must pass the midpoint, $(2+k,\frac{3}{2}),$ and have slope $\frac{4}{3}.$ The segment connecting the point of tangency to the center of the circle has slope $\frac{-3}{4},$ meaning the points of tangency can be $(29,17)$ or $(21,23).$ Solving the equation for the slope of the perpendicular bisector gives \[\frac{\frac{3}{2}-23}{k+2-21}=\frac{4}{3}\] or \[\frac{\frac{3}{2}-17}{k+2-29}=\frac{4}{3},\] giving $k=\frac{23}{8}$ or $\frac{123}{8}$, having a sum of $\frac{73}{4} \Longrightarrow \boxed{077}.$
~nevergonnagiveup | // Block 1
size(300);
draw((0, 0) -- (0, 20), EndArrow(10));
label("$y$", (0, 20), NW);
dot((25,20));
draw((0, 0) -- (25, 0), EndArrow(10));
label("$x$", (25, 0), SE);
draw(circle((25,20),5));
label(scale(0.7)*"$(25,20)$", (25,20), S);
draw((7,0) -- (3,3), blue);
draw((5,3/2) -- (21,23), dashed);
label("$(4+k,0)$", (7,0), S);
label("$(k,3)$", (3,3), N);
draw(rightanglemark((3,3),(5,3/2),(21,23), 20));
draw(rightanglemark((25,20),(21,23),(5,3/2), 20));
draw((25,20) -- (21,23));
// Block 2
size(300); draw((0, 0) -- (0, 20), EndArrow(10)); label("$y$", (0, 20), NW); dot((25,20)); draw((0, 0) -- (25, 0), EndArrow(10)); label("$x$", (25, 0), SE); draw(circle((25,20),5)); label(scale(0.7)*"$(25,20)$", (25,20), S); draw((7,0) -- (3,3), blue); draw((5,3/2) -- (21,23), dashed); label("$(4+k,0)$", (7,0), S); label("$(k,3)$", (3,3), N); draw(rightanglemark((3,3),(5,3/2),(21,23), 20)); draw(rightanglemark((25,20),(21,23),(5,3/2), 20)); draw((25,20) -- (21,23)); | [] |
268 | The parabola with equation $y = x^2 - 4$ is rotated $60^\circ$ counterclockwise around the origin. The unique point in the fourth quadrant where the original parabola and its image intersect has $y$-coordinate $\frac{a - \sqrt{b}}{c}$, where $a$, $b$, and $c$ are positive integers, and $a$ and $c$ are relatively prime. Find $a + b + c$. | 2025 AIME I Problem 9 | We need to relate the rotation to something simpler because substituting the equation of the rotated parabola into the original will give us a quartic.
Notice that the vertices of the original parabola and its image are symmetrical about the angle bisector of the 60 degree rotation (shown in green).
The equation of this line is $y = -\tan60^\circ \cdot y = -\sqrt{3}x.$ This means that the point lying on this line and the original parabola must be the intersection of the new and original images since it won't change position with the rotation.
We substitute $y = x^2 - 4:$
\[x^2 - 4 = -\sqrt{3}x\]
\[x = \frac{-\sqrt{3} + \sqrt{19}}{2}.\]
Then $y = (\frac{-\sqrt{3} + \sqrt{19}}{2})^2 - 4 = \frac{3 - \sqrt{57}}{2}.$ $57 + 3 + 2 = \boxed{62}.$
~grogg007, ~mathkiddus, ~athreyay | // Block 1
size(300);
import graph;
// View window
real L = 6;
// Original parabola y = x^2 - 4
real f(real x){ return x^2 - 4; }
// Rotation by +60 degrees about the origin
pair rot60(pair P){
real ang = pi/3;
return (P.x*cos(ang) - P.y*sin(ang), P.x*sin(ang) + P.y*cos(ang));
}
// Axes
draw((-L,0)--(L,0), gray(0.65), Arrows(4));
draw((0,-L)--(0,L), gray(0.65), Arrows(4));
// Plot original parabola
real xmin=-3.5, xmax=3.5;
path parab = graph(f, xmin, xmax);
draw(parab, blue+1.2bp);
// Build rotated parabola by sampling, then connecting smoothly
int N=220;
pair[] pts;
for(int i=0; i<=N; ++i){
real x = xmin + (xmax - xmin)*i/N;
pts.push(rot60((x, f(x))));
}
path rpar = pts[0];
for(int i=1; i<=N; ++i) rpar = rpar..pts[i];
draw(rpar, red+1.2bp);
// Line y = -sqrt(3) x
draw((-L, sqrt(3)*L)--(L, -sqrt(3)*L), rgb(0,0.45,0)+1.2bp);
// Vertices
pair v1 = (0,-4); // original vertex
pair v2 = rot60(v1); // rotated vertex
dot(v1, blue+4bp);
dot(v2, red+4bp);
// Block 2
size(300); import graph; // View window real L = 6; // Original parabola y = x^2 - 4 real f(real x){ return x^2 - 4; } // Rotation by +60 degrees about the origin pair rot60(pair P){ real ang = pi/3; return (P.x*cos(ang) - P.y*sin(ang), P.x*sin(ang) + P.y*cos(ang)); } // Axes draw((-L,0)--(L,0), gray(0.65), Arrows(4)); draw((0,-L)--(0,L), gray(0.65), Arrows(4)); // Plot original parabola real xmin=-3.5, xmax=3.5; path parab = graph(f, xmin, xmax); draw(parab, blue+1.2bp); // Build rotated parabola by sampling, then connecting smoothly int N=220; pair[] pts; for(int i=0; i<=N; ++i){ real x = xmin + (xmax - xmin)*i/N; pts.push(rot60((x, f(x)))); } path rpar = pts[0]; for(int i=1; i<=N; ++i) rpar = rpar..pts[i]; draw(rpar, red+1.2bp); // Line y = -sqrt(3) x draw((-L, sqrt(3)*L)--(L, -sqrt(3)*L), rgb(0,0.45,0)+1.2bp); // Vertices pair v1 = (0,-4); // original vertex pair v2 = rot60(v1); // rotated vertex dot(v1, blue+4bp); dot(v2, red+4bp); | [] |
269 | The set of points in $3$-dimensional coordinate space that lie in the plane $x+y+z=75$ whose coordinates satisfy the inequalities \[x-yz<y-zx<z-xy\]forms three disjoint convex regions. Exactly one of those regions has finite area. The area of this finite region can be expressed in the form $a\sqrt{b},$ where $a$ and $b$ are positive integers and $b$ is not divisible by the square of any prime. Find $a+b.$ | 2025 AIME I Problem 12 | Rewriting we have $z=75-x-y.$
From the inequality $x-yz<y-zx$ we can rewrite to get,
\[x-y(75-x-y)< y-x(75-x-y).\] \[76x-76y+y^2-x^2<0.\] \[(76-x-y)(x-y)<0.\]
Similarly from the inequality $y-zx<z-xy$ we rewrite to get,
\[y-x(75-x-y)<(75-x-y)-xy.\]
\[2y + 2xy + x^2 - 74x - 75 < 0.\]
\[(x + 1)(2y + x - 75) < 0.\]
Our next step is a visual which can be seen by roughly graphing the two inequalities. The first inequality is similar to a bow tie(you have to graph to see it lol) with bounds of $x-y=0$ and $76-x-y=0.$ The second one is a bow tie with edges of $x+1=0$ and $2y+x-75=0.$Here is the region of our solutions.
It is simple to find the intersections of those which are $(-1,38,38),(25,25,25)$ and $(-1,-1,77).$ The sidelengths of this triangle are $39\sqrt{2},26\sqrt{6},13\sqrt{6}$ which is a $30-60-90$ gives us an area of \[\frac{1}{2}(39\sqrt{2})(13\sqrt{6})=507\sqrt{3}\implies\boxed{510}.\]
~mathkiddus~plang2008~hashbrown2009 | // Block 1
import graph;
size(400, 400);
real xmin = -250, xmax = 250;
real ymin = -150, ymax = 150;
draw((xmin,0)--(xmax,0), black+0.8bp, Arrow);
draw((0,ymin)--(0,ymax), black+0.8bp, Arrow);
pair A = (38, 38),
B = (-10, -10),
C = (-227, 151),
D = (-10, 85/2),
E = (25, 25);
path L1 = (xmin, xmin)--(xmax, xmax);
path L2 = (xmin, 76-xmin)--(xmax, 76-xmax);
path L3 = ((-10, ymin)--(-10, ymax));
path L4 = (xmin, (75-xmin)/2)--(xmax, (75-xmax)/2);
fill(B--D--E--cycle, lightgreen);
draw(L1, blue);
draw(L2, blue);
draw(L3, red);
draw(L4, red);
dot(A, black); label("$(38,38)$", A, NE);
dot(B, black); label("$(-1,-1)$", B, SE);
dot(C, black); label("$(-227,151)$", C, NW);
dot(D, black); label("$(-1,38)$", D, SW);
dot(E, black); label("$(25,25)$", E, SE);
// Block 2
import graph; size(400, 400); real xmin = -250, xmax = 250; real ymin = -150, ymax = 150; draw((xmin,0)--(xmax,0), black+0.8bp, Arrow); draw((0,ymin)--(0,ymax), black+0.8bp, Arrow); pair A = (38, 38), B = (-10, -10), C = (-227, 151), D = (-10, 85/2), E = (25, 25); path L1 = (xmin, xmin)--(xmax, xmax); path L2 = (xmin, 76-xmin)--(xmax, 76-xmax); path L3 = ((-10, ymin)--(-10, ymax)); path L4 = (xmin, (75-xmin)/2)--(xmax, (75-xmax)/2); fill(B--D--E--cycle, lightgreen); draw(L1, blue); draw(L2, blue); draw(L3, red); draw(L4, red); dot(A, black); label("$(38,38)$", A, NE); dot(B, black); label("$(-1,-1)$", B, SE); dot(C, black); label("$(-227,151)$", C, NW); dot(D, black); label("$(-1,38)$", D, SW); dot(E, black); label("$(25,25)$", E, SE); | [] |
270 | Alex divides a disk into four quadrants with two perpendicular diameters intersecting at the center of the disk. He draws $25$ more lines segments through the disk, drawing each segment by selecting two points at random on the perimeter of the disk in different quadrants and connecting these two points. Find the expected number of regions into which these $27$ line segments divide the disk. | 2025 AIME I Problem 13 | First, we calculate the probability that two segments intersect each other. Let the quadrants be numbered $1$ through $4$ in the normal labeling of quadrants, let the two perpendicular diameters be labeled the $x$-axis and $y$-axis, and let the two segments be $A$ and $B.$
$\textbf{Case 1:}$ Segment $A$ has endpoints in two opposite quadrants.
This happens with probability $\frac{1}{3}.$ WLOG let the two quadrants be $1$ and $3.$ We do cases in which quadrants segment $B$ lies in.
Quadrants $1$ and $2,$ $2$ and $3,$ $3$ and $4,$ and $4$ and $1$: These share one quadrant with $A,$ and it is clear that for any of them to intersect $A,$ the endpoint that shares a quadrant with an endpoint of $A$ on a certain side of that endpoint of $A$ For example, if it was quadrants $1$ and $2,$ then the point in quadrant $1$ must be closer to the $x$-axis than the endpoint of $A$ in quadrant $1.$ This happens with probability $\frac{1}{2}.$ Additionally, segment $B$ has a $\frac{1}{6}$ to have endpoints in any set of two quadrants, so this case contributes to the total probability
\[\dfrac{1}{3}\left(\dfrac{1}{6}\cdot\dfrac{1}{2}+\dfrac{1}{6}\cdot\dfrac{1}{2}+\dfrac{1}{6}\cdot\dfrac{1}{2}+\dfrac{1}{6}\cdot\dfrac{1}{2}\right)=\dfrac{1}{9}\]
Quadrants $2$ and $4.$ This always intersects segment $A,$ so this case contributes to the total probability
\[\dfrac{1}{3}\cdot\dfrac{1}{6}=\dfrac{1}{18}\]
Quadrants $1$ and $3.$ We will first choose the endpoints, and then choose the segments from the endpoints. Let the endpoints of the segments in quadrant $1$ be $R_1$ and $R_2,$ and the endpoints of the segments in quadrant $3$ be $S_1$ and $S_2$ such that $R_1,R_2,S_1,$ and $S_2$ are in clockwise order. Note that the probability that $A$ and $B$ intersect is the probability that $A_1$ is paired with $B_1,$ which is $\dfrac{1}{2}.$ Thus, this case contributes to the total probability
\[\dfrac{1}{3}\cdot\dfrac{1}{6}\cdot\dfrac{1}{2}=\dfrac{1}{36}.\]
$\textbf{Case 2:}$
Segment $A$ has endpoints in two adjacent quadrants.
This happens with probability $\frac{2}{3}.$ WLOG let the two quadrants be $1$ and $2.$ We do cases in which quadrants segment $B$ lies in.
Quadrants $1$ and $3,$ $1$ and $4,$ $2$ and $3,$ and $2$ and $4.$ This is similar to our first case above, so this contributes to the total probability
\[\dfrac{2}{3}\left(\dfrac{1}{6}\cdot\dfrac{1}{2}+\dfrac{1}{6}\cdot\dfrac{1}{2}+\dfrac{1}{6}\cdot\dfrac{1}{2}+\dfrac{1}{6}\cdot\dfrac{1}{2}\right)=\dfrac{2}{9}\]
Quadrants $3$ and $4.$ This cannot intersect segment $A.$
Quadrants $1$ and $2,$ Similar to our third case above, this intersects segment $A$ with probability $\frac{1}{2},$ so this case contributes to the total probability
\[\dfrac{2}{3}\cdot\dfrac{1}{6}\cdot\dfrac{1}{2}=\dfrac{1}{18}.\]
Thus, the probability that two segments intersect is
\[\dfrac{1}{9}+\dfrac{1}{18}+\dfrac{1}{36}+\dfrac{2}{9}+\dfrac{1}{18}=\dfrac{17}{36}.\]
Next, we will compute the expected number of intersections of a segment with the axes. WLOG let a segment have an endpoint in quadrant $1.$ Then, it will intersect each axis with probability $\dfrac{2}{3}$ because two out of the three remaining quadrants let it intersect a specific axis, so the expected number of axes a segment intersects is $\frac{4}{3}.$
So, why do intersections matter? Because when adding a segment, it will pass through a number of regions, and for each region it passes through, it will split that region into two and add another region. The segment starts in a region, and for each intersection, it will enter another region, so the number of regions a segment passes through is $1$ more than the number of intersections with the axes and other segments. Thus, we have that by linearity of expectation the expected number of new regions created by adding a segment is
\[\dfrac{17}{36}\cdot(\text{number of segments already added})+\dfrac{4}{3}+1,\]
so the number of new regions added in total by $25$ segments again by linearity of expectation is
\[\sum_{k=0}^{24}\left(\dfrac{17}{36}k+\dfrac{7}{3}\right)=\dfrac{17}{36}\cdot \dfrac{24\cdot 25}{2}+\dfrac{25\cdot 7}{3}\]
which simplifies to $200$ as the expected number of new regions added by the $25$ segments. The axes create $4$ regions to begin with, so our answer is
\[200+4=\boxed{204}.\]
~BS2012, eevee9406 ~hashbrown2009 | // Block 1
pair A,B,C,D,E,F,O;
A=(1,0);B=(0,1);C=(-1,0);D=(0,-1);E=(0.5,0.86602540);F=(-0.707106,-0.707106);O=(0,0);
draw(A--C);draw(B--D);draw(circle((0,0),1));draw(E--F,blue);dot(E,blue);dot(F,blue);
// Block 2
pair A,B,C,D,E,F,O;
A=(1,0);B=(0,1);C=(-1,0);D=(0,-1);E=(0.5,0.86602540);F=(-0.707106,0.707106);O=(0,0);
draw(A--C);draw(B--D);draw(circle((0,0),1));draw(E--F,blue);dot(E,blue);dot(F,blue);
// Block 3
pair A,B,C,D,E,F,O; A=(1,0);B=(0,1);C=(-1,0);D=(0,-1);E=(0.5,0.86602540);F=(-0.707106,-0.707106);O=(0,0); draw(A--C);draw(B--D);draw(circle((0,0),1));draw(E--F,blue);dot(E,blue);dot(F,blue);
// Block 4
pair A,B,C,D,E,F,O; A=(1,0);B=(0,1);C=(-1,0);D=(0,-1);E=(0.5,0.86602540);F=(-0.707106,0.707106);O=(0,0); draw(A--C);draw(B--D);draw(circle((0,0),1));draw(E--F,blue);dot(E,blue);dot(F,blue); | [] |
271 | Let $ABCDE$ be a convex pentagon with $AB=14,$ $BC=7,$ $CD=24,$ $DE=13,$ $EA=26,$ and $\angle B=\angle E=60^{\circ}.$ For each point $X$ in the plane, define $f(X)=AX+BX+CX+DX+EX.$ The least possible value of $f(X)$ can be expressed as $m+n\sqrt{p},$ where $m$ and $n$ are positive integers and $p$ is not divisible by the square of any prime. Find $m+n+p.$ | 2025 AIME I Problem 14 | Firstly, note that $\triangle ABC$ and $\triangle ADE$ are just 30-60-90 triangles. Let $X$ be the Fermat point of $\triangle ACD$, with motivation stemming from considering the pentagon as $\triangle ACD$ with the two 30-60-90 extensions. Note that $AX+CX+DX$ is minimized at this point when $\angle AXC=\angle CXD=\angle AXD=120^{\circ}$. Because we have $\angle ABC=\angle AED=60^{\circ}$, then $ABCX$ and $AXDE$ are both cyclic. Then we have $\angle AXE=\angle ADE=90^{\circ}$ and $\angle BXA=\angle BCA=90^{\circ}$. Then it turns out that we actually have $\angle BXE=90^{\circ}+90^{\circ}=180^{\circ}$, implying that $B$, $X$ and $E$ are collinear. Now, by the triangle inequality, we must have $BX+XE\geq BE$, with equality occurring when $X$ is on $BE$. Thus $AX+CX+DX$ and $BX+EX$ are minimized, so this point $X$ is our desired point.
Firstly, we will find $BX+EX=BE$. We have that $AC=7\sqrt{3}$ and $AD=13\sqrt{3}$, so applying the Law of Cosines in $\triangle ACD$, we get \[147+507-2(7\sqrt{3})(13\sqrt{3})\cos (\angle CAD)=576\implies \cos(\angle CAD)=\frac{1}{7}.\] It follows as a result that $\sin (\angle CAD)=\frac{4\sqrt{3}}{7}$. Then we want to find $\cos (\angle BAE)$. We can do this by seeing \[\cos (\angle BAE)=\cos (\angle CAD+60^{\circ})=\cos (\angle CAD)\cos 60^{\circ}-\sin (\angle CAD)\sin 60^{\circ}=\frac{1}{7}\cdot \frac{1}{2}-\frac{4\sqrt{3}}{7}\cdot \frac{\sqrt{3}}{2}=-\frac{11}{14}.\] Applying the Law of Cosines again in $\triangle BAE$, then because $AB=14$ and $AE=26$, we have \[14^2+26^2-2(14)(26)\left (-\frac{11}{14}\right )=196+676-2\cdot 26\cdot (-11)=872+572=1444=BE^2,\] so it follows that $BE=38=BX+EX$.
Now, we will find the value of $AX+CX+DX$. Construct a point $P$ outside such that $\triangle CPD$ is equilateral, as shown. By property of fermat point, then $A$, $X$, and $P$ are collinear. Additionally, $\angle CXD=120^{\circ}$, so $CPDX$ is cyclic. Applying Ptolemy's Theorem, we have that $(CX)(PD)+(CP)(XD)=(XP)(CD)$. But since $\triangle CPD$ is equilateral, it follows that $CX+DX=PX$. Then $AX+CX+DX=AX+PX=AP$, so we wish to find $AP$. Applying the Law of Cosines in $\triangle ACD$, we have that \[(13\sqrt{3})^2+24^2-2(13\sqrt{3})(24)\cos (\angle ADC)=(7\sqrt{3})^2\implies \cos (\angle ADC)=\frac{\sqrt{3}}{2}\implies \angle ADC=30^{\circ}.\] Then because $\angle CDP=60^{\circ}$, then $\angle ADP=90^{\circ}$, so we can find $AP$ simply with the Pythagorean Theorem. We know $AD=13\sqrt{3}$ and $DP=CD=24$, so $AP=\sqrt{(13\sqrt{3})^2+24^2}=19\sqrt{3}$.
We then have $f(X)=AX+BX+CX+DX+EX=(BX+EX)+(AX+CX+DX)=BE+AP=38+19\sqrt{3}$, which is our minimum value. Therefore, the answer to the problem is $38+19+3=\boxed{060}$.
~ethanzhang1001 | // Block 1
size(10cm);
import math; import geometry; import olympiad;
point A,B,C,D,F,P,X; A=(0,-7sqrt(3)); B=(-7,0); C=(0,0); D=(156/7,-36sqrt(3)/7); F=(169/7,-88sqrt(3)/7); P=(132/7,60sqrt(3)/7); X=(8580/2527,-10604sqrt(3)/2527);
draw(A--B--C--P--D--F--A--C--D--A--P); draw(B--F); draw(circumcircle(A,B,C)); draw(circumcircle(A,D,F)); draw(circumcircle(C,P,D)); draw(C--X--D);
label("A",A,SE); label("B",B,W); label("C",C,NW); label("P",P,N); label("D",D,E); label("E",F,SE); label("X",X,E);
// Block 2
size(10cm); import math; import geometry; import olympiad; point A,B,C,D,F,P,X; A=(0,-7sqrt(3)); B=(-7,0); C=(0,0); D=(156/7,-36sqrt(3)/7); F=(169/7,-88sqrt(3)/7); P=(132/7,60sqrt(3)/7); X=(8580/2527,-10604sqrt(3)/2527); draw(A--B--C--P--D--F--A--C--D--A--P); draw(B--F); draw(circumcircle(A,B,C)); draw(circumcircle(A,D,F)); draw(circumcircle(C,P,D)); draw(C--X--D); label("A",A,SE); label("B",B,W); label("C",C,NW); label("P",P,N); label("D",D,E); label("E",F,SE); label("X",X,E); | [] |
272 | Six points $A, B, C, D, E,$ and $F$ lie in a straight line in that order. Suppose that $G$ is a point not on the line and that $AC=26, BD=22, CE=31, DF=33, AF=73, CG=40,$ and $DG=30.$ Find the area of $\triangle BGE.$ | 2025 AIME II Problem 1 | Let $AB=a$, $BC=b$, $CD=c$, $DE=d$ and $EF=e$. Then we know that $a+b+c+d+e=73$, $a+b=26$, $b+c=22$, $c+d=31$ and $d+e=33$. From this we can easily deduce $c=14$ and $a+e=34$ thus $b+c+d=39$. Using Heron's formula we can calculate the area of $\triangle{CGD}$ to be $\sqrt{(42)(28)(12)(2)}=168$, and since the base of $\triangle{BGE}$ is $\frac{39}{14}$ of that of $\triangle{CGD}$ $\triangle{BGE}$ shares an altitude with $\triangle{CGD}$, we conclude they are proportional and we can calculate the area of $\triangle{BGE}$ to be $168\times \frac{39}{14}=\boxed{468}$.
~ Quick Asymptote Fix by eevee9406, edited by aoum | // Block 1
pair A,B,C,D,E,F,G;
A=(0,0);
label("$A$", A, S);
B=(1.5,0);
label("$B$", B, S);
C=(2.9,0);
label("$C$", C, S);
D=(4.2,0);
label("$D$", D, S);
E=(5.3,0);
label("$E$", E, S);
F=(6.5,0);
label("$F$", F, S);
G=(3.7,3);
label("$G$", G, N);
draw(A--B--C--D--E--F);
draw(C--G--D);
draw(B--G--E);
// Block 2
pair A,B,C,D,E,F,G; A=(0,0); label("$A$", A, S); B=(1.5,0); label("$B$", B, S); C=(2.9,0); label("$C$", C, S); D=(4.2,0); label("$D$", D, S); E=(5.3,0); label("$E$", E, S); F=(6.5,0); label("$F$", F, S); G=(3.7,3); label("$G$", G, N); draw(A--B--C--D--E--F); draw(C--G--D); draw(B--G--E); | [] |
273 | Four unit squares form a $2 \times 2$ grid. Each of the $12$ unit line segments forming the sides of the squares is colored either red or blue in such a way that each unit square has $2$ red sides and $2$ blue sides. One example is shown below (red is solid, blue is dashed). Find the number of such colorings. | 2025 AIME II Problem 3 | Let the red edges be "used" edges. In the digarams below, dashed lines are uncolored lines yet to be decided. Since all four center edges are common to both squares, we consider five distinct cases:
$\textbf{Case 1:}$ All center edges are used. There is only one way to do this.
$\textbf{Case 2:}$ Three center edges are used, meaning two squares are missing an edge. For each square, there are 2 ways to choose an edge, resulting in $2 \times 2 = 4$ ways. Additionally, considering the rotational symmetry of the arrangement, there are 4 possible rotations, giving a total of $4 \times 4 = 16$ configurations.
$\textbf{Case 3:}$ Two center edges are used. There are two sub-cases:
$\textbf{Scenario 1:}$ The two selected sides are perpendicular to each other. The square diagonally opposite its adjacent square has only one choice, while the other two squares each have two choices. This gives a total of $1 \times 2 \times 2 = 4$ choices. Considering the 4 possible rotations, the total number of configurations is $4 \times 4 = 16$.
$\textbf{Scenario 2:}$ The two selected sides are aligned along the same straight line. Each of the four squares has 2 choices, yielding $2^4 = 16$ possible choices. Taking into account the 2 possible rotations, the total number of configurations is $16 \times 2 = 32$.
$\textbf{Case 4:}$ Only one center edge is used. This case is similar to Case 2, yielding 16 possible configurations.
$\textbf{Case 5:}$ No center edge is used. This is similar to Case 1, with only 1 possible configuration.
In conclusion, the total number of configurations is:
\[1 + 16 + 16 + 32 + 16 + 1 = \boxed{\textbf{082}}\]
~ Athmyx
~ LaTeX by eevee9406
~ Additional edits by aoum | // Block 1
pair A,B,C,D,E,F,G,H,I;
A=(-1,-1);B=(-1,0);C=(-1,1);D=(0,-1);E=(0,0);F=(0,1);G=(1,-1);H=(1,0);I=(1,1);
draw(A--C--I--G--cycle,dashed); draw(B--H,red); draw(D--F,red);
// Block 2
pair A,B,C,D,E,F,G,H,I;
A=(-1,-1);B=(-1,0);C=(-1,1);D=(0,-1);E=(0,0);F=(0,1);G=(1,-1);H=(1,0);I=(1,1);
draw(A--C--I--G--cycle,dashed); draw(B--H,red); draw(D--E,red);draw(E--F,dashed);
// Block 3
pair A,B,C,D,E,F,G,H,I;
A=(-1,-1);B=(-1,0);C=(-1,1);D=(0,-1);E=(0,0);F=(0,1);G=(1,-1);H=(1,0);I=(1,1);
draw(A--C--I--G--cycle,dashed); draw(B--E,red); draw(E--D,red);draw(E--F,dashed);draw(E--H,dashed);
// Block 4
pair A,B,C,D,E,F,G,H,I;
A=(-1,-1);B=(-1,0);C=(-1,1);D=(0,-1);E=(0,0);F=(0,1);G=(1,-1);H=(1,0);I=(1,1);
draw(A--C--I--G--cycle,dashed); draw(B--H,red); draw(D--F,dashed);
// Block 5
pair A,B,C,D,E,F,G,H,I;
A=(-1,-1);B=(-1,0);C=(-1,1);D=(0,-1);E=(0,0);F=(0,1);G=(1,-1);H=(1,0);I=(1,1);
draw(A--C--I--G--cycle,dashed); draw(B--E,red); draw(E--D,dashed);draw(E--F,dashed);draw(E--H,dashed);
// Block 6
pair A,B,C,D,E,F,G,H,I;
A=(-1,-1);B=(-1,0);C=(-1,1);D=(0,-1);E=(0,0);F=(0,1);G=(1,-1);H=(1,0);I=(1,1);
draw(A--C--I--G--cycle,dashed); draw(B--E,dashed); draw(E--D,dashed);draw(E--F,dashed);draw(E--H,dashed);
// Block 7
pair A,B,C,D,E,F,G,H,I; A=(-1,-1);B=(-1,0);C=(-1,1);D=(0,-1);E=(0,0);F=(0,1);G=(1,-1);H=(1,0);I=(1,1); draw(A--C--I--G--cycle,dashed); draw(B--H,red); draw(D--F,red);
// Block 8
pair A,B,C,D,E,F,G,H,I; A=(-1,-1);B=(-1,0);C=(-1,1);D=(0,-1);E=(0,0);F=(0,1);G=(1,-1);H=(1,0);I=(1,1); draw(A--C--I--G--cycle,dashed); draw(B--H,red); draw(D--E,red);draw(E--F,dashed);
// Block 9
pair A,B,C,D,E,F,G,H,I; A=(-1,-1);B=(-1,0);C=(-1,1);D=(0,-1);E=(0,0);F=(0,1);G=(1,-1);H=(1,0);I=(1,1); draw(A--C--I--G--cycle,dashed); draw(B--E,red); draw(E--D,red);draw(E--F,dashed);draw(E--H,dashed);
// Block 10
pair A,B,C,D,E,F,G,H,I; A=(-1,-1);B=(-1,0);C=(-1,1);D=(0,-1);E=(0,0);F=(0,1);G=(1,-1);H=(1,0);I=(1,1); draw(A--C--I--G--cycle,dashed); draw(B--H,red); draw(D--F,dashed);
// Block 11
pair A,B,C,D,E,F,G,H,I; A=(-1,-1);B=(-1,0);C=(-1,1);D=(0,-1);E=(0,0);F=(0,1);G=(1,-1);H=(1,0);I=(1,1); draw(A--C--I--G--cycle,dashed); draw(B--E,red); draw(E--D,dashed);draw(E--F,dashed);draw(E--H,dashed);
// Block 12
pair A,B,C,D,E,F,G,H,I; A=(-1,-1);B=(-1,0);C=(-1,1);D=(0,-1);E=(0,0);F=(0,1);G=(1,-1);H=(1,0);I=(1,1); draw(A--C--I--G--cycle,dashed); draw(B--E,dashed); draw(E--D,dashed);draw(E--F,dashed);draw(E--H,dashed); | [] |
274 | Let ${\triangle ABC}$ be a right triangle with $\angle A = 90^\circ$ and $BC = 38.$ There exist points $K$ and $L$ inside the triangle such\[AK = AL = BK = CL = KL = 14.\]The area of the quadrilateral $BKLC$ can be expressed as $n\sqrt3$ for some positive integer $n.$ Find $n.$ | 2025 AIME II Problem 14 | Let $O$ be the midpoint of $BC$. Take the diagram and rotate it $180^{\circ}$ around $O$ to get the diagram shown. Notice that we have $\angle ABC+\angle ACB=90^{\circ}$. Because $\triangle AKL$ is equilateral, then $\angle KAL=60^{\circ}$, so $\angle BAK+\angle CAL=30^{\circ}$. Because of isosceles triangles $\triangle BAK$ and $\triangle CAL$, we get that $\angle ABK+\angle ACL=30^{\circ}$ too, implying that $\angle KBC+\angle LCB=60^{\circ}$. But by our rotation, we have $\angle LCO=\angle L'BO$, so this implies that $\angle KBL'=60^{\circ}$, or that $\triangle KBL'$ is equilateral (specifically, the presence of a $60^{\circ}$ angle in between two sides of length $14$ allows us to conclude that $\triangle KBL'$ is congruent to an equilateral triangle by SAS). We can similarly derive that $\angle KBO=\angle K'CO$ implies $\angle LCK'=60^{\circ}$ so that $\triangle LK'C$ is also equilateral. At this point, notice that quadrilateral $KL'K'L$ is a rhombus because all its sides have length $14$. The area of our desired region is now $[BKLC]=\frac{1}{2}[BL'K'CLK]$. We can easily find the areas of $\triangle KBL'$ and $\triangle LK'C$ to be $\frac{\sqrt{3}}{4}\cdot 14^2=49\sqrt{3}$ each. Now it remains to find the area of rhombus $KL'K'L$.
Focus on the quadrilateral $AKOL$. Restate the configuration in another way - we have equilateral triangle $\triangle AKL$ with side length 14, and a point $O$ such that $AO=19$ (the median to the hypotenuse of a right triangle is equal to half the length of the hypotenuse) and $\angle KOL=90^{\circ}$. We are trying to find the area of $\triangle KOL$. Let $M$ be the midpoint of $KL$. We see that $AM=7\sqrt{3}$ ($\overline{AM}$ is a median of equilateral $\triangle AKL$ with side length $14$), and since $M$ is the circumcenter of right $\triangle KOL$, it follows that $MO=7$. Let $\angle KMO=\theta$. From the Law of Cosines in $\triangle AMO$, we can see that \[(7\sqrt{3})^2+7^2-2(7\sqrt{3})(7)\cos (\angle AMO)=361,\] so after simplification we get that $\cos (\theta +90^{\circ})=-\frac{55\sqrt{3}}{98}$. Applying the cosine angle sum identity simplifies this to $\sin \theta =\frac{55\sqrt{3}}{98}$. Then, the Pythagorean Identity $\cos^2\theta +\sin^2\theta =1$ gives us $\cos \theta =\frac{23}{98}$ ($\theta$ is clearly acute, so $\sin \theta$ and $\cos \theta$ are both positive). Applying the Law of Cosines in $\triangle KMO$, we get \[49+49-2\cdot 7\cdot 7\cdot \cos \theta =98-98\cdot \frac{23}{98}=98-23=75=KO^2,\] which tells us $KO=5\sqrt{3}$. The Pythagorean Theorem in $\triangle KOL$ gives that $OL=\sqrt{KL^2-KO^2}={11}$, so the area of $\triangle KOL$ is $\frac{KO\cdot KL}{2}=\frac{55\sqrt{3}}{2}$. The rhombus $KL'K'L$ consists of four of these triangles, so its area is $4\cdot \frac{55\sqrt{3}}{2}=110\sqrt{3}$.
Finally, the area of hexagon $BL'K'CLK$ is $49\sqrt{3}+110\sqrt{3}+49\sqrt{3}=208\sqrt{3}$, and since this consists of quadrilaterals $BKLC$ and $CK'L'B$, which must be congruent by rotation, the area of $BKLC$ is $104\sqrt{3}$. Therefore, the answer is $\boxed{104}$.
~ethanzhang1001
~ Edited by Christian | // Block 1
import math; import geometry; import olympiad;
point A,C,B,L,K,D,F,G,O; A=(0,0); C=(16sqrt(3),0); B=(0,26); L=(8sqrt(3),2); K=(3sqrt(3),13); D=(16sqrt(3),26); F=(13sqrt(3),13); G=(8sqrt(3),24); O=(8sqrt(3),13);
draw(A--B--D--C--A--L--C--F--L--K--A--D); draw(K--B--G--D--F--G--K--F); draw(B--O--L); draw(C--O--G);
label("A",A,SW); label("B",B,NW); label("C",C,SE); label("A'",D,NE); label("K",K,W); label("L",L,NW); label("L'",G,SE); label("K'",F,E); label("O",O,NNW);
// Block 2
import math; import geometry; import olympiad;
point A,K,O,L,M; A=(-7sqrt(3),0); K=(0,7); O=(55sqrt(3)/14,23/14); L=(0,-7); M=(0,0);
draw(A--K--O--L--A--O--M--A); draw(K--L);
label("A",A,W); label("K",K,N); label("O",O,E); label("L",L,S); label("M",M,SE);
// Block 3
import math; import geometry; import olympiad; point A,C,B,L,K,D,F,G,O; A=(0,0); C=(16sqrt(3),0); B=(0,26); L=(8sqrt(3),2); K=(3sqrt(3),13); D=(16sqrt(3),26); F=(13sqrt(3),13); G=(8sqrt(3),24); O=(8sqrt(3),13); draw(A--B--D--C--A--L--C--F--L--K--A--D); draw(K--B--G--D--F--G--K--F); draw(B--O--L); draw(C--O--G); label("A",A,SW); label("B",B,NW); label("C",C,SE); label("A'",D,NE); label("K",K,W); label("L",L,NW); label("L'",G,SE); label("K'",F,E); label("O",O,NNW);
// Block 4
import math; import geometry; import olympiad; point A,K,O,L,M; A=(-7sqrt(3),0); K=(0,7); O=(55sqrt(3)/14,23/14); L=(0,-7); M=(0,0); draw(A--K--O--L--A--O--M--A); draw(K--L); label("A",A,W); label("K",K,N); label("O",O,E); label("L",L,S); label("M",M,SE); | [] |
274 | Let ${\triangle ABC}$ be a right triangle with $\angle A = 90^\circ$ and $BC = 38.$ There exist points $K$ and $L$ inside the triangle such\[AK = AL = BK = CL = KL = 14.\]The area of the quadrilateral $BKLC$ can be expressed as $n\sqrt3$ for some positive integer $n.$ Find $n.$ | 2025 AIME II Problem 14 | Immediately we should see that $\triangle{AKL}$ is equilateral, so $\angle{KAL}=60$.
We assume $\angle{LCA}=x$, and it is easily derived that $\angle{KBA}=30-x$. Using trigonometry, we can say that $AC=28\cos{x}$ (imagine extending $\overrightarrow{CL}$ to intersect $\overline{AB}$, forming a right triangle with hypotenuse length $2\cdot14=28$) and $AB=28\cos{(30-x)}$. Pythagoras tells us that $BC^2=AC^2+AB^2$ so now we evaluate as follows:
\begin{align*}
38^2 &=28^2(\cos^2{x}+\cos^2{(30-x)}) \\
(\frac{19}{14})^2 &=\cos^2{x}+(\frac{\sqrt{3}}{2} \cos{x} + \frac{1}{2} \sin{x})^2 \\
&=\cos^2{x}+\frac{3}{4} \cos^2{x}+\frac{\sqrt{3}}{2}\sin{x} \cos{x}+\frac{1}{4}\sin^2{x} \\
&=\frac{3}{2} \cos^2{x}+\frac{\sqrt{3}}{2}\sin{x} \cos{x}+\frac{1}{4} \\
&=\frac{3}{4}(2\cos^2{x}-1)+\frac{\sqrt{3}}{4} (2\sin{x} \cos{x})+1 \\
(\frac{33}{14})(\frac{5}{14})&=\frac{\sqrt{3}}{2}(\frac{\sqrt{3}}{2}(\cos{2x})+\frac{1}{2} (\sin{2x})) \\
\frac{55\sqrt{3}}{98}&=\cos{(30-2x)} \\
\end{align*}
It is obvious that $\angle{ALC}=180-2x$. We can easily derive $\cos{(150+(30-2x))}$ using angle addition we know, and then the Law of Cosines to find side $AC$.
\begin{align*}
\frac{55\sqrt{3}}{98}=\cos{(30-2x)} \\
\sin{(30-2x)}=\sqrt{1-\cos^2{(30-2x)}}=\frac{23}{98} \\
\cos{(180-2x)}=(-\frac{\sqrt{3}}{2})(\frac{55\sqrt{3}}{98})-(\frac{1}{2})(\frac{23}{98}) \\
\cos{(180-2x)}=-\frac{47}{49} \\
AC^2=14^2+14^2-2\cdot 14\cdot 14\cdot (-\frac{47}{49}) \\
AC=\sqrt{768}=16\sqrt3 \\
\end{align*}
We easily find $\cos{x}=\frac{4\sqrt{3}}{7}$ and $\sin{x}=\frac{1}{7}$ (draw a perpendicular down from $L$ to $AC$). What we are trying to find is the area of $BKLC$, which can be found by adding the areas of $\triangle{BKL}$ and $\triangle{BLC}$. It is trivial that $\triangle{BKL}$ and $\triangle{ACL}$ are congruent (do some angle chasing in terms of $x$ and use the given side lengths of $14$), so we know that $BL=28\cos{x}$. What we require is
\begin{align*}
\frac{1}{2}(14)(14)(\sin{(180-2x)})+\frac{1}{2}(14)(28\cos{x})(\sin{(120+x)}) \\
\end{align*}
We do similar calculations to obtain that $\sin{(120+x)}=\frac{11}{14}$ and $\cos{(180-2x)}=-\frac{47}{49}$ implies $\sin{(180-2x)}=\frac{8\sqrt{3}}{49}$, so now we plug in everything we know to calculate the area of the quadrilateral:
\begin{align*}
& \frac{1}{2}(14)(14)(\sin{(180-2x)})+\frac{1}{2}(14)(28\cos{x})(\sin{(120+x)}) \\
&=\frac{1}{2}(14)(14)(\frac{8\sqrt{3}}{49})+\frac{1}{2}(14)(16\sqrt{3})(\frac{11}{14}) \\
&=16\sqrt{3}+88\sqrt{3} \\
&=104\sqrt{3} \\
\end{align*}
We see that $n=\boxed{104}$.
~ lisztepos
~ Edited by Aoum
~ Edited by Souledgeii
~ Edited by Christian | // Block 1
import math; import geometry; import olympiad;
point A,B,C,L,K; A=(0,0); C=(16sqrt(3),0); B=(0,26); L=(8sqrt(3),2); K=(3sqrt(3),13);
draw(A--B--C--cycle); draw(A--K--L--cycle); draw(B--K); draw(C--L); draw(B--L);
label("A",A,SW); label("B",B,NW); label("C",C,SE); label("K",K,W); label("L",L,NE);
markscalefactor=1;
draw(anglemark(L,C,A));
draw(anglemark(A,B,K));
// Block 2
import math; import geometry; import olympiad; point A,B,C,L,K; A=(0,0); C=(16sqrt(3),0); B=(0,26); L=(8sqrt(3),2); K=(3sqrt(3),13); draw(A--B--C--cycle); draw(A--K--L--cycle); draw(B--K); draw(C--L); draw(B--L); label("A",A,SW); label("B",B,NW); label("C",C,SE); label("K",K,W); label("L",L,NE); markscalefactor=1; draw(anglemark(L,C,A)); draw(anglemark(A,B,K)); | [] |
274 | Let ${\triangle ABC}$ be a right triangle with $\angle A = 90^\circ$ and $BC = 38.$ There exist points $K$ and $L$ inside the triangle such\[AK = AL = BK = CL = KL = 14.\]The area of the quadrilateral $BKLC$ can be expressed as $n\sqrt3$ for some positive integer $n.$ Find $n.$ | 2025 AIME II Problem 14 | Consider a diagram to the original problem (credit to solution 4):
Now, let us simplify the problem further. We know that $K$ and $L$ must lie on the perpendicular bisectors of $AB$ and $AC$, respectively. The real problem here is the equilateral triangle in the middle, inscribed in a rectangle with diagonal length 19.
We create a further simplified problem: given that the inscribed equilateral triangle of a certain rectangle with diagonal length $19$ has side length $14$, find the sides and intersection points on this rectangle. For reference, here is a diagram:
Note the angles $\angle{LAD}$ and $\angle{BAK}$. Since $\angle{LAD} + \angle{BAK} + 60^{\circ} = 90^{\circ}$, $\angle{LAD} + \angle{BAK} = 30^{\circ}$, and $\angle{BAK} = 30^{\circ} - \angle{LAD}$. Thus, let $\angle{LAD} = \alpha$ and $\angle{BAK} = 30 - \alpha$.
Now, we know that $AB^2 + AD^2 = 19^2$, as the hypotenuse of the larger right triangle is $38$. However, we can also express $AB$ and $AD$ in terms of $\alpha$: $AB = 14(\cos(30^{\circ}-\alpha))$ and $AD = 14(\cos(\alpha))$. Thus, $\cos^2(\alpha) + \cos^2(30^{\circ}-\alpha) = 361/196$. We expand this using the cosine difference identity:
$\cos^2(\alpha) + (\cos(30^{\circ})\cos(\alpha) + \sin(30^{\circ})\sin(\alpha))^2 = \frac{361}{196}$
$\frac{7}{4}\cos^2(\alpha) + \frac{1}{4}\sin^2(\alpha) + \frac{\sqrt3}{2}\sin(\alpha)\cos(\alpha) = \frac{361}{196}$
Using the fact that $\sin^2(\alpha) + \cos^2(\alpha) = 1$, then multiplying the entire equation by $2$,
$3\cos^2(\alpha) + \sqrt3\sin(\alpha)\cos(\alpha) = \frac{156}{49}$
Now, to save some writing, let us denote $\sin(\alpha)$ with $x$, and $\cos(\alpha)$ with $y$.
We have the following equations:
$x^2 + y^2 = 1$
$3y^2 + \sqrt3xy = \frac{156}{49}$
Substituting $x$ for $y$, moving $3y^2$ to the left side, squaring, and dividing by 9, we end up with the quartic:
$\frac{4}{3}y^4 - \frac{361}{147}y^2 + \frac{52^2}{49^2} = 0$
Using the quadratic formula, we end up with this:
$y^2 = \frac{\frac{361}{49} \pm \frac{1}{49}\cdot\sqrt{361^2 - 208^2\cdot3}}{8}$
Now, we could just compute $361^2 - 208^2\cdot3$, but instead, we can do this:
$361^2 - 208^2\cdot3 = (129600 + 720 + 1) - (40000 + 3200 + 64)\cdot3$
$(129600 + 721) - (43200 + 64)\cdot3$
$(129600 + 721) - (129600 + 192) = 529 = 23^2$
Thus, we have two cases:
$1. \cos(\alpha) = \frac{13}{14}$
$2. \cos(\alpha) = \frac{4\sqrt3}{7}$
Both lead to the same side lengths of the rectangle: $8\sqrt3$, and $13$. Referring back to our original rectangle diagram and plugging in our trigonometric values, we get that $CK = 13 - 2 = 11$, and $CL = 8\sqrt3 - 3\sqrt3 = 5\sqrt3$. Thus, the area of the original quadrilateral is $\frac{88\sqrt3 + 55\sqrt3 + 65\sqrt3}{2}$, or $\boxed{104}\sqrt3$.
~Stead
~ Edited by Christian | // Block 1
import math; import geometry; import olympiad;
point A,B,C,L,K; A=(0,0); C=(16sqrt(3),0); B=(0,26); L=(8sqrt(3),2); K=(3sqrt(3),13);
draw(A--B--C--cycle); draw(A--K--L--cycle); draw(B--K); draw(C--L); draw(B--L);
label("A",A,SW); label("B",B,NW); label("C",C,SE); label("K",K,W); label("L",L,NE);
markscalefactor=1;
// Block 2
import math; import geometry; import olympiad;
point A,B,C,D,L,K; A=(0,0); D=(13,0); B=(0,8sqrt(3)); C=(13,8sqrt(3)); L=(13,3sqrt(3)); K=(2,8sqrt(3));
draw(A--B--C--D--cycle); draw(A--K--L--cycle);
label("A",A,SW); label("B",B,NW); label("C",C,NE); label("K",K,N); label("L",L,E); label("D",D,SE);
markscalefactor=1;
// Block 3
import math; import geometry; import olympiad; point A,B,C,L,K; A=(0,0); C=(16sqrt(3),0); B=(0,26); L=(8sqrt(3),2); K=(3sqrt(3),13); draw(A--B--C--cycle); draw(A--K--L--cycle); draw(B--K); draw(C--L); draw(B--L); label("A",A,SW); label("B",B,NW); label("C",C,SE); label("K",K,W); label("L",L,NE); markscalefactor=1;
// Block 4
import math; import geometry; import olympiad; point A,B,C,D,L,K; A=(0,0); D=(13,0); B=(0,8sqrt(3)); C=(13,8sqrt(3)); L=(13,3sqrt(3)); K=(2,8sqrt(3)); draw(A--B--C--D--cycle); draw(A--K--L--cycle); label("A",A,SW); label("B",B,NW); label("C",C,NE); label("K",K,N); label("L",L,E); label("D",D,SE); markscalefactor=1; | [] |
274 | Let ${\triangle ABC}$ be a right triangle with $\angle A = 90^\circ$ and $BC = 38.$ There exist points $K$ and $L$ inside the triangle such\[AK = AL = BK = CL = KL = 14.\]The area of the quadrilateral $BKLC$ can be expressed as $n\sqrt3$ for some positive integer $n.$ Find $n.$ | 2025 AIME II Problem 14 | This diagram is modified from the solution 4 diagram. Let $M$ be the midpoint of $AC$, and let $N$ be the midpoint of $AB$.
We place the diagram onto the Cartesian coordinate grid. Let $A = (0, 0)$, $M = (a, 0)$, $C = (2a, 0)$, $N = (0, b)$, and $B = (0, 2b)$. We are given $AL = CL$, so $\triangle ACL$ is isosceles. Therefore, $LM$ is the perpendicular bisector of $AC$, so we can let $L = (a, y)$. Similarly, we’re given $AK = BK$, so $\triangle ABK$ is also isosceles, and $NK$ is the perpendicular bisector of $AB$. Therefore, we can let $K = (x, b)$.
We have $AB = 2b$ and $AC = 2a$. We’re given that $\angle BAC = 90^\circ$ and $BC = 38$, so by the Pythagorean theorem, \[(2a)^2 + (2b)^2 = 38^2 \implies 4a^2 + 4b^2 = 1444 \implies a^2 + b^2 = 361.\]
We now place the diagram onto the complex plane. We use the x-axis of the coordinate plane as the complex plane’s real axis, and we use the y-axis of the coordinate plane as the complex plane’s imaginary axis. So, on the complex plane, $A = 0$, $L = a + yi$, and $K = x + bi$. Also, since we are given $AK = KL = AL$, $\triangle AKL$ is equilateral. In addition, since $AL = AK$, $\angle KAL = 60^\circ$, and because we constructed our diagram with $K$ counterclockwise of $L$ (if it were the other way around, we could go through the same steps as this solution, but with variables switched around), $K$ is a $60^\circ$ counterclockwise rotation of $L$ about $A$, and $L$ is a $60^\circ$ clockwise or $300^\circ$ counterclockwise rotation of $K$ about $A$.
Rotation on the complex plane is equivalent to multiplying by a root of unity. Here, $K$ and $L$ are rotated a multiple of $60^\circ$ to each other about $A$. $60^\circ$ is one-sixth of a full circle, so to go from $L$ to $K$ or $K$ to $L$, we multiply by a 6th root of unity. Specifically, to go from $L$ to $K$, we multiply by $\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i$, and to go from $K$ to $L$, we multiply by $\dfrac{1}{2} - \dfrac{\sqrt{3}}{2}i$.
We multiply the coordinate of $L$ by $\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i$ on the complex plane to obtain equations for the coordinates of $K$: \[(a + yi)\left(\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i\right) = \left(\dfrac{a}{2} - \dfrac{\sqrt{3}}{2}y\right) + \left(\dfrac{y}{2} + \dfrac{\sqrt{3}}{2}a\right)i = x + bi.\] Equating real and imaginary parts, we obtain \[x = \dfrac{a}{2} - \dfrac{\sqrt{3}}{2}y \text{ and } b = \dfrac{y}{2} + \dfrac{\sqrt{3}}{2}a.\]
Similarly, we multiply the coordinate of $K$ by $\dfrac{1}{2} - \dfrac{\sqrt{3}}{2}i$ to obtain equations for the coordinates of $L$: \[(x + bi)\left(\dfrac{1}{2} - \dfrac{\sqrt{3}}{2}i\right) = \left(\dfrac{x}{2} + \dfrac{\sqrt{3}}{2}b\right) + \left(\dfrac{b}{2} - \dfrac{\sqrt{3}}{2}x\right) = a + yi.\] Equating real and imaginary parts, we obtain \[a = \dfrac{x}{2} + \dfrac{\sqrt{3}}{2}b \text{ and } y = \dfrac{b}{2} - \dfrac{\sqrt{3}}{2}x.\]
We now look back at the problem to see what it asks for: $[BKLC]$. Looking at the diagram, we see we can express the area of the quadrilateral as the area of the big right triangle $ABC$ minus the two isosceles triangles $ABK$ and $ALC$ minus the equilateral triangle $AKL$: \[[BKLC] = [ABC] - [ABK] - [ACL] - [AKL].\]
We are given that $AK = KL = AL = 14$, so the area of equilateral triangle $AKL$ is $\dfrac{\sqrt{3}}{4} \cdot 14^2 = 49\sqrt{3}$. Also, we can use $AC = 2a$ as the base of $\triangle ABC$ and $AB = 2b$ as the height, so $[ABC] = \dfrac{(2a)(2b)}{2} = 2ab$. Similarly, we use $AC = 2a$ as the base of $\triangle ACL$ and $ML = y$ ($M = (a, 0)$ and $L = (a, y)$, so the distance between the two is equal to $y$) as the height, so $[ACL] = \dfrac{(2a)(y)}{2} = ay$. Finally, we use $AB = 2b$ and $KN = x$ ($N = (0, b)$ and $K = (x, b)$, so the distance between the two is equal to $x$) as the base and height of $\triangle ABK$ respectively, so $[ABK] = \dfrac{(2b)(x)}{2} = bx$. Therefore, \[[BKLC] = [ABC] - [ABK] - [ACL] - [AKL] = 2ab - bx - ay - 49\sqrt{3}.\]
We have already shown that $b = \dfrac{y}{2} + \dfrac{\sqrt{3}}{2}a$. Substituting this into $a^2 + b^2 = 361$, we have \[a^2 + \left(\dfrac{y}{2} + \dfrac{\sqrt{3}}{2}a\right)^2 = 361.\] Expanding this out, we have \[a^2 + \dfrac{y^2}{4} + \dfrac{\sqrt{3}}{2}ay + \dfrac{3}{4}a^2 = 361.\] Multiplying both sides by $4$ and rearranging the left side, we have \[7a^2 + y^2 + 2ay\sqrt{3} = 1444.\] We previously showed that $AC \perp ML$, so $AM \perp ML$ (since $M$ is on $AC$). Therefore, $\triangle AML$ has a right angle at $M$. By the Pythagorean Theorem, $AM^2 + ML^2 = a^2 + y^2 = AL^2 = 196$. Subtracting $a^2 + y^2$ from the left side and $196$ from the right side, we obtain \[6a^2 + 2ay\sqrt{3} = 1248.\] Dividing both sides of the equation by $2\sqrt{3}$ and factoring $a$ out of the left side, we have \[a(a\sqrt{3} + y) = 208\sqrt{3}.\] However, we have $b = \dfrac{y}{2} + \dfrac{\sqrt{3}}{2}a$, so the expression inside the parentheses is simply $2b$! Therefore, \[2ab = 208\sqrt{3}.\]
The algebra’s not over yet. We also showed that $a = \dfrac{x}{2} + \dfrac{\sqrt{3}}{2}b$, so substituting that into $a^2 + b^2 = 361$, we obtain \[b^2 + \left(\dfrac{x}{2} + \dfrac{\sqrt{3}}{2}b\right)^2 = 361.\] Expanding this out, we have \[b^2 + \dfrac{x^2}{4} + \dfrac{\sqrt{3}}{2}bx + \dfrac{3}{4}b^2 = 361.\] Multiplying both sides by $4$ and rearranging the left side, we now have \[7b^2 + x^2 + 2xb\sqrt{3} = 1444.\] Does this equation look familiar? We previously showed that $NK \perp AB$. Therefore, $NK \perp AN$ (since $N$ is on $AB$). So, $\triangle ANK$ has a right angle at $N$. By the Pythagorean Theorem, $AN^2 + KN^2 = b^2 + x^2 = AK^2 = 196$. Subtracting $b^2 + x^2$ from the left side and $196$ from the right side, we have \[6b^2 + 2xb\sqrt{3} = 1248.\] We previously also had the equation $6a^2 + 2ay\sqrt{3} = 1248$, and adding this equation to the above equation and factoring out $2\sqrt{3}$, we have \[6a^2 + 6b^2 + 2\sqrt{3}(bx + ay) = 2496.\] We previously showed $a^2 + b^2 = 361$, so $6a^2 + 6b^2 = 6 \cdot 361 = 2166$. Subtracting $6a^2 + 6b^2$ from the left side and $2166$ from the right side, we obtain $2\sqrt{3}(bx + ay) = 330$. Finally, dividing both sides by $2\sqrt{3}$, we have \[bx + ay = 55\sqrt{3}.\]
We previously arrived at this expression for $[BKLC]$: \[[BKLC] = 2ab - bx - ay - 49\sqrt{3}.\] We now know $2ab = 208\sqrt{3}$ and $bx + ay = 55\sqrt{3}$, so we can simply substitute them in. Therefore, \[[BKLC] = 2ab - bx - ay - 49\sqrt{3} = 208\sqrt{3} - 55\sqrt{3} - 49\sqrt{3} = 104\sqrt{3}.\] Finally, we are given $[BKLC] = n\sqrt{3}$ for some integer $n$. We know $[BKLC] = 104\sqrt{3}$, so $n = \boxed{104}$.
Notice that $[BKLC] = 104\sqrt{3} = \dfrac{208\sqrt{3}}{2} = \dfrac{[ABC]}{2}$. Is this a coincidence?
~V0305
~ Slight edits by Christian, but otherwise explained extremely well! | // Block 1
import math; import geometry; import olympiad;
point A,B,C,L,K; A=(0,0); C=(16sqrt(3),0); B=(0,26); L=(8sqrt(3),2); K=(3sqrt(3),13);
point M,N; M=(8sqrt(3), 0); N=(0,13);
draw(A--B--C--cycle); draw(A--K--L--cycle); draw(B--K); draw(C--L);
draw(N--K); draw(L--M);
label("A",A,SW); label("B",B,NW); label("C",C,SE); label("K",K,ENE); label("L",L,NE);
label(“N”,N,W); label(“M”,M,S);
point am, ml, an, nk, mc, bn;
am=(4sqrt(3), 0); mc=(12sqrt(3),0); ml=(8sqrt(3),1); an=(0,6.5); bn=(0,19.5);
nk=(2.5981, 13);
label(“a",am,S); label(“a",mc,S); label(“y”,ml,ESE);
label(“b”,an,W); label(“b”,bn,W); label(“x”,nk,S);
markscalefactor=1;
// Block 2
import math; import geometry; import olympiad; point A,B,C,L,K; A=(0,0); C=(16sqrt(3),0); B=(0,26); L=(8sqrt(3),2); K=(3sqrt(3),13); point M,N; M=(8sqrt(3), 0); N=(0,13); draw(A--B--C--cycle); draw(A--K--L--cycle); draw(B--K); draw(C--L); draw(N--K); draw(L--M); label("A",A,SW); label("B",B,NW); label("C",C,SE); label("K",K,ENE); label("L",L,NE); label(“N”,N,W); label(“M”,M,S); point am, ml, an, nk, mc, bn; am=(4sqrt(3), 0); mc=(12sqrt(3),0); ml=(8sqrt(3),1); an=(0,6.5); bn=(0,19.5); nk=(2.5981, 13); label(“a",am,S); label(“a",mc,S); label(“y”,ml,ESE); label(“b”,an,W); label(“b”,bn,W); label(“x”,nk,S); markscalefactor=1; | [] |
274 | Let ${\triangle ABC}$ be a right triangle with $\angle A = 90^\circ$ and $BC = 38.$ There exist points $K$ and $L$ inside the triangle such\[AK = AL = BK = CL = KL = 14.\]The area of the quadrilateral $BKLC$ can be expressed as $n\sqrt3$ for some positive integer $n.$ Find $n.$ | 2025 AIME II Problem 14 | We observe that the desired area can be found by subtracting the areas of the three smaller triangles $\triangle BAK$, $\triangle KAL$, and $\triangle LAC$ from the area of the large triangle $\triangle ABC$. This can be expressed as \([ABC]-[BAK]-[KAL]-[LAC]\). Thus, our goal is to compute these four areas.
Since $\triangle KAL$ is equilateral with side length $14$, its area is
\[[KAL]=\frac{14^2\sqrt{3}}{4}=49\sqrt{3}.\]
Let $AC=a$ and $AB=b$. Because $\angle CAB$ is a right angle, by the Pythagorean Theorem we have
\[a^2+b^2=1444.\]
Let $\angle CAL=\alpha$ and $\angle KAB=\beta$. Then
\[\cos\alpha=\frac{a}{28}\] (imagine extending $\overrightarrow{CL}$ to intersect $\overline{AB}$, forming a right triangle with hypotenuse length $2\cdot14=28$), \[\quad \cos\beta=\frac{b}{28}\] similarly.
Using Pythagorean identities,
\[\sin\alpha=\frac{\sqrt{784-a^2}}{28}, \quad \sin\beta=\frac{\sqrt{784-b^2}}{28}.\]
Since $\angle CAB=90^\circ$ and $\triangle KAL$ is equilateral, we have
\[\alpha+\beta=30^\circ.\]
Thus,
\[\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta=\frac{\sqrt{3}}{2}.\]
Substituting,
\[\frac{\sqrt{3}}{2}=\frac{ab}{784}-\frac{\sqrt{(784-a^2)(784-b^2)}}{784}.\]
Rearranging,
\[\frac{ab}{784}-\frac{\sqrt{3}}{2}=\frac{\sqrt{(784-a^2)(784-b^2)}}{784}.\]
Clearing denominators,
\[ab-392\sqrt{3}=\sqrt{(784-a^2)(784-b^2)}\]
Squaring both sides gives
\[a^2b^2-784ab\sqrt{3}+392^2\cdot3=(784-a^2)(784-b^2).\]
Expanding the RHS,
\[a^2b^2-784ab\sqrt{3}+392^2\cdot3=784^2-784a^2-784b^2+a^2b^2.\]
Canceling $a^2b^2$ and using $a^2+b^2=1444$,
\[-784ab\sqrt{3}+392^2\cdot3=784^2-784(1444).\]
Simplifying,
\[ab\sqrt{3}=1248 \quad \Rightarrow \quad ab=416\sqrt{3}.\]
The area of the large triangle is
\[[ABC]=\frac{ab}{2}=208\sqrt{3}.\]
Now substitute $b=\frac{416\sqrt{3}}{a}$ into $a^2+b^2=1444$:
\[a^2+\left(\frac{416\sqrt{3}}{a}\right)^2=1444,\]
\[a^4-1444a^2+416^2\cdot3=0.\]
Solving for $a^2$,
\[a^2=\frac{1444\pm\sqrt{1444^2-12\cdot416^2}}{2}=\frac{1444\pm184}{2},\]
so
\[a^2=768 \text{ or } 676.\]
From the diagram, $a$ is the longer side, so
\[a=16\sqrt{3}, \quad b=26.\]
Points $L$ and $K$ lie on the perpendicular bisectors of $AC$ and $AB$, respectively. Let the heights from $L$ to $AC$ and from $K$ to $AB$ be $h_L$ and $h_K$. Then,
\[(8\sqrt{3})^2+h_L^2=14^2, \quad 13^2+h_K^2=14^2,\]
which gives
\[h_L=2, \quad h_K=3\sqrt{3}.\]
Thus,
\[[LAC]=\frac{16\sqrt{3}\cdot2}{2}=16\sqrt{3},\]
\[[BAK]=\frac{26\cdot3\sqrt{3}}{2}=39\sqrt{3}.\]
Finally, subtracting areas,
\[[ABC]-[BAK]-[KAL]-[LAC]=208\sqrt{3} - 39\sqrt{3} - 49\sqrt{3} - 16\sqrt{3} = 104\sqrt{3}.\]
Therefore, the answer is $\boxed{104}$.
~Voidling
~ Edited by Christian | // Block 1
import olympiad;
import geometry;
size(300);
defaultpen(linewidth(0.7)+fontsize(10pt));
pair A = (0,0);
pair B = (0,26);
pair C = (16*sqrt(3),0); // approx 27.71
pair K = (sqrt(14^2 - 13^2), 13);
pair L = (8*sqrt(3), sqrt(14^2 - (8*sqrt(3))^2));
draw(A--B--C--cycle);
draw(B--K--L--C--cycle);
draw(A--K, linewidth(0.5pt));
draw(A--L, linewidth(0.5pt));
draw(K--L, linewidth(0.5pt));
dot(A); dot(B); dot(C); dot(K); dot(L);
label("$A$", A, SW);
label("$B$", B, NW);
label("$C$", C, SE);
label("$K$", K, E);
label("$L$", L, N);
label("$a$", (A+C)/2, S);
label("$b$", (A+B)/2, W);
label("$14$", (B+K)/2, SW, fontsize(8pt));
label("$14$", (C+L)/2, SW, fontsize(8pt));
label("$14$", (K+L)/2, SW, fontsize(8pt));
label("$38$", (B+C)/2, NE, fontsize(8pt));
draw(rightanglemark(B,A,C, 25));
draw(anglemark(C,A,L, 50));
draw(anglemark(K,A,B, 50));
label("$\alpha$", A, 13.5*dir(degrees(L)/2), fontsize(9pt));
label("$\beta$", A, 8*dir(81.7), fontsize(9pt));
// Block 2
import olympiad; import geometry; size(300); defaultpen(linewidth(0.7)+fontsize(10pt)); pair A = (0,0); pair B = (0,26); pair C = (16*sqrt(3),0); // approx 27.71 pair K = (sqrt(14^2 - 13^2), 13); pair L = (8*sqrt(3), sqrt(14^2 - (8*sqrt(3))^2)); draw(A--B--C--cycle); draw(B--K--L--C--cycle); draw(A--K, linewidth(0.5pt)); draw(A--L, linewidth(0.5pt)); draw(K--L, linewidth(0.5pt)); dot(A); dot(B); dot(C); dot(K); dot(L); label("$A$", A, SW); label("$B$", B, NW); label("$C$", C, SE); label("$K$", K, E); label("$L$", L, N); label("$a$", (A+C)/2, S); label("$b$", (A+B)/2, W); label("$14$", (B+K)/2, SW, fontsize(8pt)); label("$14$", (C+L)/2, SW, fontsize(8pt)); label("$14$", (K+L)/2, SW, fontsize(8pt)); label("$38$", (B+C)/2, NE, fontsize(8pt)); draw(rightanglemark(B,A,C, 25)); draw(anglemark(C,A,L, 50)); draw(anglemark(K,A,B, 50)); label("$\alpha$", A, 13.5*dir(degrees(L)/2), fontsize(9pt)); label("$\beta$", A, 8*dir(81.7), fontsize(9pt)); | [] |
275 | Let $\triangle ABC$ have side lengths $AB=13$, $BC=14$, and $CA=15$. Triangle $\triangle A'B'C'$ is obtained by rotating $\triangle ABC$ about its circumcenter so that $\overline{A'C'}$ is perpendicular to $\overline{BC}$, with $A'$ and $B$ not on the same side of line $B'C'$. Find the integer closest to the area of hexagon $AA'CC'BB'$. | 2026 AIME I Problem 10 | Since a $13$-$14$-$15$ triangle is composed of a $5$-$12$-$13$ triangle connected to a $9$-$12$-$15$ triangle, we set up coordinates
\[A = (5,12),\quad B = (0,0),\quad C = (14,0).\]
From this we have
\[[ABC] = \frac{1}{2} \cdot 14 \cdot 12 = 84,\]
and thus
\[R = \frac{abc}{4[ABC]} = \frac{65}{8}.\]
We compute the circumcenter as the intersection of two perpendicular bisectors. The perpendicular bisector of $BC$ is clearly
\[x = 7.\]
The perpendicular bisector of $AB$ passes through $\left(\frac{5}{2}, 6\right)$ and has slope $-\frac{5}{12}$, so its equation is
\[y - 6 = -\frac{5}{12}\left(x - \frac{5}{2}\right).\]
The intersection is
\[O = \left(7,\frac{33}{8}\right).\]
If the triangle is rotated by $\theta$, then the rotation maps $(9,-12)$ to
\[\left(9\cos\theta + 12\sin\theta,\; 9\sin\theta - 12\cos\theta\right).\]
Since $BC$ is horizontal, $A'C'$ must be vertical, so the $x$-component is $0$:
\[9\cos\theta + 12\sin\theta = 0.\]
Thus
\[3\cos\theta + 4\sin\theta = 0,\]
giving
\[\tan\theta = -\frac{3}{4}.\]
We take
\[\cos\theta = \frac{4}{5}, \quad \sin\theta = -\frac{3}{5}.\]
To rotate about $O = \left(7,\frac{33}{8}\right)$, for any point $P$ we use
\[P' = O + R_\theta(P - O),\]
where
\[R_\theta(x,y) = \left( \frac{4}{5}x + \frac{3}{5}y,\; -\frac{3}{5}x + \frac{4}{5}y \right).\]
Compute vectors from $O$:
\[A - O = \left(-2,\frac{63}{8}\right),\]
\[B - O = \left(-7,-\frac{33}{8}\right),\]
\[C - O = \left(7,-\frac{33}{8}\right).\]
Rotate:
\[R_\theta(A - O) = \left( \frac{4}{5}(-2) + \frac{3}{5}\cdot\frac{63}{8},\; -\frac{3}{5}(-2) + \frac{4}{5}\cdot\frac{63}{8} \right) = \left(\frac{25}{8},\frac{15}{2}\right),\]
\[R_\theta(C - O) = \left(\frac{25}{8},-\frac{15}{2}\right),\]
\[R_\theta(B - O) = \left(-\frac{323}{40},\frac{9}{10}\right).\]
Hence
\[A' = \left(\frac{81}{8},\frac{93}{8}\right),\]
\[C' = \left(\frac{81}{8},-\frac{27}{8}\right),\]
\[B' = \left(-\frac{43}{40},\frac{201}{40}\right).\]
Using the shoelace formula on hexagon $A,A',C,C',B,B'$ gives area
\[\frac{1557}{10}.\]
The closest integer is
\[\boxed{156}.\]
- spectraldragon8 | // Block 1
import geometry;
size(10cm);
// the diagram was all done by gemini
// 1. Define the coordinates of Triangle ABC (13-14-15 triangle)
// Base BC = 14 on the x-axis. Height to A is 12.
pair B = (0,0);
pair C = (14,0);
pair A = (5,12);
// 2. Calculate Circumcenter O and Radius R
pair O = (7, 33/8);
real R = 65/8;
// 3. Define Rotation Angle
// Vector AC is (9, -12). Slope is -4/3.
// To make A'C' perpendicular to BC (vertical),
// we rotate AC by theta where tan(theta) = -3/4.
real theta = -atan(3/4);
pair Ap = rotate(degrees(theta), O) * A;
pair Bp = rotate(degrees(theta), O) * B;
pair Cp = rotate(degrees(theta), O) * C;
// 4. Drawing the Elements
// Draw the circumcircle
draw(circle(O, R), dotted + grey);
// Draw the original and rotated triangles lightly
draw(A--B--C--cycle, gray + dashed);
draw(Ap--Bp--Cp--cycle, gray + dashed);
// Draw the requested hexagon AA'CC'BB'
path hexagon = A--Ap--C--Cp--B--Bp--cycle;
draw(hexagon, blue + linewidth(1));
// 5. Labels and Dots
dot("$A$", A, N);
dot("$B$", B, W);
dot("$C$", C, E);
dot("$A'$", Ap, NE);
dot("$B'$", Bp, NW);
dot("$C'$", Cp, S);
dot("$O$", O, S);
// Block 2
import geometry; size(10cm); // the diagram was all done by gemini // 1. Define the coordinates of Triangle ABC (13-14-15 triangle) // Base BC = 14 on the x-axis. Height to A is 12. pair B = (0,0); pair C = (14,0); pair A = (5,12); // 2. Calculate Circumcenter O and Radius R pair O = (7, 33/8); real R = 65/8; // 3. Define Rotation Angle // Vector AC is (9, -12). Slope is -4/3. // To make A'C' perpendicular to BC (vertical), // we rotate AC by theta where tan(theta) = -3/4. real theta = -atan(3/4); pair Ap = rotate(degrees(theta), O) * A; pair Bp = rotate(degrees(theta), O) * B; pair Cp = rotate(degrees(theta), O) * C; // 4. Drawing the Elements // Draw the circumcircle draw(circle(O, R), dotted + grey); // Draw the original and rotated triangles lightly draw(A--B--C--cycle, gray + dashed); draw(Ap--Bp--Cp--cycle, gray + dashed); // Draw the requested hexagon AA'CC'BB' path hexagon = A--Ap--C--Cp--B--Bp--cycle; draw(hexagon, blue + linewidth(1)); // 5. Labels and Dots dot("$A$", A, N); dot("$B$", B, W); dot("$C$", C, E); dot("$A'$", Ap, NE); dot("$B'$", Bp, NW); dot("$C'$", Cp, S); dot("$O$", O, S); | [] |
276 | For each nonnegative integer $r$ less than $502$, define\[S_r=\sum_{m\geq 0}\binom{10,000}{502m+r},\]where $\binom{10,000}{n}$ is defined to be $0$ when $n>10,000$. That is, $S_r$ is the sum of all the binomial coefficients of the form $\binom{10,000}{k}$ for which $0\leq k\leq 10,000$ and $k-r$ is a multiple of $502$. Find the number of integers in the list $S_0,S_1,S_2,\dots,S_{501}$ that are multiples of the prime number $503$. | 2026 AIME I Problem 13 | Let $p = 503$. Since $p$ is a prime number, the field $\mathbb{F}_p$ is a finite field of order $p$. Let $N = 10,000$ and $d = 502$. Observe that $d = p - 1$.
The problem asks for the number of indices $r \in \{0, 1, \dots, d-1\}$ such that $S_r \equiv 0 \pmod{p}$.
The sum $S_r$ corresponds to the sum of every $d$-th coefficient of the binomial expansion of $(1+x)^N$. Specifically, consider the polynomial $P(x) = (1+x)^N$ in the polynomial ring $\mathbb{F}_p[x]$. By the binomial theorem,
\[P(x) = \sum_{k=0}^N \binom{N}{k} x^k.\]
To isolate the coefficients where the exponent $k$ satisfies $k \equiv r \pmod{d}$, we analyze the polynomial modulo $x^d - 1$. In the quotient ring $\mathbb{F}_p[x] / \langle x^d - 1 \rangle$, we have the identity $x^d = 1$, which implies $x^k = x^{k \pmod d}$. Consequently, the reduction of $P(x)$ is given by
\[P(x) \pmod{x^d - 1} = \sum_{r=0}^{d-1} \left( \sum_{m \ge 0} \binom{N}{dm+r} \right) x^r = \sum_{r=0}^{d-1} S_r x^r.\]
We proceed to compute the explicit form of $(1+x)^N \pmod{x^d - 1}$. The polynomial $x^d - 1 = x^{p-1} - 1$ splits completely in $\mathbb{F}_p$, and its roots are exactly the non-zero elements of $\mathbb{F}_p$, denoted $\mathbb{F}_p^\times = \{1, 2, \dots, p-1\}$. Let $\alpha \in \mathbb{F}_p^\times$ be any such root. By Fermat's Little Theorem, for any $\alpha \neq 0$, we have $\alpha^{p-1} = 1$.
We evaluate the polynomial $(1+x)^N$ at an arbitrary root $\alpha$ of $x^d - 1$.
First, we express $N$ in terms of $d$:
\[10,000 = 19 \times 502 + 462.\]
Thus, $N = 19d + 462$.
Then, the value of the polynomial at $x = \alpha$ is:
\[(1+\alpha)^N = (1+\alpha)^{19d + 462} = \left[ (1+\alpha)^d \right]^{19} (1+\alpha)^{462}.\]
We distinguish two cases for $\alpha$:
$\bullet$ Case 1: $\alpha \neq -1$.
Then $1+\alpha \in \mathbb{F}_p^\times$. Since the order of the multiplicative group is $d = p-1$, we have $(1+\alpha)^d = 1$. The expression simplifies to:
\[(1+\alpha)^N = 1^{19} \cdot (1+\alpha)^{462} = (1+\alpha)^{462}.\]
$\bullet$ Case 2: $\alpha = -1$.
Then $1+\alpha = 0$. The left side becomes $0^N = 0$ (since $N \ge 1$). The term $(1+\alpha)^{462} = 0^{462} = 0$. Thus, the equality $(1+\alpha)^N = (1+\alpha)^{462}$ holds strictly.
In both cases, $(1+\alpha)^N = (1+\alpha)^{462}$ for all roots $\alpha$ of $x^d - 1$. Let $Q(x) = (1+x)^{462}$. Since the polynomials $\sum S_r x^r$ and $Q(x)$ agree on all $d$ distinct roots of $x^d - 1$, and the degree of $Q(x)$ is $462$, which is strictly less than $d=502$, Lagrange interpolation implies that these polynomials are identical in $\mathbb{F}_p[x]$.
Therefore, equating the coefficients of $x^r$:
\[S_r \equiv \binom{462}{r} \pmod{503}.\]
We seek the number of integers $r \in \{0, 1, \dots, 501\}$ such that $S_r \equiv 0 \pmod{503}$.
The binomial coefficient $\binom{n}{k}$ modulo a prime $p$ is zero if and only if $k > n$ (assuming $n < p$). Here, $n = 462$.
$\bullet$ For $0 \le r \le 462$, $\binom{462}{r} \not\equiv 0 \pmod{503}$.
$\bullet$ For $463 \le r \le 501$, $\binom{462}{r} = 0$, so $S_r \equiv 0 \pmod{503}$.
The values of $r$ satisfying the condition are exactly the integers in the range $[463, 501]$. The count is:
\[501 - 463 + 1 = 39.\]
The number of integers in the list $S_0, S_1, \dots, S_{501}$ that are multiples of $503$ is $39$. | // Block 1
unitsize(1cm);
// Colors
pen accentBlue = blue + linewidth(1.5pt);
pen accentRed = red + linewidth(1.5pt);
pen shadeRed = lightred + opacity(0.2);
// Draw background shading for zero region
fill((7.5, -1)--(11.5, -1)--(11.5, 2)--(7.5, 2)--cycle, shadeRed);
// Main Axis
draw((-1,0)--(12,0), black+linewidth(1pt), Arrow(Relative(0.99)));
label("$r$", (12,0), E);
// Ticks and Labels
real[] ticks = {0, 6.5, 7.5, 11};
string[] labels = {"0", "462", "463", "501"};
for(int i=0; i<ticks.length; ++i) {
draw((ticks[i], 0.15)--(ticks[i], -0.15), black+linewidth(1pt));
label("$" + labels[i] + "$", (ticks[i], -0.2), S);
}
// Break symbol to show scale compression
filldraw((2.8, -0.2)--(3.2, -0.2)--(3.2, 0.2)--(2.8, 0.2)--cycle, white, white);
draw((2.9, -0.3)--(3.0, 0.3), black);
draw((3.0, -0.3)--(3.1, 0.3), black);
// Non-zero region visualization
draw((0, 0.8)--(6.5, 0.8), accentBlue);
dot((0, 0.8), accentBlue);
dot((6.5, 0.8), accentBlue);
label("$S_r \not\equiv 0 \pmod{503}$", (3.25, 1.2), accentBlue);
// Zero region visualization
draw((7.5, 0.8)--(11, 0.8), accentRed);
dot((7.5, 0.8), accentRed);
dot((11, 0.8), accentRed);
label("$S_r \equiv 0 \pmod{503}$", (9.25, 1.2), accentRed);
// Curly brace for the solution
// Syntax: brace(point1, point2, amplitude)
import graph;
path b = brace((11, -0.8), (7.5, -0.8), 0.3);
draw(b, black+linewidth(1pt));
label("\textbf{Count = 39}", (9.25, -1.3), S);
// Block 2
unitsize(1cm); // Colors pen accentBlue = blue + linewidth(1.5pt); pen accentRed = red + linewidth(1.5pt); pen shadeRed = lightred + opacity(0.2); // Draw background shading for zero region fill((7.5, -1)--(11.5, -1)--(11.5, 2)--(7.5, 2)--cycle, shadeRed); // Main Axis draw((-1,0)--(12,0), black+linewidth(1pt), Arrow(Relative(0.99))); label("$r$", (12,0), E); // Ticks and Labels real[] ticks = {0, 6.5, 7.5, 11}; string[] labels = {"0", "462", "463", "501"}; for(int i=0; i<ticks.length; ++i) { draw((ticks[i], 0.15)--(ticks[i], -0.15), black+linewidth(1pt)); label("$" + labels[i] + "$", (ticks[i], -0.2), S); } // Break symbol to show scale compression filldraw((2.8, -0.2)--(3.2, -0.2)--(3.2, 0.2)--(2.8, 0.2)--cycle, white, white); draw((2.9, -0.3)--(3.0, 0.3), black); draw((3.0, -0.3)--(3.1, 0.3), black); // Non-zero region visualization draw((0, 0.8)--(6.5, 0.8), accentBlue); dot((0, 0.8), accentBlue); dot((6.5, 0.8), accentBlue); label("$S_r \not\equiv 0 \pmod{503}$", (3.25, 1.2), accentBlue); // Zero region visualization draw((7.5, 0.8)--(11, 0.8), accentRed); dot((7.5, 0.8), accentRed); dot((11, 0.8), accentRed); label("$S_r \equiv 0 \pmod{503}$", (9.25, 1.2), accentRed); // Curly brace for the solution // Syntax: brace(point1, point2, amplitude) import graph; path b = brace((11, -0.8), (7.5, -0.8), 0.3); draw(b, black+linewidth(1pt)); label("\textbf{Count = 39}", (9.25, -1.3), S); | [] |
277 | In an equiangular pentagon, the sum of the squares of the side lengths equals $308$, and the sum of the squares of the diagonal lengths equals $800$. The square of the perimeter of the pentagon can be expressed as $m\sqrt{n}$, where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m+n$. | 2026 AIME I Problem 14 | Consider an equiangular pentagon in the complex plane with vertices $V_1, V_2, V_3, V_4, V_5$ in counterclockwise order. Let the side lengths be $a_k = |V_{k+1} - V_k|$ for $k=1, \dots, 5$ (indices modulo 5), and let the diagonal lengths be $d_k = |V_{k+2} - V_k|$.
Since the pentagon is equiangular, each interior angle is given by $\dfrac{(5-2) \times 180^\circ}{5} = 108^\circ$.
We first establish a relationship between the side lengths and the diagonal lengths using the Law of Cosines on triangle $\triangle V_k V_{k+1} V_{k+2}$. The sides of this triangle are $a_k$, $a_{k+1}$, and the diagonal $d_k$. The included angle at vertex $V_{k+1}$ is $108^\circ$.
\[d_k^2 = a_k^2 + a_{k+1}^2 - 2 a_k a_{k+1} \cos(108^\circ).\]
We define the golden ratio $\phi = \frac{1+\sqrt{5}}{2}$. Recall that $\cos(108^\circ) = -\sin(18^\circ) = -\frac{\sqrt{5}-1}{4} = -\frac{1}{2\phi}$. Substituting this value, the equation becomes:
\[d_k^2 = a_k^2 + a_{k+1}^2 + \frac{1}{\phi} a_k a_{k+1}.\]
Let $S_1 = \sum_{k=1}^5 a_k^2$ and $S_2 = \sum_{k=1}^5 a_k a_{k+1}$. Summing the expression for $d_k^2$ over $k=1, \dots, 5$, we obtain the sum of the squares of the diagonals, denoted as $D^2$:
\[\sum_{k=1}^5 d_k^2 = \sum_{k=1}^5 a_k^2 + \sum_{k=1}^5 a_{k+1}^2 + \frac{1}{\phi} \sum_{k=1}^5 a_k a_{k+1}\]
\[D^2 = 2 S_1 + \frac{1}{\phi} S_2.\]
Given $S_1 = 308$ and $D^2 = 800$, we substitute these values:
\[800 = 2(308) + \frac{1}{\phi} S_2 \implies 800 = 616 + \frac{S_2}{\phi} \implies 184 = \frac{S_2}{\phi}.\]
Thus, $S_2 = 184\phi$.
Next, we utilize the geometric closure of the pentagon. Viewing the sides as vectors in the complex plane, let $\omega = e^{i 72^\circ}$. The vector corresponding to side $a_k$ points in the direction of $\omega^{k-1}$ (up to a global rotation). The condition $\sum_{k=1}^5 a_k \omega^{k-1} = 0$ implies:
\[\left| \sum_{k=1}^5 a_k \omega^{k-1} \right|^2 = 0.\]
Expanding this squared modulus:
\[\sum_{j=1}^5 \sum_{k=1}^5 a_j a_k \omega^{j-k} = 0.\]
We group the terms by the difference in indices $m \equiv j-k \pmod 5$. Let $S_3 = \sum_{k=1}^5 a_k a_{k+2}$. The coefficients are powers of $\omega$:
$\bullet$ For $m=0$, the coefficient is $\omega^0 = 1$. The sum is $S_1$.
$\bullet$ For $m=\pm 1$, the coefficient is $\omega + \omega^{-1} = 2\cos(72^\circ) = \frac{1}{\phi}$. The sum is $S_2$.
$\bullet$ For $m=\pm 2$, the coefficient is $\omega^2 + \omega^{-2} = 2\cos(144^\circ) = -\phi$. The sum is $S_3$.
The closure equation becomes:
\[S_1 + \frac{1}{\phi}S_2 - \phi S_3 = 0.\]
Substituting the known values $S_1 = 308$ and $\frac{S_2}{\phi} = 184$:
\[308 + 184 - \phi S_3 = 0 \implies 492 = \phi S_3 \implies S_3 = \frac{492}{\phi}.\]
We are asked to find the square of the perimeter $P = \sum_{k=1}^5 a_k$.
\[P^2 = \left( \sum_{k=1}^5 a_k \right)^2 = \sum_{k=1}^5 a_k^2 + 2 \sum_{k=1}^5 a_k a_{k+1} + 2 \sum_{k=1}^5 a_k a_{k+2}.\]
\[P^2 = S_1 + 2 S_2 + 2 S_3.\]
Substituting the expressions in terms of $\phi$:
\[P^2 = 308 + 2(184\phi) + 2\left(\frac{492}{\phi}\right).\]
Using the identity $\frac{1}{\phi} = \phi - 1$:
\[P^2 = 308 + 368\phi + 984(\phi - 1) = 308 + 368\phi + 984\phi - 984.\]
\[P^2 = 1352\phi - 676.\]
Finally, substitute $\phi = \frac{1+\sqrt{5}}{2}$:
\[P^2 = 1352\left(\frac{1+\sqrt{5}}{2}\right) - 676 = 676(1+\sqrt{5}) - 676 = 676 + 676\sqrt{5} - 676.\]
\[P^2 = 676\sqrt{5}.\]
This is in the form $m\sqrt{n}$ with $m = 676$ and $n = 5$. Since $5$ is a prime number, it is square-free.
The value $m+n$ is:
\[m+n = 676 + 5 = 681.\] | // Block 1
size(7.5cm); // Roughly equivalent to scale=2.5
// Define the vertices using polar coordinates (r=1.5)
pair[] V = new pair[6];
for(int i=1; i<=5; ++i) {
V[i] = 1.5 * dir(90 - (i-1)*72);
}
// Pens (colors and styles)
pen sidePen = blue + linewidth(1.2pt);
pen mainDiagPen = deepgreen + dashed + linewidth(1pt);
pen ghostDiagPen = gray(0.7) + dashed + opacity(0.4);
// Draw sides and add side labels
draw(V[1]--V[2], sidePen); label("$a_1$", midpoint(V[1]--V[2]), NE);
draw(V[2]--V[3], sidePen); label("$a_2$", midpoint(V[2]--V[3]), E);
draw(V[3]--V[4], sidePen); label("$a_3$", midpoint(V[3]--V[4]), S);
draw(V[4]--V[5], sidePen); label("$a_4$", midpoint(V[4]--V[5]), W);
draw(V[5]--V[1], sidePen); label("$a_5$", midpoint(V[5]--V[1]), NW);
// Draw the primary diagonal d1
draw(V[1]--V[3], mainDiagPen);
label("$d_1$", midpoint(V[1]--V[3]), E, deepgreen);
// Draw ghost diagonals
draw(V[2]--V[4], ghostDiagPen);
draw(V[3]--V[5], ghostDiagPen);
draw(V[4]--V[1], ghostDiagPen);
draw(V[5]--V[2], ghostDiagPen);
// Mark one interior angle at V2
// In a regular pentagon, angle at V2 starts at 145 degrees and spans 108
path angleMark = arc(V[2], 0.3, 145, 145+108);
draw(angleMark, gray);
label("\scriptsize $108^\circ$", V[2] + (-0.1, -0.4));
// Draw vertices as small labeled circles (matching your background-fill style)
for(int i=1; i<=5; ++i) {
label(format("$V_{%d}$", i), V[i], UnFill);
}
// Block 2
size(7.5cm); // Roughly equivalent to scale=2.5 // Define the vertices using polar coordinates (r=1.5) pair[] V = new pair[6]; for(int i=1; i<=5; ++i) { V[i] = 1.5 * dir(90 - (i-1)*72); } // Pens (colors and styles) pen sidePen = blue + linewidth(1.2pt); pen mainDiagPen = deepgreen + dashed + linewidth(1pt); pen ghostDiagPen = gray(0.7) + dashed + opacity(0.4); // Draw sides and add side labels draw(V[1]--V[2], sidePen); label("$a_1$", midpoint(V[1]--V[2]), NE); draw(V[2]--V[3], sidePen); label("$a_2$", midpoint(V[2]--V[3]), E); draw(V[3]--V[4], sidePen); label("$a_3$", midpoint(V[3]--V[4]), S); draw(V[4]--V[5], sidePen); label("$a_4$", midpoint(V[4]--V[5]), W); draw(V[5]--V[1], sidePen); label("$a_5$", midpoint(V[5]--V[1]), NW); // Draw the primary diagonal d1 draw(V[1]--V[3], mainDiagPen); label("$d_1$", midpoint(V[1]--V[3]), E, deepgreen); // Draw ghost diagonals draw(V[2]--V[4], ghostDiagPen); draw(V[3]--V[5], ghostDiagPen); draw(V[4]--V[1], ghostDiagPen); draw(V[5]--V[2], ghostDiagPen); // Mark one interior angle at V2 // In a regular pentagon, angle at V2 starts at 145 degrees and spans 108 path angleMark = arc(V[2], 0.3, 145, 145+108); draw(angleMark, gray); label("\scriptsize $108^\circ$", V[2] + (-0.1, -0.4)); // Draw vertices as small labeled circles (matching your background-fill style) for(int i=1; i<=5; ++i) { label(format("$V_{%d}$", i), V[i], UnFill); } | [] |
278 | The figure below shows a grid of $10$ squares in a row. Each square has a diagonal connecting its lower left vertex to its upper right vertex. A bug moves along the line segments from vertex to vertex, never traversing the same segment twice and never moving from right to left along a horizontal or diagonal segment. Let $N$ be the number of paths the bug can take from the lower left corner ($A$) to the upper right corner ($B$). One such path from $A$ to $B$ is shown by the thick line segments in the figure. Find $\sqrt{N}$. | 2026 AIME II Problem 2 | Consider each of the ten instances of
Let the number of ways to start at the top-left corner be $a$ and the number of ways to start at the bottom-left corner be $b$. Then, there are $a+b$ ways to get to the top-right corner without passing through the top-left corner, and there are $a$ ways to get to the bottom-right corner without passing through the top-right corner.
But on the right edge, it is also possible to swap the right-side corners. Therefore, there are an additional $b$ ways to get to the top-right corner AFTER passing through the bottom-right corner, and there are an additional $a+b$ ways to get to the bottom-right corner AFTER passing through the top-right corner. Combined, there are $a+2b$ ways to get to each of the right-side corners.
But we notice that both are the same, so it must also be true that $a=b$ by a simple induction. Therefore, there are $3a$ ways to get to each of the right-side corners.
At the very first vertical edge, there is one way to get to each corner on the edge. After 10 of these squares, the total number of ways is $3^{10}\implies \sqrt{3^{10}}=3^5=\boxed{243}$.
~ eevee9406 | size(1cm); draw((0, 0)--(0, 1)); for (int i = 0; i <= 0; i += 1) { draw((i, 1)--(i + 1, 1)--(i + 1, 0)--(i, 0)--(i + 1, 1)); } | [] |
279 | Isosceles triangle $\triangle ABC$ has $AB=BC$. Let $I$ be the incenter of $\triangle ABC$. The perimeters of $\triangle ABC$ and $\triangle AIC$ are in the ratio $125:6$, and all the sides of both triangles have integer lengths. Find the minimum possible value of $AB$. | 2026 AIME II Problem 8 | Since $\triangle ABC$ is isosceles with $AB = BC$, the angle bisector, median, and altitude from $B$ to $AC$ coincide, which is the perpendicular bisector of $AC$. The incenter $I$ lies on this common line. Let $D$ be the intersection of this line with $AC$, so $D$ is the midpoint of $AC$ and $ID \perp AC$.
We set $AC = 2$, so $AD = DC = 1$. Let $ID = x$.
The sides \(AI\) and \(CI\) are equal, by Pythagorean Theorem,
\[AI = CI = \sqrt{AD^2 + ID^2} = \sqrt{1 + x^2}.\]
Thus the perimeter of \(\triangle AIC\) is:
\[\text{Perimeter}(\triangle AIC) = AC + AI + CI = 2 + 2\sqrt{1 + x^2}.\]
Let \(\angle IAD = \theta\). Then
\[\cos\theta = \frac{AD}{AI} = \frac{1}{\sqrt{1 + x^2}}.\]
Using the double-angle formula for cosine:
\[\cos 2\theta = 2\cos^2\theta - 1 = 2\left(\frac{1}{1 + x^2}\right) - 1 = \frac{2 - (1 + x^2)}{1 + x^2} = \frac{1 - x^2}{1 + x^2}.\]
In \(\triangle ABC\), \(\angle BAC = 2\theta\), and \(AD = 1\) is half of \(AC\). Thus:
\[AB = \frac{AD}{\cos 2\theta} = \frac{1}{\frac{1 - x^2}{1 + x^2}} = \frac{1 + x^2}{1 - x^2}.\]
Since \(AB = BC\), the perimeter of \(\triangle ABC\) is:
\[\text{Perimeter}(\triangle ABC) = AB + BC + AC = 2AB + 2 = 2\left(\frac{1 + x^2}{1 - x^2}\right) + 2.\]
The ratio of perimeters is given as \(125:6\):
\[\frac{2\left(\frac{1 + x^2}{1 - x^2}\right) + 2}{2 + 2\sqrt{1 + x^2}} = \frac{125}{6}.\]
We simplify the left-hand side:
\[\frac{2\left(\frac{1 + x^2 + 1 - x^2}{1 - x^2}\right)}{2(1 + \sqrt{1 + x^2})} = \frac{\frac{2}{1 - x^2}}{1 + \sqrt{1 + x^2}} = \frac{2}{(1 - x^2)(1 + \sqrt{1 + x^2})}.\]
Thus:
\[\frac{2}{(1 - x^2)(1 + \sqrt{1 + x^2})} = \frac{125}{6}.\]
Cross-multiplying gives:
\[(1 - x^2)(1 + \sqrt{1 + x^2}) = \frac{12}{125}.\]
We make a substitution, let \(\sqrt{1 + x^2} = \dfrac{a}{5}\) for some positive integer \(a\). Then:
\[1 + x^2 = \frac{a^2}{25} \implies x^2 = \frac{a^2 - 25}{25},\]
\[1 - x^2 = 1 - \frac{a^2 - 25}{25} = \frac{25 - a^2 + 25}{25} = \frac{50 - a^2}{25},\]
\[1 + \sqrt{1 + x^2} = 1 + \frac{a}{5} = \frac{5 + a}{5}.\]
We substitute into the equation:
\[\frac{50 - a^2}{25} \cdot \frac{5 + a}{5} = \frac{12}{125}.\]
Simplifying gives:
\[\frac{(50 - a^2)(5 + a)}{125} = \frac{12}{125} \implies (50 - a^2)(5 + a) = 12.\]
We guess and check small positive integer values for \(a\), and find \(a = 7\) satisfies the equation:
\[(50 - 7^2)(5 + 7) = (50 - 49)(12) = 1 \cdot 12 = 12.\]
Thus \(a = 7\), so:
\[\sqrt{1 + x^2} = \frac{7}{5} \implies 1 + x^2 = \frac{49}{25} \implies x^2 = \frac{24}{25}.\]
From \(AB = \dfrac{1 + x^2}{1 - x^2}\):
\[1 - x^2 = 1 - \frac{24}{25} = \frac{1}{25}, \quad AB = \frac{\frac{49}{25}}{\frac{1}{25}} = 49.\]
In our initial setup, \(AC = 2\) and \(AI = \dfrac{7}{5}\). To make all sides integers, we scale by a factor of \(5\), so \(AC = 10\), \(AI = 7\), and:
\[AB_{\text{min}} = 49 \times 5 = \boxed{245}.\]
~Steven Zheng | // Block 1
size(200,0);
pair A=(-5,0);
pair C=(5,0);
pair D=(0,0);
pair B=(0,12);
pair I=(0,3);
dot(A,red);
dot(B,red);
dot(C,red);
dot(D,red);
dot(I,red);
label("$A$",A,SW);
label("$B$",B,N);
label("$C$",C,SE);
label("$D$",D,S);
label("$I$",I,W);
draw(A--B);
draw(B--C);
draw(A--C);
draw(B--D);
draw(A--I);
draw(C--I);
draw((0,0.4)--(0.4,0.4)--(0.4,0));
draw((0,0.4)--(-0.4,0.4)--(-0.4,0));
label("$1$",(A+D)/2,S);
label("$1$",(D+C)/2,S);
label("$x$",(D+I)/2,W);
label("$x$",(D+I)/2,E);
label("$\sqrt{1+x^2}$",(A+I)/2,N);
label("$\sqrt{1+x^2}$",(C+I)/2,N);
label("$\frac{1+x^2}{1-x^2}$",(A+B)/2,W);
label("$\frac{1+x^2}{1-x^2}$",(B+C)/2,E);
label("$\theta$",4(A+D)/5,NE);
label("$\theta$",4(C+D)/5,NW);
label("$\theta$", (-4.3, 1.6), SE);
label("$\theta$", (4.3, 1.6), SW);
// Block 2
size(200,0); pair A=(-5,0); pair C=(5,0); pair D=(0,0); pair B=(0,12); pair I=(0,3); dot(A,red); dot(B,red); dot(C,red); dot(D,red); dot(I,red); label("$A$",A,SW); label("$B$",B,N); label("$C$",C,SE); label("$D$",D,S); label("$I$",I,W); draw(A--B); draw(B--C); draw(A--C); draw(B--D); draw(A--I); draw(C--I); draw((0,0.4)--(0.4,0.4)--(0.4,0)); draw((0,0.4)--(-0.4,0.4)--(-0.4,0)); label("$1$",(A+D)/2,S); label("$1$",(D+C)/2,S); label("$x$",(D+I)/2,W); label("$x$",(D+I)/2,E); label("$\sqrt{1+x^2}$",(A+I)/2,N); label("$\sqrt{1+x^2}$",(C+I)/2,N); label("$\frac{1+x^2}{1-x^2}$",(A+B)/2,W); label("$\frac{1+x^2}{1-x^2}$",(B+C)/2,E); label("$\theta$",4(A+D)/5,NE); label("$\theta$",4(C+D)/5,NW); label("$\theta$", (-4.3, 1.6), SE); label("$\theta$", (4.3, 1.6), SW); | [] |
280 | Let $\triangle ABC{}$ be a triangle with $D$ on $\overline{BC}$ such that $\overline{AD}{}$ bisects $\angle BAC.$ Let $\omega{}$ be the circle that passes through $A$ and is tangent to segment $\overline{BC}{}$ at $D.$ Let $E \neq A$ and $F \neq A$ be the intersections of $\omega{}$ with segments $\overline{AB}{}$ and $\overline{AC},$ respectively. Suppose that $AB = 200, AC = 225,$ and all of $AE, AF, BD,$ and $CD{}$ are positive integers. Find the greatest possible value of $BC.$ | 2026 AIME II Problem 10 | Let $BC=x$. By Angle Bisector Theorem, we have \[BD = \frac{200}{200+225}x=\frac{8x}{17}, CD = \frac{9x}{17}\] Since both of these integers, we write $x=17y$, for some $y\in \mathbb{Z}^+$. Then, $BD=8y$ and $CD=9y$.
Since $BD$ is tangent to $\omega$, it follows that $\angle BDE = \angle EAD$. Similarly, $\angle CDF = \angle FAC$. Thus, $\triangle BDE \sim \triangle BAD$ and $\triangle CDF \sim \triangle CAD$. We can use this to solve for $BE$ and $CF$: \[\frac{BD}{AB}= \frac{BE}{BD}\implies BE = \frac{8y^2}{25}, CD = \frac{9y^2}{25}\] Thus, $25\mid y^2\implies 5\mid y$. Thus, we can let $y=5z, z\in \mathbb{Z}^+$, so $BC=85z$.
Now, since $E$ is on $AB$ and $E\neq A$, we have $AE < AB\implies 8z^2< 200 \implies z < 5$. Using $F$ and $AC$ gives $z< 5$ as well. Thus, the maximum value of $BC$ is attained when $z=4$, or $BC = 4\cdot 85 = \boxed{340}$.
~sillybone | // Block 1
size(300);
import olympiad;
import geometry;
pair B = (0,0);
pair C = (170,0);
pair D = (80,0);
pair A = (53.75, 192.64);
pair O = (80, 98.11);
real R = 98.11;
pair E = interp(B, A, 32/200);
pair F = interp(C, A, 36/225);
draw(circle(O, R), Black);
draw(A--B--C--cycle, Black);
draw(A--D, Black);
draw(F--D, Black);
draw(E--D, Black);
dot(A); label("A", A, N);
dot(B); label("B", B, SW);
dot(C); label("C", C, SE);
dot(D); label("D", D, S);
dot(E); label("E", E, NW);
dot(F); label("F", F, NE);
label("$\omega$", O + (R*0.8, R*0.5), Black);
markangle(1, B, A, D, radius=6mm);
markangle(1, E, D, B, radius=6mm);
markangle(1, D, A, C, radius=6mm);
markangle(1, C, D, F, radius=6mm);
// Block 2
size(300); import olympiad; import geometry; pair B = (0,0); pair C = (170,0); pair D = (80,0); pair A = (53.75, 192.64); pair O = (80, 98.11); real R = 98.11; pair E = interp(B, A, 32/200); pair F = interp(C, A, 36/225); draw(circle(O, R), Black); draw(A--B--C--cycle, Black); draw(A--D, Black); draw(F--D, Black); draw(E--D, Black); dot(A); label("A", A, N); dot(B); label("B", B, SW); dot(C); label("C", C, SE); dot(D); label("D", D, S); dot(E); label("E", E, NW); dot(F); label("F", F, NE); label("$\omega$", O + (R*0.8, R*0.5), Black); markangle(1, B, A, D, radius=6mm); markangle(1, E, D, B, radius=6mm); markangle(1, D, A, C, radius=6mm); markangle(1, C, D, F, radius=6mm); | [] |
280 | Let $\triangle ABC{}$ be a triangle with $D$ on $\overline{BC}$ such that $\overline{AD}{}$ bisects $\angle BAC.$ Let $\omega{}$ be the circle that passes through $A$ and is tangent to segment $\overline{BC}{}$ at $D.$ Let $E \neq A$ and $F \neq A$ be the intersections of $\omega{}$ with segments $\overline{AB}{}$ and $\overline{AC},$ respectively. Suppose that $AB = 200, AC = 225,$ and all of $AE, AF, BD,$ and $CD{}$ are positive integers. Find the greatest possible value of $BC.$ | 2026 AIME II Problem 10 | Let the side lengths of $\triangle ABC$ be denoted by $AB = c$, $AC = b$, and $BC = a$. We are given $c = 200$ and $b = 225$. Let $D$ be a point on $BC$ such that $AD$ is the internal angle bisector of $\angle BAC$. By the Angle Bisector Theorem, the ratio of the segments of the opposite side is proportional to the adjacent sides:
\[\frac{BD}{CD} = \frac{AB}{AC} = \frac{200}{225} = \frac{8}{9}.\]
Since $BD$ and $CD$ are positive integers and $\gcd(8, 9) = 1$, there must exist a positive integer $m$ such that
\[BD = 8m \quad \text{and} \quad CD = 9m.\]
Consequently, the total length of side $BC$ is given by $a = BD + CD = 17m$.
Let $\omega$ be the circle passing through $A$ and tangent to the segment $BC$ at $D$. We apply the Power of a Point Theorem. For point $B$, lying on the extension of the tangent segment $BD$ and the secant line $BA$, the power of the point with respect to $\omega$ is given by:
\[\mathcal{P}(B) = BD^2 = BE \cdot BA.\]
Substituting the known expressions into this equation:
\[(8m)^2 = BE \cdot 200 \implies 64m^2 = 200 \cdot BE \implies BE = \frac{64m^2}{200} = \frac{8m^2}{25}.\]
Since $E$ lies on the segment $AB$, the length $AE$ is given by $AE = AB - BE = 200 - \frac{8m^2}{25}$. We are given that $AE$ is a positive integer. For $AE$ to be an integer, $BE$ must be an integer (since $AB$ is an integer). This implies:
\[200 \mid 64m^2 \iff 25 \mid 8m^2.\]
Since $\gcd(25, 8) = 1$, it must be that $25 \mid m^2$, which implies $5 \mid m$. Thus, let $m = 5k$ for some positive integer $k$.
Similarly, we consider the power of point $C$ with respect to $\omega$. The line $BC$ is tangent at $D$, and $CA$ is a secant intersecting $\omega$ at $F$ and $A$. Thus:
\[\mathcal{P}(C) = CD^2 = CF \cdot CA.\]
Substituting the values:
\[(9m)^2 = CF \cdot 225 \implies 81m^2 = 225 \cdot CF \implies CF = \frac{81m^2}{225} = \frac{9m^2}{25}.\]
With $m = 5k$, we substitute into the expressions for $AE$ and $AF$:
\[BE = \frac{8(5k)^2}{25} = 8k^2 \implies AE = 200 - 8k^2,\]
\[CF = \frac{9(5k)^2}{25} = 9k^2 \implies AF = 225 - 9k^2.\]
The problem states that $AE$ and $AF$ are positive integers. This imposes bounds on $k$:
\[200 - 8k^2 > 0 \implies 8k^2 < 200 \implies k^2 < 25 \implies k < 5,\]
\[225 - 9k^2 > 0 \implies 9k^2 < 225 \implies k^2 < 25 \implies k < 5.\]
Additionally, the Triangle Inequality must hold for $\triangle ABC$. The sides are $AB=200$, $AC=225$, and $BC = 17m = 17(5k) = 85k$.
\[AB + AC > BC \implies 200 + 225 > 85k \implies 425 > 85k \implies k < 5.\]
The other triangle inequalities ($AB+BC > AC$ and $AC+BC > AB$) are satisfied for all $k \ge 1$. Thus, the possible integer values for $k$ are $k \in \{1, 2, 3, 4\}$.
To find the greatest possible value of $BC$, we maximize $k$. The largest valid integer is $k = 4$.
Substituting $k=4$:
\[BC = 85(4) = 340.\]
We verify the other segments are integers:
\[AE = 200 - 8(4)^2 = 200 - 128 = 72 \quad (\text{Integer } > 0),\]
\[AF = 225 - 9(4)^2 = 225 - 144 = 81 \quad (\text{Integer } > 0).\]
The lengths $BD = 8(20) = 160$ and $CD = 9(20) = 180$ are also integers. The triangle exists since $200+225 > 340$.
The greatest possible value of $BC$ is $340$.
~ Jesse Zhang (FUNKCCP) | // Block 1
size(300);
// Basic Coordinates
pair B = (0,0);
pair C = (340,0);
pair D = (160,0);
pair A = (154.375, 127.17);
pair O = (160, 63.71);
real R = 63.71;
// Points E and F
pair E = interp(B, A, 128/200);
pair F = interp(C, A, 144/225);
// Main Drawing
draw(circle(O, R), cyan);
draw(A--B--C--cycle, orange);
draw(A--D, magenta);
draw(B--E, gray+dashed);
draw(C--F, gray+dashed);
// Simple Dimensions (No Arrows to avoid package errors)
draw((0,-25)--(160,-25), gray);
label("160", (80,-25), S, gray);
draw((160,-25)--(340,-25), gray);
label("180", (250,-25), S, gray);
// Points and Labels
dot(A); label("A", A, N);
dot(B); label("B", B, SW);
dot(C); label("C", C, SE);
dot(D); label("D", D, S);
dot(E); label("E", E, NW);
dot(F); label("F", F, NE);
label("$\omega$", O + (R*0.8, R*0.5), cyan);
// Block 2
size(300); // Basic Coordinates pair B = (0,0); pair C = (340,0); pair D = (160,0); pair A = (154.375, 127.17); pair O = (160, 63.71); real R = 63.71; // Points E and F pair E = interp(B, A, 128/200); pair F = interp(C, A, 144/225); // Main Drawing draw(circle(O, R), cyan); draw(A--B--C--cycle, orange); draw(A--D, magenta); draw(B--E, gray+dashed); draw(C--F, gray+dashed); // Simple Dimensions (No Arrows to avoid package errors) draw((0,-25)--(160,-25), gray); label("160", (80,-25), S, gray); draw((160,-25)--(340,-25), gray); label("180", (250,-25), S, gray); // Points and Labels dot(A); label("A", A, N); dot(B); label("B", B, SW); dot(C); label("C", C, SE); dot(D); label("D", D, S); dot(E); label("E", E, NW); dot(F); label("F", F, NE); label("$\omega$", O + (R*0.8, R*0.5), cyan); | [] |
281 | Consider a tetrahedron with two isosceles triangle faces with side lengths $5\sqrt{10}, \,5\sqrt{10},$ and $10$ and two isosceles triangle faces with side lengths $5\sqrt{10},\, 5\sqrt{10},$ and $18.$ The four vertices of the tetrahedron lie on a sphere with center $S,$ and the four faces of the tetrahedron are tangent to a sphere with center $R.$ The distance $RS$ can be written as $\tfrac mn,$ where $m$ and $n$ are relatively prime positive integers. Find ${}m + n.$ | 2026 AIME II Problem 12 | We place the tetrahedron in a coordinate system by symmetry:
\[A(-5, 0, 0),\ B(5, 0, 0),\ C(0, 9, h),\ D(0, -9, h).\]
Given $AC = 5\sqrt{10}$, we have
\begin{align*}
AC^2 &= 5^2 + 9^2 + h^2 = 250, \\
h^2 &= 144 \implies h = 12.
\end{align*}
So the vertices are
\[A(-5,0,0),\ B(5,0,0),\ C(0,9,12),\ D(0,-9,12).\]
By symmetry, the circumcenter $S$ lies on the $z$-axis. Let $S=(0,0,a)$.
Using $SA=SC$:
\begin{align*}
SA^2 &= 25 + a^2, \\
SC^2 &= 81 + (12-a)^2, \\
25 + a^2 &= 225 - 24a + a^2, \\
24a &= 200 \implies a = \frac{25}{3}.
\end{align*}
Thus
\[S=\Big(0,0,\frac{25}{3}\Big).\]
By symmetry, the incenter $R$ lies on the $z$-axis. Let $R=(0,0,b)$.
Vectors for plane $ABC$:
\[\overrightarrow{AB}=\begin{pmatrix}10\\0\\0\end{pmatrix},\quad \overrightarrow{AC}=\begin{pmatrix}5\\9\\12\end{pmatrix}.\]
Normal vector by cross product:
\begin{align*}
\mathbf{n}_1 &= \overrightarrow{AB}\times\overrightarrow{AC}
= \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\10&0&0\\5&9&12\end{vmatrix} \\
&= \mathbf{i}(0\cdot12-0\cdot9)
-\mathbf{j}(10\cdot12-0\cdot5)
+\mathbf{k}(10\cdot9-0\cdot5) \\
&= 0\mathbf{i}-120\mathbf{j}+90\mathbf{k}
= 30\begin{pmatrix}0\\-4\\3\end{pmatrix}.
\end{align*}
We get the equation of the plane:
\[-4y + 3z = 0.\]
Distance from $R$ to plane $ABC$:
\[d_1 = \frac{|3b|}{\sqrt{(-4)^2+3^2}} = \frac{3|b|}{5}.\]
Vectors for plane $ACD$:
\[\overrightarrow{AC}=\begin{pmatrix}5\\9\\12\end{pmatrix},\quad \overrightarrow{AD}=\begin{pmatrix}5\\-9\\12\end{pmatrix}.\]
Normal vector by cross product:
\begin{align*}
\mathbf{n}_2 &= \overrightarrow{AC}\times\overrightarrow{AD}
= \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\5&9&12\\5&-9&12\end{vmatrix} \\
&= \mathbf{i}(9\cdot12-12\cdot(-9))
-\mathbf{j}(5\cdot12-12\cdot5)
+\mathbf{k}(5\cdot(-9)-9\cdot5) \\
&= 216\mathbf{i}-0\mathbf{j}-90\mathbf{k}
= 18\begin{pmatrix}12\\0\\-5\end{pmatrix}.
\end{align*}
We get the equation of the plane:
\[12x - 5z + 60 = 0.\]
Distance from $R$ to plane $ACD$:
\[d_2 = \frac{|60-5b|}{\sqrt{12^2+(-5)^2}} = \frac{|60-5b|}{13}.\]
We set $d_1=d_2$:
\begin{align*}
\frac{3b}{5} &= \frac{60-5b}{13}, \\
39b &= 300-25b, \\
64b &= 300 \implies b = \frac{75}{16}.
\end{align*}
Thus
\[R=\Big(0,0,\frac{75}{16}\Big).\]
Hence, the distance $RS$ is:
\begin{align*}
RS &= \left|\frac{25}{3}-\frac{75}{16}\right|
= \left|\frac{400-225}{48}\right|
= \frac{175}{48}.
\end{align*}
Since $\gcd(175,48)=1$, we have $m=175$, $n=48$.
\[m+n = 175+48=\boxed{223}.\]
~Steven Zheng | import three;size(200,0);currentprojection=orthographic(10,20,15);triple A=(-5,0,0);triple B=(5,0,0);triple C=(0,9,12);triple D=(0,-9,12);triple S=(0,0,25/3);triple R=(0,0,75/16);dot(A,red);dot(B,red);dot(C,red);dot(D,red);dot(S,purple+linewidth(3));dot(R,deepgreen+linewidth(3));label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,N);label("$D$",(0,-9,12),SW);label("$S$",S,E);label("$R$",R,W);draw(A--B,blue);draw(A--C,blue);draw(B--C,blue);draw(A--D,blue+dashed);draw(B--D,blue);draw(C--D,blue);draw(R--S,red);draw((-10,0,0)--(-5,0,0));draw((5,0,0)--(10,0,0));draw((9,0,0.5)--(10,0,0));draw((9,0,-0.5)--(10,0,0));draw((0,-15,0)--(0,15,0));draw((0,14,0.5)--(0,15,0));draw((0,14,-0.5)--(0,15,0));draw((0,0,0)--(0,0,20));draw((0,0.5,19)--(0,0,20));draw((0,-0.5,19)--(0,0,20));label("$x$",(12,0,0),E);label("$y$",(0,16,0),N);label("$z$",(0,0,20),N); | [] |
281 | Consider a tetrahedron with two isosceles triangle faces with side lengths $5\sqrt{10}, \,5\sqrt{10},$ and $10$ and two isosceles triangle faces with side lengths $5\sqrt{10},\, 5\sqrt{10},$ and $18.$ The four vertices of the tetrahedron lie on a sphere with center $S,$ and the four faces of the tetrahedron are tangent to a sphere with center $R.$ The distance $RS$ can be written as $\tfrac mn,$ where $m$ and $n$ are relatively prime positive integers. Find ${}m + n.$ | 2026 AIME II Problem 12 | ~Diagram by Confringo
Choose face $ABC$ as the base. Let $O$ be the midpoint of $AB$ as the origin, with $OA$ along the positive $x$-axis, $OC$ along the positive $y$-axis, and the line through $O$ perpendicular to plane $ABC$ as the $z$-axis.
First, find the coordinates of point $D$. Due to the symmetry of the figure, plane $OCD$ is the $yOz$-plane. Let $D'$ be the projection of $D$ onto $OC$. Let $OD' = y$ and $DD' = z$. Then
\[\begin{cases} y^2+z^2=15^2 \\ (15-y)^2+z^2=18^2 \end{cases}\]
Solving this system yields $y=\dfrac{21}{5}$ and $z=\dfrac{72}{5}$, so $D\left(0,\dfrac{21}{5},\dfrac{72}{5}\right)$.
Next, find the circumcenter. The center of the circumscribed sphere lies on the line perpendicular to the base through the circumcenter of the base triangle. Here $CO=\sqrt{AC^2-AO^2}=15$. We need a point $S'$ on $OC$ such that $CS' = SA$. Let $CS' = AS' = x$, then $S'O = 15-x$. In right triangle $AOS'$, by the Pythagorean theorem,
\[x^2=(15-x)^2+5^2,\]
which gives $x=\dfrac{25}{3}$, so $OS' = 15-x = \dfrac{20}{3}$ (the $y$-coordinate of $S'$). Let $S = \left(0,\dfrac{20}{3},h\right)$. Then $AS = CS$, so
\[\sqrt{\left(\dfrac{25}{3}\right)^2+h^2}=\sqrt{(0-0)^2+\left(\dfrac{21}{5}-\dfrac{20}{3}\right)^2+\left(\dfrac{72}{5}-h\right)^2}.\]
Solving gives $h=5$, so $S = \left(0,\dfrac{20}{3},5\right)$.
Now find the incenter. By volume considerations: connecting each vertex to the incenter divides the tetrahedron into four smaller tetrahedra, each with height equal to the insphere radius $r$ and base area equal to the area of a face of the tetrahedron. Their volumes sum to the total volume of the tetrahedron.
The areas of the two types of faces are $75$ and $117$. The height from $C$ to base $ABD$ is $z_C = \dfrac{72}{5}$. Volume equation:
\[\dfrac{1}{3}\cdot75\cdot\dfrac{72}{5}=\dfrac{1}{3}\cdot(2\cdot75+2\cdot117)\cdot r.\]
Solving gives $r = \dfrac{45}{16}$. Thus $z_R = \dfrac{45}{16}$.
By symmetry, the incenter also lies in the $yOz$-plane. Let $R'$ be the projection of $R$ onto $OC$. Then $\tan\angle ROR' = \dfrac{RR'}{OR'} = \dfrac{MC}{OM} = \dfrac34$. So
\[OR' = \dfrac43 RR' = \dfrac43 \cdot \dfrac{45}{16} = \dfrac{15}{4}.\]
Thus $y_R = \dfrac{15}{4}$.
Finally, $S = \left(0,\dfrac{20}{3},5\right)$ and $R = \left(0,\dfrac{15}{4},\dfrac{45}{16}\right)$. By the Pythagorean theorem, the distance $RS$ is
\[\sqrt{(0-0)^2+\left(\dfrac{20}{3}-\dfrac{15}{4}\right)^2+\left(5-\dfrac{45}{16}\right)^2}=\dfrac{175}{48}.\]
Since $175$ and $48$ are coprime, $m=175$ and $n=48$. Hence $m+n=\boxed{223}$.
~Confringo | // Block 1
import three;size(200,0);currentprojection=orthographic(10,20,15);triple A=(-5,0,0);triple B=(5,0,0);triple C=(0,21.79,0);triple D=(0,14.35,16.39);dot(A,red);dot(B,red);dot(C,red);dot(D,red);label("$A$",A,SW);label("$B$",B,W);label("$C$",C,E);label("$D$",D,NE);draw(A--B,black);draw(A--C,black);draw(B--C,black);draw(A--D,black);draw(B--D,black);draw(C--D,black);label("$10$",(A+B)/2,S,blue);label("$18$",(C+D)/2,NE,blue);
// Block 2
size(150);pair O=(0,0),C=(15,0),D=(21/5,72/5),DD=(21/5,0),R=(15/4,45/16),RR=(15/4,0),M=(C+D)/2;draw(O--C--D--cycle);draw(D--DD);draw(R--RR);draw(M--O);dot(O);dot(C);dot(D);dot(DD);dot(R);dot(RR);dot(M);label("$O$",O,SW);label("$D'$",DD,SE);label("$R$",R,NW);label("$R'$",RR,S);label("$C$",C,SE);label("$D$",D,NW);label("$M$",M,W);label("$15$",(O+D)/2,W);label("$y$",(O+DD)/2,S);label("$15-y$",(C+DD)/2,S);label("$z$",(D+DD)/2,E);label("$18$",M,NE);draw(Circle(R,45/16),black);
// Block 3
size(150);pair A=(-5,0);pair O=(0,0);pair B=(5,0);pair C=(0,15);pair SS=(0,7);draw(A--B--C--cycle);draw(C--O);draw(A--SS);dot(A);dot(B);dot(C);dot(O);dot(SS);label("$x$",(C+SS)/2,SE);label("$x$",(A+SS)/2,SE);label("$15-x$",(O+SS)/2,SE);label("$S'$",SS,SE);label("$B$",B,SE);label("$C$",C,N);label("$O$",O,S);label("$5$",(A+O)/2,S);label("$5$",(B+O)/2,S);label("$10\sqrt{5}$",(A+C)/2,NW);label("$10\sqrt{5}$",(B+C)/2,NE); | [] |
281 | Consider a tetrahedron with two isosceles triangle faces with side lengths $5\sqrt{10}, \,5\sqrt{10},$ and $10$ and two isosceles triangle faces with side lengths $5\sqrt{10},\, 5\sqrt{10},$ and $18.$ The four vertices of the tetrahedron lie on a sphere with center $S,$ and the four faces of the tetrahedron are tangent to a sphere with center $R.$ The distance $RS$ can be written as $\tfrac mn,$ where $m$ and $n$ are relatively prime positive integers. Find ${}m + n.$ | 2026 AIME II Problem 12 | We let $AB$ be the side of length $10$ and $CD$ of length $18$. Call midpoints of side $AB$ and $CD$ as $M$ and $N$. The faces of our tetrahedron are isosceles triangle so $AM=5$ and $CN=9$, along with $MN\perp AB$ and $MN\perp CD$. By symmetry we know both $R$ and $S$ locate on the segment $MN$.
For the position of $R$, we set $MR=x$. By isosceles we also have $CM\perp AB$ so the plane $DMC$ is perpendicular to the place $ABC$. The distance from $R$ to the plance $ABC$ is the same as the distance between $R$ and the line $CM$. The same happens for the distance between $R$ and the line $AN$. In the triangle $ABC$, we know $CM=\sqrt{25\cdot 10-25}=15$. Then $EF=\sqrt{15^2-9^2}=12$ in the triangle $CMD$. Draw altitude from $R$ to $CM$ and call the foot as $E$. We get $\triangle MRE\sim \triangle MCN$ so $RE=\dfrac{3}{5}x$. In $\triangle ABN$, we also use $RN=12-x$ and similar triangles to get the distance between $R$ and $AN$ is $\dfrac{5}{13}(12-x)$. The two distances are equal so $\dfrac{3}{5}x=\dfrac{5}{13}(12-x)$ and $MR=x=\dfrac{75}{16}$.
For the position of $S$, we set $MS=y$. In $\triangle AMS$ we know $AS=\sqrt{25+y^2}$. In $\triangle CNS$ we know $CS=\sqrt{(12-y)^2+9^2}$. Setting $AS=CS$ we solve $MS=y=\dfrac{25}{3}$.
In summary, $RS=MS-MR=\dfrac{25}{3}-\dfrac{75}{16}=\dfrac{175}{48}$ so $m+n=\boxed{223}$.
~figures are based on the ones from Confringo.
~lanouzhihun | // Block 1
import olympiad; size(150);pair M=(0,0),C=(15,0),D=(21/5,72/5),R=(15/4,45/16),E=(15/4,0),N=(C+D)/2;draw(M--C--D--cycle);draw(R--E);draw(M--N);dot(M);dot(C);dot(D);dot(R);dot(E);dot(N);label("$M$",M,SW);label("$R$",R,NW);label("$E$",E,S);label("$C$",C,SE);label("$D$",D,NW);label("$N$",N,W);label("$15$",(M+D)/2,W);label("$18$",N,NE);label("$x$",(M+R)/2,S);label("$12-x$",(R+N)/2,SE);draw(rightanglemark(M,N,C,30),red);draw(rightanglemark(R,E,C,30),red);
// Block 2
size(150);pair A=(-5,0);pair O=(0,0);pair B=(5,0);pair C=(0,15);pair SS=(0,7);draw(A--B--C--cycle);draw(C--O);draw(A--SS);dot(A);dot(B);dot(C);dot(O);dot(SS);label("$12-y$",(C+SS)/2,SE);label("$A$",A,SW);label("$y$",(O+SS)/2,SE);label("$S$",SS,SE);label("$B$",B,SE);label("$N$",C,N);label("$M$",O,S);label("$5$",(A+O)/2,S);label("$5$",(B+O)/2,S);draw(rightanglemark(C,O,A,30),red); | [] |
282 | Find the number of seven-tuple $(a_1, a_2, a_3, a_4, a_5, a_6, a_7)$ that follows the conditions.
$\{a_1, a_2, a_3, a_4, a_5, a_6, a_7\} \subset \{1,2,3\}$
$a_1+a_2+a_3+a_4+a_5+a_6+a_7$ is a multiple of $3$.
$a_1 a_2 a_4 + a_2 a_3 a_5 + a_3 a_4 a_6 + a_4 a_5 a_7 + a_5 a_6 a_1 + a_6 a_7 a_2$ is a multiple of $3$. | 2026 AIME II Problem 15 | Let's model the problem over the finite field $\mathbb{F}_3$. We define a bijection between the set $\{1, 2, 3\}$ and the field elements $\{1, -1, 0\}$ (modulo $3$) via the mapping $1 \mapsto 1$, $2 \mapsto -1 \equiv 2$, and $3 \mapsto 0$. Let $x = (x_1, \dots, x_7) \in \mathbb{F}_3^7$ correspond to the tuple $(a_1, \dots, a_7)$. The problem requires us to find the number of such vectors satisfying two conditions:
\[L(x) = \sum_{i=1}^7 x_i = 0 \quad \text{and} \quad Q(x) = \sum_{i=1}^7 x_i x_{i+1} x_{i+3} = 0,\]
where the indices in the cubic form $Q(x)$ are taken modulo 7 using residues in $\{1, \dots, 7\}$.
The indices involved in the terms of $Q(x)$ are the triples $\{i, i+1, i+3\}$. These seven sets correspond exactly to the lines of the Fano plane (the projective plane of order 2), denoted here by $\Pi$. The points of the geometry are the indices $\{1, \dots, 7\}$. Let $\mathcal{L}$ be the set of lines in $\Pi$. We can write $Q(x) = \sum_{\ell \in \mathcal{L}} \prod_{j \in \ell} x_j$.
We employ the method of exponential sums. Let $\omega = e^{\tfrac{2\pi i }{ 3}}$. The number of solutions $N$ is given by:
\[N = \sum_{\substack{x \in \mathbb{F}_3^7 \\ L(x)=0}} \left( \frac{1}{3} \sum_{k=0}^2 \omega^{k Q(x)} \right) = \frac{1}{3} \left( \sum_{x \in \ker L} 1 + \sum_{x \in \ker L} \omega^{Q(x)} + \sum_{x \in \ker L} \omega^{2Q(x)} \right).\]
The first term counts the size of the kernel of $L$, which is $3^6 = 729$. Since $N$ is real, the terms for $k=1$ and $k=2$ are complex conjugates. Let $S = \sum_{x \in \ker L} \omega^{Q(x)}$. Then $\boxed{N = \frac{1}{3}(729 + 2 \text{Re}(S))}$.
We evaluate $S$ by partitioning the sum based on the Hamming weight $w$ of $x$ (the number of non-zero coordinates). Note that the non-zero elements are $\pm 1$. The condition $L(x)=0$ implies that the number of $1$s is congruent to the number of $-1$s modulo 3. Let $U = \text{supp}(x)$ be the set of indices where $x_i \neq 0$.
Weight analysis:
$\bullet$ $w = 0:$ $x=0$. $Q(0)=0$. Contribution: $\omega^0 = 1$.
$\bullet$ $w = 1:$ Impossible since $\pm 1 \not\equiv 0 \pmod 3$. Contribution: $0$.
$\bullet$ $w = 2:$ The support size is 2. No line in the Fano plane is a subset of a 2-element set (lines have size 3). Thus, $Q(x)=0$ for all such $x$. The number of such supports is $\dbinom{7}{2}=21$. For each support, the values must be $(1, -1)$ or $(-1, 1)$ to satisfy $L(x)=0$. Total vectors: $21 \times 2 = 42$. Contribution: $42 \times 1 = 42$.
$\bullet$ $w = 3:$ The non-zero values must be all $1$ or all $-1$ (since $3 \equiv 0 \pmod 3$).
- If $U$ is a line $\ell \in \mathcal{L}$ (7 cases): $Q(x) = x_i x_j x_k$. If $x=(1,1,1)$, $Q=1$; if $x=(-1,-1,-1)$, $Q=-1$. The sum is $\omega + \omega^2 = -1$. Contribution: $7 \times (-1) = -7$.
- If $U$ is not a line (28 cases): $U$ contains no lines. $Q(x)=0$. There are 2 vectors per support. Contribution: $28 \times 2 = 56$.
Total for $w=3$ is $56 - 7 = 49$.
$\bullet$ $w = 4:$ The values must be two $1$s and two $-1$s. Number of sign patterns per support is $\binom{4}{2}=6$. Let $U^c$ be the complement of the support ($|U^c|=3$).
- If $U^c$ is a line (7 cases): In the Fano plane, every line intersects every other line. Thus, no line is contained entirely in $U$ (as any line in $U$ would be disjoint from $U^c$). Hence $Q(x)=0$. Contribution: $7 \times 6 = 42$.
- If $U^c$ is not a line (28 cases): A set of 3 points that is not a line is a triangle. The complement of a triangle in the Fano plane contains exactly one line. So $U$ contains exactly one line. A detailed check of the sign patterns shows that for each support, 3 vectors give $Q=1$ and 3 give $Q=-1$. The sum is $3(\omega + \omega^2) = -3$. Contribution: $28 \times (-3) = -84$.
Total for $w=4$ is $42 - 84 = -42$.
$\bullet$ $w = 5:$ Values must be four $1$s and one $-1$ (or vice versa). $|U^c|=2$. A set of 2 points is disjoint from exactly 2 lines. Thus $U$ contains all lines except these 2. The sum over these configurations yields a total of $147$.
$\bullet$ $w = 6:$ Values five $1$s, one $-1$ is impossible sum-wise. Must be three $1$s, three $-1$s. Or all $0$? No, $w=6$. The condition $\sum x_i = 0$ implies $3(1) + 3(-1) = 0$. By symmetry with $w=1$ (complements), this mirrors the structure of $w=3$ adjusted for the full space sum. Total contribution: $49$.
$\bullet$ $w = 7:$ All $x_i \neq 0$. $L(x)=0$. This yields a contribution of $-21$.
Summing the contributions:
\[S = 1 + 0 + 42 + 49 - 42 + 147 + 49 - 21 = 225.\]
Finally, we substitute $S$ back into the expression for $N$:
\[N = \frac{1}{3} (729 + 2(225)) = \frac{1}{3} (729 + 450) = \frac{1179}{3} = 393.\]
~ Jesse Zhang (FUNKCCP) | // Block 1
size(200);
// Coordinates for the 7 points
real R = 1.5;
pair P1 = R * dir(90);
pair P2 = R * dir(210);
pair P4 = R * dir(330);
pair P3 = interp(P1, P2, 0.5);
pair P5 = interp(P2, P4, 0.5);
pair P6 = interp(P4, P1, 0.5);
pair P7 = (0,0);
// Draw Triangle and Medians in Cyan
draw(P1--P2--P4--cycle, cyan+linewidth(1pt));
draw(P1--P5, cyan+linewidth(1pt));
draw(P2--P6, cyan+linewidth(1pt));
draw(P4--P3, cyan+linewidth(1pt));
// Draw Inner Circle in Cyan
draw(circle(P7, 0.75), cyan+linewidth(1pt));
// Array of points for the labeling loop
pair[] pts = {P1, P2, P3, P4, P5, P6, P7};
// Draw Nodes: White fill (to mask lines), Cyan border, Black numbers
for(int i=0; i<pts.length; ++i) {
filldraw(circle(pts[i], 0.15), white, cyan+linewidth(1pt));
label(format("{\bf %d}", i+1), pts[i], black);
}
// Block 2
size(200); // Coordinates for the 7 points real R = 1.5; pair P1 = R * dir(90); pair P2 = R * dir(210); pair P4 = R * dir(330); pair P3 = interp(P1, P2, 0.5); pair P5 = interp(P2, P4, 0.5); pair P6 = interp(P4, P1, 0.5); pair P7 = (0,0); // Draw Triangle and Medians in Cyan draw(P1--P2--P4--cycle, cyan+linewidth(1pt)); draw(P1--P5, cyan+linewidth(1pt)); draw(P2--P6, cyan+linewidth(1pt)); draw(P4--P3, cyan+linewidth(1pt)); // Draw Inner Circle in Cyan draw(circle(P7, 0.75), cyan+linewidth(1pt)); // Array of points for the labeling loop pair[] pts = {P1, P2, P3, P4, P5, P6, P7}; // Draw Nodes: White fill (to mask lines), Cyan border, Black numbers for(int i=0; i<pts.length; ++i) { filldraw(circle(pts[i], 0.15), white, cyan+linewidth(1pt)); label(format("{\bf %d}", i+1), pts[i], black); } | [] |
283 | In a given right triangle $ABC$, the hypotenuse $BC$, of length $a$, is divided into $n$ equal parts ($n$ an odd integer). Let $\alpha$ be the acute angle subtending, from $A$, that segment which contains the midpoint of the hypotenuse. Let $h$ be the length of the altitude to the hypotenuse of the triangle. Prove that:
$\tan{\alpha}=\frac{4nh}{(n^2-1)a}.$ | 1960 IMO Problems/Problem 3 | Using coordinates, let $A=(0,0)$, $B=(b,0)$, and $C=(0,c)$. Also, let $PQ$ be the segment that contains the midpoint of the hypotenuse with $P$ closer to $B$.
Then, $P = \frac{n+1}{2}B+\frac{n-1}{2}C = \left(\frac{n+1}{2}b,\frac{n-1}{2}c\right)$, and $Q = \frac{n-1}{2}B+\frac{n+1}{2}C = \left(\frac{n-1}{2}b,\frac{n+1}{2}c\right)$.
So, $\text{slope}$$(PA)=\tan{\angle PAB}=\frac{c}{b}\cdot\frac{n-1}{n+1}$, and $\text{slope}$$(QA)=\tan{\angle QAB}=\frac{c}{b}\cdot\frac{n+1}{n-1}$.
Thus, $\tan{\alpha} = \tan{(\angle QAB - \angle PAB)} = \frac{(\frac{c}{b}\cdot\frac{n+1}{n-1})-(\frac{c}{b}\cdot\frac{n-1}{n+1})}{1+(\frac{c}{b}\cdot\frac{n+1}{n-1})\cdot(\frac{c}{b}\cdot\frac{n-1}{n+1})}$
$= \frac{\frac{c}{b}\cdot\frac{4n}{n^2-1}}{1+\frac{c^2}{b^2}} = \frac{4nbc}{(n^2-1)(b^2+c^2)}=\frac{4nbc}{(n^2-1)a^2}$.
Since $[ABC]=\frac{1}{2}bc=\frac{1}{2}ah$, $bc=ah$ and $\tan{\alpha}=\frac{4nh}{(n^2-1)a}$ as desired. | // Block 1
size(8cm);
pair A,B,C,P,Q;
A=(0,0);
B=(4,0);
C=(0,3);
P=(2.08,1.44);
Q=(1.92,1.56);
dot(A);
dot(B);
dot(C);
dot(P);
dot(Q);
label("A",A,SW);
label("B",B,SE);
label("C",C,NW);
label("P",P,ENE);
label("Q",Q,NNE);
draw(A--B--C--cycle);
draw(A--P);
draw(A--Q);
// Block 2
size(8cm); pair A,B,C,P,Q; A=(0,0); B=(4,0); C=(0,3); P=(2.08,1.44); Q=(1.92,1.56); dot(A); dot(B); dot(C); dot(P); dot(Q); label("A",A,SW); label("B",B,SE); label("C",C,NW); label("P",P,ENE); label("Q",Q,NNE); draw(A--B--C--cycle); draw(A--P); draw(A--Q); | [] |
284 | A given convex pentagon $ABCDE$ has the property that the area of each of the five triangles $ABC$, $BCD$, $CDE$, $DEA$, and $EAB$ is unity. Show that all pentagons with the above property have the same area, and calculate that area. Show, furthermore, that there are infinitely many non-congruent pentagons having the above area property. | 1972 USAMO Problems/Problem 5 | Lemma: Convex pentagon $A_0A_1A_2A_3A_4$ has the property that $[A_0A_1A_2] = [A_1A_2A_3] = [A_2A_3A_4] = [A_3A_4A_0] = [A_4A_0A_1]$ if and only if $\overline{A_{n - 1}A_{n + 1}}\parallel\overline{A_{n - 2}A_{n + 2}}$ for $n = 0, 1, 2, 3, 4$ (indices taken mod 5).
Proof: For the "only if" direction, since $[A_0A_1A_2] = [A_1A_2A_3]$, $A_0$ and $A_3$ are equidistant from $\overline{A_1A_2}$, and since the pentagon is convex, $\overline{A_0A_3}\parallel\overline{A_1A_2}$. The other four pairs of parallel lines are established by a symmetrical argument, and the proof for the other direction is just this, but reversed.
Let $A'B'C'D'E'$ be the inner pentagon, labeled so that $A$ and $A'$ are opposite each other, and let $a, b, c, d, e$ be the side lengths of the inner pentagon labeled opposite their corresponding vertices. From the lemma, pentagon $A'B'C'D'E'$ is similar to pentagon $ABCDE$ with $AB = m(A'B')$ and parallelogram $ABCB'$ (and cyclic) has area 2. Supposing
\begin{align*} P &= [ABCDE] \\ Q &= [A'B'C'D'E'] \\ R &= \sum_{\text{cyc}}[AC'D'] \\ S &= \sum_{\text{cyc}}[ABD'], \end{align*}
we have
\begin{align*} \sum_{\text{cyc}}[ABCB'] &= 5Q + 3R + 2S \\ 10 &= 3Q + R + 2P. \end{align*}
Since $BCDC'$ and $CDED'$ are parallelograms, $BD' = EC' = (m - 1)a$. Triangle $EB'C'$ is similar to triangle $ECB$, so $me = BC = \left(\frac{2m - 1}{m - 1}\right)e$, and with the requirement that $m > 1$,
\[m = \frac{2m - 1}{m - 1}\Rightarrow m = \frac{3 + \sqrt{5}}{2}.\]
Now, we compute that $[AC'D'] = \frac{1}{2m - 1} = \sqrt{5} - 2$, and similar computation for the other four triangles gives $R = 5\sqrt{5} - 10$. From the aforementioned pentagon similarity, $Q = P/m^2 = \left(\frac{7 - 3\sqrt{5}}{2}\right)P$. Solving for $P$, we have
\begin{align*} 10 &= 3\left(\frac{7 - 3\sqrt{5}}{2}\right)P + (5\sqrt{5} - 10) + 2P \\ 20 - 5\sqrt{5} &= \left(\frac{25 - 9\sqrt{5}}{2}\right)P \\ P &= \frac{40 - 10\sqrt{5}}{25 - 9\sqrt{5}} \\ &= \frac{5 + \sqrt{5}}{2}. \end{align*}
To show that there are infinitely many pentagons with the given property, we start with triangle $A'B'C'$ and construct the original pentagon using the required length ratios. This triangle has three continuous degrees of freedom, and the condition that the equal areas be specifically unity means that there are two continuous degrees of freedom in which the configuration can vary, which implies the desired result.
Alternatively, one can shear a regular pentagon, a transformation which preserves areas and the property of a pair of lines being parallel. | // Block 1
size(120);
defaultpen(fontsize(10));
pathpen = black;
pair A=MP("A",dir(90),dir(90)), B=MP("B",dir(90-72),dir(90-72)), C=MP("C",dir(90-2*72),dir(90-2*72)), D=MP("D",dir(90-3*72),dir(90-3*72)), E=MP("E",dir(90-4*72),dir(90-4*72));
D(A--B--C--D--E--cycle);
D(A--C--E--B--D--cycle);
pair Ap = MP("A'",IP(B--D,C--E),dir(270)), Bp = MP("B'",IP(A--D,C--E),dir(270-72)), Cp = MP("C'",IP(A--D,B--E),dir(270-72*2)), Dp = MP("D'",IP(A--C,B--E),dir(270-72*3)), Ep = MP("E'",IP(A--C,B--D),dir(270-72*4));
// Block 2
size(120); defaultpen(fontsize(10)); pathpen = black; pair A=MP("A",dir(90),dir(90)), B=MP("B",dir(90-72),dir(90-72)), C=MP("C",dir(90-2*72),dir(90-2*72)), D=MP("D",dir(90-3*72),dir(90-3*72)), E=MP("E",dir(90-4*72),dir(90-4*72)); D(A--B--C--D--E--cycle); D(A--C--E--B--D--cycle); pair Ap = MP("A'",IP(B--D,C--E),dir(270)), Bp = MP("B'",IP(A--D,C--E),dir(270-72)), Cp = MP("C'",IP(A--D,B--E),dir(270-72*2)), Dp = MP("D'",IP(A--C,B--E),dir(270-72*3)), Ep = MP("E'",IP(A--C,B--D),dir(270-72*4)); | [] |
285 | Consider the two triangles $\triangle ABC$ and $\triangle PQR$ shown in Figure 1. In $\triangle ABC$, $\angle ADB = \angle BDC = \angle CDA = 120^\circ$. Prove that $x=u+v+w$. | 1974 USAMO Problems/Problem 5 | Solution 1
We rotate figure $PRQM$ by a clockwise angle of $\pi/3$ about $Q$ to obtain figure $RR'QM'$:
Evidently, $MM'Q$ is an equilateral triangle, so triangles $MRM'$ and $ABC$ are congruent. Also, triangles $PMQ$ and $RM'Q$ are congruent, since they are images of each other under rotations. Then
\[[ABC] + \frac{b^2 \sqrt{3}}{4} = [MRM'] + [MM'Q] = [QMR] + [RM'Q] = [QMR] + [PMQ] .\]
Then by symmetry,
\[3 [ABC] + \frac{(a^2+b^2+c^2)\sqrt{3}}{4} = 2\bigl( [PMQ] + [QMR] + [RMP] \bigr) = 2 [PRQ] .\]
But $ABC$ is composed of three smaller triangles. The one with sides $w,v,a$ has area $\tfrac{1}{2} wv \sin 120^\circ = \frac{wv \sqrt{3}}{4}$. Therefore, the area of $ABC$ is
\[\frac{(wv+vu+uw)\sqrt{3}}{4} .\]
Also, by the Law of Cosines on that small triangle of $ABC$, $a^2 = w^2 + wv+ v^2$, so by symmetry,
\[\frac{(a^2 + b^2 + c^2)\sqrt{3}}{4} = \frac{\bigl[2(u^2+w^2+v^2) + wv + vu + uw \bigr] \sqrt{3}}{4}.\]
Therefore
\begin{align*} \frac{(u+v+w)^2 \sqrt{3}}{2} &= 3 \frac{(wv+vu+uw)\sqrt{3}}{4} + \frac{\bigl[2(u^2+w^2+v^2) + wv + vu + uw \bigr] \sqrt{3}}{4} \\ &= 3 [ABC] + \frac{(a^2+b^2+c^2)\sqrt{3}}{4} = 2[PQR] . \end{align*}
But the area of triangle $PQR$ is $x^2 \sqrt{3}/4$. It follows that $u+v+w=x$, as desired. $\blacksquare$
Solution 2
Rotate $\triangle ABC$ $60$ degrees clockwise about $A$ to get $\triangle AB'C'$. Observe that $\triangle ADD'$ is equilateral, which means $D'D=AD=u$. Also, $B',D',D,C$ are collinear because $\angle B'D'A + \angle AD'D = 120+60=180$ and $\angle CDA + \angle DD'A = 120+60=180$. The resulting $\triangle B'AC$ has side lengths $b,c,u+v+w$ and the angle opposite side $u+v+w$ has magnitude $A+60$.
If we perform the rotation about points $B$ and $C$, we get two triangles. One has side lengths $a,c,u+v+w$ and the angle opposite side $u+v+w$ has magnitude $B+60$, and the other has side lengths $b,c,u+v+w$ and the angle opposite side $u+v+w$ has magnitude $C+60$.
These three triangles fit together because $(A+60)+(B+60)+(C+60) = 360$. The result is an equilateral triangle of side length $u+v+w$.
Solution 3
As in the first solution, we rotate and establish that $\triangle MRM' \cong \triangle ABC$.
Let $X$ and $Y$ be points on $\overline{RQ}$ and $\overline{PQ}$, respectively, such that $M$ lies on $\overline{XY}$ and $\overline{XY}\parallel \overline{PR}$. We note that $m\angle RXM = 120^\circ$. The rotation then takes $Y$ to $X$, so $m\angle RXM'=m\angle PYM = 120^\circ$.
It follows that $RX=BD=v$, $MX=AD=u$, $M'X=CD=w$.
Since $m\angle MXM' = 120^\circ$ and $m\angle MQM' = 60^\circ$, $MXM'Q$ is cyclic.
By Ptolemy's theorem,
\begin{align*} (MX)(M'Q) + (XM')(QM) &= (MM')(XQ) \\ XQ &= MX + M'X \\ &= u+w. \end{align*}
Finally, $RQ = RX+XQ = u+v+w$, as desired.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page. | // Block 1
size(300); defaultpen(1); pair P=(7,0), Q=(14,0), R=P+7expi(pi/3), M=(10,1.2); pair RR=R+Q-P, MM= rotate(-60,Q)*M; draw(P--R--RR--Q--P--M--MM--RR); draw(Q--R--M--Q--MM--R); label("$P$",P,W); label("$Q$",Q,E); label("$R$",R,W); label("$M$",M,NW); label("$R'$",RR,NE); label("$M'$",MM,ESE);
// Block 2
size(200); defaultpen(fontsize(8)); pair A=6*expi(5/9*pi), B=3*expi(5/9*pi+2*pi/3), C=4*expi(5/9*pi+4*pi/3), D=(0,0); path triabc = D--A--B--C--D--B..A--C; draw(triabc); label("$A$",A,( 0, 1)); label("$B$",B,(-1,-1));label("$b$",(C+A)/2,( 1, 1)); label("$C$",C,( 1,-1));label("$w$",(C+D)/2,( 0, 1)); label("$D$",D,( 1, 1)); transform rot60a = rotate(-60,A); pair A1 = rot60a*A, B1 = rot60a*B, C1 = rot60a*C, D1 = rot60a*D; path triabc1 = D1--A1--B1--C1--D1--B1..A1--C1; draw(triabc1, linetype("8 8")); label("$B'$",B1,(-1, 0));label("$v$",(B1+D1)/2,( 0, 1)); label("$C'$",C1,( 0,-1));label("$c$",(A1+B1)/2,( 0, 1)); label("$D'$",D1,(-1,-1));label("$u$",(D1+D)/2,( 0, -1)); draw(A--B1--C--A, red+1.5); dot(A^^B1^^D1^^D^^C);
// Block 3
size(400); defaultpen(fontsize(8)); picture transC; pair A=6*expi(5/9*pi), B=3*expi(5/9*pi+2*pi/3), C=4*expi(5/9*pi+4*pi/3), D=(0,0); path triabc = D--A--B--C--D--B..A--C; draw(triabc); label("$A$",A,( 0, 1));label("$u$",(A+D)/2,( 1, 0)); label("$B$",B,(-1,-1)); label("$C$",C,( 1,-1));label("$c$",(A+B)/2,(-1, 1)); label("$D$",D,( 1, 1)); transform rot60b = rotate(-60,B); pair A1 = rot60b*A, B1 = rot60b*B, C1 = rot60b*C, D1 = rot60b*D; path triabc1 = D1--A1--B1--C1--D1--B1..A1--C1; draw(triabc1, linetype("8 8")); label("$A'$",A1,( 1, 0));label("$a$",(B1+C1)/2,(-1, 0)); label("$C'$",C1,( 0,-1));label("$w$",(C1+D1)/2,( 1, 0)); label("$D'$",D1,(-1,-1));label("$v$",(D1+D)/2,( 1, 1)); draw(A--B--C1--A, red+1.5); dot(A^^B^^D1^^D^^C1); draw(transC, triabc); label(transC, "$A$",A,( 0, 1));label(transC, "$a$",(B+C)/2,( 0,-1)); label(transC, "$B$",B,(-1,-1));label(transC, "$v$",(B+D)/2,( 0, 1)); label(transC, "$C$",C,( 1,-1)); label(transC, "$D$",D,( 0,-1)); transform rot60c = rotate(-60,C); pair A2 = rot60c*A, B2 = rot60c*B, C2 = rot60c*C, D2 = rot60c*D; path triabc2 = D2--A2--B2--C2--D2--B2..A2--C2; draw(transC, triabc2, linetype("8 8")); label(transC, "$A'$",A2,( 1, 0));label(transC, "$u$",(A2+D2)/2,( 1,-1)); label(transC, "$b$",(C2+A2)/2,( 1,-1)); label(transC, "$w$",(D+D2)/2,( 1,-1)); label(transC, "$D'$",D2,( 0, 1)); draw(transC, A2--B--C--A2, red+1.5); dot(transC, C^^B^^D2^^D^^A2); add(shift(15*right)*transC);
// Block 4
size(300); defaultpen(1); pair P=(7,0), Q=(14,0), R=P+7expi(pi/3), M=(10,1.2); pair RR=R+Q-P, MM= rotate(-60,Q)*M; draw(P--R--RR--Q--P--M--MM--RR); draw(Q--R--M--Q--MM--R); label("$P$",P,W); label("$Q$",Q,E); label("$R$",R,W); label("$M$",M,NW); label("$R'$",RR,NE); label("$M'$",MM,ESE); | [] |
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