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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512 | Regular octagon $A_1A_2A_3A_4A_5A_6A_7A_8$ is inscribed in a circle of area $1.$ Point $P$ lies inside the circle so that the region bounded by $\overline{PA_1},\overline{PA_2},$ and the minor arc $\widehat{A_1A_2}$ of the circle has area $\tfrac{1}{7},$ while the region bounded by $\overline{PA_3},\overline{PA_4},$ and the minor arc $\widehat{A_3A_4}$ of the circle has area $\tfrac{1}{9}.$ There is a positive integer $n$ such that the area of the region bounded by $\overline{PA_6},\overline{PA_7},$ and the minor arc $\widehat{A_6A_7}$ of the circle is equal to $\tfrac{1}{8}-\tfrac{\sqrt2}{n}.$ Find $n.$ | 2019 AIME II Problems/Problem 13 | Instead of considering the actual values of the areas, consider only the changes in the areas that result from moving point $P$ from the center of the circle. We will proceed by coordinates. Set the origin at the center of the circle and refer to the following diagram, where the octagon is oriented so as $A_1A_2$ is horizontal (and therefore $A_3A_4$ is vertical). Note that the area bounded by $\overline{A_iA_j}$ and the arc $\widehat{A_iA_j}$ is fixed, so we only need to consider the relevant triangles.
Define one arbitrary unit as the distance that you need to move $P$ from $A_1A_2$ to change the area of $\triangle PA_1A_2$ by $1$. We can see that $P$ was moved down by $\tfrac{1}{7}-\tfrac{1}{8}=\tfrac{1}{56}$ units to make the area defined by $P$, $A_1$, and $A_2$ $\tfrac{1}{7}$. Similarly, $P$ was moved right by $\tfrac{1}{8}-\tfrac{1}{9}=\tfrac{1}{72}$ to make the area defined by $P$, $A_3$, and $A_4$ $\tfrac{1}{9}$. This means that $P$ has coordinates $(\tfrac{1}{72},-\tfrac{1}{56})$.
Now, we need to consider how this displacement in $P$ affected the area defined by $P$, $A_6$, and $A_7$. This is equivalent to finding the shortest distance between $P$ and the blue line in the diagram (as $K=\tfrac{1}{2}bh$ and the blue line represents $h$ while $b$ is fixed). Using an isosceles right triangle, one can find the that shortest distance between $P$ and this line is $\tfrac{\sqrt{2}}{2}(\tfrac{1}{56}-\tfrac{1}{72})=\tfrac{\sqrt{2}}{504}$.
Remembering the definition of our unit, this yields a final area of
\[\frac{1}{8}-\frac{\sqrt{2}}{\boxed{504}}.\]
-Archeon | // Block 1
size(7cm);
draw(Circle((0,0),1));
pair P = (0.1,-0.15);
filldraw(P--dir(112.5)--dir(112.5-45)--cycle,yellow,red);
filldraw(P--dir(112.5-90)--dir(112.5-135)--cycle,yellow,red);
filldraw(P--dir(112.5-225)--dir(112.5-270)--cycle,green,red);
dot(P);
for(int i=0; i<8; ++i)
{
draw(dir(22.5+45i)--dir(67.5+45i));
draw((0,0)--dir(22.5+45i),gray+dashed);
}
draw(dir(135)--dir(-45),blue+linewidth(1));
label("$P$", P, dir(-75));
label("$A_1$", dir(112.5), dir(112.5));
label("$A_2$", dir(112.5-45), dir(112.5-45));
label("$A_3$", dir(112.5-90), dir(112.5-90));
label("$A_4$", dir(112.5-135), dir(112.5-135));
label("$A_5$", dir(112.5-180), dir(112.5-180));
label("$A_6$", dir(112.5-225), dir(112.5-225));
label("$A_7$", dir(112.5-270), dir(112.5-270));
label("$A_8$", dir(112.5-315), dir(112.5-315));
dot(dir(112.5)^^dir(112.5-45)^^dir(112.5-90)^^dir(112.5-135)^^dir(112.5-180)^^dir(112.5-225)^^dir(112.5-270)^^dir(112.5-315));
// Block 2
size(7cm); draw(Circle((0,0),1)); pair P = (0.1,-0.15); filldraw(P--dir(112.5)--dir(112.5-45)--cycle,yellow,red); filldraw(P--dir(112.5-90)--dir(112.5-135)--cycle,yellow,red); filldraw(P--dir(112.5-225)--dir(112.5-270)--cycle,green,red); dot(P); for(int i=0; i<8; ++i) { draw(dir(22.5+45i)--dir(67.5+45i)); draw((0,0)--dir(22.5+45i),gray+dashed); } draw(dir(135)--dir(-45),blue+linewidth(1)); label("$P$", P, dir(-75)); label("$A_1$", dir(112.5), dir(112.5)); label("$A_2$", dir(112.5-45), dir(112.5-45)); label("$A_3$", dir(112.5-90), dir(112.5-90)); label("$A_4$", dir(112.5-135), dir(112.5-135)); label("$A_5$", dir(112.5-180), dir(112.5-180)); label("$A_6$", dir(112.5-225), dir(112.5-225)); label("$A_7$", dir(112.5-270), dir(112.5-270)); label("$A_8$", dir(112.5-315), dir(112.5-315)); dot(dir(112.5)^^dir(112.5-45)^^dir(112.5-90)^^dir(112.5-135)^^dir(112.5-180)^^dir(112.5-225)^^dir(112.5-270)^^dir(112.5-315)); | [] |
513 | Triangle $ABC$ has side lengths $AB=120,BC=220$, and $AC=180$. Lines $\ell_A,\ell_B$, and $\ell_C$ are drawn parallel to $\overline{BC},\overline{AC}$, and $\overline{AB}$, respectively, such that the intersections of $\ell_A,\ell_B$, and $\ell_C$ with the interior of $\triangle ABC$ are segments of lengths $55,45$, and $15$, respectively. Find the perimeter of the triangle whose sides lie on lines $\ell_A,\ell_B$, and $\ell_C$. | 2019 AIME II Problems/Problem 7 | Let's squish a triangle with side lengths 15, 22.5, and 27.5 into a equilateral triangle with side length 1. Then, the original triangle gets turned into a equilateral triangle with side length 8. Since 15 is one eighth of 120, it has a length of one. Since 45 and 55 are one fourth of 180 and 220 respectively, they are both two long. We extend the three segments to form a big equilateral triangle shown in black. Notice it has a side length of 11. Now that our task is done, let's undo the distortion. We get 11(15+22.5+27.5)=11(65)=715.
~ Afly (talk) | // Block 1
pair A,B,C; B = (0,0); C = (1,0); A = intersectionpoints(circle(B,3/2),circle(C,11/6))[0]; draw(A--B--C--cycle); draw((3/2,3/4)--(5/2,3/4)); draw((3/2,1/4)--(5/2,1/4)); draw((9/4,1)--(11/4,1/2)--(9/4,0)); draw(shift(dir(0)*13/4)*shift(dir(30))*polygon(3));
// Block 2
for (int i=0; i<8; ++i) { for (int j=0; j<i+1; ++j) { draw(shift(dir(30))*shift(dir(0)*i*sqrt(3))*shift(dir(120)*j*sqrt(3))*polygon(3)); } } pair A = origin+2*sqrt(3)*dir(0)+7*sqrt(3)*dir(120); pair B = origin+13*sqrt(3)*dir(0)+7*sqrt(3)*dir(120); pair C = origin+2*sqrt(3)*dir(0)-4*sqrt(3)*dir(120); pair D = origin + 2*sqrt(3)*dir(0); pair E = origin + 2*sqrt(3)*dir(60); pair F = origin + 7*sqrt(3)*dir(60); pair G = origin + 7*sqrt(3)*dir(60) + 1*sqrt(3)*dir(0); pair H = origin + 6*sqrt(3)*dir(0) + 2*sqrt(3)*dir(60); pair I = origin + 6*sqrt(3)*dir(0); draw(A--B--C--cycle,linewidth(3)); draw(D--E,linewidth(3)+rgb(3/4,1/4,1/4)); draw(F--G,linewidth(3)+rgb(1/4,3/4,1/4)); draw(H--I,linewidth(3)+rgb(1/4,1/4,3/4)); | [] |
514 | 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 Problems/Problem 13 | Let $M_A$, $M_B$, $M_C$ be the midpoints of arcs $BC$, $CA$, $AB$. By Fact 5, we know that $M_AB = M_AC = M_AI$, and so by Ptolmey's theorem, we deduce that \[AB\cdot M_AC + AC\cdot M_AB = BC\cdot M_AA \implies M_AA = 2M_AI.\]
In particular, we have $AI = IM_A$.
Now the key claim is that:
Claim: $\triangle DEF$ and $\triangle M_AM_BM_C$ are homothetic at $I$ with ratio $2$.
Proof. First, we show that $D$ is the midpoint of $M_AI$. Indeed, we have
\[\frac{ID}{DM_A} = \frac{BI}{BM_A}\cdot \frac{\sin\angle IBC}{\sin \angle CBM_A} = \frac{BI}{AI}\cdot\frac{\sin \angle B/2}{\sin \angle A/2} = 1\]
by Ratio lemma and Law of Sines.
Now observe that:
$\overline{M_BM_C}$ is the perpendicular bisector of $\overline{AI}$,
$\overline{EF}$ is the perpendicular bisector of $\overline{AD}$, and
$ID = AI/2$.
Combining these facts gives that $\overline{EF}$ is a midline in $\triangle IM_BM_C$, which proves the claim. $\blacksquare$
To finish, we compute $[M_AM_BM_C]$, noting that $[AEF] = [DEF] = \tfrac{1}{4}[M_AM_BM_C]$.
By Heron's, we can calculate the circumradius $R = 8/\sqrt{7}$, and by Law of Cosines, we get
\begin{align*}\cos A &= \frac{9}{16}\implies \cos A/2 = \frac{5}{\sqrt{32}} \\ \cos B &= \frac{1}{8} \implies \cos B/2 = \frac{3}{4} \\ \cos C &= \frac{3}{4} \implies \cos C/2 = \sqrt{\frac{7}{8}}.\end{align*}
Then using $[XYZ] = 2R^2\sin X\sin Y\sin Z$, we can compute
\[[M_AM_BM_C] = 2\cdot \frac{64}{7}\cdot \frac{5}{\sqrt{32}}\cdot \frac{3}{4}\cdot \frac{\sqrt{7}}{\sqrt{8}} = \frac{30\sqrt{7}}{7}.\]
Thus $[AEF] = 15\sqrt{7}/14$, which gives a final answer of $\boxed{036}$.
~pinetree1 | // Block 1
defaultpen(fontsize(10pt));
size(200);
pair A, B, C, D, E, F, I, P, MA, MB, MC;
B = (0,0);
C = (5,0);
A = IP(Circle(B, 4), Circle(C, 6), 0);
I = incenter(A, B, C);
D = extension(A, I, B, C);
P = midpoint(A--D);
E = extension(P, rotate(90, P)*A, B, I);
F = extension(P, rotate(90, P)*A, C, I);
MA = IP(Line(A, I, 20), circumcircle(A, B, C), 1);
MB = IP(Line(B, I, 20), circumcircle(A, B, C), 1);
MC = IP(Line(C, I, 20), circumcircle(A, B, C), 1);
draw(A--B--C--cycle, orange);
draw(circumcircle(A, B, C), red);
draw(A--E--F--cycle, lightblue);
draw(E--D--F, lightblue);
draw(A--MA^^B--MB^^C--MC, heavygreen);
draw(MA--MB--MC--cycle, magenta);
dot("$A$", A, dir(120));
dot("$B$", B, dir(220));
dot("$C$", C, dir(320));
dot("$D$", D, dir(230));
dot("$E$", E, dir(330));
dot("$F$", F, dir(250));
dot("$I$", I, dir(80));
dot("$M_A$", MA, dir(270));
dot("$M_B$", MB, dir(60));
dot("$M_C$", MC, dir(150));
// Block 2
defaultpen(fontsize(10pt)); size(200); pair A, B, C, D, E, F, I, P, MA, MB, MC; B = (0,0); C = (5,0); A = IP(Circle(B, 4), Circle(C, 6), 0); I = incenter(A, B, C); D = extension(A, I, B, C); P = midpoint(A--D); E = extension(P, rotate(90, P)*A, B, I); F = extension(P, rotate(90, P)*A, C, I); MA = IP(Line(A, I, 20), circumcircle(A, B, C), 1); MB = IP(Line(B, I, 20), circumcircle(A, B, C), 1); MC = IP(Line(C, I, 20), circumcircle(A, B, C), 1); draw(A--B--C--cycle, orange); draw(circumcircle(A, B, C), red); draw(A--E--F--cycle, lightblue); draw(E--D--F, lightblue); draw(A--MA^^B--MB^^C--MC, heavygreen); draw(MA--MB--MC--cycle, magenta); dot("$A$", A, dir(120)); dot("$B$", B, dir(220)); dot("$C$", C, dir(320)); dot("$D$", D, dir(230)); dot("$E$", E, dir(330)); dot("$F$", F, dir(250)); dot("$I$", I, dir(80)); dot("$M_A$", MA, dir(270)); dot("$M_B$", MB, dir(60)); dot("$M_C$", MC, dir(150)); | [] |
514 | 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 Problems/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); | [] |
514 | 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 Problems/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); | [] |
514 | 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 Problems/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); | [] |
514 | 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 Problems/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)); | [] |
514 | 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 Problems/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); | [] |
515 | 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 Problems/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); | [] |
515 | 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 Problems/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); | [] |
515 | 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 Problems/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)); | [] |
516 | 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 Problems/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]); } | [] |
517 | 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 Problems/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); | [] |
518 | 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 Problems/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)); | [] |
519 | 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 Problems/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 */ | [] |
520 | 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 Problems/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); | [] |
520 | 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 Problems/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); | [] |
520 | 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 Problems/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 */ | [] |
521 | In square $ABCD$, points $E$ and $H$ lie on $\overline{AB}$ and $\overline{DA}$, respectively, so that $AE=AH.$ Points $F$ and $G$ lie on $\overline{BC}$ and $\overline{CD}$, respectively, and points $I$ and $J$ lie on $\overline{EH}$ so that $\overline{FI} \perp \overline{EH}$ and $\overline{GJ} \perp \overline{EH}$. See the figure below. Triangle $AEH$, quadrilateral $BFIE$, quadrilateral $DHJG$, and pentagon $FCGJI$ each has area $1.$ What is $FI^2$?
$\textbf{(A) } \frac{7}{3} \qquad \textbf{(B) } 8-4\sqrt2 \qquad \textbf{(C) } 1+\sqrt2 \qquad \textbf{(D) } \frac{7}{4}\sqrt2 \qquad \textbf{(E) } 2\sqrt2$ | 2020 AMC 10B Problems/Problem 21 | Easily, we can find that: quadrilateral $BFIE$ and $DHJG$ are congruent with each other, so we can move $DHJG$ to the shaded area ($F$ and $G$, $B$ and $D$ overlapping) to form a square $FIKJ'$ ($DG$ = $FB$, $CG$ = $FC$, ${\angle} CGF$ = ${\angle}CFG$ = $45^{\circ}$ so ${\angle} IFJ'= 90^{\circ}$). Then we can solve $AH$ = $AE$ = $\sqrt{2}$, $EB$ = $2-\sqrt{2}$, $EK$ = $2\sqrt{2}-2$.
$FI^2=\text{area of} \: BFIE+\text{area of} \:FJ'H'B+\text{area of} \:EH'K \\= 1 + 1 + \frac{1}{2}(2\sqrt{2}-2)^2=8-4\sqrt{2}\rightarrow \boxed{\textbf{(B)}\ 8-4\sqrt{2}}$
--Ryan Zhang @BRS
--Minor edit by WhySean38 | // Block 1
real x=2sqrt(2);
real y=2sqrt(16-8sqrt(2))-4+2sqrt(2);
real z=2sqrt(8-4sqrt(2));
real k= 8-2sqrt(2);
real l= 2sqrt(2)-4;
pair A, B, C, D, E, F, G, H, I, J, L, M, K;
A = (0,0);
B = (4,0);
C = (4,4);
D = (0,4);
E = (x,0);
F = (4,y);
G = (y,4);
H = (0,x);
I = F + z * dir(225);
L = (k,0);
M = F + z * dir(315);
K = (4,l);
draw(A--B--C--D--A);
draw(H--E);
draw(J--G^^F--I);
draw(F--M);
draw(M--L);
draw(E--K,dashed+linewidth(.5));
draw(K--L,dashed+linewidth(.5));
draw(B--L);
draw(rightanglemark(G, J, I), linewidth(.5));
draw(rightanglemark(F, I, E), linewidth(.5));
draw(rightanglemark(F, M, L), linewidth(.5));
fill((4,0)--(k,0)--M--(4,y)--cycle, gray);
dot("$A$", A, S);
dot("$C$", C, dir(90));
dot("$D$", D, dir(90));
dot("$E$", E, S);
dot("$G$", G, N);
dot("$H$", H, W);
dot("$I$", I, SW);
dot("$J$", J, SW);
dot("$K$", K, S);
dot("$F(G)$", F, E);
dot("$H'$", L, S);
dot("$B(D)$", B, S);
// Block 2
real x=2sqrt(2); real y=2sqrt(16-8sqrt(2))-4+2sqrt(2); real z=2sqrt(8-4sqrt(2)); real k= 8-2sqrt(2); real l= 2sqrt(2)-4; pair A, B, C, D, E, F, G, H, I, J, L, M, K; A = (0,0); B = (4,0); C = (4,4); D = (0,4); E = (x,0); F = (4,y); G = (y,4); H = (0,x); I = F + z * dir(225); L = (k,0); M = F + z * dir(315); K = (4,l); draw(A--B--C--D--A); draw(H--E); draw(J--G^^F--I); draw(F--M); draw(M--L); draw(E--K,dashed+linewidth(.5)); draw(K--L,dashed+linewidth(.5)); draw(B--L); draw(rightanglemark(G, J, I), linewidth(.5)); draw(rightanglemark(F, I, E), linewidth(.5)); draw(rightanglemark(F, M, L), linewidth(.5)); fill((4,0)--(k,0)--M--(4,y)--cycle, gray); dot("$A$", A, S); dot("$C$", C, dir(90)); dot("$D$", D, dir(90)); dot("$E$", E, S); dot("$G$", G, N); dot("$H$", H, W); dot("$I$", I, SW); dot("$J$", J, SW); dot("$K$", K, S); dot("$F(G)$", F, E); dot("$H'$", L, S); dot("$B(D)$", B, S); | [] |
521 | In square $ABCD$, points $E$ and $H$ lie on $\overline{AB}$ and $\overline{DA}$, respectively, so that $AE=AH.$ Points $F$ and $G$ lie on $\overline{BC}$ and $\overline{CD}$, respectively, and points $I$ and $J$ lie on $\overline{EH}$ so that $\overline{FI} \perp \overline{EH}$ and $\overline{GJ} \perp \overline{EH}$. See the figure below. Triangle $AEH$, quadrilateral $BFIE$, quadrilateral $DHJG$, and pentagon $FCGJI$ each has area $1.$ What is $FI^2$?
$\textbf{(A) } \frac{7}{3} \qquad \textbf{(B) } 8-4\sqrt2 \qquad \textbf{(C) } 1+\sqrt2 \qquad \textbf{(D) } \frac{7}{4}\sqrt2 \qquad \textbf{(E) } 2\sqrt2$ | 2020 AMC 10B Problems/Problem 21 | $[ABCD] = 4$, $AB = 2$, $[AHE] = 1$, $AH = AE = \sqrt{2}$, $DH = 2 - \sqrt{2}$, $JL = HK = \sqrt{2} \cdot DH = 2 \sqrt{2} - 2$
Because $ABCD$ is a square and $AH = AE$, $AC$ is the line of symmetry of pentagon $CDHEB$. Because $[DHJG] = [BFIE]$, $DHJG$ is the reflection of $BFIE$ about line $AC$
Let $FI = GJ = x$, $KL = LG = GJ - LJ = x - 2 \sqrt{2} + 2$
$[DHK] = \frac{(2 - \sqrt{2})^2}{2} = 3 - 2 \sqrt {2}$
$[GKL] = \frac{(x - 2 \sqrt{2} + 2)^2}{2} = \frac{x^2}{2} + 2x - 2x \sqrt{2} - 4 \sqrt{2} + 6$
$[HKJL] = (x - 2 \sqrt{2} + 2) \cdot (2 \sqrt{2} - 2) = 2x \sqrt{2} - 2x + 8 \sqrt{2} -12$
\[[DHK] + [GKL] + [HKLJ] = [DHJG]\]
\[3 - 2 \sqrt {2} + \frac{x^2}{2} + 2x - 2x \sqrt{2} - 4 \sqrt{2} + 6 + 2x \sqrt{2} - 2x + 8 \sqrt{2} -12= 1\]
\[\frac{x^2}{2} + 2 \sqrt{2} - 4 = 0\]
\[x^2 = 8 - 4 \sqrt{2}\]
\[FI^2 = \boxed{\textbf{(B)}\ 8-4\sqrt{2}}\]
~isabelchen | // Block 1
real x=2sqrt(2);
real y=2sqrt(16-8sqrt(2))-4+2sqrt(2);
real z=2sqrt(8-4sqrt(2));
pair A, B, C, D, E, F, G, H, I, J, K, L;
A = (0,0);
B = (4,0);
C = (4,4);
D = (0,4);
E = (x,0);
F = (4,y);
G = (y,4);
H = (0,x);
I = F + z * dir(225);
J = G + z * dir(225);
K = (4-x,4);
L = J + 1.68 * dir(45);
draw(A--B--C--D--A);
draw(H--E);
draw(J--G^^F--I);
draw(H--K,dashed+linewidth(.5));
draw(L--K,dashed+linewidth(.5));
draw(rightanglemark(G, J, I), linewidth(.5));
draw(rightanglemark(F, I, E), linewidth(.5));
draw(rightanglemark(H, K, L), linewidth(.5));
draw(rightanglemark(K, L, G), linewidth(.5));
dot("$A$", A, S);
dot("$B$", B, S);
dot("$C$", C, dir(90));
dot("$D$", D, dir(90));
dot("$E$", E, S);
dot("$F$", F, dir(0));
dot("$G$", G, N);
dot("$H$", H, W);
dot("$I$", I, SW);
dot("$J$", J, SW);
dot("$K$", K, N);
dot("$L$", L, S);
// Block 2
real x=2sqrt(2); real y=2sqrt(16-8sqrt(2))-4+2sqrt(2); real z=2sqrt(8-4sqrt(2)); pair A, B, C, D, E, F, G, H, I, J, K, L; A = (0,0); B = (4,0); C = (4,4); D = (0,4); E = (x,0); F = (4,y); G = (y,4); H = (0,x); I = F + z * dir(225); J = G + z * dir(225); K = (4-x,4); L = J + 1.68 * dir(45); draw(A--B--C--D--A); draw(H--E); draw(J--G^^F--I); draw(H--K,dashed+linewidth(.5)); draw(L--K,dashed+linewidth(.5)); draw(rightanglemark(G, J, I), linewidth(.5)); draw(rightanglemark(F, I, E), linewidth(.5)); draw(rightanglemark(H, K, L), linewidth(.5)); draw(rightanglemark(K, L, G), linewidth(.5)); dot("$A$", A, S); dot("$B$", B, S); dot("$C$", C, dir(90)); dot("$D$", D, dir(90)); dot("$E$", E, S); dot("$F$", F, dir(0)); dot("$G$", G, N); dot("$H$", H, W); dot("$I$", I, SW); dot("$J$", J, SW); dot("$K$", K, N); dot("$L$", L, S); | [] |
522 | Let $ABCD$ be a cyclic quadrilateral with $AB=6, AC=8, BD=5, CD=2$. Let $P$ be the point on $\overline{AD}$ such that $\angle APB = \angle CPD$. Then $\frac{BP}{CP}$ can be expressed in the form $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2020 CIME I Problems/Problem 9 | Let $C'$ be the reflection of $C$ over line $AD$. Since $\angle APB = \angle CPD = \angle C'PD$, $B, P, C$ are collinear. Suppose $X$ and $Y$ are the projections of $B$ and $C$ onto line $AD$, respectively. We want to find $\frac{BP}{CP}$ which by similar triangles is also equal to $\frac{BX}{C'Y}$ from $\triangle BPX \sim \triangle C'PY$. Since $C'Y=CY$, this also equals $\frac{BX}{CY}$. We know that $\triangle ABD$ and $\triangle ACD$ each share the same base, so this can also be interpreted as $\frac{[ABD]}{[ACD]}$. The sine area formula gives \[\frac{[ABD]}{[ACD]} = \frac{\frac{1}{2} \cdot 6 \cdot 5 \sin ABD}{\frac{1}{2} \cdot 8 \cdot 2 \sin ACD}.\] Quadrilateral $ABCD$ is cyclic, so $\angle ABD = \angle ACD$ because both angles subtend arc $\widehat{AD}$ on the circumcircle of Quadrilateral $ABCD$. We can then replace every $\angle ACD$ with $\angle ABD$, but realise that if we do that, the $\angle ABD$s will cancel out. The requested area ratio is thus \[\frac{\frac{1}{2} \cdot 6 \cdot 5}{\frac{1}{2} \cdot 8 \cdot 2} = \frac{15}{8}\]. The answer is $15+8=\boxed{023}$. | // Block 1
size(4cm);
/* draw figures */
draw((0,0)--(1,0));
draw(circle((0.5,-0.12046156424028276), 0.5143063177321622));
draw((0.29472641365670604,0.35110364144073225)--(0,0));
draw((0.29472641365670604,0.35110364144073225)--(0.7999672347109121,0.297307087718076));
draw((0.7999672347109121,0.297307087718076)--(1,0));
draw((0.29472641365670604,0)--(0.29472641365670604,0.35110364144073225));
draw((0.7999672347109121,0.297307087718076)--(0.7999672347109121,-0.297307087718076));
draw((0.7999672347109121,-0.297307087718076)--(0.29472641365670604,0.35110364144073225));
draw(arc((0.5683059275348462,0),0.13393442314476192,127.92567650750303,180));
draw((0.4620043965252797,0.05193209576548634)--(0.4339247468246395,0.06565000785448262));
draw(arc((0.5683059275348462,0),0.13393442314476192,307.925676507503,360));
draw((0.6746074585444126,-0.05193209576548634)--(0.7026871082450528,-0.06565000785448288));
draw((0.7999672347109121,0.297307087718076)--(0.5683059275348462,0));
draw(arc((0.5683059275348462,0),0.13393442314476192,360,412.074323492497));
draw((0.6746074585444126,0.05193209576548634)--(0.702687108245053,0.06565000785448288));
/* dots and labels */
dot((0,0));
label("$A$", (0,0), NW);
dot((1,0));
label("$D$", (1,0), NE);
dot((0.29472641365670604,0.35110364144073225));
label("$B$", (0.29472641365670604,0.35110364144073225), N);
dot((0.7999672347109121,0.297307087718076));
label("$C$", (0.7999672347109121,0.297307087718076), N);
dot((0.29472641365670604,0));
label("$X$", (0.29472641365670604,0), SW);
dot((0.7999672347109121,0));
label("$Y$", (0.7999672347109121,0), SE);
dot((0.7999672347109121,-0.297307087718076));
label("$C'$", (0.7999672347109121,-0.297307087718076), S);
dot((0.5683059275348462,0));
label("$P$", (0.5683059275348462,0), SW);
// Block 2
size(4cm); /* draw figures */ draw((0,0)--(1,0)); draw(circle((0.5,-0.12046156424028276), 0.5143063177321622)); draw((0.29472641365670604,0.35110364144073225)--(0,0)); draw((0.29472641365670604,0.35110364144073225)--(0.7999672347109121,0.297307087718076)); draw((0.7999672347109121,0.297307087718076)--(1,0)); draw((0.29472641365670604,0)--(0.29472641365670604,0.35110364144073225)); draw((0.7999672347109121,0.297307087718076)--(0.7999672347109121,-0.297307087718076)); draw((0.7999672347109121,-0.297307087718076)--(0.29472641365670604,0.35110364144073225)); draw(arc((0.5683059275348462,0),0.13393442314476192,127.92567650750303,180)); draw((0.4620043965252797,0.05193209576548634)--(0.4339247468246395,0.06565000785448262)); draw(arc((0.5683059275348462,0),0.13393442314476192,307.925676507503,360)); draw((0.6746074585444126,-0.05193209576548634)--(0.7026871082450528,-0.06565000785448288)); draw((0.7999672347109121,0.297307087718076)--(0.5683059275348462,0)); draw(arc((0.5683059275348462,0),0.13393442314476192,360,412.074323492497)); draw((0.6746074585444126,0.05193209576548634)--(0.702687108245053,0.06565000785448288)); /* dots and labels */ dot((0,0)); label("$A$", (0,0), NW); dot((1,0)); label("$D$", (1,0), NE); dot((0.29472641365670604,0.35110364144073225)); label("$B$", (0.29472641365670604,0.35110364144073225), N); dot((0.7999672347109121,0.297307087718076)); label("$C$", (0.7999672347109121,0.297307087718076), N); dot((0.29472641365670604,0)); label("$X$", (0.29472641365670604,0), SW); dot((0.7999672347109121,0)); label("$Y$", (0.7999672347109121,0), SE); dot((0.7999672347109121,-0.297307087718076)); label("$C'$", (0.7999672347109121,-0.297307087718076), S); dot((0.5683059275348462,0)); label("$P$", (0.5683059275348462,0), SW); | [] |
523 | 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 Problems/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)); | [] |
523 | 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 Problems/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)); | [] |
523 | 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 Problems/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)); | [] |
524 | 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 Problems/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); | [] |
524 | 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 Problems/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)); | [] |
525 | Semicircle $\Gamma$ has diameter $\overline{AB}$ of length $14$. Circle $\Omega$ lies tangent to $\overline{AB}$ at a point $P$ and intersects $\Gamma$ at points $Q$ and $R$. If $QR=3\sqrt3$ and $\angle QPR=60^\circ$, then the area of $\triangle PQR$ equals $\tfrac{a\sqrt{b}}{c}$, where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer not divisible by the square of any prime. What is $a+b+c$?
$\textbf{(A) }110 \qquad \textbf{(B) }114 \qquad \textbf{(C) }118 \qquad \textbf{(D) }122\qquad \textbf{(E) }126$ | 2021 AMC 12A Problems/Problem 24 | Let $O=\Gamma$ be the center of the semicircle and $X=\Omega$ be the center of the circle.
Applying the Extended Law of Sines to $\triangle PQR,$ we find the radius of $\odot X:$ \[XP=\frac{QR}{2\cdot\sin \angle QPR}=\frac{3\sqrt3}{2\cdot\frac{\sqrt3}{2}}=3.\]
Alternatively, by the Inscribed Angle Theorem, $\triangle QRX$ is a $30^\circ\text{-}30^\circ\text{-}120^\circ$ triangle with base $QR=3\sqrt3.$ Dividing $\triangle QRX$ into two congruent $30^\circ\text{-}60^\circ\text{-}90^\circ$ triangles, we get that the radius of $\odot X$ is $XQ=XR=3$ by the side-length ratios.
Let $M$ be the midpoint of $\overline{QR}.$ By the Perpendicular Chord Bisector Converse, we have $\overline{OM}\perp\overline{QR}$ and $\overline{XM}\perp\overline{QR}.$ Together, points $O, X,$ and $M$ must be collinear.
By the SAS Congruence, we have $\triangle QXM\cong\triangle RXM,$ both of which are $30^\circ\text{-}60^\circ\text{-}90^\circ$ triangles. By the side-length ratios, we obtain $RM=\frac{3\sqrt3}{2}, RX=3,$ and $XM=\frac{3}{2}.$ By the Pythagorean Theorem on right $\triangle ORM,$ we get $OM=\frac{13}{2}$ and $OX=OM-XM=5.$ By the Pythagorean Theorem on right $\triangle OXP,$ we get $OP=4.$
Let $C$ be the foot of the perpendicular from $P$ to $\overline{QR},$ and $D$ be the foot of the perpendicular from $X$ to $\overline{PC},$ as shown below:
Clearly, quadrilateral $XDCM$ is a rectangle. Since $\angle XPD=\angle OXP$ by alternate interior angles, we have $\triangle XPD\sim\triangle OXP$ by the AA Similarity, with the ratio of similitude $\frac{XP}{OX}=\frac 35.$ Therefore, we get $PD=\frac 95$ and $PC=PD+DC=PD+XM=\frac 95 + \frac 32 = \frac{33}{10}.$
The area of $\triangle PQR$ is \[\frac12\cdot QR\cdot PC=\frac12\cdot3\sqrt3\cdot\frac{33}{10}=\frac{99\sqrt3}{20},\] from which the answer is $99+3+20=\boxed{\textbf{(D) } 122}.$
~MRENTHUSIASM | // Block 1
/* Made by MRENTHUSIASM */
size(300);
pair O, X, A, B, P, Q, R, M, C, D;
O = (0,0);
X = (4,3);
A = (-7,0);
B = (7,0);
P = (4,0);
Q = intersectionpoints(Circle(O,7),Circle(X,3))[0];
R = intersectionpoints(Circle(O,7),Circle(X,3))[1];
M = midpoint(Q--R);
C = foot(P,Q,R);
D = foot(X,P,C);
fill(P--Q--R--cycle,yellow);
dot("$O$",O,S);
dot("$X$",X,N);
dot("$A$",A,SW);
dot("$B$",B,SE);
dot("$P$",P,S);
dot("$Q$",Q,E);
dot("$R$",R,N);
dot("$M$",M,dir(M));
dot("$C$",C,NE);
dot("$D$",D,SE);
markscalefactor=0.0375;
draw(rightanglemark(O,M,R),red);
draw(rightanglemark(P,C,M),red);
draw(rightanglemark(P,D,X),red);
draw(rightanglemark(O,P,X),red);
draw(P--Q--R--cycle);
draw(arc(O, 7, 0, 180)^^A--B^^Circle(X,3));
draw(O--M^^X--P);
draw(P--C^^X--D,dashed);
// Block 2
/* Made by MRENTHUSIASM */ size(300); pair O, X, A, B, P, Q, R, M, C, D; O = (0,0); X = (4,3); A = (-7,0); B = (7,0); P = (4,0); Q = intersectionpoints(Circle(O,7),Circle(X,3))[0]; R = intersectionpoints(Circle(O,7),Circle(X,3))[1]; M = midpoint(Q--R); C = foot(P,Q,R); D = foot(X,P,C); fill(P--Q--R--cycle,yellow); dot("$O$",O,S); dot("$X$",X,N); dot("$A$",A,SW); dot("$B$",B,SE); dot("$P$",P,S); dot("$Q$",Q,E); dot("$R$",R,N); dot("$M$",M,dir(M)); dot("$C$",C,NE); dot("$D$",D,SE); markscalefactor=0.0375; draw(rightanglemark(O,M,R),red); draw(rightanglemark(P,C,M),red); draw(rightanglemark(P,D,X),red); draw(rightanglemark(O,P,X),red); draw(P--Q--R--cycle); draw(arc(O, 7, 0, 180)^^A--B^^Circle(X,3)); draw(O--M^^X--P); draw(P--C^^X--D,dashed); | [] |
525 | Semicircle $\Gamma$ has diameter $\overline{AB}$ of length $14$. Circle $\Omega$ lies tangent to $\overline{AB}$ at a point $P$ and intersects $\Gamma$ at points $Q$ and $R$. If $QR=3\sqrt3$ and $\angle QPR=60^\circ$, then the area of $\triangle PQR$ equals $\tfrac{a\sqrt{b}}{c}$, where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer not divisible by the square of any prime. What is $a+b+c$?
$\textbf{(A) }110 \qquad \textbf{(B) }114 \qquad \textbf{(C) }118 \qquad \textbf{(D) }122\qquad \textbf{(E) }126$ | 2021 AMC 12A Problems/Problem 24 | Suppose we label the points as shown in the diagram above, where $C$ is the center of the semicircle and $O$ is the center of the circle tangent to $\overline{AB}$. Since $\angle QPR = 60^{\circ}$, we have $\angle QOR = 2\cdot 60^{\circ}=120^{\circ}$ and $\triangle QOR$ is a $30-30-120$ triangle, which can be split into two $30-60-90$ triangles by the altitude from $O$. Since $QR=3\sqrt{3},$ we know $OQ=OR=\tfrac{3\sqrt{3}}{\sqrt{3}}=3$ by $30-60-90$ triangles. The area of this part of $\triangle PQR$ is $\frac{1}{2}bh=\tfrac{3\sqrt{3}}{2}\cdot\tfrac{3}{2}=\tfrac{9\sqrt{3}}{4}$. We would like to add this value to the sum of the areas of the other two parts of $\triangle PQR$.
To find the areas of the other two parts of $\triangle PQR$ using the $\sin$ area formula, we need the sides and included angles. Here we know the sides but what we don't know are the angles. So it seems like we will have to use an angle from another triangle and combine them with the angles we already know to find these angles easily. We know that $\angle QOR = 120^{\circ}$ and triangles $\triangle COQ$ and $\triangle COR$ are congruent as they share a side, $CQ=CR,$ and $OQ=OR$. Therefore $\angle COQ = \angle COR = 120^{\circ}$. Suppose $CO=x$. Then $3^{2}+x^{2}-6x\cos{120^{\circ}}=7^{2}$, and since $\cos{120^{\circ}}=-\tfrac{1}{2}$, this simplifies to $x^{2}+3x=7^{2}-3^{2}\rightarrow x^{2}+3x-40=0$. This factors nicely as $(x-5)(x+8)=0$, so $x=5$ as $x$ can't be $-8$. Since $CO=5, OP=3$ and $\angle OPC=90^{\circ}$, we now know that $\triangle OPC$ is a $3-4-5$ right triangle. This may be useful info for later as we might use an angle in this triangle to find the areas of the other two parts of $\triangle PQR$.
Let $\angle POC = \alpha$. Then $\sin\alpha = \tfrac{4}{5}, \cos\alpha = \tfrac{3}{5}, \angle QOP = 120+\alpha,$ and $\angle POR = 120-\alpha$. The sum of the areas of $\triangle QOP$ and $\triangle POR$ is $3\cdot 3\cdot\tfrac{1}{2}\cdot\left[\sin(120-\alpha)+\sin(120+\alpha)\right]=\tfrac{9}{2}\left[\sin(120-\alpha)+\sin(120+\alpha)\right],$ which we will add to $\tfrac{9\sqrt{3}}{4}$ to get the area of $\triangle PQR$. Observe that \[\sin(120-\alpha) = \sin 120\cos\alpha-\sin\alpha\cos 120 = \tfrac{\sqrt{3}}{2}\cdot\tfrac{3}{5}-\tfrac{4}{5}\cdot\tfrac{-1}{2}=\tfrac{3\sqrt{3}}{10}+\tfrac{4}{10}=\tfrac{3\sqrt{3}+4}{10}\]and similarly $\sin(120+\alpha)=\tfrac{3\sqrt{3}-4}{10}$. Adding these two gives $\tfrac{3\sqrt{3}}{5}$ and multiplying that by $\tfrac{9}{2}$ gets us $\tfrac{27\sqrt{3}}{10},$ which we add to $\tfrac{9\sqrt{3}}{4}$ to get $\tfrac{54\sqrt{3}+45\sqrt{3}}{20}=\tfrac{99\sqrt{3}}{20}$. The answer is $99+3+20=102+20=\boxed{\textbf{(D)} ~122}.$
~sugar_rush | // Block 1
size(150);
draw(circle((7,0),7));
pair A = (0, 0);
pair B = (14, 0);
draw(A--B);
draw(circle((11,3),3));
label("$C$", (7, 0), S);
label("$O$", (11, 3), E);
label("$P$", (11, 0), S);
pair C = (7, 0);
pair O = (11, 3);
pair P = (11, 0);
pair Q = intersectionpoints(circle(C, 7), circle(O, 3))[1];
pair R = intersectionpoints(circle(C, 7), circle(O, 3))[0];
draw(C--O);
draw(C--Q);
draw(C--R);
draw(Q--R);
draw(O--P);
draw(O--Q);
draw(O--R);
draw(P--Q);
draw(P--R);
label("$Q$", Q, N);
label("$R$", R, E);
// Block 2
size(150); draw(circle((7,0),7)); pair A = (0, 0); pair B = (14, 0); draw(A--B); draw(circle((11,3),3)); label("$C$", (7, 0), S); label("$O$", (11, 3), E); label("$P$", (11, 0), S); pair C = (7, 0); pair O = (11, 3); pair P = (11, 0); pair Q = intersectionpoints(circle(C, 7), circle(O, 3))[1]; pair R = intersectionpoints(circle(C, 7), circle(O, 3))[0]; draw(C--O); draw(C--Q); draw(C--R); draw(Q--R); draw(O--P); draw(O--Q); draw(O--R); draw(P--Q); draw(P--R); label("$Q$", Q, N); label("$R$", R, E); | [] |
525 | Semicircle $\Gamma$ has diameter $\overline{AB}$ of length $14$. Circle $\Omega$ lies tangent to $\overline{AB}$ at a point $P$ and intersects $\Gamma$ at points $Q$ and $R$. If $QR=3\sqrt3$ and $\angle QPR=60^\circ$, then the area of $\triangle PQR$ equals $\tfrac{a\sqrt{b}}{c}$, where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer not divisible by the square of any prime. What is $a+b+c$?
$\textbf{(A) }110 \qquad \textbf{(B) }114 \qquad \textbf{(C) }118 \qquad \textbf{(D) }122\qquad \textbf{(E) }126$ | 2021 AMC 12A Problems/Problem 24 | Define points as shown above, where $N=\overleftrightarrow{PA}\cap\overleftrightarrow{QR}$. The area of $\triangle PQR$ is simply\[\dfrac{1}{2}PX\cdot QR=\dfrac{3\sqrt{3}}{2}PX;\]
it remains to compute the value of $PX$. Note that $PX$ is simply a weighted average of $BT$ and $CS;$ it is $\dfrac{CP}{BP}$ times closer to $BT$ than it is to $CS$. Observe that\[CS=\sqrt{CQ^{2}-\left(\dfrac{1}{2}QR\right)^{2}}=\sqrt{7^{2}-\left(\dfrac{3\sqrt{3}}{2}\right)^{2}}=6.5\]
since the radius of $\Gamma$ is $7$ as its diameter is $14$. Note also by the Extended Law of Sines the radius of $\Omega$ is $\dfrac{3\sqrt{3}}{2\sin 60^{\circ}}=3,$ so $OS=3\cos 60^{\circ}=1.5$. Since $C, O,$ and $S$ are collinear by symmetry we have $CO=CS-OS=5,$ so $CP=\sqrt{5^{2}-3^{2}}=4$ and $BP=7-4=3$. Therefore, $\triangle OPC$ is a $3\text{-}4\text{-}5$ right triangle; $\triangle OPC\sim\triangle NSC$ since $\angle OPC=\angle CSN=90^{\circ}$ and $\angle OCP=\angle NCS=\sin^{-1}\left(\dfrac{3}{5}\right)$. Therefore $\dfrac{CN}{CS}=\cfrac{CO}{CP}=\dfrac{5}{4}$ so $CN=\dfrac{5}{4}CS=\dfrac{65}{8}$. Since $\triangle BTN\sim\triangle CSN,$ we have $\dfrac{BT}{BN}=\dfrac{CS}{CN}=\dfrac{4}{5}$. Therefore \[BT=\dfrac{4}{5}BN=\dfrac{4}{5}\left(CN-7\right)=\dfrac{4}{5}\cdot\dfrac{9}{8}=\dfrac{36}{40}=0.9;\]
so $PX$ is $\dfrac{4}{3}$ times as close to $0.9$ as to $6.5;$ we can compute $PX=\dfrac{4}{7}BT+\dfrac{3}{7}CS=\dfrac{4}{7}\cdot0.9+\dfrac{3}{7}\cdot6.5=3.3$. The area of $\triangle PQR$ is\[\dfrac{3\sqrt{3}}{2}\cdot 3.3=\dfrac{99\sqrt{3}}{20}\]and $99+3+20=\boxed{\textbf{(D)} ~122}$.
~sugar_rush | size(300); pair C = (7, 0); draw(arc(C, 7, 0, 180)); pair A = (0, 0), B = (14, 0); draw(A--B); draw(circle((11,3),3)); label("$A$", A, SSE); label("$B$", B, SSW); label("$C$", (A+B)/2, S); label("$O$", (11, 3), E); label("$P$", (11, 0), S); pair O = (11, 3), P = (11, 0), Q = intersectionpoints(circle(C, 7), circle(O, 3))[1], R = intersectionpoints(circle(C, 7), circle(O, 3))[0], S = (Q+R)/2, N = (121/8, 0), T = (8/11)*N + (3/11)*R, X = (4/7)*T + (3/7)*S; draw(C--O, blue); draw(O--S, red); draw(C--Q); draw(C--R); draw(Q--N--B); draw(O--P); draw(O--Q); draw(O--R); draw(P--Q--R--cycle); draw(B--T); draw(P--X); label("$Q$", Q, NNE); label("$R$", R, E); label("$S$", S, ENE); label("$N$", N, SSE); label("$T$", T, ENE); label("$X$", X, NE); draw(rightanglemark(P, X, Q)); draw(rightanglemark(B, T, R)); draw(rightanglemark(C, S, Q)); | [] |
526 | Let $S$ be the set of lattice points in the coordinate plane, both of whose coordinates are integers between $1$ and $30,$ inclusive. Exactly $300$ points in $S$ lie on or below a line with equation $y=mx.$ The possible values of $m$ lie in an interval of length $\frac ab,$ where $a$ and $b$ are relatively prime positive integers. What is $a+b?$
$\textbf{(A)} ~31 \qquad \textbf{(B)} ~47 \qquad \textbf{(C)} ~62\qquad \textbf{(D)} ~72 \qquad \textbf{(E)} ~85$ | 2021 AMC 12B Problems/Problem 25 | Based on area ratios between a square of side length $30$ and a triangle with base $30$, we estimate that the slope $m$ of the line we want is approximately $\frac{2}{3}$. Following this estimate, we see if there are approximately $30 \cdot 30 - 300 = 600$ lattice points above the line $y=\frac{2}{3}x$.
Counting the number of lattice points with $x=1$ above the line, the number of lattice points with $x=2$ above the line, and so on, we find that the total number of lattice points above the line is $30+29+28+28+27+26+26 \ldots+ 10$, with the even integers repeating every third term. We see that the average of the $30$ terms is $20$, which means that exactly $20 \cdot 30 = 600$ lattice points above the line as desired. This gives a lower bound because any decrease in the slope of the line would cause points that were already on the line to shift to being above it.
To find the upper bound, notice that each lattice point less than $1$ unit above the line is either $\frac{1}{3}$ or $\frac{2}{3}$ above. Since the slope through a point is the y-coordinate divided by the x-coordinate, a shift in the slope will increase the y-value of the higher x-coordinates more than the y-value of the lower x-coordinates. So, we turn our attention to $x=28, 29, 30$ for which the line $y=\frac{2}{3}x$ intersects at $y= \frac{56}{3}, \frac{58}{3}, 20$. The point $(30,20)$ is already counted, and we can clearly see that if we slowly increase the slope of the line, we will first hit the point $(28,19)$ since $(28, \frac{56}{3})$ is the closest to it. The equation of the line which goes through both the origin and $(28,19)$ is $y=\frac{19}{28}x$. This gives an upper bound of $m=\frac{19}{28}$.
Taking the upper bound of $m$ and subtracting the lower bound yields $\frac{19}{28}-\frac{2}{3}=\frac{1}{84}$. The answer is therefore $1+84=$ $\boxed{\textbf{(E)} ~85}$.
~theAJL ~Minor edits by Eric_Zang
Diagram | /* Created by Brendanb4321 */ import graph; size(16cm); defaultpen(fontsize(9pt)); xaxis(0,30,Ticks(1.0)); yaxis(0,25,Ticks(1.0)); draw((0,0)--(30,20)); draw((0,0)--(30,30/28*19), dotted); for (int i = 1; i<=30; ++i) { for (int j = 1; j<=2/3*i+1; ++j) { dot((i,j)); } } dot((28,19), red); label("$m=2/3$", (32,20)); label("$m=19/28$", (32.3,20.8)); | [] |
526 | Let $S$ be the set of lattice points in the coordinate plane, both of whose coordinates are integers between $1$ and $30,$ inclusive. Exactly $300$ points in $S$ lie on or below a line with equation $y=mx.$ The possible values of $m$ lie in an interval of length $\frac ab,$ where $a$ and $b$ are relatively prime positive integers. What is $a+b?$
$\textbf{(A)} ~31 \qquad \textbf{(B)} ~47 \qquad \textbf{(C)} ~62\qquad \textbf{(D)} ~72 \qquad \textbf{(E)} ~85$ | 2021 AMC 12B Problems/Problem 25 | It's easier to calculate the number of lattice points inside a rectangle with vertices $(0,0)$, $(p,0)$, $(p,q)$, $(0,q)$. Those lattice points are divided by the diagonal $y = \frac{p}{q} \cdot x$ into $2$ halves. In this problem the number of lattice points on or below the diagonal and $x \ge 1$ is
$\frac{1}{2} [(p+1)(q+1) - d] + d - (p+1)$, $d$ is the number of lattice points on the diagonal, $d \ge 2$
$(p+1)(q+1)$ is the total number of lattice points inside the rectangle. Subtract the number of lattice points on the diagonal, divided by 2 is the number of lattice points below the diagonal, add the number of lattice points on the diagonal, and subtract the lattice points on the $x$ axis, then we get the total number of lattice points on or below the diagonal and $x \ge 1$.
There are $900$ lattice points in total. $300$ is $\frac{1}{3}$ of $900$. The $x$ coordinate of the top-right vertex of the rectangle is $30$, $\frac{1}{2} \cdot 30 \cdot 20 = 300$. I guess the $y$ coordinate of the top-right vertex of the rectangle is $20$. Now I am going to verify that. The slope of the diagonal is $\frac{20}{30} = \frac{2}{3}$, there are $11$ lattice points on the diagonal. Substitute $(p,q)=(30, 20)$, $d=11$ to the above formula:
$\frac{1}{2} [(30+1)(10+1) - 11] + 11 - (30+1) = 300$
Because there are $11$ lattice points on line $y = \frac{2}{3}x$, if $m < \frac{2}{3}$, then the number of lattice points on or below the line is less than $300$. So $m = \frac{2}{3}$ is the lower bound.
Now I am going to calculate the upper bound. From $\frac{b}{a} < \frac{b+1}{a+1}$,
$\frac{2}{3} = \frac{18}{27} < \frac{19}{28}$
$\frac{2}{3} = \frac{20}{30} < \frac{21}{31} < \frac{19}{28}$
If $m = \frac{21}{31}$, I will calculate by using the rectangle with blue vertex $(p,q) = (31, 21)$, then subtract lattice points on line $x = 31$, which is $21$. There are 2 lattice points on the diagonal, $d=2$.
$\frac{1}{2} [(31+1)(21+1) - 2] + 2 - (31+1) - 21 = 300$, same as that of $m = \frac{2}{3}$
If $m = \frac{19}{28}$, I will calculate by using the rectangle with red vertex $(p,q) = (28, 19)$, then add lattice points on line $x = 29$ and $x = 30$, which is $19 + 20 = 39$. There are 2 lattice points on the diagonal, $d=2$.
$\frac{1}{2} [(28+1)(19+1) - 2] + 2 - (28+1) + 39 = 301$, $1$ more than that of $m = \frac{2}{3}$
When $m$ increases, more lattice points falls below the line $y = mx$. Any value larger than $\frac{19}{28}$ has more than $301$ lattice points on or below $y = \frac{19}{28} x$. So the upper bound is $\frac{19}{28}$.
$\frac{19}{28}-\frac{2}{3}=\frac{1}{84}$, $\boxed{\textbf{(E)} ~85}$.
~isabelchen | // Block 1
/* Created by Brendanb4321, modified by isabelchen */
import graph;
size(18cm);
defaultpen(fontsize(9pt));
xaxis(0,31,Ticks(1.0));
yaxis(0,22,Ticks(1.0));
draw((0,0)--(30,20));
draw((0,0)--(30,30*19/28), dotted);
draw((0,0)--(31,31*21/31), dotted);
for (int i = 1; i<=31; ++i) {
for (int j = 1; j<=2/3*i+1; ++j) {
dot((i,j));
}
}
dot((28,19), red);
dot((31,21), blue);
label("$m=2/3$", (33,20));
label("$m=21/31$", (33,21));
label("$m=19/28$", (33,22));
// Block 2
/* Created by Brendanb4321, modified by isabelchen */ import graph; size(18cm); defaultpen(fontsize(9pt)); xaxis(0,31,Ticks(1.0)); yaxis(0,22,Ticks(1.0)); draw((0,0)--(30,20)); draw((0,0)--(30,30*19/28), dotted); draw((0,0)--(31,31*21/31), dotted); for (int i = 1; i<=31; ++i) { for (int j = 1; j<=2/3*i+1; ++j) { dot((i,j)); } } dot((28,19), red); dot((31,21), blue); label("$m=2/3$", (33,20)); label("$m=21/31$", (33,21)); label("$m=19/28$", (33,22)); | [] |
526 | Let $S$ be the set of lattice points in the coordinate plane, both of whose coordinates are integers between $1$ and $30,$ inclusive. Exactly $300$ points in $S$ lie on or below a line with equation $y=mx.$ The possible values of $m$ lie in an interval of length $\frac ab,$ where $a$ and $b$ are relatively prime positive integers. What is $a+b?$
$\textbf{(A)} ~31 \qquad \textbf{(B)} ~47 \qquad \textbf{(C)} ~62\qquad \textbf{(D)} ~72 \qquad \textbf{(E)} ~85$ | 2021 AMC 12B Problems/Problem 25 | The lower bound of $m$ is $\frac23 = \frac{20}{30}$. Inside the rectangle with vertices $(0,0)$, $(30,0)$, $(30,20)$, $(0, 20)$ and diagonal $y = \frac23 x$, there are $(30-1)(20-1) = 551$ lattice points inside, not including the edges. There are $9$ lattice points on diagonal $y = \frac23 x$ inside the rectangle, $551 + 9 = 560$. Half of the $560$ lattice points are below diagonal $y = \frac23 x$, $560 \cdot \frac12 = 280$. There are $20$ lattice points on edge $x = 30$, $280 + 20 = 300$. Once $m < \frac23$, the $9$ lattice points on diagonal $y = \frac23 x$ will be above the new diagonal, making the number of lattice points on and below the diagonal less than $300$.
Now we are going to calculate the upper bound by the following formula:
The number of lattice points inside rectangle $(0,0)$, $(a,0)$, $(a, b)$, $(0, b)$ and below diagonal $y = \frac{b}{a}x$ is $\frac{(a-1)(b-1)}{2}$, where $a$ and $b$ are relatively prime.
There are $(a-1)(b-1)$ lattice points inside the rectangle. Because $a$ and $b$ are relatively prime, the only lattice points on the diagonal are $(0,0)$ and $(a,b)$. By symmetry, half of the lattice points are below the diagonal.
$\frac{2}{3} = \frac{20}{30} < \frac{21}{31} < \frac{19}{28}$
When $a = 31$, $b = 21$, $\frac{(31-1)(21-1)}{2} = 300$.
When $a = 28$, $b = 19$, $\frac{(28-1)(19-1)}{2} = 243$. Below the line $y = \frac{19}{28} x$, there are $19$ lattice points on line $x = 28$, $19$ lattice points on line $x = 29$, $20$ lattice points on line $x = 30$, $243 + 19 + 19 + 20 = 301$.
More lattice points fall below the line $y = mx$ as $m$ increases. There are more than $301$ lattice points on and below the line for any $m$ greater than $\frac{19}{28}$. Therefore, the upper bound is $\frac{19}{28}$.
$\frac{19}{28}-\frac{2}{3}=\frac{1}{84}$, so $1+84=\boxed{\textbf{(E)} ~85}$.
~isabelchen
Remark
$\lfloor \frac{b}{a} k \rfloor$ is the number of lattice points on line $x = k$, below line $y = \frac{b}{a} x$ and above the $x$ axis, where $k$ is an integer and $0<k<a$. Therefore, $\sum_{k=1}^{a-1} \lfloor \frac{b}{a} k \rfloor$ is the number of lattice points inside the rectangle $(0,0)$, $(a,0)$, $(a, b)$, $(0, b)$, below diagonal $y = \frac{b}{a} x$. If $a$ and $b$ are relatively prime, $\sum_{k=1}^{a-1} \lfloor \frac{b}{a} k \rfloor = \frac{(a-1)(b-1)}{2}$, as explained in solution 5. This problem is about finding the upper and lower bound of $\frac{b}{a}$, given $\sum_{k=1}^{30} \lfloor \frac{b}{a} k \rfloor = 300$. The same problem can have geometric representation as stated in the original problem, or algebraic representation as stated here.
~isabelchen | // Block 1
/* Created by isabelchen */
import graph;
size(8cm);
defaultpen(fontsize(9pt));
xaxis(0,8);
yaxis(0,6);
draw((0,0)--(7,5));
draw((7,0)--(7,5));
draw((0,5)--(7,5));
dot((0,0)); dot((1,0)); dot((2,0)); dot((3,0)); dot((4,0)); dot((5,0)); dot((6,0)); dot((7,0)); dot((8,0)); dot((0,1)); dot((1,1)); dot((2,1)); dot((3,1)); dot((4,1)); dot((5,1)); dot((6,1)); dot((7,1)); dot((8,1)); dot((0,2)); dot((1,2)); dot((2,2)); dot((3,2)); dot((4,2)); dot((5,2)); dot((6,2)); dot((7,2)); dot((8,2)); dot((0,3)); dot((1,3)); dot((2,3)); dot((3,3)); dot((4,3)); dot((5,3)); dot((6,3)); dot((7,3)); dot((8,3)); dot((0,4)); dot((1,4)); dot((2,4)); dot((3,4)); dot((4,4)); dot((5,4)); dot((6,4)); dot((7,4)); dot((8,4)); dot((0,5)); dot((1,5)); dot((2,5)); dot((3,5)); dot((4,5)); dot((5,5)); dot((6,5)); dot((7,5)); dot((8,5)); dot((0,6)); dot((1,6)); dot((2,6)); dot((3,6)); dot((4,6)); dot((5,6)); dot((6,6)); dot((7,6)); dot((8,6));
label("$(0,0)$", (0,0), SW);
label("$(a, b)$", (7,5), NE);
dot((7,0));
label("$a$", (7,0), S);
dot((0,5));
label("$b$", (0,5), W);
// Block 2
/* Created by isabelchen */ import graph; size(8cm); defaultpen(fontsize(9pt)); xaxis(0,8); yaxis(0,6); draw((0,0)--(7,5)); draw((7,0)--(7,5)); draw((0,5)--(7,5)); dot((0,0)); dot((1,0)); dot((2,0)); dot((3,0)); dot((4,0)); dot((5,0)); dot((6,0)); dot((7,0)); dot((8,0)); dot((0,1)); dot((1,1)); dot((2,1)); dot((3,1)); dot((4,1)); dot((5,1)); dot((6,1)); dot((7,1)); dot((8,1)); dot((0,2)); dot((1,2)); dot((2,2)); dot((3,2)); dot((4,2)); dot((5,2)); dot((6,2)); dot((7,2)); dot((8,2)); dot((0,3)); dot((1,3)); dot((2,3)); dot((3,3)); dot((4,3)); dot((5,3)); dot((6,3)); dot((7,3)); dot((8,3)); dot((0,4)); dot((1,4)); dot((2,4)); dot((3,4)); dot((4,4)); dot((5,4)); dot((6,4)); dot((7,4)); dot((8,4)); dot((0,5)); dot((1,5)); dot((2,5)); dot((3,5)); dot((4,5)); dot((5,5)); dot((6,5)); dot((7,5)); dot((8,5)); dot((0,6)); dot((1,6)); dot((2,6)); dot((3,6)); dot((4,6)); dot((5,6)); dot((6,6)); dot((7,6)); dot((8,6)); label("$(0,0)$", (0,0), SW); label("$(a, b)$", (7,5), NE); dot((7,0)); label("$a$", (7,0), S); dot((0,5)); label("$b$", (0,5), W); | [] |
527 | Let $ABC$ be an acute triangle with orthocenter $H$ and circumcenter $O$. The tangent to the circumcircle of $\triangle ABC$ at $A$ intersects lines $BH$ and $CH$ at $X$ and $Y$, and $BY\parallel CX$. Let line $AO$ intersect $\overline{BC}$ at $D$. Suppose that $AO=25, BC=49$, and $AD=a-b\sqrt{c}$ for positive integers $a, b, c,$ where $c$ is not divisible by the square of any prime. Find $a+b+c$. | 2021 CIME I Problems/Problem 14 | Let $H$ be the orthocenter of $\triangle ABC$, and let $E$, $F$ be the feet of the altitudes from $B, C$. Also let $A'$ be the antipode of $A$ on the circumcircle and let $S=\overline{AH}\cap\overline{EF}$, as shown below:
Disregarding the condition $\overline{BY}\parallel\overline{CX}$, we contend:
$\textbf{Claim}:$ In general, $BCXY$ is cyclic.
$\textbf{Proof}.$ Recall that $\overline{AA}\parallel\overline{EF}$, so the claim follows from Reims' theorem on $BCEF, BCXY. \blacksquare$
With $\overline{BY}\parallel\overline{CX}$, it follows that $BCXY$ is an isosceles trapezoid. In particular, $HB=HY$ and $HC=HX$. Since $\overline{SF}\parallel\overline{AY}$, we have \[\frac{HS}{HA}=\frac{HF}{HY}=\frac{HF}{HB}=\cos A.\] But note that $\triangle AEF\cup H\sim\triangle ABC\cup A'$, so \[\frac{AD}{2R}=1-\frac{A'D}{2R}=1-\frac{HS}{HA}=1-\cos A,\] i.e.\ $AD=2R(1-\cos A)$. We are given $R=25$, and by the law of sines, $\sin A=\frac{49}{50}$, so $\cos A=\frac{3\sqrt{11}}{50}$, and $AD=50-3\sqrt{11}$, so $50+3+11=\boxed{064}$. | // Block 1
size(6cm); defaultpen(fontsize(10pt)); pen pri=blue; pen sec=lightblue; pen tri=heavycyan; pen qua=paleblue; pen fil=invisible; pen sfil=invisible; pen tfil=invisible;
pair O,A,B,C,H,X,Y,EE,F,Ap,SS,D; O=(0,0); A=dir(147.55); B=dir(195); C=dir(345); H=A+B+C; X=extension(B,H,A,A+rotate(90)*A); Y=extension(C,H,A,X); EE=foot(B,C,A); F=foot(C,A,B); Ap=-A; SS=extension(A,H,EE,F); D=extension(A,O,B,C);
draw(A--Ap,qua+Dotted); draw(B--Ap--C,qua); draw(B--X,tri); draw(C--Y,tri); draw(EE--F,sec+Dotted); draw(X--Y,sec); draw(B--Y,sec); draw(C--X,sec); filldraw(circumcircle(B,C,X),sfil,sec); filldraw(A--B--C--cycle,fil,pri); filldraw(circle(O,1),fil,pri);
dot("\(A\)",A,A); dot("\(B\)",B,dir(210)); dot("\(C\)",C,dir(-30)); dot("\(H\)",H,dir(300)); dot("\(X\)",X,N); dot("\(Y\)",Y,W); dot("\(E\)",EE,dir(30)); dot("\(F\)",F,dir(120)); dot("\(S\)",SS,N); dot("\(A'\)",Ap,Ap); dot("\(D\)",D,S);
// Block 2
size(6cm); defaultpen(fontsize(10pt)); pen pri=blue; pen sec=lightblue; pen tri=heavycyan; pen qua=paleblue; pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pair O,A,B,C,H,X,Y,EE,F,Ap,SS,D; O=(0,0); A=dir(147.55); B=dir(195); C=dir(345); H=A+B+C; X=extension(B,H,A,A+rotate(90)*A); Y=extension(C,H,A,X); EE=foot(B,C,A); F=foot(C,A,B); Ap=-A; SS=extension(A,H,EE,F); D=extension(A,O,B,C); draw(A--Ap,qua+Dotted); draw(B--Ap--C,qua); draw(B--X,tri); draw(C--Y,tri); draw(EE--F,sec+Dotted); draw(X--Y,sec); draw(B--Y,sec); draw(C--X,sec); filldraw(circumcircle(B,C,X),sfil,sec); filldraw(A--B--C--cycle,fil,pri); filldraw(circle(O,1),fil,pri); dot("\(A\)",A,A); dot("\(B\)",B,dir(210)); dot("\(C\)",C,dir(-30)); dot("\(H\)",H,dir(300)); dot("\(X\)",X,N); dot("\(Y\)",Y,W); dot("\(E\)",EE,dir(30)); dot("\(F\)",F,dir(120)); dot("\(S\)",SS,N); dot("\(A'\)",Ap,Ap); dot("\(D\)",D,S); | [] |
527 | Let $ABC$ be an acute triangle with orthocenter $H$ and circumcenter $O$. The tangent to the circumcircle of $\triangle ABC$ at $A$ intersects lines $BH$ and $CH$ at $X$ and $Y$, and $BY\parallel CX$. Let line $AO$ intersect $\overline{BC}$ at $D$. Suppose that $AO=25, BC=49$, and $AD=a-b\sqrt{c}$ for positive integers $a, b, c,$ where $c$ is not divisible by the square of any prime. Find $a+b+c$. | 2021 CIME I Problems/Problem 14 | Let $E$ and $F$ be the feet of the altitudes from $B$ to $\overline{AC}$ and from $C$ to $\overline{AB}$, respectively. It is clear that $\overline{EF} \parallel \overline{XY}$, as $\angle XAE = \angle AEF = \angle B$ and $\angle YAF = \angle AFE = \angle C$. We conjecture that quadrilateral $BCXY$ is cyclic because it looks cyclic. Indeed, note that for this to be true we need $\angle XYC = \angle XBC$, but $\overline{EF} \parallel \overline{XY}$ reduces this condition to $\angle EFC = \angle EBC$, which is obvious as $BCEF$ is cyclic with $\angle BEC = \angle BFC = 90^{\circ}$.
Thus, note that $BCXY$ cyclic along with $\overline{BY} \parallel \overline{CX}$ are enough to imply that it is an isosceles trapezoid and $XY = BC = 49$. Then, going back to $\overline{EF} \parallel \overline{XY}$, it is evident that $\triangle HEF$ and $\triangle HXY$ are directly similar and the ratio in which they are similar is given by \[\frac{EF}{XY} =\frac{EF}{BC} =\frac{a \cos A}{a} = \cos A.\] Consider the geometric transformation consisting of a homothety centered at $A$ with ratio $\cos A$ followed by a reflection about the bisector of $\angle A$, under which $\triangle ABC$ maps to $\triangle AEF$. Then, point $D$ (which lies on both $\overline{BC}$ and $\overline{AO}$) maps to the point $D^{\prime}$ on $\overline{EF}$ such that $\overline{AO}$ and $\overline{AD^{\prime}}$ are isogonal wrt. $\angle A$. But we know that the circumcenter and orthocenter of $\triangle ABC$ are isogonal conjugates, and this is enough to imply $D^{\prime}$ lies on $\overline{AH}$. Now \[\frac{HD^{\prime}}{HA} = \cos A \implies\frac{AD^{\prime}}{AH} = 1 - \cos A \implies AD^{\prime} = 2R \cos A(1 - \cos A) \implies AD = 2R(1 - \cos A)\] where the final step comes from undoing the homothety and reflection, thereby dividing all lengths by the ratio $\cos A$. We are given $a = 49$ and $R = 25$, thus $\sin A = \tfrac{a}{2R} = \tfrac{49}{50}$. It follows that, as $\triangle ABC$ is an acute triangle, $\cos A = \sqrt{1 - \left(\tfrac{49}{50}\right)^{2}} = \tfrac{3\sqrt{11}}{50}$. Finally, \[50\left(1 - \tfrac{3\sqrt{11}}{50}\right) = 50 - 3\sqrt{11}\] and the answer is $50 + 3 + 11 = \boxed{64}$.
~StressedPineapple | // Block 1
import olympiad;
size(7cm); defaultpen(linewidth(0.7));
usepackage("mathptmx");
pair A = (-15, 20), B = (-24, -7), C = (24, -7), O = (0, 0), H = A + B + C, Ap = -A, X = extension(B, H, A, A + rotate(90, O) * A), Y = extension(C, H, A, A + rotate(90, O) * A), D = extension(A, O, B, C), E = extension(A, C, B, X), F = extension(A, B, C, Y);
draw(A--B--C--cycle, blue);
draw(B--X^^C--Y, fuchsia);
draw(B--Y--X--C^^E--F, lightblue);
draw(H--A--D, lightblue+dotted);
draw(circle(O, 25), blue);
draw(circumcircle(B, C, X), lightblue);
dot("$A$", A, dir(127));
dot("$B$", B, dir(225));
dot("$C$", C, dir(315));
dot("$H$", H, dir(285));
dot("$X$", X, dir(80));
dot("$Y$", Y, dir(175));
dot("$D$", D, dir(270));
dot("$E$", E, dir(15));
dot("$F$", F, dir(210));
dot("$O$", O, dir(210));
dot(extension(A, H, E, F));
draw(anglemark(C, Y, X, 105), pink);
draw(anglemark(C, F, E, 105), pink);
draw(anglemark(C, B, E, 105), pink);
// Block 2
import olympiad; size(7cm); defaultpen(linewidth(0.7)); usepackage("mathptmx"); pair A = (-15, 20), B = (-24, -7), C = (24, -7), O = (0, 0), H = A + B + C, Ap = -A, X = extension(B, H, A, A + rotate(90, O) * A), Y = extension(C, H, A, A + rotate(90, O) * A), D = extension(A, O, B, C), E = extension(A, C, B, X), F = extension(A, B, C, Y); draw(A--B--C--cycle, blue); draw(B--X^^C--Y, fuchsia); draw(B--Y--X--C^^E--F, lightblue); draw(H--A--D, lightblue+dotted); draw(circle(O, 25), blue); draw(circumcircle(B, C, X), lightblue); dot("$A$", A, dir(127)); dot("$B$", B, dir(225)); dot("$C$", C, dir(315)); dot("$H$", H, dir(285)); dot("$X$", X, dir(80)); dot("$Y$", Y, dir(175)); dot("$D$", D, dir(270)); dot("$E$", E, dir(15)); dot("$F$", F, dir(210)); dot("$O$", O, dir(210)); dot(extension(A, H, E, F)); draw(anglemark(C, Y, X, 105), pink); draw(anglemark(C, F, E, 105), pink); draw(anglemark(C, B, E, 105), pink); | [] |
528 | Right triangle $ABC$ has side lengths $BC=6$, $AC=8$, and $AB=10$. A circle centered at $O$ is tangent to line $BC$ at $B$ and passes through $A$. A circle centered at $P$ is tangent to line $AC$ at $A$ and passes through $B$. What is $OP$?
$\textbf{(A)}\ \frac{23}{8} \qquad\textbf{(B)}\ \frac{29}{10} \qquad\textbf{(C)}\ \frac{35}{12} \qquad\textbf{(D)}\ \frac{73}{25} \qquad\textbf{(E)}\ 3$ | 2021 Fall AMC 12B Problems/Problem 22 | In a Cartesian plane, let $C, B,$ and $A$ be $(0,0),(0,6),(8,0)$ respectively.
~(Diagram By MaPhyCom)
By analyzing the behaviors of the two circles, we set $O$ to be $(a,6)$ and $P$ be $(b,8)$.
Hence derive the two equations:
$(x-a)^2+(y-6)^2=a^2$
$(x-8)^2+(y-b)^2=b^2$
Considering the coordinates of $A$ and $B$ for the two equations respectively, we get:
$(8-a)^2+(0-6)^2=a^2$
$(0-8)^2+(6-b)^2=b^2$
Solve to get $a=\frac{25}{4}$ and $b=\frac{25}{3}$
Through using the distance formula,
$OP=\sqrt{(8-\frac{25}{4})^2+(\frac{25}{3}-6)^2}= \boxed{\textbf{(C)}\ \frac{35}{12}}$.
~Wilhelm Z | // Block 1
size(8cm);
pair A = (0,8);
pair B = (6,0);
pair C = (0,0);
pair O = (6,6.25);
pair P = (8.3333,8);
draw(A--B--C--cycle);
draw(Circle(O, 6.2498));
draw(Circle(P, 8.3331));
draw(O--P);
draw(O--B, red+dotted+1.2bp);
draw(A--P, blue+dotted+1.2bp);
filldraw(O--(0,6.25)--A--cycle, red+opacity(0.3), invisible);
filldraw(P--B--(6,8)--cycle, blue+opacity(0.3), invisible);
dot(O);
dot(P);
label("$A$", A, W);
label("$B$", B, S);
label("$C$", C, W);
label("$O$", O, S);
label("$P$", P, E);
label("$a$", midpoint(O--B), W);
label("$b$", midpoint(P--A), N);
draw(rightanglemark(A,C,B,15));
// Block 2
size(8cm); pair A = (0,8); pair B = (6,0); pair C = (0,0); pair O = (6,6.25); pair P = (8.3333,8); draw(A--B--C--cycle); draw(Circle(O, 6.2498)); draw(Circle(P, 8.3331)); draw(O--P); draw(O--B, red+dotted+1.2bp); draw(A--P, blue+dotted+1.2bp); filldraw(O--(0,6.25)--A--cycle, red+opacity(0.3), invisible); filldraw(P--B--(6,8)--cycle, blue+opacity(0.3), invisible); dot(O); dot(P); label("$A$", A, W); label("$B$", B, S); label("$C$", C, W); label("$O$", O, S); label("$P$", P, E); label("$a$", midpoint(O--B), W); label("$b$", midpoint(P--A), N); draw(rightanglemark(A,C,B,15)); | [] |
529 | 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 Problems/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); | [] |
530 | 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 Problems/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); | [] |
531 | 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 Problems/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); | [] |
531 | 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 Problems/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); | [] |
531 | 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 Problems/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); | [] |
531 | 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 Problems/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)); | [] |
531 | 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 Problems/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); | [] |
531 | 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 Problems/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)); | [] |
532 | 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 Problems/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)); | [] |
532 | 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 Problems/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); | [] |
533 | 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 Problems/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/intermediate_geometry/2022_AIME_II_ProblemsProblem_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"] |
533 | 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 Problems/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)); | [] |
534 | Isosceles trapezoid $ABCD$ has parallel sides $\overline{AD}$ and $\overline{BC},$ with $BC < AD$ and $AB = CD.$ There is a point $P$ in the plane such that $PA=1, PB=2, PC=3,$ and $PD=4.$ What is $\tfrac{BC}{AD}?$
$\textbf{(A) }\frac{1}{4}\qquad\textbf{(B) }\frac{1}{3}\qquad\textbf{(C) }\frac{1}{2}\qquad\textbf{(D) }\frac{2}{3}\qquad\textbf{(E) }\frac{3}{4}$ | 2022 AMC 10A Problems/Problem 23 | Consider the reflection $P^{\prime}$ of $P$ over the perpendicular bisector of $\overline{BC}$, creating two new isosceles trapezoids $DAPP^{\prime}$ and $CBPP^{\prime}$. Under this reflection, $P^{\prime}A=PD=4$, $P^{\prime}D=PA=1$, $P^{\prime}C=PB=2$, and $P^{\prime}B=PC=3$.
Since $DAPP'$ and $CBPP'$ are isosceles trapezoids, they are cyclic. Using Ptolemy's theorem on $DAPP'$, we get that $(PP')(AD) + (PA)(P'D) = (AP')(PD)$, so
\[PP' \cdot AD + 1 \cdot 1 = 4 \cdot 4.\] Then, using Ptolemy's theorem again on $CBPP'$, we get that $(BC)(PP') + (BP)(CP') = (BP')(CP)$, so
\[PP' \cdot BC + 2 \cdot 2 = 3 \cdot 3.\] Thus, $PP^{\prime}\cdot AD=15$ and $PP^{\prime}\cdot BC=5$; dividing these two equations and taking the reciprocal yields $\frac{BC}{AD}=\boxed{\textbf{(B) }\frac{1}{3}}$.
(diagram by cinnamon_e) | // Block 1
size(300);
pair A = (0,0);
pair B = (1, 2);
pair C = (2,2);
pair D = (3,0);
label("$A$", A, SW);
label("$B$", B, NW);
label("$C$", C, NE);
label("$D$", D, SE);
draw(A--B--C--D--cycle, blue);
pair P = (0.8, 0.6);
dot("$P$", P, NW);
draw(P--A, magenta);
draw(P--B, magenta);
draw(P--C);
draw(P--D);
label("$1$", P--A, NW);
label("$2$", P--B, E);
label("$3$", P--C, NW);
label("$4$", P--D, S);
pair P1 = (2.2, 0.6);
dot("$P'$", P1, NE);
draw(P1--D, magenta);
draw(P1--C, magenta);
draw(P1--A);
draw(P1--B);
label("$1$", P1--D, NE);
label("$2$", P1--C, E);
label("$3$", P1--B, NE);
label("$4$", P1--A, SE);
draw(P--P1, dashed+magenta);
// Block 2
size(300); pair A = (0,0); pair B = (1, 2); pair C = (2,2); pair D = (3,0); label("$A$", A, SW); label("$B$", B, NW); label("$C$", C, NE); label("$D$", D, SE); draw(A--B--C--D--cycle, blue); pair P = (0.8, 0.6); dot("$P$", P, NW); draw(P--A, magenta); draw(P--B, magenta); draw(P--C); draw(P--D); label("$1$", P--A, NW); label("$2$", P--B, E); label("$3$", P--C, NW); label("$4$", P--D, S); pair P1 = (2.2, 0.6); dot("$P'$", P1, NE); draw(P1--D, magenta); draw(P1--C, magenta); draw(P1--A); draw(P1--B); label("$1$", P1--D, NE); label("$2$", P1--C, E); label("$3$", P1--B, NE); label("$4$", P1--A, SE); draw(P--P1, dashed+magenta); | [] |
534 | Isosceles trapezoid $ABCD$ has parallel sides $\overline{AD}$ and $\overline{BC},$ with $BC < AD$ and $AB = CD.$ There is a point $P$ in the plane such that $PA=1, PB=2, PC=3,$ and $PD=4.$ What is $\tfrac{BC}{AD}?$
$\textbf{(A) }\frac{1}{4}\qquad\textbf{(B) }\frac{1}{3}\qquad\textbf{(C) }\frac{1}{2}\qquad\textbf{(D) }\frac{2}{3}\qquad\textbf{(E) }\frac{3}{4}$ | 2022 AMC 10A Problems/Problem 23 | Extend $AB$ and $CD$ to a point $Q$ as shown, and let $PQ = s$. Then let $BQ=CQ = a$ and $AB=DC = b$. Notice that $\frac{BC}{AD} = \frac{QC}{QD} = \frac{a}{a+b}$ by similar triangles.
By Stewart's theorem on $APQ$ and $DPQ$, we have \begin{align*} ab(a+b) + 9(a+b) &= 16a + s^2b \\ ab(a+b) + 4(a+b) &= a + s^2b \\ \end{align*}
Subtracting, $5(a+b) = 15a$, and so $\frac{BC}{AD} = \frac{a}{a+b} = \frac{5}{15} = \boxed{\textbf{(B) }\frac{1}{3}}$.
~kred9 (minor edit by gwang2008) | // Block 1
size(7.5cm);
draw((0,0)--(4.2,0));
draw((0,0)--(1.4,2)--(2.8,2)--(4.2,0));
draw((1.4,2)--(2.1,3)--(2.8,2));
draw((0,0)--(1,0.5)--(1.4,2)--(1,0.5)--(2.8,2)--(1,0.5)--(4.2,0));
label("$A$",(0,0),SW);
label("$B$",(1.4,2),NW);
label("$C$",(2.8,2),NE);
label("$D$",(4.2,0),SE);
label("$P$",(1,0.5),NW);
label("$Q$",(2.1,3),N);
draw((2.1,3)--(1,0.5));
// Block 2
size(7.5cm); draw((0,0)--(4.2,0)); draw((0,0)--(1.4,2)--(2.8,2)--(4.2,0)); draw((1.4,2)--(2.1,3)--(2.8,2)); draw((0,0)--(1,0.5)--(1.4,2)--(1,0.5)--(2.8,2)--(1,0.5)--(4.2,0)); label("$A$",(0,0),SW); label("$B$",(1.4,2),NW); label("$C$",(2.8,2),NE); label("$D$",(4.2,0),SE); label("$P$",(1,0.5),NW); label("$Q$",(2.1,3),N); draw((2.1,3)--(1,0.5)); | [] |
535 | Let $c$ be a real number, and let $z_1$ and $z_2$ be the two complex numbers satisfying the equation
$z^2 - cz + 10 = 0$. Points $z_1$, $z_2$, $\frac{1}{z_1}$, and $\frac{1}{z_2}$ are the vertices of (convex) quadrilateral $\mathcal{Q}$ in the complex plane. When the area of $\mathcal{Q}$ obtains its maximum possible value, $c$ is closest to which of the following?
$\textbf{(A) }4.5 \qquad\textbf{(B) }5 \qquad\textbf{(C) }5.5 \qquad\textbf{(D) }6\qquad\textbf{(E) }6.5$ | 2022 AMC 12A Problems/Problem 22 | Note that $z=\dfrac c2\pm\dfrac{\sqrt{c^2-40}}2$, so let $z_1=\dfrac c2+\dfrac{\sqrt{c^2-40}}2$ and $z_2=\dfrac c2-\dfrac{\sqrt{c^2-40}}2$. Taking a look at the answer choices, they range between $c=4.5$ to $c=6.5$, and in that range, $c^2$ is always less than $40$. Thus, $c^2-40<0$ for our possible answer choices; we can then rewrite $z_1$ and $z_2$ as $\dfrac c2+\dfrac{\sqrt{40-c^2}}2i$ and $\dfrac c2-\dfrac{\sqrt{40-c^2}}2i$, respectively, with real coefficients.
Let us compute $\dfrac1z$:
\[\dfrac1z=\dfrac1{\frac c2\pm\frac{\sqrt{c^2-40}}2}=\dfrac{\frac c2\mp\frac{\sqrt{c^2-40}}2}{\left(\frac c2\right)^2-\left(\frac{\sqrt{c^2-40}}2\right)^2}=\dfrac{\frac c2\mp\frac{\sqrt{c^2-40}}2}{\frac{c^2}4-\frac{c^2-40}4}=\dfrac{2c\mp2\sqrt{c^2-40}}{40}=\dfrac{c\mp\sqrt{c^2-40}}{20}.\]
Then, $\dfrac1{z_1}=\dfrac{c-\sqrt{c^2-40}}{20}=\dfrac c{20}-\dfrac{\sqrt{40-c^2}}{20}i$ while $\dfrac1{z_2}=\dfrac{c+\sqrt{c^2-40}}{20}=\dfrac c{20}+\dfrac{\sqrt{40-c^2}}{20}i$.
In the complex plane, we can draw a rough sketch of $z_1,z_2,\dfrac1{z_1},\dfrac1{z_2}$:
Note that, here, our quadrilateral is an isoceles trapezoid. The shorter base length is $\left(\dfrac{\sqrt{40-c^2}}{20}-\left(-\dfrac{\sqrt{40-c^2}}{20}\right)\right)=\dfrac1{10}\sqrt{40-c^2}$.
The longer base length is $\left(\dfrac{\sqrt{40-c^2}}2-\left(-\dfrac{\sqrt{40-c^2}}2\right)\right)=\sqrt{40-c^2}$.
The average of the two bases is $\dfrac{11}{20}\sqrt{40-c^2}$.
The height of our trapezoid (which is horizontal parallel to the $x$-axis in our diagram above) is simply $\dfrac c2-\dfrac c{20}=\dfrac9{20}c$.
Since the area of a trapezoid is the product of the average of its bases and its height, we conclude that our trapezoid's area is $\dfrac{99}{400}c\sqrt{40-c^2}$, which is a function of $c$. Thus, let $A(c)=\dfrac{99}{400}c\sqrt{40-c^2}$.
Taking the derivative (FINALLY we get to the Calculus part haha, this definitely wasn't clickbait :3 trust me) with respect to $c$, we find that $\dfrac{dA}{dc}=\dfrac{99}{400}\left(\sqrt{40-c^2}+c\left(\dfrac{-2c}{2\sqrt{40-c^2}}\right)\right)=\dfrac{99}{400}\left(\sqrt{40-c^2}-\dfrac{c^2}{\sqrt{40-c^2}}\right)$.
To find an extremum, we set the derivative equal to zero:
\begin{align*}
\dfrac{dA}{dc}&=\dfrac{99}{400}\left(\sqrt{40-c^2}-\dfrac{c^2}{\sqrt{40-c^2}}\right) \\
0&=\dfrac{99}{400}\left(\sqrt{40-c^2}-\dfrac{c^2}{\sqrt{40-c^2}}\right) \\
\sqrt{40-c^2}-\dfrac{c^2}{\sqrt{40-c^2}}&=0 \\
\sqrt{40-c^2}&=\dfrac{c^2}{\sqrt{40-c^2}} \\
\left(\sqrt{40-c^2}\right)^2&=c^2 \\
c^2&=40-c^2 \\
2c^2&=40 \\
c^2&=20 \\
c&=\sqrt{20} \\
&\approx4.47 \\
\end{align*}
Clearly, this is very close to $\boxed{\textbf{(A)}~4.5}$, so we are done. QED.
~Technodoggo
(Minor Edits by dolphindesigner) | // Block 1
import graph;
unitsize(0.5cm);
/*xaxis(Ticks, xmin=-1,xmax=8);
yaxis(Ticks, ymin=-11,ymax=11);*/
draw((-1,0)--(8,0));
draw((0,-11)--(0,11));
dot((7,10));dot((7,-10));dot((0.7,1));dot((0.7,-1));
draw((7,10)--(7,-10)--(0.7,-1)--(0.7,1)--cycle);
label("$z_1$", (7,10), E);
label("$z_2$", (7,-10), E);
label("$\frac1{z_2}$", (0.7,1), N);
label("$\frac1{z_1}$", (0.7,-1), S);
// Block 2
import graph; unitsize(0.5cm); /*xaxis(Ticks, xmin=-1,xmax=8); yaxis(Ticks, ymin=-11,ymax=11);*/ draw((-1,0)--(8,0)); draw((0,-11)--(0,11)); dot((7,10));dot((7,-10));dot((0.7,1));dot((0.7,-1)); draw((7,10)--(7,-10)--(0.7,-1)--(0.7,1)--cycle); label("$z_1$", (7,10), E); label("$z_2$", (7,-10), E); label("$\frac1{z_2}$", (0.7,1), N); label("$\frac1{z_1}$", (0.7,-1), S); | [] |
536 | 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 Problems/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); | [] |
536 | 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 Problems/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); | [] |
537 | 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 Problems/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); | [] |
537 | 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 Problems/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)); | [] |
538 | 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 Problems/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)); | [] |
538 | 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 Problems/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)); | [] |
538 | 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 Problems/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); | [] |
538 | 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 Problems/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); | [] |
539 | 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 Problems/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); | [] |
539 | 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 Problems/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); | [] |
540 | 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 Problems/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)); | [] |
541 | 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 Problems/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); | [] |
542 | A regular pentagon with area $\sqrt{5}+1$ is printed on paper and cut out. The five vertices of the pentagon are folded into the center of the pentagon, creating a smaller pentagon. What is the area of the new pentagon?
$\textbf{(A)}~4-\sqrt{5}\qquad\textbf{(B)}~\sqrt{5}-1\qquad\textbf{(C)}~8-3\sqrt{5}\qquad\textbf{(D)}~\frac{\sqrt{5}+1}{2}\qquad\textbf{(E)}~\frac{2+\sqrt{5}}{3}$ | 2023 AMC 12B Problems/Problem 25 | We can find the area of the red pentagon by taking the area of the total pentagon and subtracting the area outside the red pentagon.
The area outside the red pentagon is the sum of the larger isosceles triangles, but this double counts the overlapping regions of the small isosceles triangles, so we have to subtract those out.
We have $[FGHIJ] = [ABCDE]-(5 \cdot[DKT]-5 \cdot [PFK])$
Lets focus on finding the area of each individual triangle:
Notice that we have no information about the side length, so instead we let the side length be $s$. Now we can drop an altitude from $O$ to the base of the triangle, and we know this altitude must split the base of the pentagon in half, so we can create a right triangle. Furthermore, draw a line from $O$ to $D$. This must bisect angle $D$ which is $108$ degrees, so we create $36-54-90$ triangles. Specifically, we know $\angle ODK = 54^{\circ}$, $\angle DOU = 36^{\circ}$, and $\angle DTK = 36^{\circ}$ because $\triangle DTK$ is isosceles and we know the vertex angle is $108^{\circ}$. We encode this information in the diagram below:
Since $\triangle DKT$ is isosceles, the area of $\triangle DVT$ is half the area of $\triangle DKT$. Similarly, the area of $\triangle UFK$ is half that of $\triangle PFK$. Thus:
\[[FGHIJ] = [ABCDE]-(5 \cdot[DKT]-5 \cdot [PFK]) \implies [FGHIJ] = [ABCDE]-(10 \cdot [DVT]-10 \cdot [UFK])\]
We also know that since we dropped an altitude from $O$ to $U$, the area of $\triangle ODU$ must be half of a fifth of the total area of the pentagon. Therefore we can rewrite the above equation as \[[FGHIJ]=10 \cdot ([ODU]-[DVT]+[UFK])\]
Now notice that $\triangle ODU ~ \triangle TDV ~ \triangle KFU$ by AA similarity. Therefore, if we can write the areas of the latter two triangles as a ratio of the first triangle, we can express the whole equation in terms of $[ODU]$, and by extension $[ABCDE]$, which we know. To find these ratios, we can find the side length ratios and square them because the triangles are similar.
We already know $DU = \frac{s}{2}$, so let's try to find it's analogous side for $\triangle TDV$ and $\triangle KFU$. These sides are $DV$ and $FU$, respectively.
First, $\frac{s}{2} = OD \cdot cos(54^{\circ})$, so $OD = \frac{s}{2} \cdot sec(54^{\circ})$. Then notice that $DV = \frac{OD}{2}$ because we have to fold $D$ to hit $O$, so the folding crease has to be exactly halfway between $O$ and $D$. Therefore, \[DV = \frac{s}{4} \cdot sec(54^{\circ}) \implies \frac{DU}{DV} = \frac{\frac{s}{2}}{\frac{s}{4} \cdot sec(54^{\circ})} = 2 \cdot cos(54^{\circ})\]
Now the ratio between the area of two similar triangles is the square of the ratio of their analogous side lengths. Thus
\[\frac{[ODU]}{[TDV]} = 4 \cdot cos^2(54^{\circ}) \implies \frac{10 \cdot [ODU]}{10 \cdot [TDV]} = 4 \cdot cos^2(54^{\circ})\]
\[\implies \frac{[ABCDE]}{10 \cdot [TDV]} = 4 \cdot cos^2(54^{\circ}) \implies \frac{\sqrt{5}+1}{10 \cdot [TDV]} = 4 \cdot cos^2(54^{\circ})\]
\[\implies 10 \cdot [TDV] = \frac{\sqrt{5}+1}{4 \cdot cos^2(54^{\circ})} \implies 10 \cdot [TDV] = \frac{\sqrt{5}+1}{4} \cdot sec^2(54^{\circ})\]
Now let's move on and calculate the ratio of the other side length. Calculating $FU$ is slightly tricker.
First, we find $TD$: $TD \cdot cos(54^{\circ}) = DV = \frac{s}{4} sec^54^{\circ}) \implies TD = \frac{s}{4} \cdot sec^2(54^{\circ})$. Now since $\triangle DTK$ is isosceles, $TD = DK$ and $UK = DK-DU = TD-DU = \frac{s}{4} \cdot sec^2(54^{\circ})-\frac{s}{2} = \frac{s}{4} \cdot (sec^2(54^{\circ})-2) = \frac{s}{4} \cdot (tan^2(54^{\circ})-1)$.
Now $FU = UK \cdot tan(36^{\circ}) \implies FU \frac{s}{4} \cdot (tan^2(54^{\circ})-1) \cdot tan(36^{\circ})$. Now note that $tan(x) \cdot tan(90-x) = 1$ because opposite over adjacent cancel each other out in a right triangle. Thus, $FU = \frac{s}{4} \cdot (tan(54^{\circ})-tan(36^{\circ}))$
Now, \[\frac{DU}{FU} = \frac{\frac{s}{2}}{\frac{s}{4} \cdot (tan(54^{\circ})-tan(36^{\circ}))} = \frac{2}{tan(54^{\circ})-tan(36^{\circ})}\]
\[\implies \frac{[DU]}{[FU]} = \frac{4}{(tan(54^{\circ})-tan(36^{\circ}))^2} \implies 10 \cdot [FU] = \frac{1+\sqrt{5}}{4} \cdot (tan(54^{\circ})-tan(36^{\circ}))^2\]
Now we go back to our first equation and plug in our values:
\[[FGHIJ] = [ABCDE]-(10 \cdot [DVT]-10 \cdot [UFK]) \implies [FGHIJ] = 1+\sqrt{5}-\frac{\sqrt{5}+1}{4} \cdot sec^2(54^{\circ})+\frac{1+\sqrt{5}}{4} \cdot (tan(54^{\circ})-tan(36^{\circ}))^2\]
\[\implies [FGHIJ] = \frac{1+\sqrt{5}}{4} \cdot (4-sec^2(54^{\circ})+(tan(54^{\circ})-tan(36^{\circ}))^2)\]
Note $(tan(x)-tan(90-x))^2 = tan^2(x)-2 \cdot tan(x) \cdot tan(90-x)+tan^2(90-x) = tan^2(x)-2+tan^2(90-x)$.
Also note that $tan^2(x)+1 = sec^2(x)$.
Thus \[[FGHIJ] = \frac{1+\sqrt{5}}{4} \cdot (2-sec^2(54^{\circ})+tan^2(54^{\circ})+tan^2(36^{\circ}))\]
\[\implies [FGHIJ] = \frac{1+\sqrt{5}}{4} \cdot (2+(-1)+tan^2(36^{\circ})) \implies [FGHIJ] = \frac{1+\sqrt{5}}{4} \cdot (1+tan^2(36^{\circ}))\].
Now all that remains is to find $tan^2(36^{\circ})$. We can use the tan addition formula to find the general form of $tan(5x)$ or remember question 25 from this year's AMC 12A. We have that \[tan(5x) = \frac{5tan(x)-10tan^3(x)+tan^5(x)}{1-10tan^2(x)+5tan^4(x)}\].
Plug in $x=36$. Then we have \[tan(180^{\circ}) = \frac{5tan(36^{\circ})-10tan^3(36^{\circ})+tan^5(36^{\circ})}{1-10tan^2(36^{\circ})+5tan^4(36^{\circ})}\]
Now let $y = tan(36^{\circ})$. We have the equation \[\frac{5y-10y^3+y^5}{1-10y^2+5y^4} = 0 \implies 5y-10y^3+y^5 = 0\]
\[\implies 5-10y^2+y^4 = 0 \implies 5-10z+z^2 = 0\]
Where we let $z = y^2$. Using the quadratic formula, we have \[z = \frac{10 \pm \sqrt{80}}{2} = 5 \pm 2\sqrt{5}\]
Now since $y = tan(36^{\circ})$, $z = tan^2(36^{\circ})$, which is what we were looking for. Notice that $tan(0^{\circ}) = 0$ and $tan(45^{\circ}) = 1$, so $tan(36^{\circ})$ is between $0$ and $1$, and so is it's square. Thus $z = 5 - 2\sqrt{5}$, not the other root.
Finally:
\[[FGHIJ] = \frac{1+\sqrt{5}}{4} \cdot (1+tan^2(36^{\circ})) = \frac{1+\sqrt{5}}{4} \cdot (1+5-2\sqrt{5})\]
\[\implies [FGHIJ] = \frac{1+\sqrt{5}}{4} \cdot (6-2\sqrt{5}) = \frac{4\sqrt{5}-4}{4} = \sqrt{5}-1\]
Therefore, \[[FGHIJ] = \sqrt{5}-1 = \boxed{B}\]
~KingRavi | // Block 1
unitsize(5cm);
// Define the vertices of the pentagons
pair A, B, C, D, E;
pair F, G, H, I, J;
// Calculate the vertices of the larger pentagon
A = dir(90);
B = dir(90 - 72);
C = dir(90 - 2*72);
D = dir(90 - 3*72);
E = dir(90 - 4*72);
// Draw the larger pentagon
draw(A--B--C--D--E--cycle);
pair O = (A+B+C+D+E)/5;
pair AA,OO;
real gap = 0.02;
AA = A+(0,0);
OO = O+(0,0);
label("$O$", O, S);
pair OOO, OAO;
OOO = O+(gap,0);
OAO = (O+A)/2 + (gap,0);
dot(O);
label("$A$", (0,1), E);
label("$B$", B, S);
label("$C$", C, S);
label("$D$", D, S);
label("$E$", E, W);
real scaleFactor = 1/1.618; // Adjust this value as needed
// Rotate the smaller pentagon by 180 degrees
F = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 + 180);
G = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 72 + 180);
H = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 2*72 + 180);
I = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 3*72 + 180);
J = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 4*72 + 180);
pair K, L, M, N, O, P, Q, R, S, T, U, V;
real newScaleFactor = 0.8507;
K = newScaleFactor*dir(270+18);
L = newScaleFactor*dir(270+72+18);
M = newScaleFactor*dir(270+72+72+18);
N = newScaleFactor*dir(270+72+72+72+18);
O = newScaleFactor*dir(270+72+72+72+72+18);
P = newScaleFactor*dir(270-18);
Q = newScaleFactor*dir(270+72-18);
R = newScaleFactor*dir(270+72+72-18);
S = newScaleFactor*dir(270+72+72+72-18);
T = newScaleFactor*dir(270+72+72+72+72-18);
label("$K$", K, S);
label("$L$", L, S);
label("$M$", M, S);
label("$N$", N, S);
label("$O$", O, W);
label("$P$", P, S);
label("$Q$", Q, E);
label("$R$", R, S);
label("$S$", S, S);
label("$T$", T, W);
draw(K--T, dashed);
draw(S--O, dashed);
draw(P--L, dashed);
draw(Q--M, dashed);
draw(R--N, dashed);
label("$F$", F, S);
label("$G$", G, S);
label("$H$", H, S);
label("$I$", I, S);
label("$J$", J, S);
// Draw the smaller pentagon
draw(F--G--H--I--J--cycle,red);
// Block 2
unitsize(5cm);
// Define the vertices of the pentagons
pair A, B, C, D, E;
pair F, G, H, I, J;
// Calculate the vertices of the larger pentagon
A = dir(90);
B = dir(90 - 72);
C = dir(90 - 2*72);
D = dir(90 - 3*72);
E = dir(90 - 4*72);
pair O = (A+B+C+D+E)/5;
pair AA,OO;
real gap = 0.02;
AA = A+(0,0);
OO = O+(0,0);
label("$O$", O, S);
pair OOO, OAO;
OOO = O+(gap,0);
OAO = (O+A)/2 + (gap,0);
dot(O);
label("$D$", D, S);
real scaleFactor = 1/1.618; // Adjust this value as needed
// Rotate the smaller pentagon by 180 degrees
F = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 + 180);
G = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 72 + 180);
H = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 2*72 + 180);
I = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 3*72 + 180);
J = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 4*72 + 180);
label("$F$", F, N);
pair K, L, M, N, O, P, Q, R, S, T, U, V;
real newScaleFactor = 0.8507;
K = newScaleFactor*dir(270+18);
T = newScaleFactor*dir(270+72+72+72+72-18);
label("$K$", K, E);
label("$T$", T, W);
draw(K--T);
draw(K--D);
draw(D--T);
// Block 3
unitsize(5cm);
// Define the vertices of the pentagons
pair A, B, C, D, E;
pair F, G, H, I, J;
// Calculate the vertices of the larger pentagon
A = dir(90);
B = dir(90 - 72);
C = dir(90 - 2*72);
D = dir(90 - 3*72);
E = dir(90 - 4*72);
pair O = (A+B+C+D+E)/5;
pair AA,OO;
real gap = 0.02;
AA = A+(0,0);
OO = O+(0,0);
label("$O$", O, E);
pair OOO, OAO;
OOO = O+(gap,0);
OAO = (O+A)/2 + (gap,0);
dot(O);
label("$D$", D, S);
real scaleFactor = 1/1.618; // Adjust this value as needed
// Rotate the smaller pentagon by 180 degrees
F = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 + 180);
G = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 72 + 180);
H = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 2*72 + 180);
I = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 3*72 + 180);
J = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 4*72 + 180);
label("$F$", (0.1,-1/1.618), E);
pair K, L, M, N, O, P, Q, R, S, T, U, V;
real newScaleFactor = 0.8507;
K = newScaleFactor*dir(270+18);
T = newScaleFactor*dir(270+72+72+72+72-18);
label("$K$", K, S);
label("$T$", T, W);
draw(K--T);
draw(K--D);
draw(D--T);
pair U;
U=(0,-0.809);
label("$U$",(0,-0.9), S);
draw(O--U);
draw(O--D);
pair V;
V = midpoint(O--D);
label("$V$", V+(0,0.05), N);
markscalefactor = 0.005;
draw(rightanglemark(D,U,O));
draw(rightanglemark(F,V,O));
draw(rightanglemark(K,U,F));
draw(anglemark(U,D,O));
label("$54^{\circ}$", D+(0.05,0),NE);
draw(anglemark(D,O,U));
label("$36^{\circ}$", O-(0,0.2),SW);
draw(anglemark(D,T,F));
label("$36^{\circ}$", T+(0.1,-0.17),SE);
label("$\frac{s}{2}$", D+(0.3,-0.1), S);
// Block 4
unitsize(5cm); // Define the vertices of the pentagons pair A, B, C, D, E; pair F, G, H, I, J; // Calculate the vertices of the larger pentagon A = dir(90); B = dir(90 - 72); C = dir(90 - 2*72); D = dir(90 - 3*72); E = dir(90 - 4*72); // Draw the larger pentagon draw(A--B--C--D--E--cycle); pair O = (A+B+C+D+E)/5; pair AA,OO; real gap = 0.02; AA = A+(0,0); OO = O+(0,0); label("$O$", O, S); pair OOO, OAO; OOO = O+(gap,0); OAO = (O+A)/2 + (gap,0); dot(O); label("$A$", (0,1), E); label("$B$", B, S); label("$C$", C, S); label("$D$", D, S); label("$E$", E, W); real scaleFactor = 1/1.618; // Adjust this value as needed // Rotate the smaller pentagon by 180 degrees F = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 + 180); G = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 72 + 180); H = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 2*72 + 180); I = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 3*72 + 180); J = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 4*72 + 180); pair K, L, M, N, O, P, Q, R, S, T, U, V; real newScaleFactor = 0.8507; K = newScaleFactor*dir(270+18); L = newScaleFactor*dir(270+72+18); M = newScaleFactor*dir(270+72+72+18); N = newScaleFactor*dir(270+72+72+72+18); O = newScaleFactor*dir(270+72+72+72+72+18); P = newScaleFactor*dir(270-18); Q = newScaleFactor*dir(270+72-18); R = newScaleFactor*dir(270+72+72-18); S = newScaleFactor*dir(270+72+72+72-18); T = newScaleFactor*dir(270+72+72+72+72-18); label("$K$", K, S); label("$L$", L, S); label("$M$", M, S); label("$N$", N, S); label("$O$", O, W); label("$P$", P, S); label("$Q$", Q, E); label("$R$", R, S); label("$S$", S, S); label("$T$", T, W); draw(K--T, dashed); draw(S--O, dashed); draw(P--L, dashed); draw(Q--M, dashed); draw(R--N, dashed); label("$F$", F, S); label("$G$", G, S); label("$H$", H, S); label("$I$", I, S); label("$J$", J, S); // Draw the smaller pentagon draw(F--G--H--I--J--cycle,red);
// Block 5
unitsize(5cm); // Define the vertices of the pentagons pair A, B, C, D, E; pair F, G, H, I, J; // Calculate the vertices of the larger pentagon A = dir(90); B = dir(90 - 72); C = dir(90 - 2*72); D = dir(90 - 3*72); E = dir(90 - 4*72); pair O = (A+B+C+D+E)/5; pair AA,OO; real gap = 0.02; AA = A+(0,0); OO = O+(0,0); label("$O$", O, S); pair OOO, OAO; OOO = O+(gap,0); OAO = (O+A)/2 + (gap,0); dot(O); label("$D$", D, S); real scaleFactor = 1/1.618; // Adjust this value as needed // Rotate the smaller pentagon by 180 degrees F = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 + 180); G = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 72 + 180); H = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 2*72 + 180); I = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 3*72 + 180); J = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 4*72 + 180); label("$F$", F, N); pair K, L, M, N, O, P, Q, R, S, T, U, V; real newScaleFactor = 0.8507; K = newScaleFactor*dir(270+18); T = newScaleFactor*dir(270+72+72+72+72-18); label("$K$", K, E); label("$T$", T, W); draw(K--T); draw(K--D); draw(D--T);
// Block 6
unitsize(5cm); // Define the vertices of the pentagons pair A, B, C, D, E; pair F, G, H, I, J; // Calculate the vertices of the larger pentagon A = dir(90); B = dir(90 - 72); C = dir(90 - 2*72); D = dir(90 - 3*72); E = dir(90 - 4*72); pair O = (A+B+C+D+E)/5; pair AA,OO; real gap = 0.02; AA = A+(0,0); OO = O+(0,0); label("$O$", O, E); pair OOO, OAO; OOO = O+(gap,0); OAO = (O+A)/2 + (gap,0); dot(O); label("$D$", D, S); real scaleFactor = 1/1.618; // Adjust this value as needed // Rotate the smaller pentagon by 180 degrees F = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 + 180); G = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 72 + 180); H = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 2*72 + 180); I = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 3*72 + 180); J = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 4*72 + 180); label("$F$", (0.1,-1/1.618), E); pair K, L, M, N, O, P, Q, R, S, T, U, V; real newScaleFactor = 0.8507; K = newScaleFactor*dir(270+18); T = newScaleFactor*dir(270+72+72+72+72-18); label("$K$", K, S); label("$T$", T, W); draw(K--T); draw(K--D); draw(D--T); pair U; U=(0,-0.809); label("$U$",(0,-0.9), S); draw(O--U); draw(O--D); pair V; V = midpoint(O--D); label("$V$", V+(0,0.05), N); markscalefactor = 0.005; draw(rightanglemark(D,U,O)); draw(rightanglemark(F,V,O)); draw(rightanglemark(K,U,F)); draw(anglemark(U,D,O)); label("$54^{\circ}$", D+(0.05,0),NE); draw(anglemark(D,O,U)); label("$36^{\circ}$", O-(0,0.2),SW); draw(anglemark(D,T,F)); label("$36^{\circ}$", T+(0.1,-0.17),SE); label("$\frac{s}{2}$", D+(0.3,-0.1), S); | [] |
542 | A regular pentagon with area $\sqrt{5}+1$ is printed on paper and cut out. The five vertices of the pentagon are folded into the center of the pentagon, creating a smaller pentagon. What is the area of the new pentagon?
$\textbf{(A)}~4-\sqrt{5}\qquad\textbf{(B)}~\sqrt{5}-1\qquad\textbf{(C)}~8-3\sqrt{5}\qquad\textbf{(D)}~\frac{\sqrt{5}+1}{2}\qquad\textbf{(E)}~\frac{2+\sqrt{5}}{3}$ | 2023 AMC 12B Problems/Problem 25 | Let $r_b$ and $r_s$ be the circumradius of the big and small pentagon, respectively. Let $a_s$ be the apothem of the smaller pentagon and $A_s$ and $A_b$ be the areas of the smaller and larger pentagon, respectively.
From the diagram:
\begin{align*} \cos{36^\circ} &= \dfrac{a_s}{r_s} = \dfrac{\phi}{2} = \dfrac{\sqrt{5}+1}{4}\\ a_s &= \dfrac{r_b}{2}\\ A_s &= \left(\dfrac{r_s}{r_b}\right)^2A_b\\ &=\left(\dfrac{a_s}{\cos{36^\circ} r_b}\right)^2 \left(1+\sqrt{5}\right)\\ &=\left(\dfrac{a_s}{\dfrac{\phi}{2} r_b}\right)^2 \left(1+\sqrt{5}\right)\\ &=\left(\dfrac{1}{2 \dfrac{\phi}{2}}\right)^2 \left(1+\sqrt{5}\right)\\ &=\left(\dfrac{2}{\sqrt{5}+1}\right)^2 \left(1+\sqrt{5}\right)\\ &=\dfrac{4}{\sqrt{5}+1} \\ &=\dfrac{4\left(\sqrt{5}-1\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)} \\ &=\sqrt{5}-1 \end{align*}
\[\boxed{\textbf{(B) }\sqrt{5}-1}\]
~Technodoggo | // Block 1
unitsize(5cm);
// Define the vertices of the pentagons
pair A, B, C, D, E;
pair F, G, H, I, J;
// Calculate the vertices of the larger pentagon
A = dir(90);
B = dir(90 - 72);
C = dir(90 - 2*72);
D = dir(90 - 3*72);
E = dir(90 - 4*72);
// Draw the larger pentagon
draw(A--B--C--D--E--cycle);
pair O = (A+B+C+D+E)/5;
pair AA,OO;
real gap = 0.02;
AA = A+(0,0);
OO = O+(0,0);
draw(AA--OO, blue);
pair OOO, OAO;
OOO = O+(gap,0);
OAO = (O+A)/2 + (gap,0);
draw(OOO--OAO,green);
dot(O);
dot((O+A)/2);
label("$r_b$", (O+A)*.7, E,blue);
label("$a_s$", (O+A)*.2 +(0+0.18,0.05), E,green);
label("$r_s$", O+(-0.175,0.2), E,pink);
label("$A$", (0,0), E);
real scaleFactor = 1/1.618; // Adjust this value as needed
// Rotate the smaller pentagon by 180 degrees
F = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 + 180);
G = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 72 + 180);
H = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 2*72 + 180);
I = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 3*72 + 180);
J = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 4*72 + 180);
// Draw the smaller pentagon
draw(F--G--H--I--J--cycle,red);
draw(arc(O,(H+I)*.5*.6,H*.6));
label("$36^\circ$",O+(+0.05,0.15),NW);
draw(O--H,pink);
// Block 2
unitsize(5cm); // Define the vertices of the pentagons pair A, B, C, D, E; pair F, G, H, I, J; // Calculate the vertices of the larger pentagon A = dir(90); B = dir(90 - 72); C = dir(90 - 2*72); D = dir(90 - 3*72); E = dir(90 - 4*72); // Draw the larger pentagon draw(A--B--C--D--E--cycle); pair O = (A+B+C+D+E)/5; pair AA,OO; real gap = 0.02; AA = A+(0,0); OO = O+(0,0); draw(AA--OO, blue); pair OOO, OAO; OOO = O+(gap,0); OAO = (O+A)/2 + (gap,0); draw(OOO--OAO,green); dot(O); dot((O+A)/2); label("$r_b$", (O+A)*.7, E,blue); label("$a_s$", (O+A)*.2 +(0+0.18,0.05), E,green); label("$r_s$", O+(-0.175,0.2), E,pink); label("$A$", (0,0), E); real scaleFactor = 1/1.618; // Adjust this value as needed // Rotate the smaller pentagon by 180 degrees F = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 + 180); G = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 72 + 180); H = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 2*72 + 180); I = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 3*72 + 180); J = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 4*72 + 180); // Draw the smaller pentagon draw(F--G--H--I--J--cycle,red); draw(arc(O,(H+I)*.5*.6,H*.6)); label("$36^\circ$",O+(+0.05,0.15),NW); draw(O--H,pink); | [] |
543 | 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 Problems/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); | [] |
544 | 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 Problems/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)); | [] |
544 | 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 Problems/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)); | [] |
545 | 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 Problems/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); | [] |
546 | 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 Problems/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); | [] |
547 | 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 Problems/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); | [] |
548 | $\triangle DEF$ is inscribed inside $\triangle ABC$ such that $D,E,F$ lie on $BC, AC, AB$, respectively. The circumcircles of $\triangle DEC, \triangle BFD, \triangle AFE$ have centers $O_1,O_2,O_3$, respectively. Also, $AB = 23, BC = 25, AC=24$, and $\stackrel{\frown}{BF} = \stackrel{\frown}{EC},\ \stackrel{\frown}{AF} = \stackrel{\frown}{CD},\ \stackrel{\frown}{AE} = \stackrel{\frown}{BD}$. The length of $BD$ can be written in the form $\frac mn$, where $m$ and $n$ are relatively prime integers. Find $m+n$. | Mock AIME 1 2007-2008 Problems/Problem 11 | From adjacent sides, the following relationships can be derived:
\begin{align*} DC &= EC + 1\\ AE &= AF + 1\\ BD &= BF + 2 \end{align*}
Since $BF = EC$, and $DC = BF + 1$, $BD = DC + 1$. Thus, $BC = BD + DC = BD + (BD - 1)$. $26 = 2BD$. Thus, $BD = 13/1$. Thus, the answer is $\boxed{014}$. | // Block 1
size(150); defaultpen(linewidth(0.8)); import markers;
pair B = (0,0), C = (25,0), A = (578/50,19.8838);
draw(A--B--C--cycle);
label("$B$",B,SW); label("$C$",C,SE); label("$A$",A,N);
pair D = (13,0), E = (11*A + 13*C)/24, F = (12*B + 11*A)/23;
draw(D--E--F--cycle);
label("$D$",D,dir(-90));
label("$E$",E,dir(0));
label("$F$",F,dir(180));
draw(A--E,StickIntervalMarker(1,3,size=6));draw(B--D,StickIntervalMarker(1,3,size=6));
draw(F--B,StickIntervalMarker(1,2,size=6)); draw(E--C,StickIntervalMarker(1,2,size=6));
draw(A--F,StickIntervalMarker(1,1,size=6)); draw(C--D,StickIntervalMarker(1,1,size=6));
label("24",A--C,5*dir(0)); label("25",B--C,5*dir(-90)); label("23",B--A,5*dir(180));
// Block 2
size(150); defaultpen(linewidth(0.8)); import markers; pair B = (0,0), C = (25,0), A = (578/50,19.8838); draw(A--B--C--cycle); label("$B$",B,SW); label("$C$",C,SE); label("$A$",A,N); pair D = (13,0), E = (11*A + 13*C)/24, F = (12*B + 11*A)/23; draw(D--E--F--cycle); label("$D$",D,dir(-90)); label("$E$",E,dir(0)); label("$F$",F,dir(180)); draw(A--E,StickIntervalMarker(1,3,size=6));draw(B--D,StickIntervalMarker(1,3,size=6)); draw(F--B,StickIntervalMarker(1,2,size=6)); draw(E--C,StickIntervalMarker(1,2,size=6)); draw(A--F,StickIntervalMarker(1,1,size=6)); draw(C--D,StickIntervalMarker(1,1,size=6)); label("24",A--C,5*dir(0)); label("25",B--C,5*dir(-90)); label("23",B--A,5*dir(180)); | [] |
549 | Points $A$ and $B$ lie on $\odot O$, with radius $r$, so that $\angle OAB$ is acute. Extend $AB$ to point $C$ so that $AB = BC$. Let $D$ be the intersection of $\odot O$ and $OC$ such that $CD = \frac {1}{18}$ and $\cos(2\angle OAC) = \frac 38$. If $r$ can be written as $\frac{a+b\sqrt{c}}{d}$, where $a,b,$ and $d$ are relatively prime and $c$ is not divisible by the square of any prime, find $a+b+c+d$. | Mock AIME 1 2007-2008 Problems/Problem 14 | EDIT: The following solution is wrong. See the discussion page.
By the cosine double-angle formula,
\[\cos(2\angle OAC) = 2\cos^2(\angle OAC) - 1 = \frac 38\ \Longrightarrow\ \cos \angle OAC = \sqrt{\frac{11}{16}}.\]
The Law of Cosines on $\triangle BAO$ with respect to $\angle BAO$ yields
\begin{align*}r^2 &= r^2 + AB^2 - 2 \cdot AB \cdot r \cos \angle BAO \\
AB^2 &= 2 \cdot AB \cdot r \cdot \frac{\sqrt{11}}{4}\\
AB &= \frac{r\sqrt{11}}{2} (Error compiling LaTeX. Unknown error_msg)
Now, $AC = 2AB = r\sqrt{11}$. The Law of Cosines on $\triangle CAO$ with respect to $\angle CAO$ yields
\begin{align*}\left(r+\frac{1}{18}\right)^2 &= r^2 + \left(r\sqrt{11}\right)^2 - 2 \cdot r\sqrt{11} \cdot r \cos \angle BAO \\ \frac{r}{9} + \frac{1}{324} &= 11r^2 - r\sqrt{11} \cdot \left(\frac{\sqrt{11}}{2}\right) \\ 0 &= 1782 r^2 - 36r - 1 \\ r &= \frac{36 \pm \sqrt{36^2 + 4 \cdot 1782 \cdot 1}}{2 \cdot 1782} = \frac{2 + \sqrt{26}}{198}\end{align*}
The answer is thus $a+b+c+d = 2 + 1 + 26 + 198 = \boxed{227}$. | // Block 1
size(220); defaultpen(fontsize(10));
real r = (2+26^0.5)/198;
pair O = (0,0), A=(r,0), B= A*dir(112.02432), C=2*B-A;
pair[] D = intersectionpoints(C--O,Circle(O,r));
dot(O); dot(A); dot(B); dot(C); dot(D[0]);
draw(O--A--C--cycle); draw(Circle(O,r));
draw(O--B,dashed);
draw(anglemark(B,A,O,0.15));
label("\(A\)",A,NE);
label("\(B\)",B,N);
label("\(C\)",C,N);
label("\(D\)",D[0],W);
label("\(\frac{1}{18}\)",(C+D[0])/2,W);
label("\(O\)",O,SW);
// Block 2
size(220); defaultpen(fontsize(10)); real r = (2+26^0.5)/198; pair O = (0,0), A=(r,0), B= A*dir(112.02432), C=2*B-A; pair[] D = intersectionpoints(C--O,Circle(O,r)); dot(O); dot(A); dot(B); dot(C); dot(D[0]); draw(O--A--C--cycle); draw(Circle(O,r)); draw(O--B,dashed); draw(anglemark(B,A,O,0.15)); label("\(A\)",A,NE); label("\(B\)",B,N); label("\(C\)",C,N); label("\(D\)",D[0],W); label("\(\frac{1}{18}\)",(C+D[0])/2,W); label("\(O\)",O,SW); | [] |
550 | Suppose $\triangle ABC$ is inscribed in circle $\Gamma$. $B_1$ and $C_1$ are the feet of the altitude from $B$ to $CA$ and $C$ to $AB$, respectively. Let $D$ be the intersection of lines $\overline{B_1 C_1}$ and $\overline{BC}$, let $E$ be the point of intersection of $\Gamma$ and line $\overline{DA}$ distinct from $A$, and let $F$ be the foot of the perpendicular from $E$ to $BD$. Given that $BD = 28$, $EF = \frac{20 \sqrt{159}}{7}$, and $ED^2 + EB^2 = 3050$, and that $\tan m \angle ACB$ can be expressed in the form $\frac{a \sqrt{b}}{c}$, where $a$ and $c$ are relatively prime positive integers and $b$ is an integer not divisible by the square of any prime, find the last three digits of $a + b + c$. | Mock AIME 1 2010 Problems/Problem 13 | Let $\measuredangle BED=\theta$. Because the problem gives us $ED^2+EB^2=3050$, we think to use the Law of Cosines in $\triangle BED$, which yields $BD^2=ED^2+EB^2-2ED\cdot EB\cos\theta$. Subtituting the values given by the problem, we get $28^2=3050-2ED\cdot EB\cos\theta$, which gives $ED\cdot EB=\tfrac{3050-784}{2\cos\theta}=\tfrac{1133}{\cos\theta}$.
To find another expression for $ED\cdot EB$, we think of the formula $[\triangle BED]=\tfrac12ED\cdot EB\sin\theta$. We know that the area of the triangle is $\tfrac12\cdot\tfrac{20\sqrt{159}}7\cdot28=40\sqrt{159}$. Substituting this in the previous equation for $[\triangle BED]$, we get that $40\sqrt{159}=\tfrac12ED\cdot EB\sin\theta$, so $ED\cdot EB=\tfrac{80\sqrt{159}}{\sin\theta}$.
Setting these two expressions for $ED\cdot EB$ equal to each other reveals that $\tfrac{1133}{\cos\theta}=\tfrac{80\sqrt{159}}{\sin\theta}$, so $\tan\theta=\tfrac{80\sqrt{159}}{1133}$ by the identity $\tan\theta=\tfrac{\sin\theta}{\cos\theta}.$
$\angle AEB$ is supplementary to $\angle BED$, and $\angle ACB$ is supplementary to $\angle AEB$, because $AEBC$ is a cyclic quadrilateral. Thus, $\measuredangle BED=\measuredangle ACB=\theta$, so $\tan\measuredangle ACB=\tan\theta=\tfrac{80\sqrt{159}}{1133}$. Thus, $a+b+c=80+159+1133=1372$, so our answer is $\boxed{372}$. | // Block 1
import geometry;
size(8cm);
point B = origin;
point A = (3,8);
point C = (12,0);
triangle t = triangle(A,B,C);
circle c = circumcircle(t);
point B1 = foot(t.VB);
point C1 = foot(t.VC);
point D = intersectionpoint(line(B1,C1), line(B,C));
pair[] e = intersectionpoints(line(A,D), c);
point E = e[0];
// Triangle ABC and Circumcircle
draw(t);
draw(c);
// Altitudes
draw(B--B1);
draw(C--C1);
// Segments AD,EB,BD, and B_1D
draw(A--D);
draw(E--B);
draw(B--D);
draw(B1--D);
// Point Labels
dot(A);
label("A",A,NW);
dot(B);
label("B",B,SSW);
dot(C);
label("C",C,SE);
dot(B1);
label("B$_1$",B1,NE);
dot(C1);
label("C$_1$",C1,NNW);
dot(D);
label("D",D,SW);
dot(E);
label("E",E,NW);
// Block 2
import geometry; size(8cm); point B = origin; point A = (3,8); point C = (12,0); triangle t = triangle(A,B,C); circle c = circumcircle(t); point B1 = foot(t.VB); point C1 = foot(t.VC); point D = intersectionpoint(line(B1,C1), line(B,C)); pair[] e = intersectionpoints(line(A,D), c); point E = e[0]; // Triangle ABC and Circumcircle draw(t); draw(c); // Altitudes draw(B--B1); draw(C--C1); // Segments AD,EB,BD, and B_1D draw(A--D); draw(E--B); draw(B--D); draw(B1--D); // Point Labels dot(A); label("A",A,NW); dot(B); label("B",B,SSW); dot(C); label("C",C,SE); dot(B1); label("B$_1$",B1,NE); dot(C1); label("C$_1$",C1,NNW); dot(D); label("D",D,SW); dot(E); label("E",E,NW); | [] |
551 | Let $AB$ be a line segment of length $20 \sqrt{2}$, and let $S$ be the set of all points $P$ such that $m \angle APB \geq 45^{\circ}$. Find the last three digits of the largest integer less than the area of $S$. | Mock AIME 1 2010 Problems/Problem 3 | Let point $O$ be on the perpendicular bisector of $\overline{AB}$ such that $\overline{AO} \perp \overline{OB}$. We know that $AO=OB$ (because $O$ is on the perpendicular bisector of $\overline{AB}$), so we can draw circle $O_A$. Do the same on the other side of $AB$ with point $Q$ and draw circle $Q_A$. The central angle of $\widehat{AB}$ in both circles is $90^{\circ}$. Thus, by the properties of inscribed angles, we know that the locus of all points $P$ such that $\measuredangle APB = \tfrac{90^{\circ}}2=45^{\circ}$ is the border of the area enclosed by the two circles. Going out of this area would force $\measuredangle APB<45^{\circ}$, so the inequality is not satified there. Conversely, going inside the area would force $\measuredangle APB > 45^{\circ}$, so the inequality is satisfied. Thus, we look for the area enclosed by the two circles in the diagram below.
We know that $\triangle AOB$ and $\triangle AQB$ are isosceles right triangles, so their leg lengths are $\tfrac{20\sqrt2}{\sqrt2}=20$. Thus, the radius of the circles is $20$, and so the area of the enclosed region is $2 \cdot 20^2 \cdot \pi = 800\pi$ minus the region enclosed by both circles. To find the area of this doubly enclosed region, we consider the square in the diagram below.
The entire square has an area of $(20\sqrt2)^2=800$, and the circle has area $\pi \cdot 20^2 = 400\pi$. Thus, each of the sections of the circle not inside the square has an area of $\tfrac{400\pi-800}4=100\pi-200$. Thus, looking back at the first diagram, we see that the doubly enclosed area (which we would count twice if we just added the areas of the two circles) is composed of two of these regions. Thus, it has area $2\cdot(100\pi-200)=200\pi-400$. Subtracting this from the area of the two circles, we see that the area of the enclosed region is $800\pi-(200\pi-400)=600\pi+400$. The greatest integer less than this amount is $600\cdot3.14+400=1884+400=2284$, so our answer is $\boxed{284}$. | // Block 1
import geometry;
point A = origin;
point B = A + (20*sqrt(2),0);
point O = (A+B)/2+(0,10*sqrt(2));
point Q = O-(0,20*sqrt(2));
// Segment AB and Circles
draw(A--B);
draw(circle(O,length(segment(O,A))));
draw(circle(Q,length(segment(Q,A))));
// Radii
draw(O--A);
draw(O--B);
draw(Q--A);
draw(Q--B);
// Labelling Points
dot(A);
label("A",A,W);
dot(B);
label("B",B,E);
dot(O);
label("O",O,N);
dot(Q);
label("Q",Q,S);
// Right angle marks
markscalefactor=0.35;
draw(rightanglemark(A,O,B));
draw(rightanglemark(A,Q,B));
// Block 2
import geometry;
point O = origin;
point A = dir(-135);
point B = dir(-45);
point C = dir(45);
point D = dir(135);
// Square ABCD and Circle
draw(A--B--C--D--cycle);
draw(circle(O,1));
// Diagonals
draw(A--C);
draw(B--D);
// Labelling Points
dot(A);
label("A",A,SW);
dot(B);
label("B",B,SE);
dot(C);
label("C",C,NE);
dot(D);
label("D",D,NW);
dot(O);
label("O",O,N);
// Right angle mark
markscalefactor=0.015;
draw(rightanglemark(A,O,B));
// Length Labels
label("$20\sqrt2$",midpoint(C--D),N);
label("$20$",midpoint(O--B),SW);
// Block 3
import geometry; point A = origin; point B = A + (20*sqrt(2),0); point O = (A+B)/2+(0,10*sqrt(2)); point Q = O-(0,20*sqrt(2)); // Segment AB and Circles draw(A--B); draw(circle(O,length(segment(O,A)))); draw(circle(Q,length(segment(Q,A)))); // Radii draw(O--A); draw(O--B); draw(Q--A); draw(Q--B); // Labelling Points dot(A); label("A",A,W); dot(B); label("B",B,E); dot(O); label("O",O,N); dot(Q); label("Q",Q,S); // Right angle marks markscalefactor=0.35; draw(rightanglemark(A,O,B)); draw(rightanglemark(A,Q,B));
// Block 4
import geometry; point O = origin; point A = dir(-135); point B = dir(-45); point C = dir(45); point D = dir(135); // Square ABCD and Circle draw(A--B--C--D--cycle); draw(circle(O,1)); // Diagonals draw(A--C); draw(B--D); // Labelling Points dot(A); label("A",A,SW); dot(B); label("B",B,SE); dot(C); label("C",C,NE); dot(D); label("D",D,NW); dot(O); label("O",O,N); // Right angle mark markscalefactor=0.015; draw(rightanglemark(A,O,B)); // Length Labels label("$20\sqrt2$",midpoint(C--D),N); label("$20$",midpoint(O--B),SW); | [] |
552 | Let $\omega_1$ and $\omega_2$ be circles of radii 5 and 7, respectively, and suppose that the distance between their centers is 10. There exists a circle $\omega_3$ that is internally tangent to both $\omega_1$ and $\omega_2$, and tangent to the line joining the centers of $\omega_1$ and $\omega_2$. If the radius of $\omega_3$ can be expressed in the form $a \sqrt{b} - c$, where $a$, $b$, and $c$ are integers, and $b$ is not divisible by the square if any prime, find the value of $a + b + c$. | Mock AIME 1 2010 Problems/Problem 9 | Let $\omega_1$ have center $A$, $\omega_2$ have center $B$, and $\omega_3$ have center $C$. Further, let $\overline{AB}$ intersect $\omega_1$ at $D$, $\omega_2$ at $E$, and $\omega_3$ at $T$, as in the diagram. Let $r$ be the radius of $\omega_3$ and let $TE=x$.
Because $\overline{AB}$ is tangent to $\omega_3$, $\overline{CT} \perp \overline{AB}$. Because $\omega_1$ and $\omega_3$ are tangent, we know that the line joining their centers goes through their point of tangency. Thus, because $\omega_1$ has radius $5$, $AC=5-r$. Similarly, $BC=7-r$. Because $AB=10$ with $AE=5$ and $BD=7$, $AB=2$. Thus, $DT=2-x$, $AD=5-2=3$, and $BE=7-2=5$.
By the Pythagorean Theorem in $\triangle BTC$, we have the following equation that we can solve for $r$:
\begin{align*}
BT^2+TC^2 &= BC^2 \\
(5+x)^2+r^2 &= (7-r)^2 \\
25+10x+x^2+r^2 &= 49-14r+r^2 \\
x^2+10x-24 &= -14r \\
r &= -\frac{x^2+10x-24}{14}
\end{align*}
By using Pythagoras again in $\triangle ATC$, we have the following equation:
\begin{align*}
AT^2+TC^2 &= AC^2 \\
(3+(2-x))^2+r^2 &= (5-r)^2 \\
(5-x)^2+r^2 &= r^2-10r+25 \\
x^2-10x+25 &= -10r+25 \\
x^2-10x &= -10r
\end{align*}
Substituting the expression for $r$ we found earlier, we see the following:
\begin{align*}
x^2-10x &= -10(-\frac{x^2+10x-24}{14}) \\
x^2-10x &= \frac{5x^2+50x-120}{7} \\
7x^2-70x &= 5x^2+50x-120 \\
2x^2-120x+120 &= 0 \\
x^2-60x+60 &= 0 \\
x &= \frac{60 \pm \sqrt{3600-240}}2 \\
x &= 30 \pm \sqrt{900-60} \\
x &= 30 \pm 2\sqrt{210}
\end{align*}
Because $x<2$, $x=30 - 2\sqrt{210}$. Now, we can plug this value for $x$ into our expression for $r$ to get our answer:
\begin{align*}
r &= -\frac{x^2+10x-24}{14} \\
&= -\frac{(30-2\sqrt{210})^2+10(30-2\sqrt{210})-24}{14} \\
&= -\frac{900+840-120\sqrt{210}+300-20\sqrt{210}-24}{14} \\
&= -\frac{1740+276-140\sqrt{210}}{14} \\
&= 10\sqrt{210}-\frac{2016}{14} \\
&= 10\sqrt{210}-144
\end{align*}
Thus, our answer is $10+210+144=\boxed{364}$. | // Block 1
import geometry;
size(10cm);
point A = origin;
point B = (10,0);
point C = (5-(30-2*sqrt(210)),10*sqrt(210)-144);
point D,E,T;
circle a = circle(A,5);
circle b = circle(B,7);
circle c = circle(C,10*sqrt(210)-144);
// Defining D, E, and T
pair[] d = intersectionpoints(b,A--B);
D = d[0];
pair[] e = intersectionpoints(a,A--B);
E = e[0];
pair[] t = intersectionpoints(c,A--B);
T = t[0];
// Circles
draw(a);
draw(b);
draw(c);
// Segments AB, BC, AC, and CT
draw(A--B);
draw(B--C);
draw(A--C);
draw(C--T);
// Point Labels
dot(A);
label("A",A,WSW);
dot(B);
label("B",B,ESE);
dot(C);
label("C",C,(0,0.5));
dot(D);
label("D",D,SW);
dot(E);
label("E",E,SE);
dot(T);
label("T",T,S);
// Right angle mark
markscalefactor=0.05;
draw(rightanglemark(C,T,A));
// Block 2
import geometry; size(10cm); point A = origin; point B = (10,0); point C = (5-(30-2*sqrt(210)),10*sqrt(210)-144); point D,E,T; circle a = circle(A,5); circle b = circle(B,7); circle c = circle(C,10*sqrt(210)-144); // Defining D, E, and T pair[] d = intersectionpoints(b,A--B); D = d[0]; pair[] e = intersectionpoints(a,A--B); E = e[0]; pair[] t = intersectionpoints(c,A--B); T = t[0]; // Circles draw(a); draw(b); draw(c); // Segments AB, BC, AC, and CT draw(A--B); draw(B--C); draw(A--C); draw(C--T); // Point Labels dot(A); label("A",A,WSW); dot(B); label("B",B,ESE); dot(C); label("C",C,(0,0.5)); dot(D); label("D",D,SW); dot(E); label("E",E,SE); dot(T); label("T",T,S); // Right angle mark markscalefactor=0.05; draw(rightanglemark(C,T,A)); | [] |
553 | In rectangle $ABCD$, $AD=1$, $P$ is on $\overline{AB}$, and $\overline{DB}$ and $\overline{DP}$ trisect $\angle ADC$. What is the perimeter of $\triangle BDP$?
$\textbf{(A)}\ 3+\frac{\sqrt{3}}{3} \qquad\textbf{(B)}\ 2+\frac{4\sqrt{3}}{3} \qquad\textbf{(C)}\ 2+2\sqrt{2} \qquad\textbf{(D)}\ \frac{3+3\sqrt{5}}{2} \qquad\textbf{(E)}\ 2+\frac{5\sqrt{3}}{3}$ | 2000 AMC 10 Problem 7 | $AD=1$.
Since $\angle ADC$ is trisected, $\angle ADP= \angle PDB= \angle BDC=30^\circ$.
Thus, $PD=\frac{2\sqrt{3}}{3}$
$DB=2$
$BP=\sqrt{3}-\frac{\sqrt{3}}{3}=\frac{2\sqrt{3}}{3}$.
Adding, we get $\boxed{\textbf{(B) } 2+\frac{4\sqrt{3}}{3}}$. | // Block 1
draw((0,2)--(3.4,2)--(3.4,0)--(0,0)--cycle);
draw((0,0)--(1.3,2));
draw((0,0)--(3.4,2));
dot((0,0));
dot((0,2));
dot((3.4,2));
dot((3.4,0));
dot((1.3,2));
label("$A$",(0,2),NW);
label("$B$",(3.4,2),NE);
label("$C$",(3.4,0),SE);
label("$D$",(0,0),SW);
label("$P$",(1.3,2),N);
label("$1$",(0,1),W);
label("$2$",(1.7,1),SE);
label("$\frac{\sqrt{3}}{3}$",(0.65,2),N);
label("$\frac{2\sqrt{3}}{3}$",(0.85,1),NW);
label("$\frac{2\sqrt{3}}{3}$",(2.35,2),N);
label("$\sqrt{3}$",(1.7,0),S);
label("$2$",(3,1),W);
// Block 2
draw((0,2)--(3.4,2)--(3.4,0)--(0,0)--cycle); draw((0,0)--(1.3,2)); draw((0,0)--(3.4,2)); dot((0,0)); dot((0,2)); dot((3.4,2)); dot((3.4,0)); dot((1.3,2)); label("$A$",(0,2),NW); label("$B$",(3.4,2),NE); label("$C$",(3.4,0),SE); label("$D$",(0,0),SW); label("$P$",(1.3,2),N); label("$1$",(0,1),W); label("$2$",(1.7,1),SE); label("$\frac{\sqrt{3}}{3}$",(0.65,2),N); label("$\frac{2\sqrt{3}}{3}$",(0.85,1),NW); label("$\frac{2\sqrt{3}}{3}$",(2.35,2),N); label("$\sqrt{3}$",(1.7,0),S); label("$2$",(3,1),W); | [] |
554 | Figures $0$, $1$, $2$, and $3$ consist of $1$, $5$, $13$, and $25$ nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100?
$\textbf{(A)}\ 10401 \qquad\textbf{(B)}\ 19801 \qquad\textbf{(C)}\ 20201 \qquad\textbf{(D)}\ 39801 \qquad\textbf{(E)}\ 40801$ | 2000 AMC 10 Problem 12 | By taking figure $n$, putting the centers of each square on a grid, and rotating the figure 45 degrees, we get the following shape (figure 3 is shown for reference):
There are two lattices of points (shown in red): one of side length $n$ (blue) and one of $n+1$ (green). The rest follows like in solution 1.
-Integralarefun (talk) 18:21, 29 July 2023 (EDT) | import olympiad; int fig = 3; for (int i = 0; i <= 2*fig; ++i) { for (int j = 0; j <= 2*fig; ++j) { pair p = (i, j); //do we need this point? if ((i+j)%2 == 1) { continue; } //draw squares draw((p+N)--(p+E)--(p+S)--(p+W)--cycle); //draw lattices if (i < 2*fig-1) { draw(p--p+2*E, red); } if (j < 2*fig-1) { draw(p--p+2*N, red); } //mark center points if (i % 2 == 0) { dot(p, green); } else { dot(p, blue); } } } | [] |
555 | There are 5 yellow pegs, 4 red pegs, 3 green pegs, 2 blue pegs, and 1 orange peg to be placed on a triangular peg board. In how many ways can the pegs be placed so that no (horizontal) row or (vertical) column contains two pegs of the same color?
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 5!\cdot 4!\cdot 3!\cdot 2!\cdot 1! \qquad\textbf{(D)}\ \frac{15!}{5!\cdot 4!\cdot 3!\cdot 2!\cdot 1!} \qquad\textbf{(E)}\ 15!$ | 2000 AMC 10 Problem 13 | In each column there must be one yellow peg. In particular, in the rightmost column, there is only one peg spot, therefore a yellow peg must go there.
In the second column from the right, there are two spaces for pegs. One of them is in the same row as the corner peg, so there is only one remaining choice left for the yellow peg in this column.
By similar logic, we can fill in the yellow pegs as shown:
After this we can proceed to fill in the whole pegboard, so there is only $1$ arrangement of the pegs. The answer is $\boxed{\text{B}}$ | // Block 1
unitsize(20);
dot((0,0));
dot((1,0));
dot((2,0));
dot((3,0));
label("Y",(4,-.35),N);
dot((0,1));
dot((1,1));
dot((2,1));
label("Y",(3,.6),N);
dot((0,2));
dot((1,2));
label("Y",(2,1.6),N);
dot((0,3));
label("Y",(1,2.6),N);
label("Y",(0,3.6),N);
// Block 2
unitsize(20);
label("O",(0,-.35),N);
label("B",(1,-.35),N);
label("G",(2,-.35),N);
label("R",(3,-.35),N);
label("Y",(4,-.35),N);
label("B",(0,.6),N);
label("G",(1,.6),N);
label("R",(2,.6),N);
label("Y",(3,.6),N);
label("G",(0,1.6),N);
label("R",(1,1.6),N);
label("Y",(2,1.6),N);
label("R",(0,2.6),N);
label("Y",(1,2.6),N);
label("Y",(0,3.6),N);
// Block 3
unitsize(20); dot((0,0)); dot((1,0)); dot((2,0)); dot((3,0)); label("Y",(4,-.35),N); dot((0,1)); dot((1,1)); dot((2,1)); label("Y",(3,.6),N); dot((0,2)); dot((1,2)); label("Y",(2,1.6),N); dot((0,3)); label("Y",(1,2.6),N); label("Y",(0,3.6),N);
// Block 4
unitsize(20); label("O",(0,-.35),N); label("B",(1,-.35),N); label("G",(2,-.35),N); label("R",(3,-.35),N); label("Y",(4,-.35),N); label("B",(0,.6),N); label("G",(1,.6),N); label("R",(2,.6),N); label("Y",(3,.6),N); label("G",(0,1.6),N); label("R",(1,1.6),N); label("Y",(2,1.6),N); label("R",(0,2.6),N); label("Y",(1,2.6),N); label("Y",(0,3.6),N); | [] |
556 | The diagram shows $28$ lattice points, each one unit from its nearest neighbors. Segment $AB$ meets segment $CD$ at $E$. Find the length of segment $AE$.
$\textbf{(A)}\ \frac{4\sqrt{5}}{3} \qquad\textbf{(B)}\ \frac{5\sqrt{5}}{3} \qquad\textbf{(C)}\ \frac{12\sqrt{5}}{7} \qquad\textbf{(D)}\ 2\sqrt{5} \qquad\textbf{(E)}\ \frac{5\sqrt{65}}{9}$ | 2000 AMC 10 Problem 16 | Draw the perpendiculars from $A$ and $B$ to $CD$, respectively. As it turns out, $BC \perp CD$. Let $F$ be the point on $CD$ for which $AF\perp CD$.
$m\angle AFE=m\angle BCE=90^\circ$, and $m\angle AEF=m\angle CEB$, so by AA similarity, \[\triangle AFE\sim \triangle BCE \Rightarrow \frac{AF}{AE}=\frac{BC}{BE}\]
By the Pythagorean Theorem, we have $AB=\sqrt{3^2+6^2}=3\sqrt{5}$, $AF=\sqrt{2.5^2+2.5^2}=2.5\sqrt{2}$, and $BC=\sqrt{2^2+2^2}=2\sqrt{2}$. Let $AE=x$, so $BE=3\sqrt{5}-x$, then \[\frac{2.5\sqrt{2}}{x}=\frac{2\sqrt{2}}{3\sqrt{5}-x}\] \[x=\frac{5\sqrt{5}}{3}\]
This is answer choice $\boxed{\text{B}}$ | // Block 1
path seg1, seg2;
seg1=(6,0)--(0,3);
seg2=(2,0)--(4,2);
dot((0,0));
dot((1,0));
fill(circle((2,0),0.1),black);
dot((3,0));
dot((4,0));
dot((5,0));
fill(circle((6,0),0.1),black);
dot((0,1));
dot((1,1));
dot((2,1));
dot((3,1));
dot((4,1));
dot((5,1));
dot((6,1));
dot((0,2));
dot((1,2));
dot((2,2));
dot((3,2));
fill(circle((4,2),0.1),black);
dot((5,2));
dot((6,2));
fill(circle((0,3),0.1),black);
dot((1,3));
dot((2,3));
dot((3,3));
dot((4,3));
dot((5,3));
dot((6,3));
draw(seg1);
draw(seg2);
pair [] x=intersectionpoints(seg1,seg2);
fill(circle(x[0],0.1),black);
label("$A$",(0,3),NW);
label("$B$",(6,0),SE);
label("$C$",(4,2),NE);
label("$D$",(2,0),S);
label("$E$",x[0],N);
label("$F$",(2.5,.5),E);
draw((6,0)--(4,2));
draw((0,3)--(2.5,.5));
// Block 2
path seg1, seg2; seg1=(6,0)--(0,3); seg2=(2,0)--(4,2); dot((0,0)); dot((1,0)); fill(circle((2,0),0.1),black); dot((3,0)); dot((4,0)); dot((5,0)); fill(circle((6,0),0.1),black); dot((0,1)); dot((1,1)); dot((2,1)); dot((3,1)); dot((4,1)); dot((5,1)); dot((6,1)); dot((0,2)); dot((1,2)); dot((2,2)); dot((3,2)); fill(circle((4,2),0.1),black); dot((5,2)); dot((6,2)); fill(circle((0,3),0.1),black); dot((1,3)); dot((2,3)); dot((3,3)); dot((4,3)); dot((5,3)); dot((6,3)); draw(seg1); draw(seg2); pair [] x=intersectionpoints(seg1,seg2); fill(circle(x[0],0.1),black); label("$A$",(0,3),NW); label("$B$",(6,0),SE); label("$C$",(4,2),NE); label("$D$",(2,0),S); label("$E$",x[0],N); label("$F$",(2.5,.5),E); draw((6,0)--(4,2)); draw((0,3)--(2.5,.5)); | [] |
556 | The diagram shows $28$ lattice points, each one unit from its nearest neighbors. Segment $AB$ meets segment $CD$ at $E$. Find the length of segment $AE$.
$\textbf{(A)}\ \frac{4\sqrt{5}}{3} \qquad\textbf{(B)}\ \frac{5\sqrt{5}}{3} \qquad\textbf{(C)}\ \frac{12\sqrt{5}}{7} \qquad\textbf{(D)}\ 2\sqrt{5} \qquad\textbf{(E)}\ \frac{5\sqrt{65}}{9}$ | 2000 AMC 10 Problem 16 | Drawing line $\overline{BD}$ and parallel line $\overline{CF}$, we see that $\triangle FCE \sim \triangle BDE$ by AA similarity. Thus $\frac{FE}{EB} = \frac{FC}{DB} = \frac{2}{4} = \frac{1}{2}$. Reciprocating, we know that $\frac{EB}{FE} = 2$ so $\frac{EB+FE}{FE} = 2+1 \Rightarrow \frac{FB}{FE} = 3$. Reciprocating again, we have $\frac{FE}{FB} = \frac{1}{3} \Rightarrow FE = \frac{1}{3}FB$. We know that $FD = 2$, so by the Pythagorean Theorem, $FB = \sqrt{2^{2} + 4^{2}} = 2\sqrt{5}$. Thus $FE = \frac{1}{3}FB = \frac{2\sqrt{5}}{3}$. Applying the Pythagorean Theorem again, we have $AF = \sqrt{1^{2}+2^{2}} = \sqrt{5}$. We finally have $AE = AF + FE = \sqrt{5} + \frac{2\sqrt{5}}{3} = \frac{5\sqrt{5}}{3} \Rightarrow \boxed{\text{B}}$ | // Block 1
path seg1, seg2;
seg1=(6,0)--(0,3);
seg2=(2,0)--(4,2);
dot((0,0));
dot((1,0));
fill(circle((2,0),0.1),black);
dot((3,0));
dot((4,0));
dot((5,0));
fill(circle((6,0),0.1),black);
dot((0,1));
dot((1,1));
dot((2,1));
dot((3,1));
dot((4,1));
dot((5,1));
dot((6,1));
dot((0,2));
dot((1,2));
dot((2,2));
dot((3,2));
fill(circle((4,2),0.1),black);
dot((5,2));
dot((6,2));
fill(circle((0,3),0.1),black);
dot((1,3));
dot((2,3));
dot((3,3));
dot((4,3));
dot((5,3));
dot((6,3));
draw(seg1);
draw(seg2);
pair [] x=intersectionpoints(seg1,seg2);
fill(circle(x[0],0.1),black);
label("$A$",(0,3),NW);
label("$B$",(6,0),SE);
label("$C$",(4,2),NE);
label("$D$",(2,0),S);
label("$E$",x[0],N);
label("$F$",(2,2),NE);
draw((2,2)--(4,2));
draw((6,0)--(2,0));
// Block 2
path seg1, seg2; seg1=(6,0)--(0,3); seg2=(2,0)--(4,2); dot((0,0)); dot((1,0)); fill(circle((2,0),0.1),black); dot((3,0)); dot((4,0)); dot((5,0)); fill(circle((6,0),0.1),black); dot((0,1)); dot((1,1)); dot((2,1)); dot((3,1)); dot((4,1)); dot((5,1)); dot((6,1)); dot((0,2)); dot((1,2)); dot((2,2)); dot((3,2)); fill(circle((4,2),0.1),black); dot((5,2)); dot((6,2)); fill(circle((0,3),0.1),black); dot((1,3)); dot((2,3)); dot((3,3)); dot((4,3)); dot((5,3)); dot((6,3)); draw(seg1); draw(seg2); pair [] x=intersectionpoints(seg1,seg2); fill(circle(x[0],0.1),black); label("$A$",(0,3),NW); label("$B$",(6,0),SE); label("$C$",(4,2),NE); label("$D$",(2,0),S); label("$E$",x[0],N); label("$F$",(2,2),NE); draw((2,2)--(4,2)); draw((6,0)--(2,0)); | [] |
556 | The diagram shows $28$ lattice points, each one unit from its nearest neighbors. Segment $AB$ meets segment $CD$ at $E$. Find the length of segment $AE$.
$\textbf{(A)}\ \frac{4\sqrt{5}}{3} \qquad\textbf{(B)}\ \frac{5\sqrt{5}}{3} \qquad\textbf{(C)}\ \frac{12\sqrt{5}}{7} \qquad\textbf{(D)}\ 2\sqrt{5} \qquad\textbf{(E)}\ \frac{5\sqrt{65}}{9}$ | 2000 AMC 10 Problem 16 | Extend line $\overline{DC}$ as above. This creates two similar triangles whose side lengths have the ratio $5:4$. Therefore $AE=\frac{5}{9}AB$. Using Pythagorean theorem to find $AB$ gives us:
\[AE=\frac{5}{9}AB=\frac{5}{9}\sqrt{3^2+6^2}=\frac{5}{9}\sqrt{45}= \boxed{\textbf{(B) }\frac{5\sqrt{5}}{3}}\]
~ proloto | // Block 1
// Coordinates
pair A = (0,3), B = (6,0), C = (4,2), D = (2,0);
path seg1 = B--A;
path seg2 = D--C;
pair[] intersectionPoints = intersectionpoints(seg1, seg2);
pair E = intersectionPoints[0];
for (int i = 0; i <= 6; i = i + 1) {
for (int j = 0; j <= 3; j = j + 1) {
dot((i,j));
}
}
// Draw
draw(seg1);
draw(seg2);
dot(E);
// Label
label("$A$", A, NW);
label("$B$", B, SE);
label("$C$", C, dir(0));
label("$D$", D, S);
label("$E$", E, N);
// Add extras
draw(C--(5,3), dashed);
draw(D--B, dashed);
draw((5,3)--A, dashed);
// Block 2
// Coordinates pair A = (0,3), B = (6,0), C = (4,2), D = (2,0); path seg1 = B--A; path seg2 = D--C; pair[] intersectionPoints = intersectionpoints(seg1, seg2); pair E = intersectionPoints[0]; for (int i = 0; i <= 6; i = i + 1) { for (int j = 0; j <= 3; j = j + 1) { dot((i,j)); } } // Draw draw(seg1); draw(seg2); dot(E); // Label label("$A$", A, NW); label("$B$", B, SE); label("$C$", C, dir(0)); label("$D$", D, S); label("$E$", E, N); // Add extras draw(C--(5,3), dashed); draw(D--B, dashed); draw((5,3)--A, dashed); | [] |
557 | Charlyn walks completely around the boundary of a square whose sides are each $5$ km long. From any point on her path she can see exactly $1$ km horizontally in all directions. What is the area of the region consisting of all points Charlyn can see during her walk, expressed in square kilometers and rounded to the nearest whole number?
$\textbf{(A)} 24 \qquad\textbf{(B)}\ 27 \qquad\textbf{(C)}\ 39 \qquad\textbf{(D)}\ 40 \qquad\textbf{(E)}\ 42$ | 2000 AMC 10 Problem 18 | The area she sees looks at follows:
The part inside the walk has area $5\cdot 5 - 3\cdot 3 = 16$. The part outside the walk consists of four rectangles, and four arcs. Each of the rectangles has area $5\cdot 1=5$. The four arcs together form a circle with radius $1$.
Therefore the total area she can see is $16 + 4\cdot 5 + \pi\cdot 1^2 = 36+\pi \simeq 39.14$, which rounded to the nearest integer is $39$. $\boxed{C}$ | // Block 1
unitsize(0.8cm);
path p1 = (0,0)--(5,0)--(5,5)--(0,5)--cycle;
path p2 = (1,1)--(4,1)--(4,4)--(1,4)--cycle;
path p3 = arc((0,0),1,180,270) -- arc((5,0),1,270,360) -- arc((5,5),1,0,90) -- arc((0,5),1,90,180) -- cycle;
fill(p3,lightgray);
unfill(p2);
draw(p1,linewidth(bp)); draw(p2); draw(p3);
draw( (0,0)--(-1,0), dashed ); draw( (0,0)--(0,-1), dashed );
draw( (5,0)--(6,0), dashed ); draw( (5,0)--(5,-1), dashed );
draw( (5,5)--(6,5), dashed ); // This line is correct
draw( (5,5)--(5,6), dashed ); // Add semicolon here
draw( (0,5)--(-1,5), dashed ); draw( (0,5)--(0,6), dashed );
draw( (0,-1)--(5,-1), Arrows ); label( "$5$", (2.5,-1), S );
draw( (1,1)--(4,1), Arrows ); label( "$3$", (2.5,1), N );
draw( (4,3.5)--(5,3.5), Arrows ); label( "$1$", (4.5,3.5), N );
draw( (5,3.5)--(6,3.5), Arrows ); label( "$1$", (5.5,3.5), N );
// Block 2
unitsize(0.8cm); path p1 = (0,0)--(5,0)--(5,5)--(0,5)--cycle; path p2 = (1,1)--(4,1)--(4,4)--(1,4)--cycle; path p3 = arc((0,0),1,180,270) -- arc((5,0),1,270,360) -- arc((5,5),1,0,90) -- arc((0,5),1,90,180) -- cycle; fill(p3,lightgray); unfill(p2); draw(p1,linewidth(bp)); draw(p2); draw(p3); draw( (0,0)--(-1,0), dashed ); draw( (0,0)--(0,-1), dashed ); draw( (5,0)--(6,0), dashed ); draw( (5,0)--(5,-1), dashed ); draw( (5,5)--(6,5), dashed ); // This line is correct draw( (5,5)--(5,6), dashed ); // Add semicolon here draw( (0,5)--(-1,5), dashed ); draw( (0,5)--(0,6), dashed ); draw( (0,-1)--(5,-1), Arrows ); label( "$5$", (2.5,-1), S ); draw( (1,1)--(4,1), Arrows ); label( "$3$", (2.5,1), N ); draw( (4,3.5)--(5,3.5), Arrows ); label( "$1$", (4.5,3.5), N ); draw( (5,3.5)--(6,3.5), Arrows ); label( "$1$", (5.5,3.5), N ); | [] |
558 | Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the triangle is divided into a square and two smaller right triangles. The area of one of the two small right triangles is $m$ times the area of the square. The ratio of the area of the other small right triangle to the area of the square is
$\textbf {(A)}\ \frac{1}{2m+1} \qquad \textbf {(B)}\ m \qquad \textbf {(C)}\ 1-m \qquad \textbf {(D)}\ \frac{1}{4m} \qquad \textbf {(E)}\ \frac{1}{8m^2}$ | 2000 AMC 10 Problem 19 | Solution 1
WLOG, let a side of the square be $1$. Simple angle chasing shows that the two right triangles are similar. Thus the ratio of the sides of the triangles are the same. Since $A = \frac{1}{2}bh = \frac{h}{2}$, the base of the triangle with area $m$ is $2m$. Therefore $\frac{2m}{1} = \frac{1}{x}$ where $x$ is the height of the other triangle. $x = \frac{1}{2m}$, and the area of that triangle is $\frac{1}{2} \cdot 1 \cdot \frac{1}{2m} = \frac{1}{4m}\ \text{\boxed{D}}$.
Solution 2 (Video Solution)
https://youtu.be/HTHveknJFpk
https://m.youtube.com/watch?v=TUQsHeJ6RSA&feature=youtu.be
Solution 3
From the diagram from the previous solution, we have $a$, $b$ as the legs and $c$ as the side length of the square. WLOG, let the area of triangle $A$
be $m$ times the area of square $C$.
Since triangle $A$ is similar to the large triangle, it has $h_A = a(\frac{c}{b}) = \frac{ac}{b}$, $b_A = c$ and \[[A] = \frac{bh}{2} = \frac{ac^2}{2b} = m[C] = mc^2\]
Thus $\frac{a}{2b} = m$
Now since triangle $B$ is similar to the large triangle, it has $h_B = c$, $b_B = b\frac{c}{a} = \frac{bc}{a}$ and \[[B] = \frac{bh}{2} = \frac{bc^2}{2a} = nc^2 = n[C]\]
Thus $n = \frac{b}{2a} = \frac{1}{4(\frac{a}{2b})} = \frac{1}{4m}$. $\text{\boxed{D}}$.
~ Nafer
Solution 4 (process of elimination)
Simply testing specific triangles is sufficient.
A triangle with legs of 1 and 2 gives a square of area $S=\frac{2}{3}\times\frac{2}{3}=\frac{4}{9}$. The larger sub-triangle has area $T_1=\frac{\frac{2}{3}\times\frac{4}{3}}{2}=\frac{4}{9}$, and the smaller triangle has area $T_2=\frac{\frac{2}{3}\times\frac{1}{3}}{2}=\frac{1}{9}$. Computing ratios you get $\frac{T_1}{S}=1$ and $\frac{T_2}{S}=\frac{1}{4}$. Plugging $m=1$ in shows that the only possible answer is $\text{\boxed{D}}$
~ Snacc
Solution 5
WLOG, let the length of the square be $1$ (Like Solution 1). Then the length of the larger triangle is $2m$. Let the length of the smaller triangle be $x$.
Therefore, since $A = B$ (try to prove that yourself), $1 = 2mx$ or $x = 1/2m$
The area of the other triangle is $1/4m$.
From here, the answer is $\text{\boxed{D}}$. | // Block 1
unitsize(36); draw((0,0)--(6,0)--(0,3)--cycle); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); label("$1$",(1,2),S); label("$1$",(2,1),W); label("$2m$",(4,0),S); label("$x$",(0,2.5),W);
// Block 2
unitsize(36); draw((0,0)--(6,0)--(0,3)--cycle); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); label("$b$",(2.5,0),S); label("$a$",(0,1.5),W); label("$c$",(2.5,1),W); label("$A$",(0.5,2.5),W); label("$B$",(3.5,0.75),W); label("$C$",(1,1),W);
// Block 3
unitsize(36); draw((0,0)--(6,0)--(0,3)--cycle); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); draw((0,0)--(6,0)--(6,3)--(0,3)--cycle); draw((0,0)--(6,0)--(6,2)--(0,2)--cycle); draw((0,0)--(2,0)--(2,3)--(0,3)--cycle); label("$1$",(1,2),S); label("$1$",(2,1),W); label("$2m$",(4,0),S); label("$x$",(0,2.5),W); label("$A$", (1.25,1),W); label("$B$", (4, 2.25),N); | [] |
558 | Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the triangle is divided into a square and two smaller right triangles. The area of one of the two small right triangles is $m$ times the area of the square. The ratio of the area of the other small right triangle to the area of the square is
$\textbf {(A)}\ \frac{1}{2m+1} \qquad \textbf {(B)}\ m \qquad \textbf {(C)}\ 1-m \qquad \textbf {(D)}\ \frac{1}{4m} \qquad \textbf {(E)}\ \frac{1}{8m^2}$ | 2000 AMC 10 Problem 19 | Because we know that the right triangles are always similar, in a figure
, x/y = y/a. Using cross products, we can say that xa= y^2. We are trying to solve for x because m is in terms of a and y; m= a/2y (the ratio of the area of one triangle to the area of the square). So solving for x, x= y^2/a. Then the question is asking us for the answer to x/2y (the ratio of the area of the other triangle to the area of the square). So x/2y is y^2/a2y which is simplified to y/2a- also 1/4m, the answer.
-Smartgrowth | unitsize(36); draw((0,0)--(6,0)--(0,3)--cycle); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); label("$y$",(1,2),S); label("$y$",(2,1),W); label("$a$",(4,0),S); label("$x$",(0,2.5),W); | [] |
559 | Figures $0$, $1$, $2$, and $3$ consist of $1$, $5$, $13$, and $25$ nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100?
$\textbf{(A)}\ 10401 \qquad\textbf{(B)}\ 19801 \qquad\textbf{(C)}\ 20201 \qquad\textbf{(D)}\ 39801 \qquad\textbf{(E)}\ 40801$ | 2000 AMC 12 Problem 8 | By taking figure $n$, putting the centers of each square on a grid, and rotating the figure 45 degrees, we get the following shape (figure 3 is shown for reference):
There are two lattices of points (shown in red): one of side length $n$ (blue) and one of $n+1$ (green). The rest follows like in solution 1.
-Integralarefun (talk) 18:21, 29 July 2023 (EDT) | // Block 1
import olympiad;
int fig = 3;
for (int i = 0; i <= 2*fig; ++i) {
for (int j = 0; j <= 2*fig; ++j) {
pair p = (i, j);
//do we need this point?
if ((i+j)%2 == 1) {
continue;
}
//draw squares
draw((p+N)--(p+E)--(p+S)--(p+W)--cycle);
//draw lattices
if (i < 2*fig-1) {
draw(p--p+2*E, red);
}
if (j < 2*fig-1) {
draw(p--p+2*N, red);
}
//mark center points
if (i % 2 == 0) {
dot(p, green);
} else {
dot(p, blue);
}
}
}
// Block 2
import olympiad; int fig = 3; for (int i = 0; i <= 2*fig; ++i) { for (int j = 0; j <= 2*fig; ++j) { pair p = (i, j); //do we need this point? if ((i+j)%2 == 1) { continue; } //draw squares draw((p+N)--(p+E)--(p+S)--(p+W)--cycle); //draw lattices if (i < 2*fig-1) { draw(p--p+2*E, red); } if (j < 2*fig-1) { draw(p--p+2*N, red); } //mark center points if (i % 2 == 0) { dot(p, green); } else { dot(p, blue); } } } | [] |
560 | Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the triangle is divided into a square and two smaller right triangles. The area of one of the two small right triangles is $m$ times the area of the square. The ratio of the area of the other small right triangle to the area of the square is
$\textbf {(A)}\ \frac{1}{2m+1} \qquad \textbf {(B)}\ m \qquad \textbf {(C)}\ 1-m \qquad \textbf {(D)}\ \frac{1}{4m} \qquad \textbf {(E)}\ \frac{1}{8m^2}$ | 2000 AMC 12 Problem 21 | Solution 1
WLOG, let a side of the square be $1$. Simple angle chasing shows that the two right triangles are similar. Thus the ratio of the sides of the triangles are the same. Since $A = \frac{1}{2}bh = \frac{h}{2}$, the base of the triangle with area $m$ is $2m$. Therefore $\frac{2m}{1} = \frac{1}{x}$ where $x$ is the height of the other triangle. $x = \frac{1}{2m}$, and the area of that triangle is $\frac{1}{2} \cdot 1 \cdot \frac{1}{2m} = \frac{1}{4m}\ \text{\boxed{D}}$.
Solution 2 (Video Solution)
https://youtu.be/HTHveknJFpk
https://m.youtube.com/watch?v=TUQsHeJ6RSA&feature=youtu.be
Solution 3
From the diagram from the previous solution, we have $a$, $b$ as the legs and $c$ as the side length of the square. WLOG, let the area of triangle $A$
be $m$ times the area of square $C$.
Since triangle $A$ is similar to the large triangle, it has $h_A = a(\frac{c}{b}) = \frac{ac}{b}$, $b_A = c$ and \[[A] = \frac{bh}{2} = \frac{ac^2}{2b} = m[C] = mc^2\]
Thus $\frac{a}{2b} = m$
Now since triangle $B$ is similar to the large triangle, it has $h_B = c$, $b_B = b\frac{c}{a} = \frac{bc}{a}$ and \[[B] = \frac{bh}{2} = \frac{bc^2}{2a} = nc^2 = n[C]\]
Thus $n = \frac{b}{2a} = \frac{1}{4(\frac{a}{2b})} = \frac{1}{4m}$. $\text{\boxed{D}}$.
~ Nafer
Solution 4 (process of elimination)
Simply testing specific triangles is sufficient.
A triangle with legs of 1 and 2 gives a square of area $S=\frac{2}{3}\times\frac{2}{3}=\frac{4}{9}$. The larger sub-triangle has area $T_1=\frac{\frac{2}{3}\times\frac{4}{3}}{2}=\frac{4}{9}$, and the smaller triangle has area $T_2=\frac{\frac{2}{3}\times\frac{1}{3}}{2}=\frac{1}{9}$. Computing ratios you get $\frac{T_1}{S}=1$ and $\frac{T_2}{S}=\frac{1}{4}$. Plugging $m=1$ in shows that the only possible answer is $\text{\boxed{D}}$
~ Snacc
Solution 5
WLOG, let the length of the square be $1$ (Like Solution 1). Then the length of the larger triangle is $2m$. Let the length of the smaller triangle be $x$.
Therefore, since $A = B$ (try to prove that yourself), $1 = 2mx$ or $x = 1/2m$
The area of the other triangle is $1/4m$.
From here, the answer is $\text{\boxed{D}}$. | // Block 1
unitsize(36); draw((0,0)--(6,0)--(0,3)--cycle); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); label("$1$",(1,2),S); label("$1$",(2,1),W); label("$2m$",(4,0),S); label("$x$",(0,2.5),W);
// Block 2
unitsize(36); draw((0,0)--(6,0)--(0,3)--cycle); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); label("$b$",(2.5,0),S); label("$a$",(0,1.5),W); label("$c$",(2.5,1),W); label("$A$",(0.5,2.5),W); label("$B$",(3.5,0.75),W); label("$C$",(1,1),W);
// Block 3
unitsize(36); draw((0,0)--(6,0)--(0,3)--cycle); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); draw((0,0)--(6,0)--(6,3)--(0,3)--cycle); draw((0,0)--(6,0)--(6,2)--(0,2)--cycle); draw((0,0)--(2,0)--(2,3)--(0,3)--cycle); label("$1$",(1,2),S); label("$1$",(2,1),W); label("$2m$",(4,0),S); label("$x$",(0,2.5),W); label("$A$", (1.25,1),W); label("$B$", (4, 2.25),N); | [] |
560 | Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the triangle is divided into a square and two smaller right triangles. The area of one of the two small right triangles is $m$ times the area of the square. The ratio of the area of the other small right triangle to the area of the square is
$\textbf {(A)}\ \frac{1}{2m+1} \qquad \textbf {(B)}\ m \qquad \textbf {(C)}\ 1-m \qquad \textbf {(D)}\ \frac{1}{4m} \qquad \textbf {(E)}\ \frac{1}{8m^2}$ | 2000 AMC 12 Problem 21 | Because we know that the right triangles are always similar, in a figure
, x/y = y/a. Using cross products, we can say that xa= y^2. We are trying to solve for x because m is in terms of a and y; m= a/2y (the ratio of the area of one triangle to the area of the square). So solving for x, x= y^2/a. Then the question is asking us for the answer to x/2y (the ratio of the area of the other triangle to the area of the square). So x/2y is y^2/a2y which is simplified to y/2a- also 1/4m, the answer.
-Smartgrowth | // Block 1
unitsize(36);
draw((0,0)--(6,0)--(0,3)--cycle);
draw((0,0)--(2,0)--(2,2)--(0,2)--cycle);
label("$y$",(1,2),S);
label("$y$",(2,1),W);
label("$a$",(4,0),S);
label("$x$",(0,2.5),W);
// Block 2
unitsize(36); draw((0,0)--(6,0)--(0,3)--cycle); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); label("$y$",(1,2),S); label("$y$",(2,1),W); label("$a$",(4,0),S); label("$x$",(0,2.5),W); | [] |
561 | Eight congruent equilateral triangles, each of a different color, are used to construct a regular octahedron. How many distinguishable ways are there to construct the octahedron? (Two colored octahedrons are distinguishable if neither can be rotated to look just like the other.)
$\textbf {(A)}\ 210 \qquad \textbf {(B)}\ 560 \qquad \textbf {(C)}\ 840 \qquad \textbf {(D)}\ 1260 \qquad \textbf {(E)}\ 1680$ | 2000 AMC 12 Problem 25 | Since the octahedron is indistinguishable by rotations, without loss of generality fix a face to be red.
There are $7!$ ways to arrange the remaining seven colors, but there still are three possible rotations about the fixed face, so the answer is $7!/3 = 1680$. | // Block 1
size(8cm);
defaultpen(0.5);
import three;
import math;
currentprojection=orthographic(2,0.2,1);
triple A=(0,0,1);
triple B=(sqrt(2)/2,sqrt(2)/2,0);
triple C=(sqrt(2)/2,-sqrt(2)/2,0);
triple D=(-sqrt(2)/2,-sqrt(2)/2,0);
triple E=(-sqrt(2)/2,sqrt(2)/2,0);
triple F=(0,0,-1);
draw(A--B--E--cycle);
draw(A--C--D--cycle);
draw(F--C--B--cycle);
draw(F--D--E--cycle,dotted+linewidth(0.7));
draw(surface(A--B--C--cycle),rgb(1,.6,.6),nolight);
// Block 2
size(8cm);
defaultpen(0.5);
import three;
import math;
currentprojection=orthographic(2,0,1);
triple A=(0,0,1);
triple B=(sqrt(2)/2,sqrt(2)/2,0);
triple C=(sqrt(2)/2,-sqrt(2)/2,0);
triple D=(-sqrt(2)/2,-sqrt(2)/2,0);
triple E=(-sqrt(2)/2,sqrt(2)/2,0);
triple F=(0,0,-1);
triple right=(0,1,0);
picture p = new picture, r = new picture, s = new picture;
draw(p,A--B--E--cycle);
draw(p,A--C--D--cycle);
draw(p,F--C--B--cycle);
draw(p,F--D--E--cycle,dotted+linewidth(0.7));
draw(p,surface(A--B--C--cycle),rgb(1,.6,.6),nolight);
draw(p,surface(A--B--E--cycle),rgb(1,1,.6),nolight);
add(scale3(2.2)*p);
draw(r,A--B--E--cycle);
draw(r,A--C--D--cycle);
draw(r,F--C--B--cycle);
draw(r,F--D--E--cycle,dotted+linewidth(0.7));
draw(r,surface(A--B--C--cycle),rgb(1,.6,.6),nolight);
draw(r,surface(A--C--D--cycle),rgb(1,1,.6),nolight);
add(scale3(2.2)*shift(2*right)*r);
draw(s,A--B--E--cycle);
draw(s,A--C--D--cycle);
draw(s,F--C--B--cycle);
draw(s,F--D--E--cycle,dotted+linewidth(0.7));
draw(s,surface(A--B--C--cycle),rgb(1,.6,.6),nolight);
draw(s,surface(B--C--F--cycle),rgb(1,1,.6),nolight);
add(scale3(2.2)*shift(4*right)*s);
// Block 3
size(8cm); defaultpen(0.5); import three; import math; currentprojection=orthographic(2,0.2,1); triple A=(0,0,1); triple B=(sqrt(2)/2,sqrt(2)/2,0); triple C=(sqrt(2)/2,-sqrt(2)/2,0); triple D=(-sqrt(2)/2,-sqrt(2)/2,0); triple E=(-sqrt(2)/2,sqrt(2)/2,0); triple F=(0,0,-1); draw(A--B--E--cycle); draw(A--C--D--cycle); draw(F--C--B--cycle); draw(F--D--E--cycle,dotted+linewidth(0.7)); draw(surface(A--B--C--cycle),rgb(1,.6,.6),nolight);
// Block 4
size(8cm); defaultpen(0.5); import three; import math; currentprojection=orthographic(2,0,1); triple A=(0,0,1); triple B=(sqrt(2)/2,sqrt(2)/2,0); triple C=(sqrt(2)/2,-sqrt(2)/2,0); triple D=(-sqrt(2)/2,-sqrt(2)/2,0); triple E=(-sqrt(2)/2,sqrt(2)/2,0); triple F=(0,0,-1); triple right=(0,1,0); picture p = new picture, r = new picture, s = new picture; draw(p,A--B--E--cycle); draw(p,A--C--D--cycle); draw(p,F--C--B--cycle); draw(p,F--D--E--cycle,dotted+linewidth(0.7)); draw(p,surface(A--B--C--cycle),rgb(1,.6,.6),nolight); draw(p,surface(A--B--E--cycle),rgb(1,1,.6),nolight); add(scale3(2.2)*p); draw(r,A--B--E--cycle); draw(r,A--C--D--cycle); draw(r,F--C--B--cycle); draw(r,F--D--E--cycle,dotted+linewidth(0.7)); draw(r,surface(A--B--C--cycle),rgb(1,.6,.6),nolight); draw(r,surface(A--C--D--cycle),rgb(1,1,.6),nolight); add(scale3(2.2)*shift(2*right)*r); draw(s,A--B--E--cycle); draw(s,A--C--D--cycle); draw(s,F--C--B--cycle); draw(s,F--D--E--cycle,dotted+linewidth(0.7)); draw(s,surface(A--B--C--cycle),rgb(1,.6,.6),nolight); draw(s,surface(B--C--F--cycle),rgb(1,1,.6),nolight); add(scale3(2.2)*shift(4*right)*s); | [] |
562 | A street has parallel curbs $40$ feet apart. A crosswalk bounded by two parallel stripes crosses the street at an angle. The length of the curb between the stripes is $15$ feet and each stripe is $50$ feet long. Find the distance, in feet, between the stripes.
$\textbf{(A)}\ 9 \qquad \textbf{(B)}\ 10 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)}\ 25$ | 2001 AMC 10 Problem 15 | Drawing the problem out, we see we get a parallelogram with a height of $40$ and a base of $15$, giving an area of $600$.
If we look at it the other way, we see the distance between the stripes is the height and the base is $50$.
The area is still the same, so the distance between the stripes is $600/50 = \boxed{\textbf{(C)}\ 12}$. | // Block 1
draw((0,0)--(5,0));
draw((2.5,5)--(7.5,5));
draw((0,0)--(2.5,5));
draw((5,0)--(7.5,5),linewidth(2));
draw((2,4)--(6,2),dashed);
// Block 2
draw((0,0)--(5,0)); draw((2.5,5)--(7.5,5)); draw((0,0)--(2.5,5)); draw((5,0)--(7.5,5),linewidth(2)); draw((2,4)--(6,2),dashed); | [] |
563 | Which of the cones listed below can be formed from a $252^\circ$ sector of a circle of radius $10$ by aligning the two straight sides?
$\textbf{(A)} \text{ A cone with slant height of } 10 \text{ and radius } 6$
$\textbf{(B)} \text{ A cone with height of } 10 \text{ and radius } 6$
$\textbf{(C)} \text{ A cone with slant height of } 10 \text{ and radius } 7$
$\textbf{(D)} \text{ A cone with height of } 10 \text{ and radius } 7$
$\textbf{(E)} \text{ A cone with slant height of } 10 \text{ and radius } 8$ | 2001 AMC 10 Problem 17 | The blue lines will be joined together to form a single blue line on the surface of the cone, so $10$ will be the slant height of the cone.
The red line will form the circumference of the base. We can compute its length and use it to determine the radius.
The length of the red line is $\dfrac{252^{\circ}}{360^{\circ}}\cdot 2\pi \cdot 10 = \frac{7}{10} \cdot 2\pi \cdot 10 = 14\pi$. This is the circumference of a circle with radius $\frac{14\pi}{2\pi} = 7$.
Therefore the correct answer is $\boxed{\textbf{(C)} \text{ A cone with slant height of } 10 \text{ and radius } 7}$. | import graph; unitsize(1.5cm); defaultpen(fontsize(8pt)); draw(Arc((0,0),1,-72,180),linewidth(.8pt) + red); draw(dir(288)--(0,0)--(-1,0),linewidth(.8pt) + blue); label("$10$",(-0.5,0),S); draw(Arc((0,0),0.1,-72,180)); label("$252^{\circ}$",(0.05,0.05),NE); | [] |
564 | The plane is tiled by congruent squares and congruent pentagons as indicated. The percent of the plane that is enclosed by the pentagons is closest to
$\textbf{(A) }50 \qquad \textbf{(B) }52 \qquad \textbf{(C) }54 \qquad \textbf{(D) }56 \qquad \textbf{(E) }58$ | 2001 AMC 10 Problem 18 | Consider any single tile:
If the side of the small square is $a$, then the area of the tile is $\left(3a\right)^2 = 9a^2$, with $4a^2$ covered by squares, and therefore $9a^2-4a^2 = 5a^2$ by pentagons.
Hence exactly $\frac{5}{9}$ of any tile is covered by pentagons, and therefore pentagons cover $\frac{5}{9}$ of the plane. When expressed as a percentage, this is $55.\overline{5}\%$, and the closest integer to this value is $\boxed{\textbf{(D) }56}$. | unitsize(1cm); defaultpen(linewidth(0.8pt)); path p1=(0,0)--(3,0)--(3,3)--(0,3)--(0,0); path p2=(0,1)--(1,1)--(1,0); path p3=(2,0)--(2,1)--(3,1); path p4=(3,2)--(2,2)--(2,3); path p5=(1,3)--(1,2)--(0,2); path p6=(1,1)--(2,2); path p7=(2,1)--(1,2); path[] p=p1^^p2^^p3^^p4^^p5^^p6^^p7; draw(p); | [] |
565 | A right circular cylinder with its diameter equal to its height is inscribed in a right circular cone. The cone has diameter $10$ and altitude $12$, and the axes of the cylinder and cone coincide. Find the radius of the cylinder.
$\textbf{(A)}\ \frac{8}3\qquad\textbf{(B)}\ \frac{30}{11}\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ \frac{25}{8}\qquad\textbf{(E)}\ \frac{7}{2}$ | 2001 AMC 10 Problem 21 | Let the diameter of the cylinder be $2r$. Examining the cross section of the cone and cylinder, we find two similar triangles. Hence, $\frac{12-2r}{12}=\frac{2r}{10}$ which we solve to find $r=\frac{30}{11}$. Our answer is $\boxed{\textbf{(B)}\ \frac{30}{11}}$. | // Block 1
draw((5,0)--(-5,0)--(0,12)--cycle);
unitsize(.75cm);
draw((-30/11,0)--(-30/11,60/11));
draw((-30/11,60/11)--(30/11,60/11));
draw((30/11,60/11)--(30/11,0));
draw((0,0)--(0,12));
label("$2r$",(0,30/11),E);
label("$12-2r$",(0,80/11),E);
label("$2r$",(0,60/11),S);
label("$10$",(0,0),S);
label("$A$",(0,12),N);
label("$B$",(-5,0),SW);
label("$C$",(5,0),SE);
label("$D$",(-30/11,60/11),W);
label("$E$",(30/11,60/11),E);
// Block 2
draw((5,0)--(-5,0)--(0,12)--cycle); unitsize(.75cm); draw((-30/11,0)--(-30/11,60/11)); draw((-30/11,60/11)--(30/11,60/11)); draw((30/11,60/11)--(30/11,0)); draw((0,0)--(0,12)); label("$2r$",(0,30/11),E); label("$12-2r$",(0,80/11),E); label("$2r$",(0,60/11),S); label("$10$",(0,0),S); label("$A$",(0,12),N); label("$B$",(-5,0),SW); label("$C$",(5,0),SE); label("$D$",(-30/11,60/11),W); label("$E$",(30/11,60/11),E); | [] |
565 | A right circular cylinder with its diameter equal to its height is inscribed in a right circular cone. The cone has diameter $10$ and altitude $12$, and the axes of the cylinder and cone coincide. Find the radius of the cylinder.
$\textbf{(A)}\ \frac{8}3\qquad\textbf{(B)}\ \frac{30}{11}\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ \frac{25}{8}\qquad\textbf{(E)}\ \frac{7}{2}$ | 2001 AMC 10 Problem 21 | Like in Solution 2, we draw a diagram.
It is known that $\overline{AH}$ has length $12$ and $\overline{BC}$ has length $10$, so triangle $\triangle ABC$ has area $60$. Also, let $x$ be equal to the radius of the cylinder.
Triangles $\triangle DBG$ and $\triangle ECF$ can be combined into one triangle with base $10-2x$ and height $2x$. The area of this new triangle is $\frac{2x(10-2x)}{2} = 2x(5-x)$.
Triangle $\triangle ADE$ has base $2x$ and height $12-2x$, so its area is$\frac{2x(12-2x)}{2} = 2x(6-x)$.
Finally, square $DEFG$ has area $4x^2$.
Now we can construct an equation to find $x$:
\[2x(5-x) + 2x(6-x) + 4x^2 = 60\]
\[\Rightarrow 2x(5-x+6-x+2x) = 60\]
\[\Rightarrow 2x(11) = 60\]
\[\Rightarrow 22x = 60\]
\[\Rightarrow x = \frac{60}{22} = \boxed{(B) \frac{30}{11}}\]
~Dreamer1297 | // Block 1
draw((5,0)--(-5,0)--(0,12)--cycle);
unitsize(.75cm);
draw((-30/11,0)--(-30/11,60/11));
draw((-30/11,60/11)--(30/11,60/11));
draw((30/11,60/11)--(30/11,0));
draw((0,0)--(0,12));
label("$2x$",(0,30/11),E);
label("$2x$",(0,60/11),S);
label("$H$",(0,0),S);
label("$A$",(0,12),N);
label("$B$",(-5,0),SW);
label("$C$",(5,0),SE);
label("$D$",(-30/11,60/11),W);
label("$E$",(30/11,60/11),E);
label("$F$",(30/11,0),S);
label("$G$",(-30/11,0),S);
// Block 2
draw((5,0)--(-5,0)--(0,12)--cycle); unitsize(.75cm); draw((-30/11,0)--(-30/11,60/11)); draw((-30/11,60/11)--(30/11,60/11)); draw((30/11,60/11)--(30/11,0)); draw((0,0)--(0,12)); label("$2x$",(0,30/11),E); label("$2x$",(0,60/11),S); label("$H$",(0,0),S); label("$A$",(0,12),N); label("$B$",(-5,0),SW); label("$C$",(5,0),SE); label("$D$",(-30/11,60/11),W); label("$E$",(30/11,60/11),E); label("$F$",(30/11,0),S); label("$G$",(-30/11,0),S); | [] |
566 | In the magic square shown, the sums of the numbers in each row, column, and diagonal are the same. Five of these numbers are represented by $v$, $w$, $x$, $y$, and $z$. Find $y + z$.
$\textbf{(A)}\ 43 \qquad \textbf{(B)}\ 44 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 46 \qquad \textbf{(E)}\ 47$ | 2001 AMC 10 Problem 22 | We know that $y+z=2v$, so we could find one variable rather than two.
$v+24+w=43+v$
$24+w=43$
$w=19$
$44+x=24+x+z \implies z=20$
The sum per row is $25+21+20=66$.
Thus $66-18-25=66-43=v=23$.
Since we needed $2v$ and we know $v=23$, $23 \times 2 = \boxed{\textbf{(D)}\ 46}$. | // Block 1
unitsize(10mm);
defaultpen(linewidth(1pt));
for(int i=0; i<=3; ++i)
{
draw((0,i)--(3,i));
draw((i,0)--(i,3));
}
label("$25$",(0.5,0.5));
label("$z$",(1.5,0.5));
label("$21$",(2.5,0.5));
label("$18$",(0.5,1.5));
label("$x$",(1.5,1.5));
label("$y$",(2.5,1.5));
label("$v$",(0.5,2.5));
label("$24$",(1.5,2.5));
label("$19$",(2.5,2.5));
// Block 2
unitsize(10mm);
defaultpen(linewidth(1pt));
for(int i=0; i<=3; ++i)
{
draw((0,i)--(3,i));
draw((i,0)--(i,3));
}
label("$25$",(0.5,0.5));
label("$20$",(1.5,0.5));
label("$21$",(2.5,0.5));
label("$18$",(0.5,1.5));
label("$x$",(1.5,1.5));
label("$y$",(2.5,1.5));
label("$v$",(0.5,2.5));
label("$24$",(1.5,2.5));
label("$19$",(2.5,2.5));
// Block 3
unitsize(10mm); defaultpen(linewidth(1pt)); for(int i=0; i<=3; ++i) { draw((0,i)--(3,i)); draw((i,0)--(i,3)); } label("$25$",(0.5,0.5)); label("$z$",(1.5,0.5)); label("$21$",(2.5,0.5)); label("$18$",(0.5,1.5)); label("$x$",(1.5,1.5)); label("$y$",(2.5,1.5)); label("$v$",(0.5,2.5)); label("$24$",(1.5,2.5)); label("$19$",(2.5,2.5));
// Block 4
unitsize(10mm); defaultpen(linewidth(1pt)); for(int i=0; i<=3; ++i) { draw((0,i)--(3,i)); draw((i,0)--(i,3)); } label("$25$",(0.5,0.5)); label("$20$",(1.5,0.5)); label("$21$",(2.5,0.5)); label("$18$",(0.5,1.5)); label("$x$",(1.5,1.5)); label("$y$",(2.5,1.5)); label("$v$",(0.5,2.5)); label("$24$",(1.5,2.5)); label("$19$",(2.5,2.5)); | [] |
566 | In the magic square shown, the sums of the numbers in each row, column, and diagonal are the same. Five of these numbers are represented by $v$, $w$, $x$, $y$, and $z$. Find $y + z$.
$\textbf{(A)}\ 43 \qquad \textbf{(B)}\ 44 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 46 \qquad \textbf{(E)}\ 47$ | 2001 AMC 10 Problem 22 | $v+24+w=43+v$
$24+w=43$
$w=19$
$44+x=24+x+z \implies z=20$
The magic sum is determined by the bottom row. $25+20+21=66$.
Solving for $y$:
$y=66-19-21=66-40=26$.
To find our answer, we need to find $y+z$. $y+z=20+26 = \boxed{\textbf{(D)}\ 46}$. | // Block 1
unitsize(10mm);
defaultpen(linewidth(1pt));
for(int i=0; i<=3; ++i)
{
draw((0,i)--(3,i));
draw((i,0)--(i,3));
}
label("$25$",(0.5,0.5));
label("$z$",(1.5,0.5));
label("$21$",(2.5,0.5));
label("$18$",(0.5,1.5));
label("$x$",(1.5,1.5));
label("$y$",(2.5,1.5));
label("$v$",(0.5,2.5));
label("$24$",(1.5,2.5));
label("$19$",(2.5,2.5));
// Block 2
unitsize(10mm);
defaultpen(linewidth(1pt));
for(int i=0; i<=3; ++i)
{
draw((0,i)--(3,i));
draw((i,0)--(i,3));
}
label("$25$",(0.5,0.5));
label("$20$",(1.5,0.5));
label("$21$",(2.5,0.5));
label("$18$",(0.5,1.5));
label("$x$",(1.5,1.5));
label("$y$",(2.5,1.5));
label("$v$",(0.5,2.5));
label("$24$",(1.5,2.5));
label("$19$",(2.5,2.5));
// Block 3
unitsize(10mm); defaultpen(linewidth(1pt)); for(int i=0; i<=3; ++i) { draw((0,i)--(3,i)); draw((i,0)--(i,3)); } label("$25$",(0.5,0.5)); label("$z$",(1.5,0.5)); label("$21$",(2.5,0.5)); label("$18$",(0.5,1.5)); label("$x$",(1.5,1.5)); label("$y$",(2.5,1.5)); label("$v$",(0.5,2.5)); label("$24$",(1.5,2.5)); label("$19$",(2.5,2.5));
// Block 4
unitsize(10mm); defaultpen(linewidth(1pt)); for(int i=0; i<=3; ++i) { draw((0,i)--(3,i)); draw((i,0)--(i,3)); } label("$25$",(0.5,0.5)); label("$20$",(1.5,0.5)); label("$21$",(2.5,0.5)); label("$18$",(0.5,1.5)); label("$x$",(1.5,1.5)); label("$y$",(2.5,1.5)); label("$v$",(0.5,2.5)); label("$24$",(1.5,2.5)); label("$19$",(2.5,2.5)); | [] |
567 | In trapezoid $ABCD$, $\overline{AB}$ and $\overline{CD}$ are perpendicular to $\overline{AD}$, with $AB+CD=BC$, $AB<CD$, and $AD=7$. What is $AB\cdot CD$?
$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 12.25 \qquad \textbf{(C)}\ 12.5 \qquad \textbf{(D)}\ 12.75 \qquad \textbf{(E)}\ 13$ | 2001 AMC 10 Problem 24 | If $AB=x$ and $CD=y$, then $BC=x+y$. By the Pythagorean theorem, we have $(x+y)^2=(y-x)^2+49.$
Solving the equation, we get $4xy=49 \implies xy = \boxed{\textbf{(B)}\ 12.25}$. | // Block 1
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(7cm);
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 = -4.3, xmax = 7.3, ymin = -3.16, ymax = 6.3; /* image dimensions */
/* draw figures */
draw(circle((0.2,4.92), 1.3));
draw(circle((1.04,1.58), 2.14));
draw((-1.1,4.92)--(0.2,4.92));
draw((0.2,4.92)--(1.04,1.58));
draw((1.04,1.58)--(-1.1,1.58));
draw((-1.1,1.58)--(-1.1,4.92));
/* dots and labels */
dot((-1.1,4.92),dotstyle);
label("$A$", (-1.02,5.12), NE * labelscalefactor);
dot((0.2,4.92),dotstyle);
label("$B$", (0.28,5.12), NE * labelscalefactor);
dot((-1.1,1.58),dotstyle);
label("$D$", (-1.02,1.78), NE * labelscalefactor);
dot((1.04,1.58),dotstyle);
label("$C$", (1.12,1.78), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
// Block 2
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(7cm); 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 = -4.3, xmax = 7.3, ymin = -3.16, ymax = 6.3; /* image dimensions */ /* draw figures */ draw(circle((0.2,4.92), 1.3)); draw(circle((1.04,1.58), 2.14)); draw((-1.1,4.92)--(0.2,4.92)); draw((0.2,4.92)--(1.04,1.58)); draw((1.04,1.58)--(-1.1,1.58)); draw((-1.1,1.58)--(-1.1,4.92)); /* dots and labels */ dot((-1.1,4.92),dotstyle); label("$A$", (-1.02,5.12), NE * labelscalefactor); dot((0.2,4.92),dotstyle); label("$B$", (0.28,5.12), NE * labelscalefactor); dot((-1.1,1.58),dotstyle); label("$D$", (-1.02,1.78), NE * labelscalefactor); dot((1.04,1.58),dotstyle); label("$C$", (1.12,1.78), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ | [] |
568 | How many positive integers not exceeding $2001$ are multiples of $3$ or $4$ but not $5$?
$\textbf{(A) }768 \qquad \textbf{(B) }801 \qquad \textbf{(C) }934 \qquad \textbf{(D) }1067 \qquad \textbf{(E) }1167$ | 2001 AMC 10 Problem 25 | We have the following Venn diagram (sorry for the bad asymptote). To calculate this Venn diagram we can just use PIE to find the numbers or just work our way out from the inside. We just need to find the numbers that are divisible by $3$ or $4$ but not by $5$, and looking back at the diagram we just have the answer of \[401+133+267 = \boxed{\textbf{(B)}\ 801}\].
-jb2015007
Solution 6
Similar to @above:
Let the function $M_{2001}(n)$ return how many multiples of $n$ are there not exceeding $2001$. Then we have that the desired number is:
\[M_{2001}(3)+M_{2001}(4)-M_{2001}(3\cdot 4)-M_{2001}(3 \cdot 5) - M_{2001}(4 \cdot 5)+M_{2001}(3 \cdot 4 \cdot 5)\]
Evaluating each of these we get:
\[667+500-166-133-100+33 = 1100-299 = 801.\]
Thus, the answer is $\boxed{\textbf{(B)}\ 801}.$
-ConfidentKoala4
Solution 7
The smallest common multiple of 3, 4, and 5 is 30.
There are 12 integers between 1~30 that are multiples of 3 and 4 but not 5.
2001/30 = 66 R 21.
Between 1~21 there are 9 integers that are multiples of 3 and 4 but not 5.
66 * 12 + 9 = $\boxed{\textbf{(B)}\ 801}.$
- mathlover1205
- AH2025 (math and number corrections) | size(10cm); // Draw circles draw(circle((0,0),2), blue); // Circle for multiples of 3 draw(circle((2,0),2), red); // Circle for multiples of 4 draw(circle((1,-1.5),2), green); // Circle for multiples of 5 // Labels for the sets label("$3$", (-2.5,1.5), blue); label("$4$", (4.5,1.5), red); label("$5$", (1,-3.8), green); // Numbers in regions (precisely adjusted) label("$401$", (-1.5,0.5), black); // Only 3 label("$267$", (3.5,0.5), black); // Only 4 label("$200$", (1,-3), black); // Only 5 label("$133$", (1,1.1), black); // 3 and 4 (not 5) label("$100$", (0.2,-1.8), black); // 3 and 5 (not 4) label("$67$", (1.8,-1.8), black); // 4 and 5 (not 3) label("$33$", (1,-0.4), black); // 3, 4, and 5 | [] |
569 | Which of the cones listed below can be formed from a $252^\circ$ sector of a circle of radius $10$ by aligning the two straight sides?
$\textbf{(A)} \text{ A cone with slant height of } 10 \text{ and radius } 6$
$\textbf{(B)} \text{ A cone with height of } 10 \text{ and radius } 6$
$\textbf{(C)} \text{ A cone with slant height of } 10 \text{ and radius } 7$
$\textbf{(D)} \text{ A cone with height of } 10 \text{ and radius } 7$
$\textbf{(E)} \text{ A cone with slant height of } 10 \text{ and radius } 8$ | 2001 AMC 12 Problem 8 | The blue lines will be joined together to form a single blue line on the surface of the cone, so $10$ will be the slant height of the cone.
The red line will form the circumference of the base. We can compute its length and use it to determine the radius.
The length of the red line is $\dfrac{252^{\circ}}{360^{\circ}}\cdot 2\pi \cdot 10 = \frac{7}{10} \cdot 2\pi \cdot 10 = 14\pi$. This is the circumference of a circle with radius $\frac{14\pi}{2\pi} = 7$.
Therefore the correct answer is $\boxed{\textbf{(C)} \text{ A cone with slant height of } 10 \text{ and radius } 7}$. | // Block 1
import graph;
unitsize(1.5cm);
defaultpen(fontsize(8pt));
draw(Arc((0,0),1,-72,180),linewidth(.8pt) + red);
draw(dir(288)--(0,0)--(-1,0),linewidth(.8pt) + blue);
label("$10$",(-0.5,0),S);
draw(Arc((0,0),0.1,-72,180));
label("$252^{\circ}$",(0.05,0.05),NE);
// Block 2
import graph; unitsize(1.5cm); defaultpen(fontsize(8pt)); draw(Arc((0,0),1,-72,180),linewidth(.8pt) + red); draw(dir(288)--(0,0)--(-1,0),linewidth(.8pt) + blue); label("$10$",(-0.5,0),S); draw(Arc((0,0),0.1,-72,180)); label("$252^{\circ}$",(0.05,0.05),NE); | [] |
570 | The plane is tiled by congruent squares and congruent pentagons as indicated. The percent of the plane that is enclosed by the pentagons is closest to
$\textbf{(A) }50 \qquad \textbf{(B) }52 \qquad \textbf{(C) }54 \qquad \textbf{(D) }56 \qquad \textbf{(E) }58$ | 2001 AMC 12 Problem 10 | Consider any single tile:
If the side of the small square is $a$, then the area of the tile is $\left(3a\right)^2 = 9a^2$, with $4a^2$ covered by squares, and therefore $9a^2-4a^2 = 5a^2$ by pentagons.
Hence exactly $\frac{5}{9}$ of any tile is covered by pentagons, and therefore pentagons cover $\frac{5}{9}$ of the plane. When expressed as a percentage, this is $55.\overline{5}\%$, and the closest integer to this value is $\boxed{\textbf{(D) }56}$. | // Block 1
unitsize(1cm);
defaultpen(linewidth(0.8pt));
path p1=(0,0)--(3,0)--(3,3)--(0,3)--(0,0);
path p2=(0,1)--(1,1)--(1,0);
path p3=(2,0)--(2,1)--(3,1);
path p4=(3,2)--(2,2)--(2,3);
path p5=(1,3)--(1,2)--(0,2);
path p6=(1,1)--(2,2);
path p7=(2,1)--(1,2);
path[] p=p1^^p2^^p3^^p4^^p5^^p6^^p7;
draw(p);
// Block 2
unitsize(1cm); defaultpen(linewidth(0.8pt)); path p1=(0,0)--(3,0)--(3,3)--(0,3)--(0,0); path p2=(0,1)--(1,1)--(1,0); path p3=(2,0)--(2,1)--(3,1); path p4=(3,2)--(2,2)--(2,3); path p5=(1,3)--(1,2)--(0,2); path p6=(1,1)--(2,2); path p7=(2,1)--(1,2); path[] p=p1^^p2^^p3^^p4^^p5^^p6^^p7; draw(p); | [] |
571 | How many positive integers not exceeding $2001$ are multiples of $3$ or $4$ but not $5$?
$\textbf{(A) }768 \qquad \textbf{(B) }801 \qquad \textbf{(C) }934 \qquad \textbf{(D) }1067 \qquad \textbf{(E) }1167$ | 2001 AMC 12 Problem 12 | We have the following Venn diagram (sorry for the bad asymptote). To calculate this Venn diagram we can just use PIE to find the numbers or just work our way out from the inside. We just need to find the numbers that are divisible by $3$ or $4$ but not by $5$, and looking back at the diagram we just have the answer of \[401+133+267 = \boxed{\textbf{(B)}\ 801}\].
-jb2015007
Solution 6
Similar to @above:
Let the function $M_{2001}(n)$ return how many multiples of $n$ are there not exceeding $2001$. Then we have that the desired number is:
\[M_{2001}(3)+M_{2001}(4)-M_{2001}(3\cdot 4)-M_{2001}(3 \cdot 5) - M_{2001}(4 \cdot 5)+M_{2001}(3 \cdot 4 \cdot 5)\]
Evaluating each of these we get:
\[667+500-166-133-100+33 = 1100-299 = 801.\]
Thus, the answer is $\boxed{\textbf{(B)}\ 801}.$
-ConfidentKoala4
Solution 7
The smallest common multiple of 3, 4, and 5 is 30.
There are 12 integers between 1~30 that are multiples of 3 and 4 but not 5.
2001/30 = 66 R 21.
Between 1~21 there are 9 integers that are multiples of 3 and 4 but not 5.
66 * 12 + 9 = $\boxed{\textbf{(B)}\ 801}.$
- mathlover1205
- AH2025 (math and number corrections) | // Block 1
size(10cm);
// Draw circles
draw(circle((0,0),2), blue); // Circle for multiples of 3
draw(circle((2,0),2), red); // Circle for multiples of 4
draw(circle((1,-1.5),2), green); // Circle for multiples of 5
// Labels for the sets
label("$3$", (-2.5,1.5), blue);
label("$4$", (4.5,1.5), red);
label("$5$", (1,-3.8), green);
// Numbers in regions (precisely adjusted)
label("$401$", (-1.5,0.5), black); // Only 3
label("$267$", (3.5,0.5), black); // Only 4
label("$200$", (1,-3), black); // Only 5
label("$133$", (1,1.1), black); // 3 and 4 (not 5)
label("$100$", (0.2,-1.8), black); // 3 and 5 (not 4)
label("$67$", (1.8,-1.8), black); // 4 and 5 (not 3)
label("$33$", (1,-0.4), black); // 3, 4, and 5
// Block 2
size(10cm); // Draw circles draw(circle((0,0),2), blue); // Circle for multiples of 3 draw(circle((2,0),2), red); // Circle for multiples of 4 draw(circle((1,-1.5),2), green); // Circle for multiples of 5 // Labels for the sets label("$3$", (-2.5,1.5), blue); label("$4$", (4.5,1.5), red); label("$5$", (1,-3.8), green); // Numbers in regions (precisely adjusted) label("$401$", (-1.5,0.5), black); // Only 3 label("$267$", (3.5,0.5), black); // Only 4 label("$200$", (1,-3), black); // Only 5 label("$133$", (1,1.1), black); // 3 and 4 (not 5) label("$100$", (0.2,-1.8), black); // 3 and 5 (not 4) label("$67$", (1.8,-1.8), black); // 4 and 5 (not 3) label("$33$", (1,-0.4), black); // 3, 4, and 5 | [] |
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