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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448 | In triangle $ABC$, $AB = AC = 100$, and $BC = 56$. Circle $P$ has radius $16$ and is tangent to $\overline{AC}$ and $\overline{BC}$. Circle $Q$ is externally tangent to $P$ and is tangent to $\overline{AB}$ and $\overline{BC}$. No point of circle $Q$ lies outside of $\triangle ABC$. The radius of circle $Q$ can be expressed in the form $m - n\sqrt {k}$, where $m$, $n$, and $k$ are positive integers and $k$ is the product of distinct primes. Find $m + nk$. | 2008 AIME II Problems/Problem 11 | Let $X$ and $Y$ be the feet of the perpendiculars from $P$ and $Q$ to $BC$, respectively. Let the radius of $\odot Q$ be $r$. We know that $PQ = r + 16$. From $Q$ draw segment $\overline{QM} \parallel \overline{BC}$ such that $M$ is on $PX$. Clearly, $QM = XY$ and $PM = 16-r$. Also, we know $QPM$ is a right triangle.
To find $XC$, consider the right triangle $PCX$. Since $\odot P$ is tangent to $\overline{AC},\overline{BC}$, then $PC$ bisects $\angle ACB$. Let $\angle ACB = 2\theta$; then $\angle PCX = \angle QBX = \theta$. Dropping the altitude from $A$ to $BC$, we recognize the $7 - 24 - 25$ right triangle, except scaled by $4$.
So we get that $\tan(2\theta) = 24/7$. From the half-angle identity, we find that $\tan(\theta) = \frac {3}{4}$. Therefore, $XC = \frac {64}{3}$. By similar reasoning in triangle $QBY$, we see that $BY = \frac {4r}{3}$.
We conclude that $XY = 56 - \frac {4r + 64}{3} = \frac {104 - 4r}{3}$.
So our right triangle $QPM$ has sides $r + 16$, $r - 16$, and $\frac {104 - 4r}{3}$.
By the Pythagorean Theorem, simplification, and the quadratic formula, we can get $r = 44 - 6\sqrt {35}$, for a final answer of $\fbox{254}$. | // Block 1
size(200);
pathpen=black;pointpen=black;pen f=fontsize(9);
real r=44-6*35^.5;
pair A=(0,96),B=(-28,0),C=(28,0),X=C-(64/3,0),Y=B+(4*r/3,0),P=X+(0,16),Q=Y+(0,r),M=foot(Q,X,P);
path PC=CR(P,16),QC=CR(Q,r);
D(A--B--C--cycle); D(Y--Q--P--X); D(Q--M); D(P--C,dashed);
D(PC); D(QC);
MP("A",A,N,f);MP("B",B,f);MP("C",C,f);MP("X",X,f);MP("Y",Y,f);D(MP("P",P,NW,f));D(MP("Q",Q,NW,f));
// Block 2
size(200); pathpen=black;pointpen=black;pen f=fontsize(9); real r=44-6*35^.5; pair A=(0,96),B=(-28,0),C=(28,0),X=C-(64/3,0),Y=B+(4*r/3,0),P=X+(0,16),Q=Y+(0,r),M=foot(Q,X,P); path PC=CR(P,16),QC=CR(Q,r); D(A--B--C--cycle); D(Y--Q--P--X); D(Q--M); D(P--C,dashed); D(PC); D(QC); MP("A",A,N,f);MP("B",B,f);MP("C",C,f);MP("X",X,f);MP("Y",Y,f);D(MP("P",P,NW,f));D(MP("Q",Q,NW,f)); | [] |
449 | A regular hexagon with center at the origin in the complex plane has opposite pairs of sides one unit apart. One pair of sides is parallel to the imaginary axis. Let $R$ be the region outside the hexagon, and let $S = \left\lbrace\frac{1}{z}|z \in R\right\rbrace$. Then the area of $S$ has the form $a\pi + \sqrt{b}$, where $a$ and $b$ are positive integers. Find $a + b$. | 2008 AIME II Problems/Problem 13 | If a point $z = r\text{cis}\,\theta$ is in $R$, then the point $\frac{1}{z} = \frac{1}{r} \text{cis}\, \left(-\theta\right)$ is in $S$ (where cis denotes $\text{cis}\, \theta = \cos \theta + i \sin \theta$). Since $R$ is symmetric every $60^{\circ}$ about the origin, it suffices to consider the area of the result of the transformation when $-30 \le \theta \le 30$, and then to multiply by $6$ to account for the entire area.
We note that if the region $S_2 = \left\lbrace\frac{1}{z}|z \in R_2\right\rbrace$, where $R_2$ is the region (in green below) outside the circle of radius $1/\sqrt{3}$ centered at the origin, then $S_2$ is simply the region inside a circle of radius $\sqrt{3}$ centered at the origin. It now suffices to find what happens to the mapping of the region $R-R_2$ (in blue below).
The equation of the hexagon side in that region is $x = r \cos \theta = \frac{1}{2}$, which is transformed to $\frac{1}{r} \cos (-\theta) = \frac{1}{r} \cos \theta =$2 . Let $r\text{cis}\,\theta = a+bi$ where $a,b \in \mathbb{R}$; then $r = \sqrt{a^2 + b^2}, \cos \theta = \frac{a}{\sqrt{a^2 + b^2}}$, so the equation becomes $a^2 - 2a + b^2 = 0 \Longrightarrow (a-1)^2 + b^2 = 1$. Hence the side is sent to an arc of the unit circle centered at $(1,0)$, after considering the restriction that the side of the hexagon is a segment of length $1/\sqrt{3}$.
Including $S_2$, we find that $S$ is the union of six unit circles centered at $\text{cis}\, \frac{k\pi}{6}$, $k = 0,1,2,3,4,5$, as shown below.
$\Longrightarrow$
The area of the regular hexagon is $6 \cdot \left( \frac{\left(\sqrt{3}\right)^2 \sqrt{3}}{4} \right) = \frac{9}{2}\sqrt{3}$. The total area of the six $120^{\circ}$ sectors is $6\left(\frac{1}{3}\pi - \frac{1}{2} \cdot \frac{1}{2} \cdot \sqrt{3}\right) = 2\pi - \frac{3}{2}\sqrt{3}$. Their sum is $2\pi + \sqrt{27}$, and $a+b = \boxed{029}$.
- Th3Numb3rThr33 | // Block 1
unitsize(1.5cm);
defaultpen(linewidth(0.7)); picture p; real max = .5 + 1/3^.5; pen d = linetype("4 4"); fill(1.5*expi(-pi/6)--arc((0,0),1,-30,30)--1.5*expi(pi/6)--cycle,rgb(0.5,1,0.5));fill(arc((0,0),1,-30,30)--cycle,rgb(0.5,0.5,1));
draw(p,expi(pi/6)--expi(-pi/6)--(0,0)--cycle);for(int i = 0; i < 6; ++i) add(rotate(i*60)*p); draw((0,max)--(0,-max),d,Arrows(4));draw((max,0)--(-max,0),d,Arrows(4)); draw(Circle((0,0),1),d); draw(expi(pi/6)--1.5*expi(pi/6),EndArrow(4)); draw(expi(-pi/6)--1.5*expi(-pi/6),EndArrow(4)); label("$1/\sqrt{3}$",(0,-0.5),W,fontsize(8));
// Block 2
unitsize(4.5cm);
defaultpen(linewidth(0.7)); picture p; fill((0,0)--arc((0,0),1,-30,30)--cycle,rgb(0.5,1,0.5));fill(arc((0,0),1,-30,30)--arc(1/3^.5,1/3^.5,60,-60)--cycle,rgb(0.5,0.5,1));
draw(p,expi(pi/6)--expi(-pi/6)--(0,0)--cycle);draw(p,arc(1/3^.5,1/3^.5,-60,60)); draw(arc(1/3^.5*expi(pi/3),1/3^.5,120,359.99),linetype("4 4")); draw(expi(pi/2)--1/3^.5*expi(pi/3)--expi(pi/6),linetype("4 4")); draw(Circle((0,0),1),linetype("4 4")); label("$\sqrt{3}$",(0,-0.5),W,fontsize(8));
add(p);add(rotate(60)*p);add(rotate(120)*p);add(rotate(180)*p);add(rotate(240)*p);add(rotate(300)*p);
// Block 3
unitsize(1.5cm); defaultpen(linewidth(0.7)); picture p; real max = .5 + 1/3^.5; pen d = linetype("4 4"); fill(1.5*expi(-pi/6)--arc((0,0),1,-30,30)--1.5*expi(pi/6)--cycle,rgb(0.5,1,0.5));fill(arc((0,0),1,-30,30)--cycle,rgb(0.5,0.5,1)); draw(p,expi(pi/6)--expi(-pi/6)--(0,0)--cycle);for(int i = 0; i < 6; ++i) add(rotate(i*60)*p); draw((0,max)--(0,-max),d,Arrows(4));draw((max,0)--(-max,0),d,Arrows(4)); draw(Circle((0,0),1),d); draw(expi(pi/6)--1.5*expi(pi/6),EndArrow(4)); draw(expi(-pi/6)--1.5*expi(-pi/6),EndArrow(4)); label("$1/\sqrt{3}$",(0,-0.5),W,fontsize(8));
// Block 4
unitsize(4.5cm); defaultpen(linewidth(0.7)); picture p; fill((0,0)--arc((0,0),1,-30,30)--cycle,rgb(0.5,1,0.5));fill(arc((0,0),1,-30,30)--arc(1/3^.5,1/3^.5,60,-60)--cycle,rgb(0.5,0.5,1)); draw(p,expi(pi/6)--expi(-pi/6)--(0,0)--cycle);draw(p,arc(1/3^.5,1/3^.5,-60,60)); draw(arc(1/3^.5*expi(pi/3),1/3^.5,120,359.99),linetype("4 4")); draw(expi(pi/2)--1/3^.5*expi(pi/3)--expi(pi/6),linetype("4 4")); draw(Circle((0,0),1),linetype("4 4")); label("$\sqrt{3}$",(0,-0.5),W,fontsize(8)); add(p);add(rotate(60)*p);add(rotate(120)*p);add(rotate(180)*p);add(rotate(240)*p);add(rotate(300)*p); | [] |
450 | In trapezoid $ABCD$ with $\overline{BC}\parallel\overline{AD}$, let $BC = 1000$ and $AD = 2008$. Let $\angle A = 37^\circ$, $\angle D = 53^\circ$, and $M$ and $N$ be the midpoints of $\overline{BC}$ and $\overline{AD}$, respectively. Find the length $MN$. | 2008 AIME II Problems/Problem 5 | Solution 1
Extend $\overline{AB}$ and $\overline{CD}$ to meet at a point $E$. Then $\angle AED = 180 - 53 - 37 = 90^{\circ}$.
As $\angle AED = 90^{\circ}$, note that the midpoint of $\overline{AD}$, $N$, is the center of the circumcircle of $\triangle AED$. We can do the same with the circumcircle about $\triangle BEC$ and $M$ (or we could apply the homothety to find $ME$ in terms of $NE$). It follows that
\[NE = ND = \frac {AD}{2} = 1004, \quad ME = MC = \frac {BC}{2} = 500.\]
Thus $MN = NE - ME = \boxed{504}$.
For purposes of rigor we will show that $E,M,N$ are collinear. Since $\overline{BC} \parallel \overline{AD}$, then $BC$ and $AD$ are homothetic with respect to point $E$ by a ratio of $\frac{BC}{AD} = \frac{125}{251}$. Since the homothety carries the midpoint of $\overline{BC}$, $M$, to the midpoint of $\overline{AD}$, which is $N$, then $E,M,N$ are collinear.
Solution 2
Let $F,G,H$ be the feet of the perpendiculars from $B,C,M$ onto $\overline{AD}$, respectively. Let $x = NH$, so $DG = 1004 - 500 - x = 504 - x$ and $AF = 1004 - (500 - x) = 504 + x$. Also, let $h = BF = CG = HM$.
By AA~, we have that $\triangle AFB \sim \triangle CGD$, and so
\[\frac{BF}{AF} = \frac {DG}{CG} \Longleftrightarrow \frac{h}{504+x} = \frac{504-x}{h} \Longrightarrow x^2 + h^2 = 504^2.\]
By the Pythagorean Theorem on $\triangle MHN$,
\[MN^{2} = x^2 + h^2 = 504^2,\] so $MN = \boxed{504}$.
Solution 3
If you drop perpendiculars from $B$ and $C$ to $AD$, and call the points where they meet $\overline{AD}$, $E$ and $F$ respectively, then $FD = x$ and $EA = 1008-x$ , and so you can solve an equation in tangents. Since $\angle{A} = 37$ and $\angle{D} = 53$, you can solve the equation [by cross-multiplication]:
\begin{align*}\tan{37}\times (1008-x) &= \tan{53} \times x\\ \frac{(1008-x)}{x} &= \frac{\tan{53}}{\tan{37}} = \frac{\sin{53}}{\cos{53}} \times\frac{\cos{37}}{\sin{37}}\end{align*}
However, we know that $\cos{90-x} = \sin{x}$ and $\sin{90-x} = \cos{x}$ are co-functions. Applying this,
\begin{align*}\frac{(1008-x)}{x} &= \frac{\sin^2{53}}{\cos^2{53}} \\ x\sin^2{53} &= 1008\cos^2{53} - x\cos^2{53}\\ x(\sin^2{53} + \cos^2{53}) &= 1008\cos^2{53}\\ x = 1008\cos^2{53} &\Longrightarrow 1008-x = 1008\sin^2{53} \end{align*}
Now, if we can find $1004 - (EA + 500)$, and the height of the trapezoid, we can create a right triangle and use the Pythagorean Theorem to find $MN$.
The leg of the right triangle along the horizontal is:
\[1004 - 1008\sin^2{53} - 500 = 504 - 1008\sin^2{53}.\]
Now to find the other leg of the right triangle (also the height of the trapezoid), we can simplify the following expression:
\begin{align*}\tan{37} \times 1008 \sin^2{53} = \tan{37} \times 1008 \cos^2{37} = 1008\cos{37}\sin{37} = 504\sin74\end{align*}
Now we used Pythagorean Theorem and get that $MN$ is equal to:
\begin{align*}&\sqrt{(1008\sin^2{53} + 500 -1004)^2 + (504\sin{74})^2} = 504\sqrt{1-2\sin^2{53} + \sin^2{74}} \end{align*}
However, $1-2\sin^2{53} = \cos^2{106}$ and $\sin^2{74} = \sin^2{106}$ so now we end up with:
\[504\sqrt{\cos^2{106} + \sin^2{106}} =\fbox{504}.\]
Solution 4
Plot the trapezoid such that $B=\left(1000\cos 37^\circ, 0\right)$, $C=\left(0, 1000\sin 37^\circ\right)$, $A=\left(2008\cos 37^\circ, 0\right)$, and $D=\left(0, 2008\sin 37^\circ\right)$.
The midpoints of the requested sides are $\left(500\cos 37^\circ, 500\sin 37^\circ\right)$ and $\left(1004\cos 37^\circ, 1004\sin 37^\circ\right)$.
To find the distance from $M$ to $N$, we simply apply the distance formula and the Pythagorean identity $\sin^2 x + \cos^2 x = 1$ to get $MN=\boxed{504}$.
Solution 5
Similar to solution 1; Notice that it forms a right triangle. Remembering that the median to the hypotenuse is simply half the length of the hypotenuse, we quickly see that the length is $\frac{2008}{2}-\frac{1000}{2}=504$.
Solution 6
Obviously, these angles are random--the only special thing about them is that they add up to 90. So we might as well let the given angles equal 45 and 45, and now the answer is trivially $\boxed{504}$. (The trapezoid is isosceles, and you see two 45-45-90 triangles;from there you can get the answer.) | // Block 1
size(220); defaultpen(0.7+fontsize(10)); real f=100, r=1004/f; pair A=(0,0), D=(2*r, 0), N=(r,0), E=N+r*expi(74*pi/180); pair B=(126*A+125*E)/251, C=(126*D + 125*E)/251; pair[] M = intersectionpoints(N--E,B--C); draw(A--B--C--D--cycle); draw(B--E--C,dashed); draw(M[0]--N); draw(N--E,dashed); draw(rightanglemark(D,E,A,2)); picture p = new picture; draw(p,Circle(N,r),dashed+linewidth(0.5)); clip(p,A--D--D+(0,20)--A+(0,20)--cycle); add(p); label("\(A\)",A,SW); label("\(B\)",B,NW); label("\(C\)",C,NE); label("\(D\)",D,SE); label("\(E\)",E,NE); label("\(M\)",M[0],SW); label("\(N\)",N,S); label("\(1004\)",(N+D)/2,S); label("\(500\)",(M[0]+C)/2,S);
// Block 2
size(220); defaultpen(0.7+fontsize(10)); real f=100, r=1004/f; pair A=(0,0), D=(2*r, 0), N=(r,0), E=N+r*expi(74*pi/180); pair B=(126*A+125*E)/251, C=(126*D + 125*E)/251; pair[] M = intersectionpoints(N--E,B--C); pair F = foot(B,A,D), G=foot(C,A,D), H=foot(M[0],A,D); draw(A--B--C--D--cycle); draw(M[0]--N); draw(B--F,dashed); draw(C--G,dashed); draw(M[0]--H,dashed); label("\(A\)",A,SW); label("\(B\)",B,NW); label("\(C\)",C,NE); label("\(D\)",D,NE); label("\(F\)",F,S); label("\(G\)",G,SW); label("\(M\)",M[0],SW); label("\(N\)",N,S); label("\(H\)",H,S); label("\(x\)",(N+H)/2+(0,1),S); label("\(h\)",(B+F)/2,W); label("\(h\)",(C+G)/2,W); label("\(1000\)",(B+C)/2,NE); label("\(504-x\)",(G+D)/2,S); label("\(504+x\)",(A+F)/2,S); label("\(h\)",(M[0]+H)/2,(1,0)); | [] |
451 | Triangle $ABC$ has $\angle C = 60^{\circ}$ and $BC = 4$. Point $D$ is the midpoint of $BC$. What is the largest possible value of $\tan{\angle BAD}$?
$\mathrm{(A)}\ \frac{\sqrt{3}}{6}\qquad\mathrm{(B)}\ \frac{\sqrt{3}}{3}\qquad\mathrm{(C)}\ \frac{\sqrt{3}}{2\sqrt{2}}\qquad\mathrm{(D)}\ \frac{\sqrt{3}}{4\sqrt{2}-3}\qquad\mathrm{(E)}\ 1$ | 2008 AMC 12A Problems/Problem 24 | Let $x = CA$. Then $\tan\theta = \tan(\angle BAF - \angle DAE)$, and since $\tan\angle BAF = \frac{2\sqrt{3}}{x-2}$ and $\tan\angle DAE = \frac{\sqrt{3}}{x-1}$, we have
\[\tan\theta = \frac{\frac{2\sqrt{3}}{x-2} - \frac{\sqrt{3}}{x-1}}{1 + \frac{2\sqrt{3}}{x-2}\cdot\frac{\sqrt{3}}{x-1}}= \frac{x\sqrt{3}}{x^2-3x+8}\]
With calculus, taking the derivative and setting equal to zero will give the maximum value of $\tan \theta$. Otherwise, we can apply AM-GM:
\begin{align*} \frac{x^2 - 3x + 8}{x} = \left(x + \frac{8}{x}\right) -3 &\geq 2\sqrt{x \cdot \frac 8x} - 3 = 4\sqrt{2} - 3\\ \frac{x}{x^2 - 3x + 8} &\leq \frac{1}{4\sqrt{2}-3}\\ \frac{x\sqrt{3}}{x^2 - 3x + 8} = \tan \theta &\leq \frac{\sqrt{3}}{4\sqrt{2}-3}\end{align*}
Thus, the maximum is at
$\frac{\sqrt{3}}{4\sqrt{2}-3} \Rightarrow \mathbf{(D)}$. | // Block 1
unitsize(12mm);
pair C=(0,0), B=(4 * dir(60)), A = (8,0), D=(2 * dir(60));
pair E=(1,0), F=(2,0);
draw(C--B--A--C);
draw(A--D);draw(D--E);draw(B--F);
dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);
label("\(C\)",C,SW);
label("\(B\)",B,N);
label("\(A\)",A,SE);
label("\(D\)",D,NW);
label("\(E\)",E,S);
label("\(F\)",F,S);
label("\(60^\circ\)",C+(.1,.1),ENE);
label("\(2\)",1*dir(60),NW);
label("\(2\)",3*dir(60),NW);
label("\(\theta\)",(7,.4));
label("\(1\)",(.5,0),S);
label("\(1\)",(1.5,0),S);
label("\(x-2\)",(5,0),S);
// Block 2
unitsize(12mm); pair C=(0,0), B=(4 * dir(60)), A = (8,0), D=(2 * dir(60)); pair E=(1,0), F=(2,0); draw(C--B--A--C); draw(A--D);draw(D--E);draw(B--F); dot(A);dot(B);dot(C);dot(D);dot(E);dot(F); label("\(C\)",C,SW); label("\(B\)",B,N); label("\(A\)",A,SE); label("\(D\)",D,NW); label("\(E\)",E,S); label("\(F\)",F,S); label("\(60^\circ\)",C+(.1,.1),ENE); label("\(2\)",1*dir(60),NW); label("\(2\)",3*dir(60),NW); label("\(\theta\)",(7,.4)); label("\(1\)",(.5,0),S); label("\(1\)",(1.5,0),S); label("\(x-2\)",(5,0),S); | [] |
452 | In order to earn her vacation spending money, Alexis helped her mother remove weeds from the garden. When she was done, she came into the house to put away her gardening gloves and change into clean clothes.
On her way to her room she notices Joshua with his face to the floor in the family room, looking pretty silly. "Josh, did you know you lose IQ points for sniffing the carpet?"
"Shut up. I'm not sniffing the carpet. I'm doing something."
"Sure, if sniffing the carpet counts as doing something." At this point Alexis stands over her twin brother grinning, trying to see how silly she can make him feel.
Joshua climbs to his feet and stands on his toes to make himself a half inch taller than his sister, who is ordinarily a half inch taller than Joshua. "I'm measuring something. I'm designing something."
Alexis stands on her toes too, reminding her brother that she is still taller than he. "When you're done, can you design me a dress?"
"Very funny." Joshua walks to the table and points to some drawings. "I'm designing the sand castle I want to build at the beach. Everything needs to be measured out so that I can build something awesome."
"And this requires sniffing carpet?" inquires Alexis, who is just a little intrigued by her brother's project.
"I was imagining where to put the base of a spiral staircase. Everything needs to be measured out correctly. See, the castle walls will be in the shape of a rectangle, like this room. The center of the staircase will be $9$ inches from one of the corners, $15$ inches from another, 16 inches from another, and some whole number of inches from the furthest corner." Joshua shoots Alexis a wry smile. The twins liked to challenge each other, and Alexis knew she had to find the distance from the center of the staircase to the fourth corner of the castle on her own, or face Joshua's pestering, which might last for hours or days.
Find the distance from the center of the staircase to the furthest corner of the rectangular castle, assuming all four of the distances to the corners are described as distances on the same plane (the ground). | 2008 iTest Problems/Problem 24 | Note that because the unknown length is the furthest diagonal, draw the diagram as shown because it maximizes the unknown length. By the Pythagorean Theorem,
\[a^2 + b^2 = 81\]
\[a^2 + d^2 = 225\]
\[b^2 + c^2 = 256\]
Adding the second and third equations results in
\[a^2 + b^2 + c^2 + d^2 = 481\]
By substitution,
\[81 + c^2 + d^2 = 481\]
\[c^2 + d^2 = 400\]
Thus, the distance from the center of the staircase to the furthest corner of the castle is $\boxed{20}$ inches. | // Block 1
draw((0,0)--(50,0)--(50,60)--(0,60)--(0,0));
dot((16,32));
draw((16,32)--(0,60));
label("9",(8,46),SW);
draw((16,32)--(0,0));
label("15",(8,16),NW);
draw((16,32)--(50,60));
label("16",(33,46),NW);
draw((16,32)--(50,0));
draw((0,32)--(50,32),dotted);
label("$a$",(8,32),S);
label("$c$",(33,32),S);
draw((16,0)--(16,60),dotted);
label("$b$",(16,46),E);
label("$d$",(16,16),E);
// Block 2
draw((0,0)--(50,0)--(50,60)--(0,60)--(0,0)); dot((16,32)); draw((16,32)--(0,60)); label("9",(8,46),SW); draw((16,32)--(0,0)); label("15",(8,16),NW); draw((16,32)--(50,60)); label("16",(33,46),NW); draw((16,32)--(50,0)); draw((0,32)--(50,32),dotted); label("$a$",(8,32),S); label("$c$",(33,32),S); draw((16,0)--(16,60),dotted); label("$b$",(16,46),E); label("$d$",(16,16),E); | [] |
453 | Wendy takes Honors Biology at school, a smallish class with only fourteen students (including Wendy) who sit around a circular table. Wendy’s friends Lucy, Starling, and Erin are also in that class. Last Monday none of the fourteen students were absent from class. Before the teacher arrived, Lucy and Starling stretched out a blue piece of yarn between them. Then Wendy and Erin stretched out a red piece of yarn between them at about the same height so that the yarns would intersect if possible. If all possible positions of the students around the table are equally likely, let $m/n$ be the probability that the yarns intersect, where $m$ and $n$ are relatively prime positive integers. Compute $m + n$. | 2008 iTest Problems/Problem 49 | First, let Wendy pick a seat, and let that point be point $W$. Also, pick three arbitrary points clockwise from Wendy. The four points form a cyclic quadrilateral, and the only way for two lines to intersect strictly inside the circle is if the two lines are diagonals. Given one point, only one of the three points make a diagonal of the cyclic quadrilateral, so $\tfrac{m}{n} = \tfrac{1}{3}.$ Thus, $m+n = \boxed{4}.$ | // Block 1
draw(circle((0,0),65));
dot((39,52));
dot((-52,39));
dot((25,-60));
dot((-60,-25));
label("W",(-52,39),NW);
// Block 2
draw(circle((0,0),65)); dot((39,52)); dot((-52,39)); dot((25,-60)); dot((-60,-25)); label("W",(-52,39),NW); | [] |
454 | Michael draws $\triangle ABC$ in the sand such that $\angle ACB=90^\circ$ and $\angle CBA=15^\circ$. He then picks a point at random from within the triangle and labels it point $M$. Next, he draws a segment from $A$ to $BC$ that passes through $M$, hitting $BC$ at a point he labels $D$. Just then, a wave washes over his work, so Michael redraws the exact same diagram farther from the water, labeling all the points the same way as before. If hypotenuse $AB$ is $4$ feet in length, let $p$ be the probability that the number of feet in the length of $AD$ is less than $2\sqrt3-2$. Compute $\lfloor1000p\rfloor$. | 2008 iTest Problems/Problem 66 | Let $D'$ be a point on $BC$ such that $AD' = 2\sqrt3-2$ feet. By using the definitions of sine and cosine, $AC = \sqrt6 - \sqrt2$ feet and $BC = \sqrt6 + \sqrt2$ feet. From the Pythagorean Theorem, $D'C = \sqrt6 - \sqrt2$ feet.
If the number of feet in the length of $AD$ is less than $2\sqrt3-2,$ then point $D$ is within the line segment $AD',$ making point $M$ inside $\triangle ACD'.$ Since $[ACD'] = 4-2\sqrt3$ and $[ACB] = 2,$ the probability $p$ is equal to $2-\sqrt3.$
That means $1000p = 2000 - 1000\sqrt{3}.$ We can find $\lfloor 1000p \rfloor$ easily by using a calculator, but if we do not, note that $1732 < 1000\sqrt{3} < 1733,$ making $\lfloor 1000p \rfloor = 2000 - 1733 = \boxed{267}.$ | // Block 1
pair C=(0,0), B=(3.864,0), A=(0,1.035);
draw(C--B--A--C);
dot(C);
label("C",C,SW);
dot(B);
label("B",B,SE);
dot(A);
label("A",A,NW);
dot((1.035,0));
label("$D'$",(1.035,0),S);
draw((1.035,0)--A);
label("4",(1.932,0.518),NE);
draw(anglemark(A,B,C,20));
label("$15^\circ$",(2.6,0),NW);
draw((0,0.25)--(0.25,0.25)--(0.25,0));
// Block 2
pair C=(0,0), B=(3.864,0), A=(0,1.035); draw(C--B--A--C); dot(C); label("C",C,SW); dot(B); label("B",B,SE); dot(A); label("A",A,NW); dot((1.035,0)); label("$D'$",(1.035,0),S); draw((1.035,0)--A); label("4",(1.932,0.518),NE); draw(anglemark(A,B,C,20)); label("$15^\circ$",(2.6,0),NW); draw((0,0.25)--(0.25,0.25)--(0.25,0)); | [] |
455 | As the Kubiks head homeward, away from the beach in the family van, Jerry decides to take a different route away from the beach than the one they took to get there. The route involves lots of twists and turns, prompting Hannah to wonder aloud if Jerry's "shortcut" will save any time at all.
Michael offers up a problem as an analogy to his father's meandering: "Suppose dad drives around, making right-angled turns after $\textit{every}$ mile. What is the farthest he could get us from our starting point after driving us $500$ miles assuming that he makes exactly $300$ right turns?"
"Sounds almost like an energy efficiency problem," notes Hannah only half jokingly. Hannah is always encouraging her children to think along these lines.
Let $d$ be the answer to Michael's problem. Compute $\lfloor d\rfloor$. | 2008 iTest Problems/Problem 73 | First, we'll show that the farthest distance Jerry can travel with $n$ left turns and $n$ right turns is $n\sqrt{2}$.
Lemma 1: Farthest distance with $n$ left turns and $n$ right turns is $n\sqrt{2}$
Note that all of Jared's turns are perpendicular and occur at every mile. Thus, he travels north or south for $n$ miles and east or west for $n$ miles.
That means the maximum distance can not be more than the diagonal distance of a square with side lengths $n$. The diagonal case can be achieved by alternating left turns with right turns. An example can be seen when $n = 2$ in the picture above. $\blacktriangleright$
Assume Jerry travels his first mile north. If Jerry takes a turns after every mile and $300$ of these turns are right turns, then he takes $199$ left turns. From the Lemma, after turning left $199$ times and turning right $199$ times, he is $199\sqrt{2}$ miles from the start.
However, Jerry still has $101$ right turns to use. By only using right turns, Jerry can only be at most $\sqrt{2}$ miles from the original spot. If Jerry uses four right turns, then he ends up in the original spot. That means Jerry can use $100$ right turns to end in the original location, resulting in one left-over right turn.
If Jerry ends up being $199\sqrt{2}$ miles from the start after using a right turn, then his final destination is one unit south of the diagonal's endpoint, so Jerry is $\sqrt{199^2 + 198^2} \approx 280.72$ miles from the start. If Jerry ends up being $199\sqrt{2}$ miles from the start after using a left turn, then his final destination is one unit east of the diagonal's endpoint, so Jerry is $\sqrt{200^2 + 199^2} \approx 282.14$ miles from the start.
Thus, the maximum distance from the start is around $282.14$ miles, so $\lfloor d\rfloor = \boxed{282}$.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page. | // Block 1
draw((0,0)--(0,2)--(2,2)--(2,0)--(0,0),dotted);
draw((0,0)--(0,1)--(1,1)--(1,2)--(2,2));
draw((0,0)--(2,2));
// Block 2
draw((0,0)--(0,2)--(2,2)--(2,0)--(0,0),dotted); draw((0,0)--(0,1)--(1,1)--(1,2)--(2,2)); draw((0,0)--(2,2)); | [] |
456 | Points $C$ and $D$ lie on opposite sides of line $\overline{AB}$. Let $M$ and $N$ be the centroids of $\triangle ABC$ and $\triangle ABD$ respectively. If $AB=841$, $BC=840$, $AC=41$, $AD=609$, and $BD=580$, find the sum of the numerator and denominator of the value of $MN$ when expressed as a fraction in lowest terms. | 2008 iTest Problems/Problem 74 | Notice that $41^2 + 840^2 = 841^2$ and $580^2 + 609^2 = 841^2$, so $\triangle ABC$ and $\triangle ABD$ are right triangles. Since $\angle ACB = \angle ADB = 90^\circ$, quadrilateral $ACBD$ is cyclic.
Let $O$ be the midpoint of $AB$. Since $M$ and $N$ are centroids, $CM : MO = DN : DO = 2 : 1$. Thus, by SAS Similarity, $\triangle COD \sim \triangle MON$, so $MN = \tfrac13 CD$.
By Ptolemy's Theorem, $AC \cdot BD + BC \cdot AD = CD \cdot AB$, so
\begin{align*} 41 \cdot 580 + 840 \cdot 609 &= 841 \cdot CD \\ 535340 &= 841 \cdot CD \\ CD &= \frac{18460}{29} \end{align*}
That means $MN = \tfrac13 \cdot \tfrac{18460}{29} = \tfrac{18460}{87}$, and the sum of the numerator and denominator in lowest terms is $18460 + 87 = \boxed{18547}$. | // Block 1
pair A=(0,0),C=(1681/841,34440/841),B=(841,0),D=(370881/841,-353220/841);
draw(A--C--B--D--A);
dot(A);
label("A",A,W);
dot(C);
label("C",C,NW);
dot(B);
label("B",B,E);
dot(D);
label("D",D,S);
draw(circle((841/2,0),841/2));
dot((841/2,0));
draw(C--(841/2,0)--D);
pair M=(1417924/5046,11480/841),n=(1282/3,-140);
dot(M);
dot(n);
label("M",M,N);
label("N",n,E);
draw(C--D,dotted);
draw(M--n,dotted);
// Block 2
pair A=(0,0),C=(1681/841,34440/841),B=(841,0),D=(370881/841,-353220/841); draw(A--C--B--D--A); dot(A); label("A",A,W); dot(C); label("C",C,NW); dot(B); label("B",B,E); dot(D); label("D",D,S); draw(circle((841/2,0),841/2)); dot((841/2,0)); draw(C--(841/2,0)--D); pair M=(1417924/5046,11480/841),n=(1282/3,-140); dot(M); dot(n); label("M",M,N); label("N",n,E); draw(C--D,dotted); draw(M--n,dotted); | [] |
457 | Two perpendicular planes intersect a sphere in two circles. These circles intersect in two points, $A$ and $B$, such that $AB=42$. If the radii of the two circles are $54$ and $66$, find $R^2$, where $R$ is the radius of the sphere. | 2008 iTest Problems/Problem 89 | Let $O$ be the center of the sphere, $X$ be the center of the circle radius $54$, $Y$ be the center of the circle radius $66$, and $M$ be the midpoint of $AB$. Since $XM \perp AB$ and $YM \perp AB$, by the Pythagorean Theorem, $XM = \sqrt{54^2 - 21^2} = 15\sqrt{11}$ and $YM = \sqrt{66^2 - 21^2} = 3\sqrt{435}$.
Additionally, by symmetry, the plane containing $X, M, Y$ must also contain $O$. Since the two planes are perpendicular, $XM \perp YM$. Because $OX \perp XM$ and $OY \perp YM$, $OXMY$ is a rectangle, so $OX = 3\sqrt{435}$.
Thus, by the Pythagorean Theorem the radius of the circle is $\sqrt{54^2 + 3915} = \sqrt{6831}$, so $R^2 = \boxed{6831}$. | // Block 1
draw(circle((0,0),100));
draw((-100,0)--(10,99.499));
draw((95,-31.225)--(-20,97.980));
dot((0,0));
label("O",(0,0),S);
pair X=(-45,49.750),Y=(37.5,33.376);
draw(X--(0,0)--Y);
dot(X);
label("X",X,NW);
dot(Y);
label("Y",Y,NE);
draw((0,0)--(-100,0),dotted);
// Block 2
draw(circle((0,0),100)); draw((-100,0)--(10,99.499)); draw((95,-31.225)--(-20,97.980)); dot((0,0)); label("O",(0,0),S); pair X=(-45,49.750),Y=(37.5,33.376); draw(X--(0,0)--Y); dot(X); label("X",X,NW); dot(Y); label("Y",Y,NE); draw((0,0)--(-100,0),dotted); | [] |
458 | In regular hexagon $ABCDEF$ with side length $1$, $AD$ intersects $BF$ at $G$, and $BD$ intersects $EC$ at $H$. Compute the length of $GH$. | 2008 Mock ARML 1 Problems/Problem 3 | Let $H'$ be the foot of the perpendicular from $H$ to $\overline{AD}$. Since $\angle CDA$ is an inscribed angle with measure $\frac{120}{2} = 60^{\circ}$, it follows that $\triangle CDH'$ is a $30-60-90 \triangle$, and $DH' = \frac{1}{2}$ and $CH' = BG = \frac{\sqrt{3}}{2}$. Also, $H'G = CB = 1$. Note that $\triangle DH'H \sim \triangle DGB$ by ratio $1/3$. Thus $HH' = BG/3 = \frac{\sqrt{3}}{6}$.
By the Pythagorean Theorem, $HG^2 = H'H^2 + H'H^2 = \left(\frac{\sqrt{3}}{6}\right)^2 + 1 = \frac{13}{12}$. Thus $HG = \boxed{\frac{\sqrt{39}}{6}}$. | // Block 1
pointpen = black; pathpen = black + linewidth(0.62);
pair v(int n){ return dir(n * 60); }
D(MP("A",v(0))--MP("B",v(1),N)--MP("C",v(2),N)--MP("D",v(3),SW)--MP("E",v(4))--MP("F",v(5))--cycle);
D(v(0)--v(3));D(v(1)--v(5));D(v(1)--v(3));D(v(2)--v(4));
D(D(MP("G",IP(v(0)--v(3),v(1)--v(5)),NE))--D(MP("H",IP(v(1)--v(3),v(2)--v(4)),NW)),dashed);
D(MP("H'",IP(v(2)--v(4),v(0)--v(3)),SW));
// Block 2
pointpen = black; pathpen = black + linewidth(0.62); pair v(int n){ return dir(n * 60); } D(MP("A",v(0))--MP("B",v(1),N)--MP("C",v(2),N)--MP("D",v(3),SW)--MP("E",v(4))--MP("F",v(5))--cycle); D(v(0)--v(3));D(v(1)--v(5));D(v(1)--v(3));D(v(2)--v(4)); D(D(MP("G",IP(v(0)--v(3),v(1)--v(5)),NE))--D(MP("H",IP(v(1)--v(3),v(2)--v(4)),NW)),dashed); D(MP("H'",IP(v(2)--v(4),v(0)--v(3)),SW)); | [] |
459 | Square $ABCD$ has side length $2$. $M$ is the midpoint of $CD$, and $N$ is the midpoint of $BC$. $P$ is on $MN$ such that $N$ is between $M$ and $P$, and $m\angle MAN = m\angle NAP$. Compute the length of $AP$. | 2008 Mock ARML 1 Problems/Problem 6 | By the Pythagorean Theorem, $MA = NA = \sqrt{5}$ and $MN = \sqrt{2}$. Let $\theta = \angle MAN = \angle NAP$. By the Law of Cosines on $\triangle MAN$,
\[\left(\sqrt{2}\right)^2 = \left(\sqrt{5}\right)^2 + \left(\sqrt{5}\right)^2 - 2 \cdot \left(\sqrt{5}\right) \cdot \left(\sqrt{5}\right) \cos \theta \Longrightarrow \cos \theta = \frac{4}{5}.\]
The Law of Cosines on $\triangle NAP$ yields
$NP^2 = AP^2 + \left(\sqrt{5}\right)^2 - 2 \cdot AP \cdot \left(\sqrt{5}\right) \cos \theta = AP^2 - \frac{8}{\sqrt{5}}AP + 5.$
The Angle Bisector Theorem on $\triangle MAP$ yields
\[\frac{AP}{NP} = \frac{AM}{MN} = \sqrt{\frac{5}{2}} \Longrightarrow NP = \sqrt{\frac{2}{5}}AP.\]
Substituting,
\begin{align*} 0 &= 3AP^2 - 8\sqrt{5}AP + 25\\ AP &= \frac{8\sqrt{5} \pm \sqrt{20}}{6} = \boxed{\frac{5\sqrt{5}}{3}}, \sqrt{5}. \end{align*} | // Block 1
pointpen=black;pathpen=black+linewidth(0.7);
pair A=(0,2), D=(0,0), C=(2,0), B=(2,2), M=(C+D)/2, N=(B+C)/2, P=8/3*(N-M)+M;
D(MP("A",A,(0,1))--MP("B",B,(0,1))--MP("C",C)--MP("D",D)--cycle); D(D(MP("M",M))--D(MP("N",N,E))--D(MP("P",P,E)));
D(M--A--N--A--P); D(anglemark(M,A,P)); MP("2",(A+D)/2,W);
// Block 2
pointpen=black;pathpen=black+linewidth(0.7); pair A=(0,2), D=(0,0), C=(2,0), B=(2,2), M=(C+D)/2, N=(B+C)/2, P=8/3*(N-M)+M; D(MP("A",A,(0,1))--MP("B",B,(0,1))--MP("C",C)--MP("D",D)--cycle); D(D(MP("M",M))--D(MP("N",N,E))--D(MP("P",P,E))); D(M--A--N--A--P); D(anglemark(M,A,P)); MP("2",(A+D)/2,W); | [] |
460 | $ABCD$ is a convex quadrilateral such that $|AB| = 5$, $|BC| = 17$, $|CD| = 7$, and $|DA| = 25$. Given that $m\angle{ABC} + m\angle{BCD} = 270^{\circ}$, find the area of $ABCD$. | 2008 Mock ARML 2 Problems/Problem 1 | Note that $m\angle{BAD} + m\angle{ADC} = 360 - 270 = 90^{\circ}$. Thus, if we let $E$ be the intersection of the extensions of $\overline{AB}$ and $\overline{CD}$, it follows that $\triangle EAD$ is a right triangle. Immediately we notice that $\triangle ECB$ is a $8-15-17,\,\triangle$ and that $\triangle EAD$ is a $15-20-25\, \triangle$; otherwise we can determine these lengths through the Pythagorean Theorem.
The answer is $[ABCD] = [EAD] - [ECB] = \frac{1}{2}(15)(20) - \frac{1}{2}(8)(15) = \boxed{90}$. | // Block 1
pointpen = black; pathpen = black + linewidth(0.7);
pair A=(0,0), D=(25,0), F=(16,12), B=(3*A + F)/4, C=(8*D+7*F)/15;
D(MP("A",A,SW)--MP("B",B,NW)--MP("C",C,NE)--MP("D",D,SE)--cycle); D(B--MP("E",F,N)--C,linetype("4 4")); D(rightanglemark(A,F,D,30),linetype("4 4"));
// Block 2
pointpen = black; pathpen = black + linewidth(0.7); pair A=(0,0), D=(25,0), F=(16,12), B=(3*A + F)/4, C=(8*D+7*F)/15; D(MP("A",A,SW)--MP("B",B,NW)--MP("C",C,NE)--MP("D",D,SE)--cycle); D(B--MP("E",F,N)--C,linetype("4 4")); D(rightanglemark(A,F,D,30),linetype("4 4")); | [] |
461 | Equilateral triangle $ABC$ has a side length of $7$. A ball begins at vertex $A$, rolls through the interior of the triangle, bounces off side $BC$, and settles at point $P$. Given that $BP = 3$ and $CP = 5$, find the total distance that the ball travels. | 2008 Mock ARML 2 Problems/Problem 4 | Reflect $P$ across $\overline{BC}$ to point $P'$; since the ball travels in a straight path, it follows that the distance the ball traveled is $AP'$. By symmetry, $BP' = 3, CP' = 5$. By the Law of Cosines on $\triangle BCP'$,
$7^2 = 3^2 + 5^2 - 2 \cdot 3 \cdot 5 \cos \theta \Longrightarrow \theta = 120^{\circ}.$
Since $\angle BAC + \angle BP'C = 60 + 120 = 180$, it follows that quadrilateral $ABP'C$ is a cyclic quadrilateral. By Ptolemy's Theorem,
$3\cdot 7 + 5\cdot 7 = 7 \cdot AP' \Longrightarrow \boxed{AP' = 8}.$ | // Block 1
pointpen = black; pathpen = black + linewidth(0.7);
pair A=(0,0),B=(7,0),C=7*expi(pi/3);
pair Q=IP(CR(B,3),CR(C,5)), P=OP(CR(B,3),CR(C,5)),D=IP(A--Q,B--C);
D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle);
D(A--D--D(MP("P",P)));D(D--D(MP("P'",Q,NE))--C--Q--B,linetype("5 5"));
MP("7",(A+B)/2); MP("3",(B+Q)/2,SE); MP("5",(C+Q)/2,NE);
// Block 2
pointpen = black; pathpen = black + linewidth(0.7); pair A=(0,0),B=(7,0),C=7*expi(pi/3); pair Q=IP(CR(B,3),CR(C,5)), P=OP(CR(B,3),CR(C,5)),D=IP(A--Q,B--C); D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle); D(A--D--D(MP("P",P)));D(D--D(MP("P'",Q,NE))--C--Q--B,linetype("5 5")); MP("7",(A+B)/2); MP("3",(B+Q)/2,SE); MP("5",(C+Q)/2,NE); | [] |
462 | In right $\triangle ABC$ with hypotenuse $\overline{AB}$, $AC = 12$, $BC = 35$, and $\overline{CD}$ is the altitude to $\overline{AB}$. Let $\omega$ be the circle having $\overline{CD}$ as a diameter. Let $I$ be a point outside $\triangle ABC$ such that $\overline{AI}$ and $\overline{BI}$ are both tangent to circle $\omega$. The ratio of the perimeter of $\triangle ABI$ to the length $AB$ can be expressed in the form $\frac {m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$. | 2009 AIME I Problems/Problem 12 | Let $AP=AD=x$, let $BQ=BD=y$, and let $IP=IQ=z$. Let $OD=r$. We find $AB=37$. Let $\alpha$, $\beta$, and $\theta$ be the angles $OAD$, $OBD$, and $OPI$ respectively. Then $\alpha + \beta + \theta = 90^\circ$, so \[\theta = 90^\circ - (\alpha+\beta).\]
The perimeter of $\triangle ABI$ is $2(x+y+z)=2(37+z)$. The desired ratio is then
\[\rho = 2\left(1+\frac z{37}\right)\]
We need to find $z$. In $\triangle OPI$, $z=r\cot\theta = r\tan (\alpha+\beta)$. We get \[\tan\alpha = \frac{OD}{AD} = \frac 12 \frac{CD}{AD} = \frac 12 \tan A = \frac 12 \frac{BC}{AC} = \frac{35}{24}.\] Similarly, $\tan\beta = \tfrac 6{35}$. Then \[z = r\cdot \tan (\alpha+\beta) = r\cdot \frac{\tan\alpha + \tan\beta}{1-\tan\alpha\tan\beta}= \frac{37^2\cdot r}{18\cdot 35}\]
Computing $[ABC]$ in two ways we get $CD = \tfrac{12\cdot 35}{37}$, so $r=\tfrac{6\cdot 35}{37}$. Using this value of $r$ we get $z=\tfrac {37}3$. Thus \[\rho = 2\left(1+\frac 1{3}\right) = \frac 8{3},\]
and $8+3=\boxed{011}$. | // Block 1
size(300);
defaultpen(linewidth(0.4)+fontsize(10));
pen s = linewidth(0.8)+fontsize(8);
pair A,B,C,D,O,X;
C=origin;
A=(0,12);
B=(18,0);
D=foot(C,A,B);
O = (C+D)/2;
real r = length(D-C)/2;
path c = CR(O, r);
pair OA = (O+A)/2;
real rA = length(A-O)/2;
pair Ap = OP(CR(OA,rA), c);
pair OB = (O+B)/2;
real rB = length(B-O)/2;
pair Bp = OP(CR(OB,rB), c);
X=extension(A,Ap,B,Bp);
draw(A--B--C--A, s);
draw(C--D^^B--O--A^^Ap--O--X, gray+0.25);
draw(c^^A--X--B);
dot("$A$", A, N);
dot("$B$", B, SE);
dot("$C$", C, SW);
dot("$D$", D, 0.2*(D-C));
dot("$I$", X, 0.5*(X-C));
dot("$P$", Ap, 0.3*(Ap-O));
dot("$Q$", Bp, 0.3*(Bp-O));
dot("$O$", O, W);
label("$\beta$",B,10*dir(157));
label("$\alpha$",A,5*dir(-55));
label("$\theta$",X,5*dir(55));
// Block 2
size(300); defaultpen(linewidth(0.4)+fontsize(10)); pen s = linewidth(0.8)+fontsize(8); pair A,B,C,D,O,X; C=origin; A=(0,12); B=(18,0); D=foot(C,A,B); O = (C+D)/2; real r = length(D-C)/2; path c = CR(O, r); pair OA = (O+A)/2; real rA = length(A-O)/2; pair Ap = OP(CR(OA,rA), c); pair OB = (O+B)/2; real rB = length(B-O)/2; pair Bp = OP(CR(OB,rB), c); X=extension(A,Ap,B,Bp); draw(A--B--C--A, s); draw(C--D^^B--O--A^^Ap--O--X, gray+0.25); draw(c^^A--X--B); dot("$A$", A, N); dot("$B$", B, SE); dot("$C$", C, SW); dot("$D$", D, 0.2*(D-C)); dot("$I$", X, 0.5*(X-C)); dot("$P$", Ap, 0.3*(Ap-O)); dot("$Q$", Bp, 0.3*(Bp-O)); dot("$O$", O, W); label("$\beta$",B,10*dir(157)); label("$\alpha$",A,5*dir(-55)); label("$\theta$",X,5*dir(55)); | [] |
463 | In parallelogram $ABCD$, point $M$ is on $\overline{AB}$ so that $\frac {AM}{AB} = \frac {17}{1000}$ and point $N$ is on $\overline{AD}$ so that $\frac {AN}{AD} = \frac {17}{2009}$. Let $P$ be the point of intersection of $\overline{AC}$ and $\overline{MN}$. Find $\frac {AC}{AP}$. | 2009 AIME I Problems/Problem 4 | We approach the problem using mass points on triangle $ABD$ as displayed below.
But as $MN$ does not protrude from a vertex, we will have to "split the mass" at point $A$. First, we know that $DO$ is congruent to $BO$ because diagonals of parallelograms bisect each other. Therefore, we can assign equal masses to points $B$ and $D$. In this case, we assign $B$ and $D$ a mass of 17 each. Now we split the mass at $A$, so we balance segments $AB$ and $AD$ separately, and then the mass of $A$ is the sum of those masses. A mass of 983 is required to balance segment $AB$, while a mass of 1992 is required to balance segment $AD$. Therefore, $A$ has a mass of $1992+983=2975$. Also, $O$ has a mass of 34. Therefore, $\frac{AO}{AP}=\frac{2975+34}{34}=\frac{3009}{34}$, so $\frac{AC}{AP}=\frac{2 (3009)}{34}=177$. | // Block 1
pair A=(0,0),B=(50,0),D=(10,40),C=B+D,M=(8,0),NN=(2,8);
draw(A--B--C--D--cycle);
draw(B--D^^A--C^^M--NN);
pair O=extension(A,C,B,D);
pair P=extension(A,C,M,NN);
dot(A);dot(B);dot(C);dot(D);dot(O);dot(M);dot(NN);dot(P);
label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,NE);
label("$D$",D,NW);
label("$M$",M,S);
label("$N$",NN,NW);
label("$P$",P,NNE);
label("$O$",O,N);
// Block 2
pair A=(0,0),B=(50,0),D=(10,40),C=B+D,M=(8,0),NN=(2,8); draw(A--B--C--D--cycle); draw(B--D^^A--C^^M--NN); pair O=extension(A,C,B,D); pair P=extension(A,C,M,NN); dot(A);dot(B);dot(C);dot(D);dot(O);dot(M);dot(NN);dot(P); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,NW); label("$M$",M,S); label("$N$",NN,NW); label("$P$",P,NNE); label("$O$",O,N); | [] |
464 | Triangle $ABC$ has $AC = 450$ and $BC = 300$. Points $K$ and $L$ are located on $\overline{AC}$ and $\overline{AB}$ respectively so that $AK = CK$, and $\overline{CL}$ is the angle bisector of angle $C$. Let $P$ be the point of intersection of $\overline{BK}$ and $\overline{CL}$, and let $M$ be the point on line $BK$ for which $K$ is the midpoint of $\overline{PM}$. If $AM = 180$, find $LP$. | 2009 AIME I Problems/Problem 5 | Since $K$ is the midpoint of $\overline{PM}$ and $\overline{AC}$, quadrilateral $AMCP$ is a parallelogram, which implies $AM||LP$ and $\bigtriangleup{AMB}$ is similar to $\bigtriangleup{LPB}$
Thus,
\[\frac {AM}{LP}=\frac {AB}{LB}=\frac {AL+LB}{LB}=\frac {AL}{LB}+1\]
Now let's apply the angle bisector theorem.
\[\frac {AL}{LB}=\frac {AC}{BC}=\frac {450}{300}=\frac {3}{2}\]
\[\frac {AM}{LP}=\frac {AL}{LB}+1=\frac {5}{2}\]
\[\frac {180}{LP}=\frac {5}{2}\]
\[LP=\boxed {072}\]
Note: Another way to realize that $\bigtriangleup{AMB}$ is similar to $\bigtriangleup{LPB}$ is through angle chasing and congruent triangles. Angle $AKM$ is congruent to angle $CKP$. Segments $MK$ and $PK$ are congruent as stated in the problem. Segments $AK$ and $CK$ are congruent as stated in the problem. Thus, by Side-Angle-Side congruency $\bigtriangleup{AMK}$ is similar to $\bigtriangleup{CPK}$. Then angle $AMK = AMB$ is congruent to angle $CPK = BPL$. Notice triangles $\bigtriangleup{AMB}$ and $\bigtriangleup{LPB}$ both share angle $MBA$. Thus the two triangles are similar because of AA similarity with angles [$AMB$ and $LPB$] and [$MBA = PBL$]. *Solution in Beauty of Math video written out | // Block 1
import markers;
defaultpen(fontsize(8));
size(300);
pair A=(0,0), B=(30*sqrt(331),0), C, K, L, M, P;
C = intersectionpoints(Circle(A,450), Circle(B,300))[0];
K = midpoint(A--C);
L = (3*B+2*A)/5;
P = extension(B,K,C,L);
M = 2*K-P;
draw(A--B--C--cycle);
draw(C--L);draw(B--M--A);
markangle(n=1,radius=15,A,C,L,marker(markinterval(stickframe(n=1),true)));
markangle(n=1,radius=15,L,C,B,marker(markinterval(stickframe(n=1),true)));
dot(A^^B^^C^^K^^L^^M^^P);
label("$A$",A,(-1,-1));label("$B$",B,(1,-1));label("$C$",C,(1,1));
label("$K$",K,(0,2));label("$L$",L,(0,-2));label("$M$",M,(-1,1));
label("$P$",P,(1,1));
label("$180$",(A+M)/2,(-1,0));label("$180$",(P+C)/2,(-1,0));label("$225$",(A+K)/2,(0,2));label("$225$",(K+C)/2,(0,2));
label("$300$",(B+C)/2,(1,1));
// Block 2
import markers; defaultpen(fontsize(8)); size(300); pair A=(0,0), B=(30*sqrt(331),0), C, K, L, M, P; C = intersectionpoints(Circle(A,450), Circle(B,300))[0]; K = midpoint(A--C); L = (3*B+2*A)/5; P = extension(B,K,C,L); M = 2*K-P; draw(A--B--C--cycle); draw(C--L);draw(B--M--A); markangle(n=1,radius=15,A,C,L,marker(markinterval(stickframe(n=1),true))); markangle(n=1,radius=15,L,C,B,marker(markinterval(stickframe(n=1),true))); dot(A^^B^^C^^K^^L^^M^^P); label("$A$",A,(-1,-1));label("$B$",B,(1,-1));label("$C$",C,(1,1)); label("$K$",K,(0,2));label("$L$",L,(0,-2));label("$M$",M,(-1,1)); label("$P$",P,(1,1)); label("$180$",(A+M)/2,(-1,0));label("$180$",(P+C)/2,(-1,0));label("$225$",(A+K)/2,(0,2));label("$225$",(K+C)/2,(0,2)); label("$300$",(B+C)/2,(1,1)); | [] |
465 | Let $\overline{MN}$ be a diameter of a circle with diameter 1. Let $A$ and $B$ be points on one of the semicircular arcs determined by $\overline{MN}$ such that $A$ is the midpoint of the semicircle and $MB=\frac{3}5$. Point $C$ lies on the other semicircular arc. Let $d$ be the length of the line segment whose endpoints are the intersections of diameter $\overline{MN}$ with chords $\overline{AC}$ and $\overline{BC}$. The largest possible value of $d$ can be written in the form $r-s\sqrt{t}$, where $r, s$ and $t$ are positive integers and $t$ is not divisible by the square of any prime. Find $r+s+t$. | 2009 AIME II Problems/Problem 15 | Solution 1 (Quick Calculus)
Let $V = \overline{NM} \cap \overline{AC}$ and $W = \overline{NM} \cap \overline{BC}$. Further more let $\angle NMC = \alpha$ and $\angle MNC = 90^\circ - \alpha$. Angle chasing reveals $\angle NBC = \angle NAC = \alpha$ and $\angle MBC = \angle MAC = 90^\circ - \alpha$. Additionally $NB = \frac{4}{5}$ and $AN = AM$ by the Pythagorean Theorem.
By the Angle Bisector Formula,
\[\frac{NV}{MV} = \frac{\sin (\alpha)}{\sin (90^\circ - \alpha)} = \tan (\alpha)\]
\[\frac{MW}{NW} = \frac{3\sin (90^\circ - \alpha)}{4\sin (\alpha)} = \frac{3}{4} \cot (\alpha)\]
As $NV + MV =MW + NW = 1$ we compute $NW = \frac{1}{1+\frac{3}{4}\cot(\alpha)}$ and $MV = \frac{1}{1+\tan (\alpha)}$, and finally $VW = NW + MV - 1 = \frac{1}{1+\frac{3}{4}\cot(\alpha)} + \frac{1}{1+\tan (\alpha)} - 1$. Taking the derivative of $VW$ with respect to $\alpha$, we arrive at
\[VW' = \frac{7\cos^2 (\alpha) - 4}{(\sin(\alpha) + \cos(\alpha))^2(4\sin(\alpha)+3\cos(\alpha))^2}\]
Clearly the maximum occurs when $\alpha = \cos^{-1}\left(\frac{2}{\sqrt{7}}\right)$. Plugging this back in, using the fact that $\tan(\cos^{-1}(x)) = \frac{\sqrt{1-x^2}}{x}$ and $\cot(\cos^{-1}(x)) = \frac{x}{\sqrt{1-x^2}}$, we get
$VW = 7 - 4\sqrt{3}$
with $7 + 4 + 3 = \boxed{014}$
~always_correct
Solution 2 (Projective)
Since $MA = \frac{\sqrt{2}}{2} \approx 0.707 > \frac{3}{5}$, point $B$ lies between $M$ and $A$ on the semicircular arc. We will first compute the length of $\overline{AB}$. By the law of cosines, $\cos \angle MOB = \frac{-(3/5)^2 + 2(1/2)^2}{2(1/2)^2} = \frac{7}{25}$, so $\cos \angle AOB = \sin \angle MOB = \frac{24}{25}$. Then $AB^2 = 2\left(\frac{1}{2}\right)^2 - 2\left(\frac{1}{2}\right)^2 \cdot \frac{24}{25} = \frac{1}{50}$, so $AB = \frac{1}{5\sqrt{2}}$.
Let $P = AC \cap MN$ and $Q = BC \cap MN$, and let $MQ = x$, $PQ = d$, $PN = y$. Note that\[(M, P; Q, N) \stackrel{C}{=} (M, A; B, N),\]that is,\[\frac{QP}{QM} \div \frac{NP}{NM} = \frac{BA}{BM} \div \frac{NA}{NM}\]or\[\frac{d}{xy} = \frac{1/(5\sqrt{2})}{(3/5) \cdot (\sqrt{2}/2)} = \frac{1}{3}.\]Hence $d = \frac{1}{3}xy$, and we also know $d+x+y=1$. Now AM-GM gives\[\frac{x+y}{2} \ge \sqrt{xy} \implies \frac{1-d}{2} \ge \sqrt{3d}.\]This gives the quadratic inequality $d^2 - 14d + 1 \ge 0$, which solves as\[d \in \left(-\infty, 7-4\sqrt3\right] \cup \left[7+4\sqrt3, \infty\right).\]But $d \le 1$, so the greatest possible value of $d$ is $7-4\sqrt3$. The answer is $7+4+3=\boxed{014}$.
~MSTang
Solution 3 (Calculus)
Let $O$ be the center of the circle. Define $\angle{MOC}=t$, $\angle{BOA}=2a$, and let $BC$ and $AC$ intersect $MN$ at points $X$ and $Y$, respectively. We will express the length of $XY$ as a function of $t$ and maximize that function in the interval $[0, \pi]$.
Let $C'$ be the foot of the perpendicular from $C$ to $MN$. We compute $XY$ as follows.
(a) By the Extended Law of Sines in triangle $ABC$, we have
\[CA\]
\[= \sin\angle{ABC}\]
\[= \sin\left(\frac{\widehat{AN} + \widehat{NC}}{2}\right)\]
\[= \sin\left(\frac{\frac{\pi}{2} + (\pi-t)}{2}\right)\]
\[= \sin\left(\frac{3\pi}{4} - \frac{t}{2}\right)\]
\[= \sin\left(\frac{\pi}{4} + \frac{t}{2}\right)\]
(b) Note that $CC' = CO\sin(t) = \left(\frac{1}{2}\right)\sin(t)$ and $AO = \frac{1}{2}$. Since $CC'Y$ and $AOY$ are similar right triangles, we have $CY/AY = CC'/AO = \sin(t)$, and hence,
\[CY/CA\]
\[= \frac{CY}{CY + AY}\]
\[= \frac{\sin(t)}{1 + \sin(t)}\]
\[= \frac{\sin(t)}{\sin\left(\frac{\pi}{2}\right) + \sin(t)}\]
\[= \frac{\sin(t)}{2\sin\left(\frac{\pi}{4} + \frac{t}{2}\right)\cos\left(\frac{\pi}{4} - \frac{t}{2}\right)}\]
(c) We have $\angle{XCY} = \frac{\widehat{AB}}{2}=a$ and $\angle{CXY} = \frac{\widehat{MB}+\widehat{CN}}{2} = \frac{\left(\frac{\pi}{2} - 2a\right) + (\pi - t)}{2} = \frac{3\pi}{4} - a - \frac{t}{2}$, and hence by the Law of Sines,
\[XY/CY\]
\[= \frac{\sin\angle{XCY}}{\sin\angle{CXY}}\]
\[= \frac{\sin(a)}{\sin\left(\frac{3\pi}{4} - a - \frac{t}{2}\right)}\]
\[= \frac{\sin(a)}{\sin\left(\frac{\pi}{4} + a + \frac{t}{2}\right)}\]
(d) Multiplying (a), (b), and (c), we have
\[XY\]
\[= CA * (CY/CA) * (XY/CY)\]
\[= \frac{\sin(t)\sin(a)}{2\cos\left(\frac{\pi}{4} - \frac{t}{2}\right)\sin\left(\frac{\pi}{4} + a + \frac{t}{2}\right)}\]
\[= \frac{\sin(t)\sin(a)}{\sin\left(\frac{\pi}{2} + a\right) + \sin(a + t)}\]
\[= \sin(a)\times\frac{\sin(t)}{\sin(t + a) + \cos(a)}\],
which is a function of $t$ (and the constant $a$). Differentiating this with respect to $t$ yields
\[\sin(a)\times\frac{\cos(t)(\sin(t + a) + \cos(a)) - \sin(t)\cos(t + a)}{(\sin(t + a) + \cos(a))^2}\],
and the numerator of this is
\[\sin(a) \times(\sin(t + a)\cos(t) - \cos(t + a)\sin(t) + \cos(a)\cos(t))\]
\[= \sin(a) \times (\sin(a) + \cos(a)\cos(t))\],
which vanishes when $\sin(a) + \cos(a)\cos(t) = 0$. Therefore, the length of $XY$ is maximized when $t=t'$, where $t'$ is the value in $[0, \pi]$ that satisfies $\cos(t') = -\tan(a)$.
Note that
\[\frac{1 - \tan(a)}{1 + \tan(a)} = \tan\left(\frac{\pi}{4} - a\right) = \tan((\widehat{MB})/2) = \tan\angle{MNB} = \frac{3}{4}\],
so $\tan(a) = \frac{1}{7}$. We compute
\[\sin(a) = \frac{\sqrt{2}}{10}\]
\[\cos(a) = \frac{7\sqrt{2}}{10}\]
\[\cos(t') = -\tan(a) = -\frac{1}{7}\]
\[\sin(t') = \frac{4\sqrt{3}}{7}\]
\[\sin(t' + a)=\sin(t')\cos(a) + \cos(t')\sin(a) = \frac{28\sqrt{6} - \sqrt{2}}{70}\],
so the maximum length of $XY$ is $\sin(a)\times\frac{\sin(t')}{\sin(t' + a) + \cos(a)} = 7 - 4\sqrt{3}$, and the answer is $7 + 4 + 3 = \boxed{014}$.
Solution 4
Suppose $\overline{AC}$ and $\overline{BC}$ intersect $\overline{MN}$ at $D$ and $E$, respectively, and let $MC = x$ and $NC = y$. Since $A$ is the midpoint of arc $MN$, $\overline{CA}$ bisects $\angle MCN$, and we get
\[\frac{MC}{MD} = \frac{NC}{ND}\Rightarrow MD = \frac{x}{x + y}.\]
To find $ME$, we note that $\triangle BNE\sim\triangle MCE$ and $\triangle BME\sim\triangle NCE$, so
\begin{align*} \frac{BN}{NE} &= \frac{MC}{CE} \\ \frac{ME}{BM} &= \frac{CE}{NC}. \end{align*}
Writing $NE = 1 - ME$, we can substitute known values and multiply the equations to get
\[\frac{4(ME)}{3 - 3(ME)} = \frac{x}{y}\Rightarrow ME = \frac{3x}{3x + 4y}.\]
The value we wish to maximize is
\begin{align*} DE &= MD - ME \\ &= \frac{x}{x + y} - \frac{3x}{3x + 4y} \\ &= \frac{xy}{3x^2 + 7xy + 4y^2} \\ &= \frac{1}{3(x/y) + 4(y/x) + 7}. \end{align*}
By the AM-GM inequality, $3(x/y) + 4(y/x)\geq 2\sqrt{12} = 4\sqrt{3}$, so
\[DE\leq \frac{1}{4\sqrt{3} + 7} = 7 - 4\sqrt{3},\]
giving the answer of $7 + 4 + 3 = \boxed{014}$. Equality is achieved when $3(x/y) = 4(y/x)$ subject to the condition $x^2 + y^2 = 1$, which occurs for $x = \frac{2\sqrt{7}}{7}$ and $y = \frac{\sqrt{21}}{7}$.
Solution 5 (Projective)
By Pythagoras in $\triangle BMN,$ we get $BN=\dfrac{4}{5}.$
Since cross ratios are preserved upon projecting, note that $(M,Y;X,N)\stackrel{C}{=}(M,B;A,N).$ By definition of a cross ratio, this becomes \[\dfrac{XM}{NY}:\dfrac{NM}{NY}=\dfrac{AM}{AB}:\dfrac{MN}{NB}.\] Let $MY=a,YX=b,XN=c$ such that $a+b+c=1.$ We know that $XM=a+b,XY=b,NM=1,NY=b+c,$ so the LHS becomes $\dfrac{(a+b)(b+c)}{b}.$
In the RHS, we are given every value except for $AB.$ However, Ptolemy's Theorem on $MBAN$ gives $AB\cdot MN+AN\cdot BM=AM\cdot BN\implies AB+\dfrac{3}{5\sqrt{2}}=\dfrac{4}{5\sqrt{2}}\implies AB=\dfrac{1}{5\sqrt{2}}.$ Substituting, we get $\dfrac{(a+b)(b+c)}{b}=4\implies b(a+b+c)+ac=4b, b=\dfrac{ac}{3}$ where we use $a+b+c=1.$
Again using $a+b+c=1,$ we have $a+b+c=1\implies a+\dfrac{ac}{3}+c=1\implies a=3\dfrac{1-c}{c+3}.$ Then $b=\dfrac{ac}{3}=\dfrac{c-c^2}{c+3}.$ Since this is a function in $c,$ we differentiate WRT $c$ to find its maximum. By quotient rule, it suffices to solve \[(-2c+1)(c+3)-(c-c^2)=0 \implies c^2+6c-3,c=-3+2\sqrt{3}.\] Substituting back yields $b=7-4\sqrt{3},$ so $7+4+3=\boxed{014}$ is the answer.
~Generic_Username | unitsize(144); pair A, B, C, M, n; A = (0,1); B = (-7/25, 24/25); C=(1/7,-4*sqrt(3)/7); M = (-1,0); n = (1,0); pair [] D = intersectionpoints(A--C,M--n); pair [] e = intersectionpoints(B--C,M--n); draw(circle((0,0),1)); draw(M--n--B--M--A--n--C--A--B--C--cycle); label("$A$",A,N); label("$B$",B,NNW); label("$M$",M,W); label("$C$",C,SSE); label("$N$",n,E); label("$D$",D[0],SE); label("$E$",e[0],SW); label("$x$",(M+C)/2,SW); label("$y$",(n+C)/2,SE); | [] |
466 | In rectangle $ABCD$, $AB=100$. Let $E$ be the midpoint of $\overline{AD}$. Given that line $AC$ and line $BE$ are perpendicular, find the greatest integer less than $AD$. | 2009 AIME II Problems/Problem 3 | Solution 1
From the problem, $AB=100$ and triangle $FBA$ is a right triangle. As $ABCD$ is a rectangle, triangles $BCA$, and $ABE$ are also right triangles. By $AA$, $\triangle FBA \sim \triangle BCA$, and $\triangle FBA \sim \triangle ABE$, so $\triangle ABE \sim \triangle BCA$. This gives $\frac {AE}{AB}= \frac {AB}{BC}$. $AE=\frac{AD}{2}$ and $BC=AD$, so $\frac {AD}{2AB}= \frac {AB}{AD}$, or $(AD)^2=2(AB)^2$, so $AD=AB \sqrt{2}$, or $100 \sqrt{2}$, so the answer is $\boxed{141}$.
Solution 2
Let $x$ be the ratio of $BC$ to $AB$. On the coordinate plane, plot $A=(0,0)$, $B=(100,0)$, $C=(100,100x)$, and $D=(0,100x)$. Then $E=(0,50x)$. Furthermore, the slope of $\overline{AC}$ is $x$ and the slope of $\overline{BE}$ is $-x/2$. They are perpendicular, so they multiply to $-1$, that is,
\[x\cdot-\frac{x}{2}=-1,\]
which implies that $-x^2=-2$ or $x=\sqrt 2$. Therefore $AD=100\sqrt 2\approx 141.42$ so $\lfloor AD\rfloor=\boxed{141}$.
Solution 3
Similarly to Solution 2, let the positive x-axis be in the direction of ray $BC$ and let the positive y-axis be in the direction of ray $BA$. Thus, the vector $BE=(x,100)$ and the vector $AC=(2x,-100)$ are perpendicular and thus have a dot product of 0. Thus, calculating the dot product:
\[x\cdot2x+(100)\cdot(-100)=2x^2-10000=0\]
\[2x^2-10000=0\rightarrow x^2=5000\]
Substituting AD/2 for x:
\[(AD/2)^2=5000\rightarrow AD^2=20000\]
\[AD=100\sqrt2\]
Solution 4
Draw $CX$ and $EX$ to form a parallelogram $AEXC$. Since $EX \parallel AC$, $\angle BEX=90^\circ$ by the problem statement, so $\triangle BEX$ is right.
Letting $AE=y$, we have $BE=\sqrt{100^2+y^2}$ and $AC=EX=\sqrt{100^2+(2y)^2}$. Since $CX=EA$, $\sqrt{100^2+y^2}^2+\sqrt{100^2+(2y)^2}=(3y)^2$. Solving this, we have
\[100^2+ 100^2 + y^2 + 4y^2 = 9y^2\]
\[2\cdot 100^2 = 4y^2\]
\[\frac{100^2}{2}=y^2\]
\[\frac{100}{\sqrt{2}}=y\]
\[\frac{100\sqrt{2}}{2}=y\]
\[100\sqrt{2}=2y=AD\], so the answer is $\boxed{141}$. | // Block 1
pair A=(0,10), B=(0,0), C=(14,0), D=(14,10), Q=(0,5); draw (A--B--C--D--cycle); pair E=(7,10); draw (B--E); draw (A--C); pair F=(6.7,6.7); label("\(E\)",E,N); label("\(A\)",A,NW); label("\(B\)",B,SW); label("\(C\)",C,SE); label("\(D\)",D,NE); label("\(F\)",F,W); label("\(100\)",Q,W);
// Block 2
pair A=(0,10), B=(0,0), C=(14,0), D=(14,10), Q=(0,5),X=(21,0); draw (A--B--C--D--cycle); pair E=(7,10); draw (B--E); draw (A--C); draw(C--X--E); label("\(E\)",E,N); label("\(A\)",A,NW); label("\(B\)",B,SW); label("\(C\)",C,SE); label("\(D\)",D,NE); label("\(X\)",X,S); label("\(100\)",Q,W); | [] |
467 | A square is divided into three pieces of
equal area by two parallel lines as shown.
If the distance between the two parallel
lines is $8$ what is the area of the square? | 2009 UNCO Math Contest II Problems/Problem 9 | In $\triangle$ADC, $\frac {AE}{ED}$ = $\frac {1}{2}$ because $\frac {[ACE]}{[CED]}$ = $\frac {1}{2}$ (According to question)
$[ECD]$ = $2[EFC]$ (AECF is a parallelogram and $2[EFC]$ = $[AECF]$)
$2[EFC]$ = $[EDC]$ (same reason as before)
$[EFC]$ = $\frac {1}{2}$ * 8 * EC
$[EDC]$ = $\frac{1}{2}$ * EC * (Perpendicular from D to EC)
2 * $\frac {1}{2}$ * 8 * EC = $\frac{1}{2}$ * EC * (Perpendicular from D to EC)
(Perpendicular from D to EC) = 16
Let AE = $x$, then ED = $2x$. So the side length of square is $3x$.
(Perpendicular from D to EC) = $\frac {ED * DC}{EC}$
EC = $\sqrt{13}x$ (By Pythagoras Theorem)
16 = $\frac {2x * 3x}{\sqrt{13}x}$
16 = $\frac {6x^{2}}{\sqrt{13}x}$
$x$ = $\frac {8\sqrt{13}}{3}$
$3x$ = $8\sqrt{13}$
Area of square ABCD = $9x^{2}$ = $(8\sqrt{13})^{2}$ = 832 | // Block 1
unitsize(135);
defaultpen(linewidth(.8pt)+fontsize(10pt));
pair A, B, C, D, E, F;
A=(0, 0); B=(1, 0); C=(1, -1); D=(0,-1); E=(0,-1/3); F=(1,-2/3);
draw(A--B--C--D--cycle);
label("$A$",A,N); label("$B$",B,N); label("$C$",C,E); label("$D$",D,S); label("$E$",E,W); label("$F$",F,SE);
draw(A--F);
draw(C--E);
draw(A--C);
draw(E--F);
// Block 2
unitsize(135); defaultpen(linewidth(.8pt)+fontsize(10pt)); pair A, B, C, D, E, F; A=(0, 0); B=(1, 0); C=(1, -1); D=(0,-1); E=(0,-1/3); F=(1,-2/3); draw(A--B--C--D--cycle); label("$A$",A,N); label("$B$",B,N); label("$C$",C,E); label("$D$",D,S); label("$E$",E,W); label("$F$",F,SE); draw(A--F); draw(C--E); draw(A--C); draw(E--F); | [] |
468 | Let $\mathcal{R}$ be the region consisting of the set of points in the coordinate plane that satisfy both $|8 - x| + y \le 10$ and $3y - x \ge 15$. When $\mathcal{R}$ is revolved around the line whose equation is $3y - x = 15$, the volume of the resulting solid is $\frac {m\pi}{n\sqrt {p}}$, where $m$, $n$, and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m + n + p$. | 2010 AIME I Problems/Problem 11 | The inequalities are equivalent to $y \ge x/3 + 5, y \le 10 - |x - 8|$. We can set them equal to find the two points of intersection, $x/3 + 5 = 10 - |x - 8| \Longrightarrow |x - 8| = 5 - x/3$. This implies that one of $x - 8, 8 - x = 5 - x/3$, from which we find that $(x,y) = \left(\frac 92, \frac {13}2\right), \left(\frac{39}{4}, \frac{33}{4}\right)$. The region $\mathcal{R}$ is a triangle, as shown above. When revolved about the line $y = x/3+5$, the resulting solid is the union of two right cones that share the same base and axis.
Let $h_1,h_2$ denote the height of the left and right cones, respectively (so $h_1 > h_2$), and let $r$ denote their common radius. The volume of a cone is given by $\frac 13 Bh$; since both cones share the same base, then the desired volume is $\frac 13 \cdot \pi r^2 \cdot (h_1 + h_2)$. The distance from the point $(8,10)$ to the line $x - 3y + 15 = 0$ is given by $\left|\frac{(8) - 3(10) + 15}{\sqrt{1^2 + (-3)^2}}\right| = \frac{7}{\sqrt{10}}$. The distance between $\left(\frac 92, \frac {13}2\right)$ and $\left(\frac{39}{4}, \frac{33}{4}\right)$ is given by $h_1 + h_2 = \sqrt{\left(\frac{18}{4} - \frac{39}{4}\right)^2 + \left(\frac{26}{4} - \frac{33}{4}\right)^2} = \frac{7\sqrt{10}}{4}$. Thus, the answer is $\frac{343\sqrt{10}\pi}{120} = \frac{343\pi}{12\sqrt{10}} \Longrightarrow 343 + 12 + 10 = \boxed{365}$. | // Block 1
size(280);
import graph; real min = 2, max = 12; pen dark = linewidth(1);
real P(real x) { return x/3 + 5; }
real Q(real x) { return 10 - abs(x - 8); }
path p = (2,P(2))--(8,P(8))--(12,P(12)), q = (2,Q(2))--(12,Q(12));
pair A = (8,10), B = (4.5,6.5), C= (9.75,8.25), F=foot(A,B,C), G=2*F-A; fill(A--B--C--cycle,rgb(0.9,0.9,0.9));
draw(graph(P,min,max),dark);
draw(graph(Q,min,max),dark);
draw(Arc((8,7.67),A,G,CW),dark,EndArrow(8)); draw(B--C--G--cycle,linetype("4 4"));
label("$y \ge x/3 + 5$",(max,P(max)),E,fontsize(10));
label("$y \le 10 - |x-8|$",(max,Q(max)),E,fontsize(10));
label("$\mathcal{R}$",(6,Q(6)),NW);
/* axes */
Label f; f.p=fontsize(8);
xaxis(0, max, Ticks(f, 6, 1));
yaxis(0, 10, Ticks(f, 5, 1));
// Block 2
size(200);
import three; currentprojection = perspective(0,0,10); defaultpen(linewidth(0.7)); pen dark=linewidth(1.3);
pair Fxy = foot((8,10),(4.5,6.5),(9.75,8.25));
triple A = (8,10,0), B = (4.5,6.5,0), C= (9.75,8.25,0), F=(Fxy.x,Fxy.y,0), G=2*F-A, H=(F.x,F.y,abs(F-A)),I=(F.x,F.y,-abs(F-A));
real theta1 = 1.2, theta2 = -1.7,theta3= abs(F-A),theta4=-2.2;
triple J=F+theta1*unit(A-F)+(0,0,((abs(F-A))^2-(theta1)^2)^.5 ),K=F+theta2*unit(A-F)+(0,0,((abs(F-A))^2-(theta2)^2)^.5 ),L=F+theta3*unit(A-F)+(0,0,((abs(F-A))^2-(theta3)^2)^.5 ),M=F+theta4*unit(A-F)-(0,0,((abs(F-A))^2-(theta4)^2)^.5 );
draw(C--A--B--G--cycle,linetype("4 4")+dark); draw(A..H..G..I..A); draw(C--B^^A--G,linetype("4 4")); draw(J--C--K); draw(L--B--M);
dot(B);dot(C);dot(F);
label("$h_1$",(B+F)/2,SE,fontsize(10));
label("$h_2$",(C+F)/2,S,fontsize(10));
label("$r$",(A+F)/2,E,fontsize(10));
// Block 3
size(280); import graph; real min = 2, max = 12; pen dark = linewidth(1); real P(real x) { return x/3 + 5; } real Q(real x) { return 10 - abs(x - 8); } path p = (2,P(2))--(8,P(8))--(12,P(12)), q = (2,Q(2))--(12,Q(12)); pair A = (8,10), B = (4.5,6.5), C= (9.75,8.25), F=foot(A,B,C), G=2*F-A; fill(A--B--C--cycle,rgb(0.9,0.9,0.9)); draw(graph(P,min,max),dark); draw(graph(Q,min,max),dark); draw(Arc((8,7.67),A,G,CW),dark,EndArrow(8)); draw(B--C--G--cycle,linetype("4 4")); label("$y \ge x/3 + 5$",(max,P(max)),E,fontsize(10)); label("$y \le 10 - |x-8|$",(max,Q(max)),E,fontsize(10)); label("$\mathcal{R}$",(6,Q(6)),NW); /* axes */ Label f; f.p=fontsize(8); xaxis(0, max, Ticks(f, 6, 1)); yaxis(0, 10, Ticks(f, 5, 1));
// Block 4
size(200); import three; currentprojection = perspective(0,0,10); defaultpen(linewidth(0.7)); pen dark=linewidth(1.3); pair Fxy = foot((8,10),(4.5,6.5),(9.75,8.25)); triple A = (8,10,0), B = (4.5,6.5,0), C= (9.75,8.25,0), F=(Fxy.x,Fxy.y,0), G=2*F-A, H=(F.x,F.y,abs(F-A)),I=(F.x,F.y,-abs(F-A)); real theta1 = 1.2, theta2 = -1.7,theta3= abs(F-A),theta4=-2.2; triple J=F+theta1*unit(A-F)+(0,0,((abs(F-A))^2-(theta1)^2)^.5 ),K=F+theta2*unit(A-F)+(0,0,((abs(F-A))^2-(theta2)^2)^.5 ),L=F+theta3*unit(A-F)+(0,0,((abs(F-A))^2-(theta3)^2)^.5 ),M=F+theta4*unit(A-F)-(0,0,((abs(F-A))^2-(theta4)^2)^.5 ); draw(C--A--B--G--cycle,linetype("4 4")+dark); draw(A..H..G..I..A); draw(C--B^^A--G,linetype("4 4")); draw(J--C--K); draw(L--B--M); dot(B);dot(C);dot(F); label("$h_1$",(B+F)/2,SE,fontsize(10)); label("$h_2$",(C+F)/2,S,fontsize(10)); label("$r$",(A+F)/2,E,fontsize(10)); | [] |
469 | Rectangle $ABCD$ and a semicircle with diameter $AB$ are coplanar and have nonoverlapping interiors. Let $\mathcal{R}$ denote the region enclosed by the semicircle and the rectangle. Line $\ell$ meets the semicircle, segment $AB$, and segment $CD$ at distinct points $N$, $U$, and $T$, respectively. Line $\ell$ divides region $\mathcal{R}$ into two regions with areas in the ratio $1: 2$. Suppose that $AU = 84$, $AN = 126$, and $UB = 168$. Then $DA$ can be represented as $m\sqrt {n}$, where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m + n$. | 2010 AIME I Problems/Problem 13 | Diagram
Solution 1
The center of the semicircle is also the midpoint of $AB$. Let this point be O. Let $h$ be the length of $AD$.
Rescale everything by 42, so $AU = 2, AN = 3, UB = 4$. Then $AB = 6$ so $OA = OB = 3$.
Since $ON$ is a radius of the semicircle, $ON = 3$. Thus $OAN$ is an equilateral triangle.
Let $X$, $Y$, and $Z$ be the areas of triangle $OUN$, sector $ONB$, and trapezoid $UBCT$ respectively.
$X = \frac {1}{2}(UO)(NO)\sin{O} = \frac {1}{2}(1)(3)\sin{60^\circ} = \frac {3}{4}\sqrt {3}$
$Y = \frac {1}{3}\pi(3)^2 = 3\pi$
To find $Z$ we have to find the length of $TC$. Project $T$ and $N$ onto $AB$ to get points $T'$ and $N'$. Notice that $UNN'$ and $TUT'$ are similar. Thus:
$\frac {TT'}{UT'} = \frac {UN'}{NN'} \implies \frac {TT'}{h} = \frac {1/2}{3\sqrt {3}/2} \implies TT' = \frac {\sqrt {3}}{9}h$.
Then $TC = T'C - T'T = UB - TT' = 4 - \frac {\sqrt {3}}{9}h$. So:
$Z = \frac {1}{2}(BU + TC)(CB) = \frac {1}{2}\left(8 - \frac {\sqrt {3}}{9}h\right)h = 4h - \frac {\sqrt {3}}{18}h^2$
Let $L$ be the area of the side of line $l$ containing regions $X, Y, Z$. Then
$L = X + Y + Z = \frac {3}{4}\sqrt {3} + 3\pi + 4h - \frac {\sqrt {3}}{18}h^2$
Obviously, the $L$ is greater than the area on the other side of line $l$. This other area is equal to the total area minus $L$. Thus:
$\frac {2}{1} = \frac {L}{6h + \frac {9}{2}{\pi} - L} \implies 12h + 9\pi = 3L$.
Now just solve for $h$.
\begin{align*} 12h + 9\pi & = \frac {9}{4}\sqrt {3} + 9\pi + 12h - \frac {\sqrt {3}}{6}h^2 \\ 0 & = \frac {9}{4}\sqrt {3} - \frac {\sqrt {3}}{6}h^2 \\ h^2 & = \frac {9}{4}(6) \\ h & = \frac {3}{2}\sqrt {6} \end{align*}
Don't forget to un-rescale at the end to get $AD = \frac {3}{2}\sqrt {6} \cdot 42 = 63\sqrt {6}$.
Finally, the answer is $63 + 6 = \boxed{069}$. | /* geogebra conversion, see azjps userscripts.org/scripts/show/72997 */ import graph; defaultpen(linewidth(0.7)+fontsize(10)); size(500); pen zzttqq = rgb(0.6,0.2,0); pen xdxdff = rgb(0.4902,0.4902,1); /* segments and figures */ draw((0,-154.31785)--(0,0)); draw((0,0)--(252,0)); draw((0,0)--(126,0),zzttqq); draw((126,0)--(63,109.1192),zzttqq); draw((63,109.1192)--(0,0),zzttqq); draw((-71.4052,(+9166.01287-109.1192*-71.4052)/21)--(504.60925,(+9166.01287-109.1192*504.60925)/21)); draw((0,-154.31785)--(252,-154.31785)); draw((252,-154.31785)--(252,0)); draw((0,0)--(84,0)); draw((84,0)--(252,0)); draw((63,109.1192)--(63,0)); draw((84,0)--(84,-154.31785)); draw(arc((126,0),126,0,180)); /* points and labels */ dot((0,0)); label("$A$",(-16.43287,-9.3374),NE/2); dot((252,0)); label("$B$",(255.242,5.00321),NE/2); dot((0,-154.31785)); label("$D$",(3.48464,-149.55669),NE/2); dot((252,-154.31785)); label("$C$",(255.242,-149.55669),NE/2); dot((126,0)); label("$O$",(129.36332,5.00321),NE/2); dot((63,109.1192)); label("$N$",(44.91307,108.57427),NE/2); label("$126$",(28.18236,40.85473),NE/2); dot((84,0)); label("$U$",(87.13819,5.00321),NE/2); dot((113.69848,-154.31785)); label("$T$",(116.61611,-149.55669),NE/2); dot((63,0)); label("$84$",(41.72627,-12.5242),NE/2); label("$168$",(167.60494,-12.5242),NE/2); dot((84,-154.31785)); dot((252,0)); label("$I$",(255.242,5.00321),NE/2); clip((-71.4052,-225.24323)--(-71.4052,171.51361)--(504.60925,171.51361)--(504.60925,-225.24323)--cycle); | [] |
470 | In $\triangle{ABC}$ with $AB = 12$, $BC = 13$, and $AC = 15$, let $M$ be a point on $\overline{AC}$ such that the incircles of $\triangle{ABM}$ and $\triangle{BCM}$ have equal radii. Then $\frac{AM}{CM} = \frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p + q$. | 2010 AIME I Problems/Problem 15 | Solution 1
Let $AM = x$, then $CM = 15 - x$. Also let $BM = d$ Clearly, $\frac {[ABM]}{[CBM]} = \frac {x}{15 - x}$. We can also express each area by the rs formula. Then $\frac {[ABM]}{[CBM]} = \frac {p(ABM)}{p(CBM)} = \frac {12 + d + x}{28 + d - x}$. Equating and cross-multiplying yields $25x + 2dx = 15d + 180$ or $d = \frac {25x - 180}{15 - 2x}.$ Note that for $d$ to be positive, we must have $7.2 < x < 7.5$.
By Stewart's Theorem, we have $12^2(15 - x) + 13^2x = d^215 + 15x(15 - x)$ or $432 = 3d^2 + 40x - 3x^2.$ Brute forcing by plugging in our previous result for $d$, we have $432 = \frac {3(25x - 180)^2}{(15 - 2x)^2} + 40x - 3x^2.$ Clearing the fraction and gathering like terms, we get $0 = 12x^4 - 340x^3 + 2928x^2 - 7920x.$
Aside: Since $x$ must be rational in order for our answer to be in the desired form, we can use the Rational Root Theorem to reveal that $6x$ is an integer because we can divide the polynomial by $2$. The only such $x$ in the above-stated range is $\frac {22}3$.
Legitimately solving that quartic, note that $x = 0$ and $x = 15$ should clearly be solutions, corresponding to the sides of the triangle and thus degenerate cevians. Factoring those out, we get $0 = 4x(x - 15)(3x^2 - 40x + 132) = x(x - 15)(x - 6)(3x - 22).$ The only solution in the desired range is thus $\frac {22}3$. Then $CM = \frac {23}3$, and our desired ratio $\frac {AM}{CM} = \frac {22}{23}$, giving us an answer of $\boxed{045}$.
Solution 2
Let $AM = 2x$ and $BM = 2y$ so $CM = 15 - 2x$. Let the incenters of $\triangle ABM$ and $\triangle BCM$ be $I_1$ and $I_2$ respectively, and their equal inradii be $r$. From $r = \sqrt {(s - a)(s - b)(s - c)}/s$, we find that
\begin{align*}r^2 & = \frac {(x + y - 6)( - x + y + 6)(x - y + 6)}{x + y + 6} & (1) \\ & = \frac {( - x + y + 1)(x + y - 1)( - x - y + 14)}{ - x + y + 14}. & (2) \end{align*}
Let the incircle of $\triangle ABM$ meet $AM$ at $P$ and the incircle of $\triangle BCM$ meet $CM$ at $Q$. Then note that $I_1 P Q I_2$ is a rectangle. Also, $\angle I_1 M I_2$ is right because $MI_1$ and $MI_2$ are the angle bisectors of $\angle AMB$ and $\angle CMB$ respectively and $\angle AMB + \angle CMB = 180^\circ$. By properties of tangents to circles $MP = (MA + MB - AB)/2 = x + y - 6$ and $MQ = (MB + MC - BC)/2 = - x + y + 1$. Now notice that the altitude of $M$ to $I_1 I_2$ is of length $r$, so by similar triangles we find that $r^2 = MP \cdot MQ = (x + y - 6)( - x + y + 1)$ (3). Equating (3) with (1) and (2) separately yields
\begin{align*} 2y^2 - 30 = 2xy + 5x - 7y \\ 2y^2 - 70 = - 2xy - 5x + 7y, \end{align*}
and adding these we have
\[4y^2 - 100 = 0\implies y = 5\implies x = 11/3 \\ \implies AM/MC = (22/3)/(15 - 22/3) = 22/23 \implies \boxed{045}.\]
Solution 3
Let the incircle of $ABM$ hit $AM$, $AB$, $BM$ at $X_{1},Y_{1},Z_{1}$, and let the incircle of $CBM$ hit $MC$, $BC$, $BM$ at $X_{2},Y_{2},Z_{2}$. Draw the incircle of $ABC$, and let it be tangent to $AC$ at $X$. Observe that we have a homothety centered at A sending the incircle of $ABM$ to that of $ABC$, and one centered at $C$ taking the incircle of $BCM$ to that of $ABC$. These have the same power, since they take incircles of the same size to the same circle. Also, the power of the homothety is $AX_{1}/AX=CX_{2}/CX$.
By standard computations, $AX=\dfrac{AB+AC-BC}{2}=7$ and $CX=\dfrac{BC+AC-AB}{2}=8$. Now, let $AX_{1}=7x$ and $CX_{2}=8x$. We will now go around and chase lengths. Observe that $BY_{1}=BA-AY_{1}=BA-AX_{1}=12-7x$. Then, $BZ_{1}=12-7x$. We also have $CY_{2}=CX_{2}=8x$, so $BY_{2}=13-8x$ and $BZ_{2}=13-8x$.
Observe now that $X_{1}M+MX_{2}=AC-15x=15(1-x)$. Also,$X_{1}M-MX_{2}=MZ_{1}-MZ_{2}=BZ_{2}-BZ_{1}=BY_{2}-BY_{1}=(1-x)$. Solving, we get $X_{1}M=8-8x$ and $MX_{2}=7-7x$ (as a side note, note that $AX_{1}+MX_{2}=X_{1}M+X_{2}C$, a result that I actually believe appears in Mandelbrot 1995-2003, or some book in that time-range).
Now, we get $BM=BZ_{2}+Z_{2}M=BZ_{2}+MX_{2}=20-15x$. To finish, we will compute area ratios. $\dfrac{[ABM]}{[CBM]}=\dfrac{AM}{MC}=\dfrac{8-x}{7+x}$. Also, since their inradii are equal, we get $\dfrac{[ABM]}{[CBM]}=\dfrac{40-16x}{40-14x}$. Equating and cross multiplying yields the quadratic $3x^{2}-8x+4=0$, so $x=2/3,2$. However, observe that $AX_{1}+CX_{2}=15x<15$, so we take $x=2/3$. Our ratio is therefore $\dfrac{8-2/3}{7+2/3}=\dfrac{22}{23}$, giving the answer $\boxed{045}$.
Note: Once we have $MX_1=8-8x$ and $MX_2=7-7x$, it's bit easier to use use the right triangle of $O_1MO_2$ than chasing the area ratio. The inradius of $\triangle{ABC}$ can be calculated to be $r=\sqrt{14}$, and the inradius of $ABM$ and $ACM$ are $r_1=r_2= xr$, so,
\[O_1O_2^2 = O_1M^2 + O_2M^2 = r_1^2+X_1M^2 + r_2^2 + X_2M^2\]
or,
\[(15(1-x))^2 = 2(\sqrt{14}x)^2 + (7(1-x))^2 + (8(1-x))^2\]
\[112(1-x)^2 = 28x^2\]
\[4(1-x)^2 = x^2\]
We get $x=\frac{2}{3}$ or $x=2$.
Solution 4
Suppose the incircle of $ABM$ touches $AM$ at $X$, and the incircle of $CBM$ touches $CM$ at $Y$. Then
\[r = AX \tan(A/2) = CY \tan(C/2)\]
We have $\cos A = \frac{12^2+15^2-13^2}{2\cdot 12\cdot 15} = \frac{200}{30\cdot 12}=\frac{5}{9}$, $\tan(A/2) = \sqrt{\frac{1-\cos A}{1+\cos A}} = \sqrt{\frac{9-5}{9+5}} = \frac{2}{\sqrt{14}}$
$\cos C = \frac{13^2+15^2-12^2}{2\cdot 13\cdot 15} = \frac{250}{30\cdot 13} = \frac{25}{39}$, $\tan(C/2) = \sqrt{\frac{39-25}{39+25}}=\frac{\sqrt{14}}{8}$,
Therefore $AX/CY = \tan(C/2)/\tan(A/2) = \frac{14}{2\cdot 8}= \frac{7}{8}.$
And since $AX=\frac{1}{2}(12+AM-BM)$, $CY = \frac{1}{2}(13+CM-BM)$,
\[\frac{12+AM-BM}{13+CM-BM} = \frac{7}{8}\]
\[96+8AM-8BM = 91 +7CM-7BM\]
\[BM= 5 + 8AM-7CM = 5 + 15AM - 7(CM+AM) = 5+15(AM-7) \dots\dots (1)\]
Now,
$\frac{AM}{CM} = \frac{[ABM]}{[CBM]} = \frac{\frac{1}{2}(12+AM+BM)r}{\frac{1}{2}(13+CM+BM)r}=\frac{12+AM+BM}{13+CM+BM}= \frac{12+BM}{13+BM} = \frac{17+15(AM-7)}{18+15(AM-7)}$
\[\frac{AM}{15} = \frac{17+15(AM-7)}{35+30(AM-7)} = \frac{15AM-88}{30AM-175}\]
\[6AM^2 - 35AM = 45AM-264\]
\[3AM^2 -40AM+132=0\]
\[(3AM-22)(AM-6)=0\]
So $AM=22/3$ or $6$. But from (1) we know that $5+15(AM-7)>0$, or $AM>7-1/3>6$, so $AM=22/3$, $CM=15-22/3=23/3$, $AM/CM=22/23$.
Solution 5
Let the common inradius equal r, $BM = x$, $AM = y$, $MC = z$
From the prespective of $\triangle{ABM}$ and $\triangle{BMC}$ we get:
$S_{ABM} = rs = r \cdot (\frac{12+x+y}{2})$ $S_{BMC} = rs = r \cdot (\frac{13+x+z}{2})$
Add two triangles up, we get $\triangle{ABC}$ :
$S_{ABC} = S_{ABM} + S_{BMC} = r \cdot \frac{25+2x+y+z}{2}$
Since $y + z = 15$, we get:
$r = \frac{S_{ABC}}{20 + x}$
By drawing an altitude from $I_1$ down to a point $H_1$ and from $I_2$ to $H_2$, we can get:
$r \cdot cot(\frac{\angle A}{2}) =r \cdot A H_1 = r \cdot \frac{12+y-x}{2}$ and
$r \cdot cot(\frac{\angle C}{2}) = r \cdot H_2 C = r \cdot \frac{13+z-x}{2}$
Adding these up, we get:
$r \cdot (cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})) = \frac{25+y+z-2x}{2} = \frac{40-2x}{2} = 20-x$
$r = \frac{20-x}{cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})}$
Now, we have 2 values equal to r, we can set them equal to each other:
$\frac{S_{ABC}}{20 + x} = \frac{20-x}{cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})}$
If we let R denote the incircle of ABC, note:
AC = $(cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})) \cdot R = 15$ and
$S_{ABC} = \frac{12+13+15}{2} \cdot R = 20 \cdot R$.
By cross multiplying the equation above, we get:
$400 - x^2 = (cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})) \cdot S_{ABC} = (cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})) \cdot R \cdot 20 = 15 \cdot 20 = 300$
We can find out x:
$x = 10$.
Now, we can find ratio of y and z:
$\frac{AM}{CM} = \frac{y}{z} = \frac{S_{ABM}}{S_{BCM}} = \frac{r \cdot \frac{22+y}{2} }{r \cdot \frac{23+z}{2}} = \frac{22+y}{23+z} = \frac{22}{23}$
The answer is $\boxed{045}$.
-Alexlikemath
Solution 6 (Similar to Solution 1 with easier computation)
Let $CM=x, AM=rx, BM=d$. $x+rx=15\Rightarrow x=\frac{15}{1+r}$.
Similar to Solution 1, we have
\[r=\frac{[AMB]}{[CMB]}=\frac{12+rx+d}{13+x+d} \Rightarrow d=\frac{13r-12}{1-r}\]
as well as
\[12^2\cdot x + 13^2 rx=15x\cdot rx+15d^2 (\text{via Stewart's Theorem})\]
\[\frac{(12^2 + 13^2r) \cdot 15}{1+r} - \frac{15r\cdot 15^2}{(1+r)^2}=\frac{15(13r-12)^2}{(1-r)^2}\]
\[\frac{169r^2+88r+144}{(1+r)^2}=\frac{(13r-12)^2}{(1-r)^2} =\frac{169r^2-312r+144}{(1-r)^2} =\frac{400r}{4r}=100\]
(here we used the fact that if $\frac{a}{b} = \frac{c}{d} = k,$ then $\frac{a-c}{b-d}=k$ as well.)
Notice $\frac{12}{13} < r < 1$, so $\frac{13r-12}{1-r} = 10$ and $\boxed{r = \frac{22}{23}}.$
~ asops
Solution 7 (No Stewart's or trig, fast + clever)
Let $BM = d, AM = x, CM = 15 - x$. Observe that we have the equation by the incircle formula:
\[\frac{[ABM]}{12 + AM + MB} = \frac{[CBM]}{13 + CM + MB} \implies \frac{AM}{CM} = \frac{12 + MB}{13 + MB} \implies \frac{x}{15 - x} = \frac{12 + d}{13 + d}.\]
Now let $X$ be the point of tangency between the incircle of $\triangle ABC$ and $AC$. Additionally, let $P$ and $Q$ be the points of tangency between the incircles of $\triangle ABM$ and $\triangle CBM$ with $AC$ respectively. Some easy calculation yields $AX = 7, CX = 8$. By homothety we have
\[\frac{AP}{7} = \frac{CQ}{8} \implies 8(AP) = 7(CQ) \implies 8(12 + x - d) = 7(13 + 15 - x - d) \implies d = 15x - 100.\]
Substituting into the first equation derived earlier it is left to solve
\[\frac{x}{15 - x} = \frac{15x - 88}{15x - 87} \implies 3x^2 - 40x + 132 \implies (x - 6)(3x - 22) = 0.\]
Now $x = 6$ yields $d = -10$ which is invalid, hence $x = \frac{22}{3}$ so $\frac{AM}{CM} = \frac{\frac{22}{3}}{15 - \frac{22}{3}} = \frac{22}{23}.$ The requested sum is $22 + 23 = \boxed{45}$. ~blueprimes | // Block 1
/* from geogebra: see azjps, userscripts.org/scripts/show/72997 */
import graph; defaultpen(linewidth(0.7)+fontsize(10)); size(200);
/* segments and figures */
draw((0,0)--(15,0));
draw((15,0)--(6.66667,9.97775));
draw((6.66667,9.97775)--(0,0));
draw((7.33333,0)--(6.66667,9.97775));
draw(circle((4.66667,2.49444),2.49444));
draw(circle((9.66667,2.49444),2.49444));
draw((4.66667,0)--(4.66667,2.49444));
draw((9.66667,2.49444)--(9.66667,0));
/* points and labels */
label("r",(10.19662,1.92704),SE);
label("r",(5.02391,1.8773),SE);
dot((0,0));
label("$A$",(-1.04408,-0.60958),NE);
dot((15,0));
label("$C$",(15.41907,-0.46037),NE);
dot((6.66667,9.97775));
label("$B$",(6.66525,10.23322),NE);
label("$15$",(6.01866,-1.15669),NE);
label("$13$",(11.44006,5.50815),NE);
label("$12$",(2.28834,5.75684),NE);
dot((7.33333,0));
label("$M$",(7.56053,-1.000),NE);
label("$H_1$",(3.97942,-1.200),NE);
label("$H_2$",(9.54741,-1.200),NE);
dot((4.66667,2.49444));
label("$I_1$",(3.97942,2.92179),NE);
dot((9.66667,2.49444));
label("$I_2$",(9.54741,2.92179),NE);
clip((-3.72991,-6.47862)--(-3.72991,17.44518)--(32.23039,17.44518)--(32.23039,-6.47862)--cycle);
// Block 2
/* from geogebra: see azjps, userscripts.org/scripts/show/72997 */ import graph; defaultpen(linewidth(0.7)+fontsize(10)); size(200); /* segments and figures */ draw((0,0)--(15,0)); draw((15,0)--(6.66667,9.97775)); draw((6.66667,9.97775)--(0,0)); draw((7.33333,0)--(6.66667,9.97775)); draw(circle((4.66667,2.49444),2.49444)); draw(circle((9.66667,2.49444),2.49444)); draw((4.66667,0)--(4.66667,2.49444)); draw((9.66667,2.49444)--(9.66667,0)); /* points and labels */ label("r",(10.19662,1.92704),SE); label("r",(5.02391,1.8773),SE); dot((0,0)); label("$A$",(-1.04408,-0.60958),NE); dot((15,0)); label("$C$",(15.41907,-0.46037),NE); dot((6.66667,9.97775)); label("$B$",(6.66525,10.23322),NE); label("$15$",(6.01866,-1.15669),NE); label("$13$",(11.44006,5.50815),NE); label("$12$",(2.28834,5.75684),NE); dot((7.33333,0)); label("$M$",(7.56053,-1.000),NE); label("$H_1$",(3.97942,-1.200),NE); label("$H_2$",(9.54741,-1.200),NE); dot((4.66667,2.49444)); label("$I_1$",(3.97942,2.92179),NE); dot((9.66667,2.49444)); label("$I_2$",(9.54741,2.92179),NE); clip((-3.72991,-6.47862)--(-3.72991,17.44518)--(32.23039,17.44518)--(32.23039,-6.47862)--cycle); | [] |
471 | Triangle $ABC$ with right angle at $C$, $\angle BAC < 45^\circ$ and $AB = 4$. Point $P$ on $\overline{AB}$ is chosen such that $\angle APC = 2\angle ACP$ and $CP = 1$. The ratio $\frac{AP}{BP}$ can be represented in the form $p + q\sqrt{r}$, where $p$, $q$, $r$ are positive integers and $r$ is not divisible by the square of any prime. Find $p+q+r$. | 2010 AIME II Problems/Problem 14 | Let $O$ be the circumcenter of $ABC$ and let the intersection of $CP$ with the circumcircle be $D$. It now follows that $\angle{DOA} = 2\angle ACP = \angle{APC} = \angle{DPB}$. Hence $ODP$ is isosceles and $OD = DP = 2$.
Denote $E$ the projection of $O$ onto $CD$. Now $CD = CP + DP = 3$. By the Pythagorean Theorem, $OE = \sqrt {2^2 - \frac {3^2}{2^2}} = \sqrt {\frac {7}{4}}$. Now note that $EP = \frac {1}{2}$. By the Pythagorean Theorem, $OP = \sqrt {\frac {7}{4} + \frac {1^2}{2^2}} = \sqrt {2}$. Hence it now follows that,
\[\frac {AP}{BP} = \frac {AO + OP}{BO - OP} = \frac {2 + \sqrt {2}}{2 - \sqrt {2}} = 3 + 2\sqrt {2}\]
This gives that the answer is $\boxed{007}$.
An alternate finish for this problem would be to use Power of a Point on $BA$ and $CD$. By Power of a Point Theorem, $CP\cdot PD=1\cdot 2=BP\cdot PA$. Since $BP+PA=4$, we can solve for $BP$ and $PA$, giving the same values and answers as above. | // Block 1
/* geogebra conversion, see azjps userscripts.org/scripts/show/72997 */
import graph; defaultpen(linewidth(0.7)+fontsize(10)); size(250); real lsf = 0.5; /* changes label-to-point distance */
pen xdxdff = rgb(0.49,0.49,1); pen qqwuqq = rgb(0,0.39,0); pen fftttt = rgb(1,0.2,0.2);
/* segments and figures */
draw((0.2,0.81)--(0.33,0.78)--(0.36,0.9)--(0.23,0.94)--cycle,qqwuqq); draw((0.81,-0.59)--(0.93,-0.54)--(0.89,-0.42)--(0.76,-0.47)--cycle,qqwuqq); draw(circle((2,0),2)); draw((0,0)--(0.23,0.94),linewidth(1.6pt)); draw((0.23,0.94)--(4,0),linewidth(1.6pt)); draw((0,0)--(4,0),linewidth(1.6pt)); draw((0.23,(+0.55-0.94*0.23)/0.35)--(4.67,(+0.55-0.94*4.67)/0.35));
/* points and labels */
label("$1$", (0.26,0.42), SE*lsf); draw((1.29,-1.87)--(2,0)); label("$2$", (2.91,-0.11), SE*lsf); label("$2$", (1.78,-0.82), SE*lsf); pair parametricplot0_cus(real t){
return (0.28*cos(t)+0.23,0.28*sin(t)+0.94);
}
draw(graph(parametricplot0_cus,-1.209429202888189,-0.24334747753738661)--(0.23,0.94)--cycle,fftttt); pair parametricplot1_cus(real t){
return (0.28*cos(t)+0.59,0.28*sin(t)+0);
}
draw(graph(parametricplot1_cus,0.0,1.9321634507016043)--(0.59,0)--cycle,fftttt); label("$\theta$", (0.42,0.77), SE*lsf); label("$2\theta$", (0.88,0.38), SE*lsf); draw((2,0)--(0.76,-0.47)); pair parametricplot2_cus(real t){
return (0.28*cos(t)+2,0.28*sin(t)+0);
}
draw(graph(parametricplot2_cus,-1.9321634507016048,0.0)--(2,0)--cycle,fftttt); label("$2\theta$", (2.18,-0.3), SE*lsf); dot((0,0)); label("$B$", (-0.21,-0.2),NE*lsf); dot((4,0)); label("$A$", (4.03,0.06),NE*lsf); dot((2,0)); label("$O$", (2.04,0.06),NE*lsf); dot((0.59,0)); label("$P$", (0.28,-0.27),NE*lsf); dot((0.23,0.94)); label("$C$", (0.07,1.02),NE*lsf); dot((1.29,-1.87)); label("$D$", (1.03,-2.12),NE*lsf); dot((0.76,-0.47)); label("$E$", (0.56,-0.79),NE*lsf); clip((-0.92,-2.46)--(-0.92,2.26)--(4.67,2.26)--(4.67,-2.46)--cycle);
// Block 2
/* geogebra conversion, see azjps userscripts.org/scripts/show/72997 */ import graph; defaultpen(linewidth(0.7)+fontsize(10)); size(250); real lsf = 0.5; /* changes label-to-point distance */ pen xdxdff = rgb(0.49,0.49,1); pen qqwuqq = rgb(0,0.39,0); pen fftttt = rgb(1,0.2,0.2); /* segments and figures */ draw((0.2,0.81)--(0.33,0.78)--(0.36,0.9)--(0.23,0.94)--cycle,qqwuqq); draw((0.81,-0.59)--(0.93,-0.54)--(0.89,-0.42)--(0.76,-0.47)--cycle,qqwuqq); draw(circle((2,0),2)); draw((0,0)--(0.23,0.94),linewidth(1.6pt)); draw((0.23,0.94)--(4,0),linewidth(1.6pt)); draw((0,0)--(4,0),linewidth(1.6pt)); draw((0.23,(+0.55-0.94*0.23)/0.35)--(4.67,(+0.55-0.94*4.67)/0.35)); /* points and labels */ label("$1$", (0.26,0.42), SE*lsf); draw((1.29,-1.87)--(2,0)); label("$2$", (2.91,-0.11), SE*lsf); label("$2$", (1.78,-0.82), SE*lsf); pair parametricplot0_cus(real t){ return (0.28*cos(t)+0.23,0.28*sin(t)+0.94); } draw(graph(parametricplot0_cus,-1.209429202888189,-0.24334747753738661)--(0.23,0.94)--cycle,fftttt); pair parametricplot1_cus(real t){ return (0.28*cos(t)+0.59,0.28*sin(t)+0); } draw(graph(parametricplot1_cus,0.0,1.9321634507016043)--(0.59,0)--cycle,fftttt); label("$\theta$", (0.42,0.77), SE*lsf); label("$2\theta$", (0.88,0.38), SE*lsf); draw((2,0)--(0.76,-0.47)); pair parametricplot2_cus(real t){ return (0.28*cos(t)+2,0.28*sin(t)+0); } draw(graph(parametricplot2_cus,-1.9321634507016048,0.0)--(2,0)--cycle,fftttt); label("$2\theta$", (2.18,-0.3), SE*lsf); dot((0,0)); label("$B$", (-0.21,-0.2),NE*lsf); dot((4,0)); label("$A$", (4.03,0.06),NE*lsf); dot((2,0)); label("$O$", (2.04,0.06),NE*lsf); dot((0.59,0)); label("$P$", (0.28,-0.27),NE*lsf); dot((0.23,0.94)); label("$C$", (0.07,1.02),NE*lsf); dot((1.29,-1.87)); label("$D$", (1.03,-2.12),NE*lsf); dot((0.76,-0.47)); label("$E$", (0.56,-0.79),NE*lsf); clip((-0.92,-2.46)--(-0.92,2.26)--(4.67,2.26)--(4.67,-2.46)--cycle); | [] |
472 | A cube with side length 10 is suspended above a plane. The vertex closest to the plane is labeled $A$. The three vertices adjacent to vertex $A$ are at heights 10, 11, and 12 above the plane. The distance from vertex $A$ to the plane can be expressed as $\frac{r-\sqrt{s}}{t}$, where $r$, $s$, and $t$ are positive integers. Find $r+s+t$. | 2011 AIME I Problems/Problem 13 | First, shift the cube down by 11 so the vertices adjacent to $A$ are $-1$, $0$, and $1$ above the plane.
Consider the two points 1 above and 1 below the plane:
Now we rotate the cube so that all 3 vertices adjacent to $A$ are on the plane. We can calculate $A$ to be $\frac{10\sqrt3}{3}$ below the plane.
Notice that when $A$ is rotated, it gets slightly closer to the plane.
The triangle shown in the diagram has legs of ration $1:7$, because it is similar to the triangle in the first diagram, since we are rotating. (kind of bad explanation)
So, we can compute $h$, where $h^2 + \left(\dfrac{h}{7}\right)^2 = \left(\dfrac{10\sqrt3}{3}\right)^2$.
Solving yields $h=\frac{7\sqrt{6}}{3}$.
Remember, this is the distance that $A$ is below the plane, so when we shift up by 11, $A$ is $11-\frac{7\sqrt{6}}{3}$ above the plane. Answer extraction yields $\boxed{330}$. | // Block 1
defaultpen(fontsize(10)+0.8); size(300);
pen p=fontsize(9)+linewidth(3);
// Define points
pair A=(-7,0), B=(0,0), C=(7,0);
pair D=(-7,-1), E=(7,1);
// Draw segments
draw(A--B--C, linewidth(1.2));
draw(A--D, linewidth(1.2));
draw(C--E, linewidth(1.2));
draw(D--E, linewidth(1.2));
// Label lengths
label("$7$", A--B, dir(90));
label("$7$", B--C, dir(270));
label("$1$", A--D, dir(180));
label("$1$", C--E, dir(0));
label("$5\sqrt{2}$", D--B, dir(225));
label("$5\sqrt{2}$", B--E, dir(45));
// Block 2
defaultpen(fontsize(10)+0.8); size(250);
pen p=fontsize(9)+linewidth(1.2);
// Define points
pair A=(-5*sqrt(2),0), B=(0,0), C=(5*sqrt(2),0);
pair D=(0,-10*sqrt(3)/3);
pair E=(7*sqrt(3)/15, -49*sqrt(3)/15);
// Draw triangle ABC base
draw(A--B--C, linewidth(1));
// Draw vertical line from B to D
draw(B--D, linewidth(1));
// Draw right triangle BDE
draw(B--E--D--B, linewidth(1));
// Label points
dot(D, p); label("$A$", D, S, p);
// Label lengths (approximate positions)
label("$\frac{10\sqrt{3}}{3}$", B--D, W);
label("$h$", B--E, dir(0));
// Block 3
defaultpen(fontsize(10)+0.8); size(300); pen p=fontsize(9)+linewidth(3); // Define points pair A=(-7,0), B=(0,0), C=(7,0); pair D=(-7,-1), E=(7,1); // Draw segments draw(A--B--C, linewidth(1.2)); draw(A--D, linewidth(1.2)); draw(C--E, linewidth(1.2)); draw(D--E, linewidth(1.2)); // Label lengths label("$7$", A--B, dir(90)); label("$7$", B--C, dir(270)); label("$1$", A--D, dir(180)); label("$1$", C--E, dir(0)); label("$5\sqrt{2}$", D--B, dir(225)); label("$5\sqrt{2}$", B--E, dir(45));
// Block 4
defaultpen(fontsize(10)+0.8); size(250); pen p=fontsize(9)+linewidth(1.2); // Define points pair A=(-5*sqrt(2),0), B=(0,0), C=(5*sqrt(2),0); pair D=(0,-10*sqrt(3)/3); pair E=(7*sqrt(3)/15, -49*sqrt(3)/15); // Draw triangle ABC base draw(A--B--C, linewidth(1)); // Draw vertical line from B to D draw(B--D, linewidth(1)); // Draw right triangle BDE draw(B--E--D--B, linewidth(1)); // Label points dot(D, p); label("$A$", D, S, p); // Label lengths (approximate positions) label("$\frac{10\sqrt{3}}{3}$", B--D, W); label("$h$", B--E, dir(0)); | [] |
473 | In triangle $ABC$, $AB=125$, $AC=117$ and $BC=120$. The angle bisector of angle $A$ intersects $\overline{BC}$ at point $L$, and the angle bisector of angle $B$ intersects $\overline{AC}$ at point $K$. Let $M$ and $N$ be the feet of the perpendiculars from $C$ to $\overline{BK}$ and $\overline{AL}$, respectively. Find $MN$. | 2011 AIME I Problems/Problem 4 | Extend ${CM}$ and ${CN}$ such that they intersect line ${AB}$ at points $P$ and $Q$, respectively.
Since ${BM}$ is the angle bisector of angle $B$ and ${CM}$ is perpendicular to ${BM}$, $\triangle BCP$ must be an isoceles triangle, so $BP=BC=120$, and $M$ is the midpoint of ${CP}$. For the same reason, $AQ=AC=117$, and $N$ is the midpoint of ${CQ}$.
Hence $MN=\tfrac 12 PQ$. Since \[PQ=BP+AQ-AB=120+117-125=112,\] so $MN=\boxed{056}$. | // Block 1
defaultpen(fontsize(10)+0.8); size(200);
pen p=fontsize(9)+linewidth(3);
pair A,B,C,D,K,L,M,N,P,Q;
A=origin; B=(125,0); C=IP(CR(A,117),CR(B,120)); L=extension(B,C,A,bisectorpoint(B,A,C)); K=extension(A,C,B,bisectorpoint(C,B,A)); M=foot(C,B,K); N=foot(C,A,L);
draw(A--B--C--A); draw(A--L^^B--K, gray+dashed+0.5); draw(M--C--N^^N--extension(A,B,C,N)^^M--extension(A,B,C,M), gray+0.5);
dot("$A$",A,dir(200),p); dot("$B$",B,right,p); dot("$C$",C,up,p); dot("$L$",L,2*dir(70),p); dot("$N$",N,2*dir(-90),p); dot("$M$",M,2*dir(-90),p); dot("$P$",extension(A,B,C,M),2*down,p); dot("$Q$",extension(A,B,C,N),2*down,p);
label("$125$",A--B,down,fontsize(10)); label("$117$",A--C,2*dir(130),fontsize(10)); label("$120$",B--C,1.5*dir(30),fontsize(10));
// Block 2
defaultpen(fontsize(10)+0.8); size(200); pen p=fontsize(9)+linewidth(3); pair A,B,C,D,K,L,M,N,P,Q; A=origin; B=(125,0); C=IP(CR(A,117),CR(B,120)); L=extension(B,C,A,bisectorpoint(B,A,C)); K=extension(A,C,B,bisectorpoint(C,B,A)); M=foot(C,B,K); N=foot(C,A,L); draw(A--B--C--A); draw(A--L^^B--K, gray+dashed+0.5); draw(M--C--N^^N--extension(A,B,C,N)^^M--extension(A,B,C,M), gray+0.5); dot("$A$",A,dir(200),p); dot("$B$",B,right,p); dot("$C$",C,up,p); dot("$L$",L,2*dir(70),p); dot("$N$",N,2*dir(-90),p); dot("$M$",M,2*dir(-90),p); dot("$P$",extension(A,B,C,M),2*down,p); dot("$Q$",extension(A,B,C,N),2*down,p); label("$125$",A--B,down,fontsize(10)); label("$117$",A--C,2*dir(130),fontsize(10)); label("$120$",B--C,1.5*dir(30),fontsize(10)); | [] |
474 | On square $ABCD$, point $E$ lies on side $AD$ and point $F$ lies on side $BC$, so that $BE=EF=FD=30$. Find the area of the square $ABCD$. | 2011 AIME II Problems/Problem 2 | Drawing the square and examining the given lengths,
you find that the three segments cut the square into three equal horizontal sections. Therefore, ($x$ being the side length), $\sqrt{x^2+(x/3)^2}=30$, or $x^2+(x/3)^2=900$. Multiplying both sides by $9$ and simplifying, we find that $10x^2=8100$. Dividing by ten gives $x^2=810.$
Area of the square is $\fbox{810}$. | // Block 1
size(2inch, 2inch);
currentpen = fontsize(8pt);
pair A = (0, 0); dot(A); label("$A$", A, plain.SW);
pair B = (3, 0); dot(B); label("$B$", B, plain.SE);
pair C = (3, 3); dot(C); label("$C$", C, plain.NE);
pair D = (0, 3); dot(D); label("$D$", D, plain.NW);
pair E = (0, 1); dot(E); label("$E$", E, plain.W);
pair F = (3, 2); dot(F); label("$F$", F, plain.E);
label("$\frac x3$", E--A);
label("$\frac x3$", F--C);
label("$x$", A--B);
label("$x$", C--D);
label("$\frac {2x}3$", B--F);
label("$\frac {2x}3$", D--E);
label("$30$", B--E);
label("$30$", F--E);
label("$30$", F--D);
draw(B--C--D--F--E--B--A--D);
// Block 2
size(2inch, 2inch); currentpen = fontsize(8pt); pair A = (0, 0); dot(A); label("$A$", A, plain.SW); pair B = (3, 0); dot(B); label("$B$", B, plain.SE); pair C = (3, 3); dot(C); label("$C$", C, plain.NE); pair D = (0, 3); dot(D); label("$D$", D, plain.NW); pair E = (0, 1); dot(E); label("$E$", E, plain.W); pair F = (3, 2); dot(F); label("$F$", F, plain.E); label("$\frac x3$", E--A); label("$\frac x3$", F--C); label("$x$", A--B); label("$x$", C--D); label("$\frac {2x}3$", B--F); label("$\frac {2x}3$", D--E); label("$30$", B--E); label("$30$", F--E); label("$30$", F--D); draw(B--C--D--F--E--B--A--D); | [] |
475 | In triangle $ABC$, $AB=20$ and $AC=11$. The angle bisector of $\angle A$ intersects $BC$ at point $D$, and point $M$ is the midpoint of $AD$. Let $P$ be the point of the intersection of $AC$ and $BM$. The ratio of $CP$ to $PA$ can be expressed in the form $\dfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2011 AIME II Problems/Problem 4 | Let $D'$ be on $\overline{AC}$ such that $BP \parallel DD'$. It follows that $\triangle BPC \sim \triangle DD'C$, so \[\frac{PC}{D'C} = 1 + \frac{BD}{DC} = 1 + \frac{AB}{AC} = \frac{31}{11}\] by the Angle Bisector Theorem. Similarly, we see by the Midline Theorem that $AP = PD'$. Thus, \[\frac{CP}{PA} = \frac{1}{\frac{PD'}{PC}} = \frac{1}{1 - \frac{D'C}{PC}} = \frac{31}{20},\] and $m+n = \boxed{51}$. | // Block 1
pointpen = black; pathpen = linewidth(0.7);
pair A = (0,0), C= (11,0), B=IP(CR(A,20),CR(C,18)), D = IP(B--C,CR(B,20/31*abs(B-C))), M = (A+D)/2, P = IP(M--2*M-B, A--C), D2 = IP(D--D+P-B, A--C);
D(MP("A",D(A))--MP("B",D(B),N)--MP("C",D(C))--cycle); D(A--MP("D",D(D),NE)--MP("D'",D(D2))); D(B--MP("P",D(P))); D(MP("M",M,NW)); MP("20",(B+D)/2,ENE); MP("11",(C+D)/2,ENE);
// Block 2
pointpen = black; pathpen = linewidth(0.7); pair A = (0,0), C= (11,0), B=IP(CR(A,20),CR(C,18)), D = IP(B--C,CR(B,20/31*abs(B-C))), M = (A+D)/2, P = IP(M--2*M-B, A--C), D2 = IP(D--D+P-B, A--C); D(MP("A",D(A))--MP("B",D(B),N)--MP("C",D(C))--cycle); D(A--MP("D",D(D),NE)--MP("D'",D(D2))); D(B--MP("P",D(P))); D(MP("M",M,NW)); MP("20",(B+D)/2,ENE); MP("11",(C+D)/2,ENE); | [] |
476 | Triangle $ABC$ is inscribed in circle $\omega$ with $AB=5$, $BC=7$, and $AC=3$. The bisector of angle $A$ meets side $\overline{BC}$ at $D$ and circle $\omega$ at a second point $E$. Let $\gamma$ be the circle with diameter $\overline{DE}$. Circles $\omega$ and $\gamma$ meet at $E$ and a second point $F$. Then $AF^2 = \frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2012 AIME II Problems/Problem 15 | Use the angle bisector theorem to find $CD=\tfrac{21}{8}$, $BD=\tfrac{35}{8}$, and use Stewart's Theorem to find $AD=\tfrac{15}{8}$. Use Power of Point $D$ to find $DE=\tfrac{49}{8}$, and so $AE=8$. Use law of cosines to find $\angle CAD = \tfrac{\pi} {3}$, hence $\angle BAD = \tfrac{\pi}{3}$ as well, and $\triangle BCE$ is equilateral, so $BC=CE=BE=7$.
In triangle $AEF$, let $X$ be the foot of the altitude from $A$; then $EF=EX+XF$, where we use signed lengths. Writing $EX=AE \cdot \cos \angle AEF$ and $XF=AF \cdot \cos \angle AFE$, we get
\begin{align}\tag{1} EF = AE \cdot \cos \angle AEF + AF \cdot \cos \angle AFE. \end{align}
Note $\angle AFE = \angle ACE$, and the Law of Cosines in $\triangle ACE$ gives $\cos \angle ACE = -\tfrac 17$.
Also, $\angle AEF = \angle DEF$, and $\angle DFE = \tfrac{\pi}{2}$ ($DE$ is a diameter), so $\cos \angle AEF = \tfrac{EF}{DE} = \tfrac{8}{49}\cdot EF$.
Plugging in all our values into equation $(1)$, we get:
\[EF = \tfrac{64}{49} EF -\tfrac{1}{7} AF \quad \Longrightarrow \quad EF = \tfrac{7}{15} AF.\]
The Law of Cosines in $\triangle AEF$, with $EF=\tfrac 7{15}AF$ and $\cos\angle AFE = -\tfrac 17$ gives
\[8^2 = AF^2 + \tfrac{49}{225} AF^2 + \tfrac 2{15} AF^2 = \tfrac{225+49+30}{225}\cdot AF^2\]
Thus $AF^2 = \frac{900}{19}$. The answer is $\boxed{919}$.
~Shen Kislay Kai | // Block 1
size(150);
defaultpen(fontsize(9pt));
picture pic;
pair A,B,C,D,E,F,W;
B=MP("B",origin,dir(180)); C=MP("C",(7,0),dir(0)); A=MP("A",IP(CR(B,5),CR(C,3)),N); D=MP("D",extension(B,C,A,bisectorpoint(C,A,B)),dir(220)); path omega=circumcircle(A,B,C); E=MP("E",OP(omega,A--(A+20*(D-A))),S); path gamma=CR(midpoint(D--E),length(D-E)/2); F=MP("F",OP(omega,gamma),SE);
draw(omega^^A--B--C--cycle^^gamma); draw(pic, A--E--F--cycle, gray); add(pic);
dot("$W$",circumcenter(A,B,C),dir(180)); label("$\gamma$",gamma,dir(180));
// Block 2
size(150); defaultpen(fontsize(9pt)); picture pic; pair A,B,C,D,E,F,W; B=MP("B",origin,dir(180)); C=MP("C",(7,0),dir(0)); A=MP("A",IP(CR(B,5),CR(C,3)),N); D=MP("D",extension(B,C,A,bisectorpoint(C,A,B)),dir(220)); path omega=circumcircle(A,B,C); E=MP("E",OP(omega,A--(A+20*(D-A))),S); path gamma=CR(midpoint(D--E),length(D-E)/2); F=MP("F",OP(omega,gamma),SE); draw(omega^^A--B--C--cycle^^gamma); draw(pic, A--E--F--cycle, gray); add(pic); dot("$W$",circumcenter(A,B,C),dir(180)); label("$\gamma$",gamma,dir(180)); | [] |
476 | Triangle $ABC$ is inscribed in circle $\omega$ with $AB=5$, $BC=7$, and $AC=3$. The bisector of angle $A$ meets side $\overline{BC}$ at $D$ and circle $\omega$ at a second point $E$. Let $\gamma$ be the circle with diameter $\overline{DE}$. Circles $\omega$ and $\gamma$ meet at $E$ and a second point $F$. Then $AF^2 = \frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2012 AIME II Problems/Problem 15 | Let $a = BC$, $b = CA$, $c = AB$ for convenience. Let $M$ be the midpoint of segment $BC$. We claim that $\angle MAD=\angle DAF$.
$\textit{Proof}$. Since $AE$ is the angle bisector, it follows that $EB = EC$ and consequently $EM\perp BC$. Therefore, $M\in \gamma$. Now let $X = FD\cap \omega$. Since $\angle EFX=90^\circ$, $EX$ is a diameter, so $X$ lies on the perpendicular bisector of $BC$; hence $E$, $M$, $X$ are collinear. From $\angle DAG = \angle DMX = 90^\circ$, quadrilateral $ADMX$ is cyclic. Therefore, $\angle MAD = \angle MXD$. But $\angle MXD$ and $\angle EAF$ are both subtended by arc $EF$ in $\omega$, so they are equal. Thus $\angle MAD=\angle DAF$, as claimed.
As a result, $\angle CAM = \angle FAB$. Combined with $\angle BFA=\angle MCA$, we get $\triangle ABF\sim\triangle AMC$ and therefore\[\frac c{AM}=\frac {AF}b\qquad \Longrightarrow \qquad AF^2=\frac{b^2c^2}{AM^2} = \frac{15^2}{AM^2}\]
By Stewart's Theorem on $\triangle ABC$ (with cevian $AM$), we get \[AM^2 = \tfrac 12 (b^2+c^2)-\tfrac 14 a^2 = \tfrac{19}{4},\] so $AF^2 = \tfrac{900}{19}$, so the answer is $900+19=\boxed{919}$.
-Solution by thecmd999 | // Block 1
size(175);
defaultpen(fontsize(10pt));
picture pic;
pair A,B,C,D,E,F,W;
B=MP("B",origin,dir(180)); C=MP("C",(7,0),dir(0)); A=MP("A",IP(CR(B,5),CR(C,3)),N); D=MP("D",extension(B,C,A,bisectorpoint(C,A,B)),dir(220)); path omega=circumcircle(A,B,C); E=MP("E",OP(omega,A--(A+20*(D-A))),S); path gamma=CR(midpoint(D--E),length(D-E)/2); F=MP("F",OP(omega,gamma),SE); pair X=MP("X",IP(omega,F--(F+2*(D-F))),N); pair M=MP("M",midpoint(B--C),dir(220));
draw(omega^^A--B--C--cycle^^gamma); draw(A--E--F--cycle, gray); draw(E--X--F, gray); draw(pic, A--M--C--cycle^^A--B--F--cycle); draw(A--M, royalblue);
dot("$W$",circumcenter(A,B,C),dir(180)); dot(circumcenter(D,E,F)); label("$\gamma$",gamma,dir(180));
draw(A--B--F--cycle, black+1);
// Block 2
size(175); defaultpen(fontsize(10pt)); picture pic; pair A,B,C,D,E,F,W; B=MP("B",origin,dir(180)); C=MP("C",(7,0),dir(0)); A=MP("A",IP(CR(B,5),CR(C,3)),N); D=MP("D",extension(B,C,A,bisectorpoint(C,A,B)),dir(220)); path omega=circumcircle(A,B,C); E=MP("E",OP(omega,A--(A+20*(D-A))),S); path gamma=CR(midpoint(D--E),length(D-E)/2); F=MP("F",OP(omega,gamma),SE); pair X=MP("X",IP(omega,F--(F+2*(D-F))),N); pair M=MP("M",midpoint(B--C),dir(220)); draw(omega^^A--B--C--cycle^^gamma); draw(A--E--F--cycle, gray); draw(E--X--F, gray); draw(pic, A--M--C--cycle^^A--B--F--cycle); draw(A--M, royalblue); dot("$W$",circumcenter(A,B,C),dir(180)); dot(circumcenter(D,E,F)); label("$\gamma$",gamma,dir(180)); draw(A--B--F--cycle, black+1); | [] |
476 | Triangle $ABC$ is inscribed in circle $\omega$ with $AB=5$, $BC=7$, and $AC=3$. The bisector of angle $A$ meets side $\overline{BC}$ at $D$ and circle $\omega$ at a second point $E$. Let $\gamma$ be the circle with diameter $\overline{DE}$. Circles $\omega$ and $\gamma$ meet at $E$ and a second point $F$. Then $AF^2 = \frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2012 AIME II Problems/Problem 15 | Use the angle bisector theorem to find $CD=\tfrac{21}{8}$, $BD=\tfrac{35}{8}$, and use Stewart's Theorem to find $AD=\tfrac{15}{8}$. Use Power of Point $D$ to find $DE=\tfrac{49}{8}$, and so $AE=8$. Then use the Extended Law of Sine to find that the length of the circumradius of $\triangle ABC$ is $\tfrac{7\sqrt{3}}{3}$.
Since $DE$ is the diameter of circle $\gamma$, $\angle DFE$ is $90^\circ$. Extending $DF$ to intersect circle $\omega$ at $X$, we find that $XE$ is the diameter of $\omega$ (since $\angle DFE$ is $90^\circ$). Therefore, $XE=\tfrac{14\sqrt{3}}{3}$.
Let $EF=x$, $XD=u$, and $DF=v$. Then $XE^2-XF^2=EF^2=DE^2-DF^2$, so we get
\[(u+v)^2-v^2=\frac{196}{3}-\frac{2401}{64}\]
which simplifies to
\[u^2+2uv = \frac{5341}{192}.\]
By Power of Point $D$, $uv=BD \cdot DC=735/64$. Combining with above, we get
\[XD^2=u^2=\frac{931}{192}.\]
Note that $\triangle XDE\sim \triangle ADF$ and the ratio of similarity is $\rho = AD : XD = \tfrac{15}{8}:u$. Then $AF=\rho\cdot XE = \tfrac{15}{8u}\cdot R$ and \[AF^2 = \frac{225}{64}\cdot \frac{R^2}{u^2} = \frac{900}{19}.\]
The answer is $900+19=\boxed{919}$.
-Solution by TheBoomBox77 | // Block 1
size(175);
defaultpen(fontsize(9pt));
pair A,B,C,D,E,F,W;
B=MP("B",origin,dir(180)); C=MP("C",(7,0),dir(0)); A=MP("A",IP(CR(B,5),CR(C,3)),N); D=MP("D",extension(B,C,A,bisectorpoint(C,A,B)),dir(220)); path omega=circumcircle(A,B,C); E=MP("E",OP(omega,A--(A+20*(D-A))),S); path gamma=CR(midpoint(D--E),length(D-E)/2); F=MP("F",OP(omega,gamma),SE); pair X=MP("X",IP(omega,F--(F+2*(D-F))),N);
draw(omega^^A--B--C--cycle^^gamma); draw(A--E--F--cycle, gray); draw(E--X--F, royalblue);
dot("$W$",circumcenter(A,B,C),dir(180)); dot(circumcenter(D,E,F)); label("$\gamma$",gamma,dir(180)); label("$u$",X--D,dir(60)); label("$v$",D--F,dir(70));
// Block 2
size(175); defaultpen(fontsize(9pt)); pair A,B,C,D,E,F,W; B=MP("B",origin,dir(180)); C=MP("C",(7,0),dir(0)); A=MP("A",IP(CR(B,5),CR(C,3)),N); D=MP("D",extension(B,C,A,bisectorpoint(C,A,B)),dir(220)); path omega=circumcircle(A,B,C); E=MP("E",OP(omega,A--(A+20*(D-A))),S); path gamma=CR(midpoint(D--E),length(D-E)/2); F=MP("F",OP(omega,gamma),SE); pair X=MP("X",IP(omega,F--(F+2*(D-F))),N); draw(omega^^A--B--C--cycle^^gamma); draw(A--E--F--cycle, gray); draw(E--X--F, royalblue); dot("$W$",circumcenter(A,B,C),dir(180)); dot(circumcenter(D,E,F)); label("$\gamma$",gamma,dir(180)); label("$u$",X--D,dir(60)); label("$v$",D--F,dir(70)); | [] |
477 | A circle of radius $1$ is externally tangent to a circle of radius $3$
and both circles are tangent to a line. Find the area of the shaded
region that lies between the two circles and the line. | 2012 UNCO Math Contest II Problems/Problem 7 | If we draw in the second tangent between these circles, we get a point where the tangents intersect. Drawing a line through the centers of the circle and the point where the tangents meet, we get two similar triangles in ratio $1:3$. Let the hypotenuse of the smaller triangle be $x$. Since the distance between the centers of the circles is $1+3=4$, we can write the ratio $\frac{x}{x+4} = \frac{1}{3}$. Solving for $x$, we get $x=2$. Since the hypotenuse of the smaller triangle is $2$, and one of the legs is $1$, we see that it is a $30-60-90$ triangle. So the side lengths of the smaller triangle is $\sqrt{3},1,2$ and the side lengths of the larger triangle is $3\sqrt{3},3,6$. Finding the difference between the areas of both triangles, we get $4\sqrt3$ which is the area of the trapezoid. The trapezoid is the area of a $120^\circ$ sector of the smaller circle, a $60^\circ$ sector of the larger circle, and the shaded region. Subtracting the areas of both sectors from the area of the trapezoid, we get $4\sqrt{3} - \frac{120}{360} \cdot \pi - \frac{60}{360} \cdot 9\pi = 4\sqrt{3} - \frac{1}{3} \cdot \pi - \frac{1}{6} \cdot 9\pi = \boxed {4\sqrt{3} - \frac{11\pi}{6}}$
~Ultraman | // Block 1
draw((-1,-1)--(7,-1),black);
filldraw((0,-1)--(0,0)--(2*sqrt(3),2)--(2*sqrt(3),-1)--cycle,grey);
filldraw(circle((0,0),1),white);
filldraw(circle((2*sqrt(3),2),3),white);
// Block 2
draw((-1,-1)--(7,-1),black); filldraw((0,-1)--(0,0)--(2*sqrt(3),2)--(2*sqrt(3),-1)--cycle,grey); filldraw(circle((0,0),1),white); filldraw(circle((2*sqrt(3),2),3),white); | [] |
478 | In $\triangle ABC$, $AC = BC$, and point $D$ is on $\overline{BC}$ so that $CD = 3\cdot BD$. Let $E$ be the midpoint of $\overline{AD}$. Given that $CE = \sqrt{7}$ and $BE = 3$, the area of $\triangle ABC$ 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 $m+n$. | 2013 AIME II Problems/Problem 13 | Let $X$ be the foot of the altitude from $C$ with other points labelled as shown below.
Now we proceed using mass points. To balance along the segment $BC$, we assign $B$ a mass of $3$ and $C$ a mass of $1$. Therefore, $D$ has a mass of $4$. As $E$ is the midpoint of $AD$, we must assign $A$ a mass of $4$ as well. This gives $L$ a mass of $5$ and $M$ a mass of $7$.
Now let $AB=b$ be the base of the triangle, and let $CX=h$ be the height. Then as $AM:MB=3:4$, and as $AX=\frac{b}{2}$, we know that \[MX=\frac{b}{2}-\frac{3b}{7}=\frac{b}{14}.\] Also, as $CE:EM=7:1$, we know that $EM=\frac{1}{\sqrt{7}}$. Therefore, by the Pythagorean Theorem on $\triangle {XCM}$, we know that \[\frac{b^2}{196}+h^2=\left(\sqrt{7}+\frac{1}{\sqrt{7}}\right)^2=\frac{64}{7}.\]
Also, as $LE:BE=5:3$, we know that $BL=\frac{8}{5}\cdot 3=\frac{24}{5}$. Furthermore, as $\triangle YLA\sim \triangle XCA$, and as $AL:LC=1:4$, we know that $LY=\frac{h}{5}$ and $AY=\frac{b}{10}$, so $YB=\frac{9b}{10}$. Therefore, by the Pythagorean Theorem on $\triangle BLY$, we get \[\frac{81b^2}{100}+\frac{h^2}{25}=\frac{576}{25}.\]
Solving this system of equations yields $b=2\sqrt{7}$ and $h=3$. Therefore, the area of the triangle is $3\sqrt{7}$, giving us an answer of $\boxed{010}$. | // Block 1
size(200);
pair A=(0,0),B=(2*sqrt(7),0),C=(sqrt(7),3),D=(3*B+C)/4,L=C/5,M=3*B/7;
draw(A--B--C--cycle);draw(A--D^^B--L^^C--M);
label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,N);label("$D$",D,NE);label("$L$",L,NW);label("$M$",M,S);
pair X=foot(C,A,B), Y=foot(L,A,B);
pair EE=D/2;
label("$X$",X,S);label("$E$",EE,NW);label("$Y$",Y,S);
draw(C--X^^L--Y,dotted);
draw(rightanglemark(B,X,C)^^rightanglemark(B,Y,L));
// Block 2
size(200); pair A=(0,0),B=(2*sqrt(7),0),C=(sqrt(7),3),D=(3*B+C)/4,L=C/5,M=3*B/7; draw(A--B--C--cycle);draw(A--D^^B--L^^C--M); label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,N);label("$D$",D,NE);label("$L$",L,NW);label("$M$",M,S); pair X=foot(C,A,B), Y=foot(L,A,B); pair EE=D/2; label("$X$",X,S);label("$E$",EE,NW);label("$Y$",Y,S); draw(C--X^^L--Y,dotted); draw(rightanglemark(B,X,C)^^rightanglemark(B,Y,L)); | [] |
478 | In $\triangle ABC$, $AC = BC$, and point $D$ is on $\overline{BC}$ so that $CD = 3\cdot BD$. Let $E$ be the midpoint of $\overline{AD}$. Given that $CE = \sqrt{7}$ and $BE = 3$, the area of $\triangle ABC$ 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 $m+n$. | 2013 AIME II Problems/Problem 13 | Let $BD = x$. Then $CD = 3x$ and $AC = 4x$. Also, let $AE = ED = l$. Using Stewart's Theorem on $\bigtriangleup CEB$ gives us the equation $(x)(3x)(4x) + (4x)(l^2) = 27x + 7x$ or, after simplifying, $4l^2 = 34 - 12x^2$. We use Stewart's again on $\bigtriangleup CAD$: $(l)(l)(2l) + 7(2l) = (16x^2)(l) + (9x^2)(l)$, which becomes $2l^2 = 25x^2 - 14$. Substituting $2l^2 = 17 - 6x^2$, we see that $31x^2 = 31$, or $x = 1$. Then $l^2 = \frac{11}{2}$.
We now use Law of Cosines on $\bigtriangleup CAD$. $(2l)^2 = (4x)^2 + (3x)^2 - 2(4x)(3x)\cos C$. Plugging in for $x$ and $l$, $22 = 16 + 9 - 2(4)(3)\cos C$, so $\cos C = \frac{1}{8}$. Using the Pythagorean trig identity $\sin^2 + \cos^2 = 1$, $\sin^2 C = 1 - \frac{1}{64}$, so $\sin C = \frac{3\sqrt{7}}{8}$.
$[ABC] = \frac{1}{2} AC \cdot BC \sin C = (\frac{1}{2})(4)(4)(\frac{3\sqrt{7}}{8}) = 3\sqrt{7}$, and our answer is $3 + 7 = \boxed{010}$.
Note to writter: Couldn't we just use Heron's formula for $[CEB]$ after $x$ is solved then noticing that $[ABC] = 2 \times [CEB]$? | // Block 1
size(200);
pair A=(0,0),B=(2*sqrt(7),0),C=(sqrt(7),3),D=(3*B+C)/4,L=C/5,M=3*B/7;
draw(A--B--C--cycle);draw(A--D^^B--L^^C--M);
label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,N);label("$D$",D,NE);label("$L$",L,NW);label("$M$",M,S);
pair EE=D/2;
label("$\sqrt{7}$", C--EE, W); label("$x$", D--B, E); label("$3x$", C--D, E); label("$l$", EE--D, N); label("$3$", EE--B, N);
label("$E$",EE,NW);
// Block 2
size(200); pair A=(0,0),B=(2*sqrt(7),0),C=(sqrt(7),3),D=(3*B+C)/4,L=C/5,M=3*B/7; draw(A--B--C--cycle);draw(A--D^^B--L^^C--M); label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,N);label("$D$",D,NE);label("$L$",L,NW);label("$M$",M,S); pair EE=D/2; label("$\sqrt{7}$", C--EE, W); label("$x$", D--B, E); label("$3x$", C--D, E); label("$l$", EE--D, N); label("$3$", EE--B, N); label("$E$",EE,NW); | [] |
478 | In $\triangle ABC$, $AC = BC$, and point $D$ is on $\overline{BC}$ so that $CD = 3\cdot BD$. Let $E$ be the midpoint of $\overline{AD}$. Given that $CE = \sqrt{7}$ and $BE = 3$, the area of $\triangle ABC$ 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 $m+n$. | 2013 AIME II Problems/Problem 13 | Let $AB = 2x$ and let $y = BD.$ Then $CD = 3y$ and $AC = 4y.$
By the Law of Cosines on triangle $ABC,$
\[\cos C = \frac{16y^2 + 16y^2 - 4x^2}{2 \cdot 4y \cdot 4y} = \frac{32y^2 - 4x^2}{32y^2} = \frac{8y^2 - x^2}{8y^2}.\]Then by the Law of Cosines on triangle $ACD,$
\begin{align*}
AD^2 &= 16y^2 + 9y^2 - 2 \cdot 4y \cdot 3y \cdot \cos C \\
&= 25y^2 - 24y^2 \cdot \frac{8y^2 - x^2}{8y^2} \\
&= 3x^2 + y^2.
\end{align*}Applying Stewart's Theorem to median $\overline{BE}$ in triangle $ABD,$ we get
\[BE^2 + AE \cdot DE = \frac{AB^2 + BD^2}{2}.\]Thus,
\[9 + \frac{3x^2 + y^2}{4} = \frac{4x^2 + y^2}{2}.\]This simplifies to $5x^2 + y^2 = 36.$
Applying Stewart's Theorem to median $\overline{CE}$ in triangle $ACD,$ we get
\[CE^2 + AE \cdot DE = \frac{AC^2 + CD^2}{2}.\]Thus,
\[7 + \frac{3x^2 + y^2}{4} = \frac{16y^2 + 9y^2}{2}.\]This simplifies to $3x^2 + 28 = 49y^2.$
Solving the system $5x^2 + y^2 = 36$ and $3x^2 + 28 = 49y^2,$ we find $x^2 = 7$ and $y^2 = 1,$ so $x = \sqrt{7}$ and $y = 1.$
Plugging this back in for our equation for $\cos C$ gives us $\frac{1}{8}$, so $\sin C = \frac{3\sqrt{7}}{8}.$ We can apply the alternative area of a triangle formula, where $AC \cdot BC \cdot \sin C \cdot \frac{1}{2} = 3\sqrt{7}.$ Therefore, our answer is $\boxed{010}$. | // Block 1
unitsize(1.5 cm);
pair A, B, C, D, E;
A = (-sqrt(7),0);
B = (sqrt(7),0);
C = (0,3);
D = interp(B,C,1/4);
E = (A + D)/2;
draw(A--B--C--cycle);
draw(A--D);
draw(B--E--C);
label("$A$", A, SW);
label("$B$", B, SE);
label("$C$", C, N);
label("$D$", D, NE);
label("$E$", E, NW);
label("$2x$", (A + B)/2, S);
label("$y$", (B + D)/2, NE);
label("$3y$", (C + D)/2, NE);
label("$4y$", (A + C)/2, NW);
label("$3$", (B + E)/2, N);
label("$\sqrt{7}$", (C + E)/2, W);
// Block 2
unitsize(1.5 cm); pair A, B, C, D, E; A = (-sqrt(7),0); B = (sqrt(7),0); C = (0,3); D = interp(B,C,1/4); E = (A + D)/2; draw(A--B--C--cycle); draw(A--D); draw(B--E--C); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, N); label("$D$", D, NE); label("$E$", E, NW); label("$2x$", (A + B)/2, S); label("$y$", (B + D)/2, NE); label("$3y$", (C + D)/2, NE); label("$4y$", (A + C)/2, NW); label("$3$", (B + E)/2, N); label("$\sqrt{7}$", (C + E)/2, W); | [] |
479 | In acute triangle $ABC$, the orthocenter $H$ lies on the line connecting the midpoint of segment $AB$ to the midpoint of segment $BC$. If $AC=24$, and the altitude from $B$ has length $14$, find $AB\cdot BC$. | 2013 Mock AIME I Problems/Problem 12 | Toss on the coordinate plane with $A=(0,0)$, $B=(t,14)$, and $C=(24,0)$, where $t$ is a real number and $0<t<24$.
Then, the line connecting the midpoints of $AB$ and $BC$ runs from $\left(\frac{t}{2}, 7\right)$ to $\left(\frac{24+t}{2}, 7\right)$, or more simply the line $y=7$.
The orthocenter of $\triangle ABC$ will be at the intersection of the altitudes from $A$ and $B$.
The slope of the altitude from $A$ is the negative reciprocal of the slope of $\overline{BC}$. The slope of $\overline{BC}$ is $-\frac{14}{24-t}$, and its negative reciprocal is $\frac{24-t}{14}$. Since the altitude from $A$ passes through the origin, its equation is $y=\frac{24-t}{14}x$.
The altitude from $B$ is the vertical line running through $B=(t, 14)$ which has equation $x=t$.
Thus the lines $x=t$ and $y=\frac{24-t}{14}x$ meet on the line $y=7$. Substituting the first equation into the second, $\frac{t(24-t)}{14}=7$.
Multiplying both sides by $14$, we have $t(24-t)=98$.
This rearranges to the quadratic $t^{2}-24t+98=0$, and completing the square by adding $46$ to each side gives us $(t-12)^{2}=46$. Thus $t=12\pm\sqrt{46}$.
The cases where $t=12-\sqrt{46}$ and $t=12+\sqrt{46}$ are similar; they merely correspond to two triangles that can each be obtained by reflecting the other across the perpendicular bisector of $\overline{AC}$, so we consider the case where $t=12-\sqrt{46}$.
So $A=(0,0),B=(12-\sqrt{46},14),C=(24,0)$.
Thus \[AB\cdot BC=\sqrt{14^{2}+\left(12-\sqrt{46}\right)^{2}}\sqrt{14^{2}+\left(12+\sqrt{46}\right)^{2}}=\sqrt{\left(386-24\sqrt{46}\right)\left(386+24\sqrt{46}\right)}=\sqrt{386^{2}-46\cdot 24^{2}}=\boxed{350}.\]
The cases where $t=12-\sqrt{46}$ and $t=12+\sqrt{46}$ are shown below, labeled $\triangle ABC$ and $\triangle AB^{\prime}C$, respectively, where the dotted line is a midline in both triangles. As you can see, the orthocenter falls perfectly on that line for both triangles, and the value of $AB\cdot BC$ is the same for both triangles. | // Block 1
real t = 12 - sqrt(46);
pair A = (0, 0);
pair B = (t, 14);
pair C = (24, 0);
draw(A--B--C--cycle);
draw((A+B)/2--(B+C)/2, dashed);
draw(B--(t, 0), blue);
draw(A--(8.572, 11.55), red);
draw(C--(2.927, 7.854), green);
label("$A$", A, W);
label("$B$", B, N);
label("$C$", C, E);
real t = 12 + sqrt(46);
pair A = (0, 0);
pair B = (t, 14);
pair C = (24, 0);
draw(A--B--C--cycle);
draw((A+B)/2--(B+C)/2, dashed);
draw(B--(t, 0), blue);
draw(A--(21.073, 7.854), red);
draw(C--(15.428, 11.55), green);
label("$A$", A, W);
label("$B^{\prime}$", B, N);
label("$C$", C, E);
// Block 2
real t = 12 - sqrt(46); pair A = (0, 0); pair B = (t, 14); pair C = (24, 0); draw(A--B--C--cycle); draw((A+B)/2--(B+C)/2, dashed); draw(B--(t, 0), blue); draw(A--(8.572, 11.55), red); draw(C--(2.927, 7.854), green); label("$A$", A, W); label("$B$", B, N); label("$C$", C, E); real t = 12 + sqrt(46); pair A = (0, 0); pair B = (t, 14); pair C = (24, 0); draw(A--B--C--cycle); draw((A+B)/2--(B+C)/2, dashed); draw(B--(t, 0), blue); draw(A--(21.073, 7.854), red); draw(C--(15.428, 11.55), green); label("$A$", A, W); label("$B^{\prime}$", B, N); label("$C$", C, E); | [] |
479 | In acute triangle $ABC$, the orthocenter $H$ lies on the line connecting the midpoint of segment $AB$ to the midpoint of segment $BC$. If $AC=24$, and the altitude from $B$ has length $14$, find $AB\cdot BC$. | 2013 Mock AIME I Problems/Problem 12 | Let $M$ be the midpoint of $\overline{AB}$ and $N$ be the midpoint of $\overline{BC}$. Further let $D$ be the foot of the altitude from $A$, $E$ from $B$, and $F$ from $C$, as in the diagram.
Because $\overline{MN}$ is a midpoint connector of $\triangle ABC$ amd $H$ is on $\overline{MN}$ and $\overline{BE}$, we know that $H$ is the midpoint of altitude $BE$. Thus, because, from the problem, $BE=14$, $BH=HE=7$. Now we see that $\overline{MH}$ is a midpoint connector of $\triangle BAE$, so $MH=\tfrac12 AE$.
Now, let $\measuredangle CHE = \theta$. We know that $\measuredangle BHF = \measuredangle CHE = \theta$, because they are vertical angles. Because $\triangle BHF$ is right (by the definition of an altitude), we know that $\measuredangle MBH = 90^{\circ}-\theta$. $\triangle BHM$ is also right, so $\measuredangle BMH = 90^{\circ} -(90^{\circ}-\theta)=\theta$.
From $\triangle HEC$, we know that $\tan\theta = \tfrac{CE}7$. From $\triangle BHM$, we know that $\tan\theta = \tfrac7{AE/2} = \tfrac{14}{AE}$. Equating these two expressions for $\tan\theta$, we see that $AE \cdot CE = 7 \cdot 14 = 98$. From the problem, we know that $AE+CE=AC=24$.
Now, we can proceed as in Solution 1 by using the quadratic formula to solve for $AE$ and the Pythagorean Theorem to find $AB$ and $BC$. We do this to obtain our answer $AB \cdot BC = \boxed{350}$. | // Block 1
import geometry;
point A = origin;
point B = (12-sqrt(46),14);
point C = (24,0);
point M = midpoint(A--B);
point N = midpoint(B--C);
triangle t = triangle(A,B,C);
point H = orthocentercenter(t);
point D, E, F;
line a = altitude(t.BC);
line b = altitude(t.AC);
line c = altitude(t.AB);
// Triangle ABC and Segment MN
draw(t);
draw(M--N);
// Altitudes
pair[] e = intersectionpoints(b,A--C);
E = e[0];
draw(B--E);
pair[] d = intersectionpoints(a,B--C);
D = d[0];
draw(A--D);
pair[] f = intersectionpoints(c,A--B);
F = f[0];
draw(C--F);
// Labeling Points
dot(A);
label("A",A,SW);
dot(B);
label("B",B,NW);
dot(C);
label("C",C,SE);
dot(M);
label("M",M,WSW);
dot(N);
label("N",N,ENE);
dot(H);
label("H",H,SSE);
dot(D);
label("D",D,NE);
dot(E);
label("E",E,S);
dot(F);
label("F",F,NW);
// Right angle marks
markscalefactor = 0.15;
draw(rightanglemark(B,H,M));
draw(rightanglemark(B,D,C));
// Block 2
import geometry; point A = origin; point B = (12-sqrt(46),14); point C = (24,0); point M = midpoint(A--B); point N = midpoint(B--C); triangle t = triangle(A,B,C); point H = orthocentercenter(t); point D, E, F; line a = altitude(t.BC); line b = altitude(t.AC); line c = altitude(t.AB); // Triangle ABC and Segment MN draw(t); draw(M--N); // Altitudes pair[] e = intersectionpoints(b,A--C); E = e[0]; draw(B--E); pair[] d = intersectionpoints(a,B--C); D = d[0]; draw(A--D); pair[] f = intersectionpoints(c,A--B); F = f[0]; draw(C--F); // Labeling Points dot(A); label("A",A,SW); dot(B); label("B",B,NW); dot(C); label("C",C,SE); dot(M); label("M",M,WSW); dot(N); label("N",N,ENE); dot(H); label("H",H,SSE); dot(D); label("D",D,NE); dot(E); label("E",E,S); dot(F); label("F",F,NW); // Right angle marks markscalefactor = 0.15; draw(rightanglemark(B,H,M)); draw(rightanglemark(B,D,C)); | [] |
480 | In acute $\triangle ABC$, $H$ is the orthocenter, $G$ is the centroid, and $M$ is the midpoint of $BC$. It is obvious that $AM \ge GM$, but $GM \ge HM$ does not always hold. If $[ABC] = 162$, $BC=18$, then the value of $GM$ which produces the smallest value of $AB$ such that $GM \ge HM$ can be expressed in the form $a+b\sqrt{c}$, for $b$ squarefree. Compute $a+b+c$. | 2013 Mock AIME I Problems/Problem 13 | Because $[\triangle ABC] = 168$ and $BC=18$, we know that the height from $A$ to $BC$ must be $18$. Thus, because the perpendicular is the shortest segment from a line to a point not on the line, we know that $AB \geq 18$. Thus, the minimum value of $AB$ is $18$, when $\overline{AB} \perp \overline{BC}$. This is technically not possible, because $\triangle ABC$ is acute, but it may be helpful in getting a better sense of the problem. This scenario is shown below:
Because the orthocenter of $\triangle ABC$ is at $B$, $HM=BM=9$. Now, by the Pythagorean Theorem, $AM=9\sqrt5$. Because the centrod is $\tfrac23$ of the way along the median from the vertex, we know that $GM=\tfrac{9\sqrt5}3=3\sqrt5<9$. Thus, we have $GM<HM$, so the given inequality does not hold. Now, let us look at the example where $A$ is collinear with $O$ and $M$:
By centroid properties, we know that $AG=\tfrac23 AM = \tfrac23 \cdot 18 = 12$. Let $R$ be the circumradius of $\triangle ABC$. Then, by Pythagoras in $\triangle OBM$, $OM=\sqrt{R^2-81}$. Because $OA=R$ and $AM=18$, we have the equation $\sqrt{R^2-81}+R=18$, which yields $AO=R=\tfrac{45}4<12=AG$, so $G$ is between $O$ and $M$. Because $\triangle ABC$ is acute, we know that $H$ is in the interior of $\triangle ABC$. This fact, combined with the properties of the Euler Line, show that $H$ must be closer to $M$ than $G$ is, so $GM>HM$, and the inequality is thereby satisfied.
As in the above diagram, let $C^{\prime}$ be the point on the circle such that $\overline{CC^{\prime}}$ is a diameter of the circle. Because $\triangle ABC$ is acute, $A$ and $B$ must be on the opposite sides of $\overline{CC^{\prime}}$ (so that the circumcenter lies inside the triangle). We want $A$ to be as close to $C^{\prime}$ as possible to minimize $AB$, but, from the first example we explored, we know that when $A=C^{\prime}$, $HM>GM$. We would reasonably expect the difference $GM-HM$ to vary continuously as we move $A$ towards $C^{\prime}$, and this difference is positive in the second example and negative in the first example. Thus, by the Intermediate Value Theorem, there should be a point on the circumcircle between these two locations for $A$ such that this difference is zero, or $GM=HM$. This point should be as close as we can get to $C^{\prime}$ while still satisfying the inequality.
Now, let $GM=HM$ and $D$ be the foot of the altitude from $A$ to $\overline{BC}$. Further, let $MD=x$, so $BD=9-x$, as shown below:
By Pythagoras, we know that $AM = \sqrt{AD^2+DM^2} = \sqrt{18^2+x^2} = \sqrt{x^2+324}$. Thus, by centroid properties, $GM = \tfrac13 AM = \tfrac{\sqrt{x^2+324}}3$. Now, we desire to find another expression for $GM=HM$. By using Pythagoras again, we see that $AB = \sqrt{(9-x)^2+18^2} = \sqrt{x^2-18x+405}$ and $AC = \sqrt{(9+x)^2+18^2} = \sqrt{x^2+18x+405}$. Now, let $\measuredangle BAC = \alpha$. Also let $AC=b$ and $AB=c$. By the Law of Cosines in $\triangle ABC$, we have the following equation:
\begin{align*}
BC^2 &= AB^2 + AC^2 - 2AB \cdot AC \cos\alpha \\
18^2 &= (x^2+18x+405)+(x^2-18x+405)-2bc\cos\alpha \\
\frac{324}2 &= x^2+405-bc\cos\alpha \\
bc\cos\alpha &= x^2+405-162 \\
\cos\alpha &= \frac{x^2+243}{bc}
\end{align*}
By rearranging the formula for the area of a triangle $\tfrac{abc}{4R}$ and recalling that, from the problem, $[\triangle ABC] = 162$, we see that $R=\tfrac{abc}{4 \cdot 162}$. Because $BC=18$, this expression equates to $\tfrac{18bc}{4 \cdot 162}=\tfrac{bc}{36}$. By the formula for the distance from a vertex to the orthocenter and substitution, we know that $AH=2R\cos\alpha=2(\tfrac{bc}{36})(\tfrac{x^2+243}{bc})=\tfrac{x^2+243}{18}$. Thus, because $AD=18$, $HD=18-\tfrac{x^2+243}{18}=\tfrac{81-x^2}{18}$. By Pythagoras in $\triangle HDM$, $HM = \sqrt{\tfrac{(81-x^2)^2}{324}+x^2} = \tfrac{\sqrt{81^2-162x^2+x^4+324x^2}}{18} = \tfrac{\sqrt{x^4+162x^2+81^2}}{18} = \tfrac{x^2+81}{18}$. Equating this with our earlier expression for $GM$, we get the following equation:
\begin{align*}
\frac{x^2+81}{18} &= \frac{\sqrt{x^2+324}}3 \\
\frac{x^2+81}6 &= \sqrt{x^2+324} \\
\frac{x^4+162x^2+81^2}{36} &= x^2+18^2 \\
x^4+162x^2+81^2 &= 36x^2+6^2 \cdot 18^2 \\
x^4+126x^2+81^2-108^2 &= 0 \\
x^4+126x^2-(108-81)(108+81) &= 0 \\
x^4+126x^2-27 \cdot 189 &= 0 \\
x^2 &= \frac{-126 \pm \sqrt{126^2+4\cdot27\cdot189}}2 \\
x^2 &= -63 \pm 36\sqrt7
\end{align*}
Because $x^2>0$, $x^2=-63+36\sqrt7$. Plugging this into $GM=HM=\tfrac{x^2+81}{18}$ yields $\tfrac{-63+36\sqrt7+81}{18}=\tfrac{18+36\sqrt7}{18}=1+2\sqrt7$. Thus, our answer is $1+2+7=\boxed{010}$. | // Block 1
import geometry;
point A = origin;
point B = (0,18);
point C = (18,18);
triangle t = triangle(A,B,C);
point M = midpoint(B--C);
circle c = circumcircle(t);
point O = circumcenter(t);
point G = centroid(t);
// Triangle and Circumcircle
draw(t);
draw(c);
// Median AM
draw(A--M);
// Labelling Points
dot(A);
label("A",A,SW);
dot(B);
label("H=B",B,NW);
dot(C);
label("C",C,NE);
dot(O);
label("O",O,SE);
dot(M);
label("M",M,N);
dot(G);
label("G",G,WSW);
// Length Labels
label("$9$",midpoint(B--M),N);
label("$9$",midpoint(M--C),N);
label("$18$",midpoint(A--B),W);
// Right Angle Mark
markscalefactor = 0.18;
draw(rightanglemark(A,B,C));
// Block 2
import geometry;
point A = origin;
point B = (-9,18);
point C = (9,18);
triangle t = triangle(A,B,C);
point M = midpoint(B--C);
circle c = circumcircle(t);
point O = circumcenter(t);
point G = centroid(t);
point H = orthocentercenter(t);
point Cp;
// Triangle and Circumcircle
draw(t);
draw(c);
// Median AM, Segment OB
draw(A--M);
draw(O--B);
// Defining C'
pair[] cp = intersectionpoints(line(O,C),c);
Cp = cp[0];
// Labelling Points
dot(A);
label("A",A,S);
dot(B);
label("B",B,NW);
dot(C);
label("C",C,NE);
dot(O);
label("O",O,SW);
dot(M);
label("M",M,N);
dot(G);
label("G",G,E);
dot(H);
label("H",H,ENE);
dot(Cp);
label("C$^{\prime}$",Cp,SW);
// Length Labels
label("$9$",midpoint(B--M),N);
label("$9$",midpoint(M--C),N);
// Right Angle Mark
markscalefactor = 0.18;
draw(rightanglemark(A,M,B));
// Block 3
import geometry;
point A = (-5.678648, 1.454964);
point B = (-9,18);
point C = (9,18);
triangle t = triangle(A,B,C);
line a = altitude(t.BC);
point D;
point M = midpoint(B--C);
circle c = circumcircle(t);
point O = circumcenter(t);
point G = centroid(t);
point H = orthocentercenter(t);
// Triangle and Circumcircle
draw(t);
draw(c);
// Median AM, Segment HM
draw(A--M);
draw(H--M);
// Altitude AD
pair[] d = intersectionpoints(a,B--C);
D = d[0];
draw(A--D);
// Labelling Points
dot(A);
label("A",A,S);
dot(B);
label("B",B,NW);
dot(C);
label("C",C,NE);
dot(O);
label("O",O,S);
dot(M);
label("M",M,N);
dot(G);
label("G",G,WSW);
dot(H);
label("H",H,ESE);
dot(D);
label("D",D,N);
// Length Labels
label("$x$",midpoint(D--M),N);
label("$9$",midpoint(M--C),N);
// Right Angle Mark
markscalefactor = 0.15;
draw(rightanglemark(A,D,B));
// Block 4
import geometry; point A = origin; point B = (0,18); point C = (18,18); triangle t = triangle(A,B,C); point M = midpoint(B--C); circle c = circumcircle(t); point O = circumcenter(t); point G = centroid(t); // Triangle and Circumcircle draw(t); draw(c); // Median AM draw(A--M); // Labelling Points dot(A); label("A",A,SW); dot(B); label("H=B",B,NW); dot(C); label("C",C,NE); dot(O); label("O",O,SE); dot(M); label("M",M,N); dot(G); label("G",G,WSW); // Length Labels label("$9$",midpoint(B--M),N); label("$9$",midpoint(M--C),N); label("$18$",midpoint(A--B),W); // Right Angle Mark markscalefactor = 0.18; draw(rightanglemark(A,B,C));
// Block 5
import geometry; point A = origin; point B = (-9,18); point C = (9,18); triangle t = triangle(A,B,C); point M = midpoint(B--C); circle c = circumcircle(t); point O = circumcenter(t); point G = centroid(t); point H = orthocentercenter(t); point Cp; // Triangle and Circumcircle draw(t); draw(c); // Median AM, Segment OB draw(A--M); draw(O--B); // Defining C' pair[] cp = intersectionpoints(line(O,C),c); Cp = cp[0]; // Labelling Points dot(A); label("A",A,S); dot(B); label("B",B,NW); dot(C); label("C",C,NE); dot(O); label("O",O,SW); dot(M); label("M",M,N); dot(G); label("G",G,E); dot(H); label("H",H,ENE); dot(Cp); label("C$^{\prime}$",Cp,SW); // Length Labels label("$9$",midpoint(B--M),N); label("$9$",midpoint(M--C),N); // Right Angle Mark markscalefactor = 0.18; draw(rightanglemark(A,M,B));
// Block 6
import geometry; point A = (-5.678648, 1.454964); point B = (-9,18); point C = (9,18); triangle t = triangle(A,B,C); line a = altitude(t.BC); point D; point M = midpoint(B--C); circle c = circumcircle(t); point O = circumcenter(t); point G = centroid(t); point H = orthocentercenter(t); // Triangle and Circumcircle draw(t); draw(c); // Median AM, Segment HM draw(A--M); draw(H--M); // Altitude AD pair[] d = intersectionpoints(a,B--C); D = d[0]; draw(A--D); // Labelling Points dot(A); label("A",A,S); dot(B); label("B",B,NW); dot(C); label("C",C,NE); dot(O); label("O",O,S); dot(M); label("M",M,N); dot(G); label("G",G,WSW); dot(H); label("H",H,ESE); dot(D); label("D",D,N); // Length Labels label("$x$",midpoint(D--M),N); label("$9$",midpoint(M--C),N); // Right Angle Mark markscalefactor = 0.15; draw(rightanglemark(A,D,B)); | [] |
481 | In quadrilateral $ABCD$, $AC\cap BD=M$. Also, $MA=6, MB=8, MC=4, MD=3$, and $BC=2CD$. The perimeter of $ABCD$ can be expressed in the form $\frac{p\sqrt{q}}{r}$ where $p$ and $r$ are relatively prime, and $q$ is not divisible by the square of any prime number. Find $p+q+r$. | 2013 Mock AIME I Problems/Problem 5 | Let $CD=x$, as in the diagram. Thus, from the problem, $BC=2x$. Because $AM \cdot MC = DM \cdot MB = 24$, by Power of a Point, we know that $ABCD$ is cyclic. Thus, we know that $\measuredangle DAC = \measuredangle DBC$, so, by the congruency of vertical angles and subsequently AA Similarity, we know that $\triangle AMD \sim \triangle BMC$. Thus, we have the proportion $\tfrac{AM}{AD} = \tfrac{BM}{BC}$, or, by substitution, $\tfrac6{AD}=\tfrac8{2x}$. Solving this equation for $AD$ yields $AD=\tfrac3 2 x$. Similarly, we know that $\measuredangle ABD = \measuredangle ACD$, so, like before, we can see that $\triangle AMB \sim \triangle DMC$. Thus, we have the proportion $\tfrac{AM}{AB} = \tfrac{DM}{DC}$, or, by substitution, $\tfrac6{AB} = \tfrac3 x$. Solving for $AB$ yields $AB=2x$.
Now, we can use Ptolemy's Theorem on cyclic $ABCD$ and solve for $x$:
\begin{align*}
x \cdot 2x + 2x \cdot \frac3 2 x &= (6+4)(8+3) \\
5x^2 &= 110 \\
x^2 &= 22 \\
x &= \pm \sqrt{22}
\end{align*}
Because $x>0$, $x=\sqrt{22}$. Thus, the perimeter of $ABCD$ is $2x+2x+\tfrac3 2 x+x = \tfrac{13}2 x = \tfrac{13\sqrt{22}}2$. Thus, $p+q+r=13+22+2=\boxed{037}$. | // Block 1
import geometry;
// Defining Points
point O = origin;
point B = (1,0);
point A = dir(115.583);
point C = dir(-115.583);
point D = dir(-165.638);
point M;
// Circle
draw(circle(O, 1));
// Quadrilateral and Diagonals
draw(A--B--C--D--cycle);
draw(A--C);
draw(B--D);
// Defining M
pair[] m = intersectionpoints((A--C),(B--D));
M = m[0];
// Labelling Points
dot(A);
label("A",A,NW);
dot(B);
label("B",B,E);
dot(C);
label("C",C,SW);
dot(D);
label("D",D,WSW);
dot(M);
label("M",M,NE);
// Length Labels
label("$3$", midpoint(D--M), NNW);
label("$8$", midpoint(M--B), NNW);
label("$6$", midpoint(A--M), E);
label("$4$", midpoint(C--M), E);
label("$2x$", midpoint(B--C), SE);
label("$x$", midpoint(C--D), NE);
// Block 2
import geometry; // Defining Points point O = origin; point B = (1,0); point A = dir(115.583); point C = dir(-115.583); point D = dir(-165.638); point M; // Circle draw(circle(O, 1)); // Quadrilateral and Diagonals draw(A--B--C--D--cycle); draw(A--C); draw(B--D); // Defining M pair[] m = intersectionpoints((A--C),(B--D)); M = m[0]; // Labelling Points dot(A); label("A",A,NW); dot(B); label("B",B,E); dot(C); label("C",C,SW); dot(D); label("D",D,WSW); dot(M); label("M",M,NE); // Length Labels label("$3$", midpoint(D--M), NNW); label("$8$", midpoint(M--B), NNW); label("$6$", midpoint(A--M), E); label("$4$", midpoint(C--M), E); label("$2x$", midpoint(B--C), SE); label("$x$", midpoint(C--D), NE); | [] |
482 | In $\triangle ABC$, $AB = 3$, $BC = 4$, and $CA = 5$. Circle $\omega$ intersects $\overline{AB}$ at $E$ and $B$, $\overline{BC}$ at $B$ and $D$, and $\overline{AC}$ at $F$ and $G$. Given that $EF=DF$ and $\frac{DG}{EG} = \frac{3}{4}$, length $DE=\frac{a\sqrt{b}}{c}$, where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer not divisible by the square of any prime. Find $a+b+c$. | 2014 AIME I Problems/Problem 15 | First we note that $\triangle DEF$ is an isosceles right triangle with hypotenuse $\overline{DE}$ the same as the diameter of $\omega$. We also note that $\triangle DGE \sim \triangle ABC$ since $\angle EGD$ is a right angle and the ratios of the sides are $3:4:5$.
From congruent arc intersections, we know that $\angle GED \cong \angle GBC$, and that from similar triangles $\angle GED$ is also congruent to $\angle GCB$. Thus, $\triangle BGC$ is an isosceles triangle with $BG = GC$, so $G$ is the midpoint of $\overline{AC}$ and $AG = GC = 5/2$. Similarly, we can find from angle chasing that $\angle ABF = \angle EDF = \frac{\pi}4$. Therefore, $\overline{BF}$ is the angle bisector of $\angle B$. From the angle bisector theorem, we have $\frac{AF}{AB} = \frac{CF}{CB}$, so $AF = 15/7$ and $CF = 20/7$.
Lastly, we apply power of a point from points $A$ and $C$ with respect to $\omega$ and have $AE \times AB=AF \times AG$ and $CD \times CB=CG \times CF$, so we can compute that $EB = \frac{17}{14}$ and $DB = \frac{31}{14}$. From the Pythagorean Theorem, we result in $DE = \frac{25 \sqrt{2}}{14}$, so $a+b+c=25+2+14= \boxed{041}$
Also: $FG=\frac{20}{7}-\frac{5}{2}=\frac{5}{2}-\frac{15}{7}=\frac{5}{14}$. We can also use Ptolemy's Theorem on quadrilateral $DEFG$ to figure what $FG$ is in terms of $d$:
\[DE\cdot FG+DG\cdot EF=DF\cdot EG\]
\[d\cdot FG+\frac{3d}{5}\cdot \frac{d}{\sqrt{2}}=\frac{4d}{5}\cdot \frac{d}{\sqrt{2}}\]
\[d\cdot FG+\frac{3d^2}{5\sqrt{2}}=\frac{4d^2}{5\sqrt{2}}\implies FG=\frac{d}{5\sqrt{2}}\]
Thus $\frac{d}{5\sqrt{2}}=\frac{5}{14}\rightarrow d=5\sqrt{2}\cdot\frac{5}{14}=\frac{25\sqrt{2}}{14}$. $a+b+c=25+2+14= \boxed{041}$
Solution 3
Call $DE=x$ and as a result $DF=EF=\frac{x\sqrt{2}}{2}, EG=\frac{4x}{5}, GD=\frac{3x}{5}$. Since $EFGD$ is cyclic we just need to get $DG$ and using LoS(for more detail see the $2$nd paragraph of Solution $2$) we get $AG=\frac{5}{2}$ and using a similar argument(use LoS again) and subtracting you get $FG=\frac{5}{14}$ so you can use Ptolemy to get $x=\frac{25\sqrt{2}}{14} \implies \boxed{041}$.
~First | // Block 1
pair A = (0,3);
pair B = (0,0);
pair C = (4,0);
draw(A--B--C--cycle);
dotfactor = 3;
dot("$A$",A,dir(135));
dot("$B$",B,dir(215));
dot("$C$",C,dir(305));
pair D = (2.21, 0);
pair E = (0, 1.21);
pair F = (1.71, 1.71);
pair G = (2, 1.5);
dot("$D$",D,dir(270));
dot("$E$",E,dir(180));
dot("$F$",F,dir(90));
dot("$G$",G,dir(0));
draw(Circle((1.109, 0.609), 1.28));
draw(D--E);
draw(E--F);
draw(D--F);
draw(E--G);
draw(D--G);
draw(B--F);
draw(B--G);
// Block 2
pair A = (0,3); pair B = (0,0); pair C = (4,0); draw(A--B--C--cycle); dotfactor = 3; dot("$A$",A,dir(135)); dot("$B$",B,dir(215)); dot("$C$",C,dir(305)); pair D = (2.21, 0); pair E = (0, 1.21); pair F = (1.71, 1.71); pair G = (2, 1.5); dot("$D$",D,dir(270)); dot("$E$",E,dir(180)); dot("$F$",F,dir(90)); dot("$G$",G,dir(0)); draw(Circle((1.109, 0.609), 1.28)); draw(D--E); draw(E--F); draw(D--F); draw(E--G); draw(D--G); draw(B--F); draw(B--G); | [] |
483 | In $\triangle RED$, $\measuredangle DRE=75^{\circ}$ and $\measuredangle RED=45^{\circ}$. $RD=1$. Let $M$ be the midpoint of segment $\overline{RD}$. Point $C$ lies on side $\overline{ED}$ such that $\overline{RC}\perp\overline{EM}$. Extend segment $\overline{DE}$ through $E$ to point $A$ such that $CA=AR$. Then $AE=\frac{a-\sqrt{b}}{c}$, where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer. Find $a+b+c$. | 2014 AIME II Problems/Problem 11 | Let $P$ be the foot of the perpendicular from $A$ to $\overline{CR}$, so $\overline{AP}\parallel\overline{EM}$. Since triangle $ARC$ is isosceles, $P$ is the midpoint of $\overline{CR}$, and $\overline{PM}\parallel\overline{CD}$. Thus, $APME$ is a parallelogram and $AE = PM = \frac{CD}{2}$. We can then use coordinates. Let $O$ be the foot of altitude $RO$ and set $O$ as the origin. Now we notice special right triangles! In particular, $DO = \frac{1}{2}$ and $EO = RO = \frac{\sqrt{3}}{2}$, so $D\left(\frac{1}{2}, 0\right)$, $E\left(-\frac{\sqrt{3}}{2}, 0\right)$, and $R\left(0, \frac{\sqrt{3}}{2}\right).$ $M =$ midpoint$(D, R) = \left(\frac{1}{4}, \frac{\sqrt{3}}{4}\right)$ and the slope of $ME = \frac{\frac{\sqrt{3}}{4}}{\frac{1}{4} + \frac{\sqrt{3}}{2}} = \frac{\sqrt{3}}{1 + 2\sqrt{3}}$, so the slope of $RC = -\frac{1 + 2\sqrt{3}}{\sqrt{3}}.$ Instead of finding the equation of the line, we use the definition of slope: for every $CO = x$ to the left, we go $\frac{x(1 + 2\sqrt{3})}{\sqrt{3}} = \frac{\sqrt{3}}{2}$ up. Thus, $x = \frac{\frac{3}{2}}{1 + 2\sqrt{3}} = \frac{3}{4\sqrt{3} + 2} = \frac{3(4\sqrt{3} - 2)}{44} = \frac{6\sqrt{3} - 3}{22}.$ $DC = \frac{1}{2} - x = \frac{1}{2} - \frac{6\sqrt{3} - 3}{22} = \frac{14 - 6\sqrt{3}}{22}$, and $AE = \frac{7 - \sqrt{27}}{22}$, so the answer is $\boxed{056}$. | // Block 1
unitsize(8cm);
pair a, o, d, r, e, m, cm, c,p;
o =(0,0);
d = (0.5, 0);
r = (0,sqrt(3)/2);
e = (-sqrt(3)/2,0);
m = midpoint(d--r);
draw(e--m);
cm = foot(r, e, m);
draw(L(r, cm,1, 1));
c = IP(L(r, cm, 1, 1), e--d);
clip(r--d--e--cycle);
draw(r--d--e--cycle);
draw(rightanglemark(e, cm, c, 1.5));
a = -(4sqrt(3)+9)/11+0.5;
dot(a);
draw(a--r, dashed);
draw(a--c, dashed);
pair[] PPAP = {a, o, d, r, e, m, c};
for(int i = 0; i<7; ++i) {
dot(PPAP[i]);
}
label("$A$", a, W);
label("$E$", e, SW);
label("$C$", c, S);
label("$O$", o, S);
label("$D$", d, SE);
label("$M$", m, NE);
label("$R$", r, N);
p = foot(a, r, c);
label("$P$", p, NE);
draw(p--m, dashed);
draw(a--p, dashed);
dot(p);
// Block 2
unitsize(8cm); pair a, o, d, r, e, m, cm, c,p; o =(0,0); d = (0.5, 0); r = (0,sqrt(3)/2); e = (-sqrt(3)/2,0); m = midpoint(d--r); draw(e--m); cm = foot(r, e, m); draw(L(r, cm,1, 1)); c = IP(L(r, cm, 1, 1), e--d); clip(r--d--e--cycle); draw(r--d--e--cycle); draw(rightanglemark(e, cm, c, 1.5)); a = -(4sqrt(3)+9)/11+0.5; dot(a); draw(a--r, dashed); draw(a--c, dashed); pair[] PPAP = {a, o, d, r, e, m, c}; for(int i = 0; i<7; ++i) { dot(PPAP[i]); } label("$A$", a, W); label("$E$", e, SW); label("$C$", c, S); label("$O$", o, S); label("$D$", d, SE); label("$M$", m, NE); label("$R$", r, N); p = foot(a, r, c); label("$P$", p, NE); draw(p--m, dashed); draw(a--p, dashed); dot(p); | [] |
483 | In $\triangle RED$, $\measuredangle DRE=75^{\circ}$ and $\measuredangle RED=45^{\circ}$. $RD=1$. Let $M$ be the midpoint of segment $\overline{RD}$. Point $C$ lies on side $\overline{ED}$ such that $\overline{RC}\perp\overline{EM}$. Extend segment $\overline{DE}$ through $E$ to point $A$ such that $CA=AR$. Then $AE=\frac{a-\sqrt{b}}{c}$, where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer. Find $a+b+c$. | 2014 AIME II Problems/Problem 11 | Let $P$ be the foot of the perpendicular from $A$ to $\overline{CR}$, so $\overline{AP}\parallel\overline{EM}$. Since $\triangle ARC$ is isosceles, $P$ is the midpoint of $\overline{CR}$, and by midpoint theorem $\overline{PM}\parallel\overline{CD}$. Thus, $APME$ is a parallelogram and therefore $AE = PM = \tfrac 12 CD$.
We can now use coordinates with $D(0,0)$ as origin and $DE$ along the $x$-axis.
Let $RD=4$ instead of $1$ (in the end we will scale down by $4$). Since $\angle D = 60^\circ$, we get $R(2,2\sqrt{3})$, and therefore $M(1, \sqrt{3})$.
We use sine-law in $\triangle RED$ to find the coordinates $E(2+2\sqrt{3}, 0)$:\[DE =4\cdot \frac{\sin 75^\circ}{\sin 45^\circ} = 4(\sin 30^\circ + \cos 30^\circ) = 2+2\sqrt{3}.\]
Since slope$(ME)= -\sqrt{3}/(1+2\sqrt{3})$, and $RC\perp ME$, it follows that slope$(RC)=(1+2\sqrt{3})/\sqrt{3}$. If $C(c,0)$ then we have\[\frac{2\sqrt{3}}{2-c}=\frac{1+2\sqrt{3}}{\sqrt{3}}\qquad \Longrightarrow\qquad c=\frac{4\sqrt{3}-4}{1+2\sqrt{3}}\]
Now $\tfrac 12 CD = \tfrac 12c =(2\sqrt{3}-2)/(1+2\sqrt{3})= \tfrac 1{11}(14-6\sqrt{3})$.
Scaling down by $4$, we get $AE=\tfrac 1{22}(7-3\sqrt{3})$, so our answer is $7+27+22=056$. | // Block 1
unitsize(8cm); pair a, d, r, e, m, cm, c,p;
d=origin; r=dir(60); e=extension(d,left,r,r+dir(75)*(d-r)); m = midpoint(d--r); cm = foot(r, e, m); c=extension(r,cm,d,e); p=midpoint(r--c); a=p+(e-m);
draw(e--m); draw(L(r, cm,1, 1)); clip(r--d--e--cycle); draw(r--d--e--cycle); draw(rightanglemark(e, cm, c, 1.5)); draw(a--r, dashed); draw(a--c, dashed); draw(p--m, dashed); draw(a--p, dashed);
pair[] PPAP = {a, d, r, e, m, c, p};
for(int i = 0; i<7; ++i) { dot(PPAP[i]); }
label("$A$", a, E); label("$E$", e, S); label("$C$", c, S); label("$D$", d, SW); label("$M$", m, NW); label("$R$", r, N); label("$P$", p, NW);
MA("60^\circ",black,c,d,m,0.07, black);
// Block 2
unitsize(8cm); pair a, d, r, e, m, cm, c,p; d=origin; r=dir(60); e=extension(d,left,r,r+dir(75)*(d-r)); m = midpoint(d--r); cm = foot(r, e, m); c=extension(r,cm,d,e); p=midpoint(r--c); a=p+(e-m); draw(e--m); draw(L(r, cm,1, 1)); clip(r--d--e--cycle); draw(r--d--e--cycle); draw(rightanglemark(e, cm, c, 1.5)); draw(a--r, dashed); draw(a--c, dashed); draw(p--m, dashed); draw(a--p, dashed); pair[] PPAP = {a, d, r, e, m, c, p}; for(int i = 0; i<7; ++i) { dot(PPAP[i]); } label("$A$", a, E); label("$E$", e, S); label("$C$", c, S); label("$D$", d, SW); label("$M$", m, NW); label("$R$", r, N); label("$P$", p, NW); MA("60^\circ",black,c,d,m,0.07, black); | [] |
484 | In $\triangle{ABC}, AB=10, \angle{A}=30^\circ$ , and $\angle{C=45^\circ}$. Let $H, D,$ and $M$ be points on the line $BC$ such that $AH\perp{BC}$, $\angle{BAD}=\angle{CAD}$, and $BM=CM$. Point $N$ is the midpoint of the segment $HM$, and point $P$ is on ray $AD$ such that $PN\perp{BC}$. Then $AP^2=\dfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2014 AIME II Problems/Problem 14 | Draw the $45-45-90 \triangle AHC$. Now, take the perpendicular bisector of $BC$ to intersect the circumcircle of $\triangle ABC$ and $AC$ at $F, L, G$ as shown, and denote $O$ to be the circumcenter of $\triangle ABC$. It is not difficult to see by angle chasing that $AHBGO$ is cyclic, namely with diameter $AB$. Then, by symmetry, $EH = HB$ and as $HB, OG$ are both subtended by equal arcs they are equal. Hence, $EH = GO$. Now, draw line $HL$ and intersect it at $AC$ at point $K$ in the diagram. It is not hard to use angle chase to arrive at $AEOL$ a parallelogram, and from our length condition derived earlier, $AL \parallel HG$. From here, it is clear that $AK = KG$; that is, $P$ is just the intersection of the perpendicular from $K$ down to $BC$ and $AD$! After this point, note that $AP = PF$. It is easily derived that the circumradius of $\triangle ABC$ is $\frac{10}{\sqrt{2}}$. Now, $APO$ is a $30-60-90$ triangle, and from here it is easy to arrive at the final answer of $\boxed{077}$. ~awang11's sol | // Block 1
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(15cm);
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 = -8.455641974276588, xmax = 26.731282460265, ymin = -10.92318356252699, ymax = 9.023689834456471; /* image dimensions */
pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882);
draw((-1.4934334172297545,2.6953043701763835)--(-1.459546107520503,-6.96389444957376)--(1.1286284157632023,-6.954814372303504)--(4.651736947776926,-3.406898607850789)--(4.642656870506668,-0.8187240845670819)--cycle, linewidth(2) + rvwvcq);
/* draw figures */
draw((-1.4934334172297545,2.6953043701763835)--(1.1286284157632023,-6.954814372303504), linewidth(2) + wrwrwr);
draw((xmin, -0.9930079421029264*xmin + 1.2123131258653241)--(xmax, -0.9930079421029264*xmax + 1.2123131258653241), linewidth(2) + wrwrwr); /* line */
draw((xmin, 0.0035082940460819836*xmin-6.958773932654766)--(xmax, 0.0035082940460819836*xmax-6.958773932654766), linewidth(2) + wrwrwr); /* line */
draw((xmin, -285.03882139434313*xmin-422.9911967079192)--(xmax, -285.03882139434313*xmax-422.9911967079192), linewidth(2) + wrwrwr); /* line */
draw((xmin, -1.7181023895538718*xmin + 0.12943284739433739)--(xmax, -1.7181023895538718*xmax + 0.12943284739433739), linewidth(2) + wrwrwr); /* line */
draw(circle((4.642656870506668,-0.8187240845670819), 7.071067811865476), linewidth(2) + wrwrwr);
draw((xmin, -285.0388213943529*xmin + 1322.5187184230485)--(xmax, -285.0388213943529*xmax + 1322.5187184230485), linewidth(2) + wrwrwr); /* line */
draw((-1.4934334172297545,2.6953043701763835)--(4.617849638067675,6.252300211899359), linewidth(2) + wrwrwr);
draw((4.617849638067675,6.252300211899359)--(-1.459546107520503,-6.96389444957376), linewidth(2) + wrwrwr);
draw(circle((-0.18240250073327363,-2.12975500106356), 5), linewidth(2) + wrwrwr);
draw((xmin, -285.0388213943432*xmin + 449.7637608575419)--(xmax, -285.0388213943432*xmax + 449.7637608575419), linewidth(2) + wrwrwr); /* line */
draw((1.1286284157632023,-6.954814372303504)--(4.651736947776926,-3.406898607850789), linewidth(2) + wrwrwr);
draw((-1.4934334172297545,2.6953043701763835)--(4.642656870506668,-0.8187240845670819), linewidth(2) + wrwrwr);
draw((-1.4934334172297545,2.6953043701763835)--(-1.459546107520503,-6.96389444957376), linewidth(2) + rvwvcq);
draw((-1.459546107520503,-6.96389444957376)--(1.1286284157632023,-6.954814372303504), linewidth(2) + rvwvcq);
draw((1.1286284157632023,-6.954814372303504)--(4.651736947776926,-3.406898607850789), linewidth(2) + rvwvcq);
draw((4.651736947776926,-3.406898607850789)--(4.642656870506668,-0.8187240845670819), linewidth(2) + rvwvcq);
draw((4.642656870506668,-0.8187240845670819)--(-1.4934334172297545,2.6953043701763835), linewidth(2) + rvwvcq);
/* dots and labels */
dot((-1.4934334172297545,2.6953043701763835),dotstyle);
label("$A$", (-1.3954084351380491,2.9230996889873015), NE * labelscalefactor);
dot((1.1286284157632023,-6.954814372303504),dotstyle);
label("$B$", (1.2093379191072373,-6.719031552166216), NE * labelscalefactor);
dot((8.199652712229643,-6.930007139864511),linewidth(4pt) + dotstyle);
label("$C$", (8.292420110475998,-6.741880204396438), NE * labelscalefactor);
dot((-1.459546107520503,-6.96389444957376),linewidth(4pt) + dotstyle);
label("$H$", (-1.3725597829078273,-6.787577508856881), NE * labelscalefactor);
dot((-1.4686261847907602,-4.375719926290057),linewidth(4pt) + dotstyle);
label("$E$", (-1.3725597829078273,-4.182831154611618), NE * labelscalefactor);
dot((4.617849638067675,6.252300211899359),linewidth(4pt) + dotstyle);
label("$L$", (4.705181710331174,6.441792132441429), NE * labelscalefactor);
dot((4.642656870506668,-0.8187240845670819),linewidth(4pt) + dotstyle);
label("$O$", (4.728030362561396,-0.6412900589272691), NE * labelscalefactor);
dot((4.117194931218359,-6.944329602191013),linewidth(4pt) + dotstyle);
label("$D$", (4.2025113612662945,-6.764728856626659), NE * labelscalefactor);
dot((4.6674641029456625,-7.889748381033524),linewidth(4pt) + dotstyle);
label("$F$", (4.750879014791618,-7.701523598065745), NE * labelscalefactor);
dot((4.651736947776926,-3.406898607850789),linewidth(4pt) + dotstyle);
label("$G$", (4.750879014791618,-3.2231877609423107), NE * labelscalefactor);
dot((1.5791517652735851,-0.3557971188372022),linewidth(4pt) + dotstyle);
label("$K$", (1.6663109637116735,-0.18431701432283698), NE * labelscalefactor);
dot((1.5870153428579534,-2.5972220054285695),linewidth(4pt) + dotstyle);
label("$P$", (1.6891596159418953,-2.4234849328845542), 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(15cm); 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 = -8.455641974276588, xmax = 26.731282460265, ymin = -10.92318356252699, ymax = 9.023689834456471; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); draw((-1.4934334172297545,2.6953043701763835)--(-1.459546107520503,-6.96389444957376)--(1.1286284157632023,-6.954814372303504)--(4.651736947776926,-3.406898607850789)--(4.642656870506668,-0.8187240845670819)--cycle, linewidth(2) + rvwvcq); /* draw figures */ draw((-1.4934334172297545,2.6953043701763835)--(1.1286284157632023,-6.954814372303504), linewidth(2) + wrwrwr); draw((xmin, -0.9930079421029264*xmin + 1.2123131258653241)--(xmax, -0.9930079421029264*xmax + 1.2123131258653241), linewidth(2) + wrwrwr); /* line */ draw((xmin, 0.0035082940460819836*xmin-6.958773932654766)--(xmax, 0.0035082940460819836*xmax-6.958773932654766), linewidth(2) + wrwrwr); /* line */ draw((xmin, -285.03882139434313*xmin-422.9911967079192)--(xmax, -285.03882139434313*xmax-422.9911967079192), linewidth(2) + wrwrwr); /* line */ draw((xmin, -1.7181023895538718*xmin + 0.12943284739433739)--(xmax, -1.7181023895538718*xmax + 0.12943284739433739), linewidth(2) + wrwrwr); /* line */ draw(circle((4.642656870506668,-0.8187240845670819), 7.071067811865476), linewidth(2) + wrwrwr); draw((xmin, -285.0388213943529*xmin + 1322.5187184230485)--(xmax, -285.0388213943529*xmax + 1322.5187184230485), linewidth(2) + wrwrwr); /* line */ draw((-1.4934334172297545,2.6953043701763835)--(4.617849638067675,6.252300211899359), linewidth(2) + wrwrwr); draw((4.617849638067675,6.252300211899359)--(-1.459546107520503,-6.96389444957376), linewidth(2) + wrwrwr); draw(circle((-0.18240250073327363,-2.12975500106356), 5), linewidth(2) + wrwrwr); draw((xmin, -285.0388213943432*xmin + 449.7637608575419)--(xmax, -285.0388213943432*xmax + 449.7637608575419), linewidth(2) + wrwrwr); /* line */ draw((1.1286284157632023,-6.954814372303504)--(4.651736947776926,-3.406898607850789), linewidth(2) + wrwrwr); draw((-1.4934334172297545,2.6953043701763835)--(4.642656870506668,-0.8187240845670819), linewidth(2) + wrwrwr); draw((-1.4934334172297545,2.6953043701763835)--(-1.459546107520503,-6.96389444957376), linewidth(2) + rvwvcq); draw((-1.459546107520503,-6.96389444957376)--(1.1286284157632023,-6.954814372303504), linewidth(2) + rvwvcq); draw((1.1286284157632023,-6.954814372303504)--(4.651736947776926,-3.406898607850789), linewidth(2) + rvwvcq); draw((4.651736947776926,-3.406898607850789)--(4.642656870506668,-0.8187240845670819), linewidth(2) + rvwvcq); draw((4.642656870506668,-0.8187240845670819)--(-1.4934334172297545,2.6953043701763835), linewidth(2) + rvwvcq); /* dots and labels */ dot((-1.4934334172297545,2.6953043701763835),dotstyle); label("$A$", (-1.3954084351380491,2.9230996889873015), NE * labelscalefactor); dot((1.1286284157632023,-6.954814372303504),dotstyle); label("$B$", (1.2093379191072373,-6.719031552166216), NE * labelscalefactor); dot((8.199652712229643,-6.930007139864511),linewidth(4pt) + dotstyle); label("$C$", (8.292420110475998,-6.741880204396438), NE * labelscalefactor); dot((-1.459546107520503,-6.96389444957376),linewidth(4pt) + dotstyle); label("$H$", (-1.3725597829078273,-6.787577508856881), NE * labelscalefactor); dot((-1.4686261847907602,-4.375719926290057),linewidth(4pt) + dotstyle); label("$E$", (-1.3725597829078273,-4.182831154611618), NE * labelscalefactor); dot((4.617849638067675,6.252300211899359),linewidth(4pt) + dotstyle); label("$L$", (4.705181710331174,6.441792132441429), NE * labelscalefactor); dot((4.642656870506668,-0.8187240845670819),linewidth(4pt) + dotstyle); label("$O$", (4.728030362561396,-0.6412900589272691), NE * labelscalefactor); dot((4.117194931218359,-6.944329602191013),linewidth(4pt) + dotstyle); label("$D$", (4.2025113612662945,-6.764728856626659), NE * labelscalefactor); dot((4.6674641029456625,-7.889748381033524),linewidth(4pt) + dotstyle); label("$F$", (4.750879014791618,-7.701523598065745), NE * labelscalefactor); dot((4.651736947776926,-3.406898607850789),linewidth(4pt) + dotstyle); label("$G$", (4.750879014791618,-3.2231877609423107), NE * labelscalefactor); dot((1.5791517652735851,-0.3557971188372022),linewidth(4pt) + dotstyle); label("$K$", (1.6663109637116735,-0.18431701432283698), NE * labelscalefactor); dot((1.5870153428579534,-2.5972220054285695),linewidth(4pt) + dotstyle); label("$P$", (1.6891596159418953,-2.4234849328845542), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ | [] |
485 | Circle $C$ with radius 2 has diameter $\overline{AB}$. Circle D is internally tangent to circle $C$ at $A$. Circle $E$ is internally tangent to circle $C$, externally tangent to circle $D$, and tangent to $\overline{AB}$. The radius of circle $D$ is three times the radius of circle $E$, and can be written in the form $\sqrt{m}-n$, where $m$ and $n$ are positive integers. Find $m+n$. | 2014 AIME II Problems/Problem 8 | Using the diagram above, let the radius of $D$ be $3r$, and the radius of $E$ be $r$. Then, $EF=r$, and $CE=2-r$, so the Pythagorean theorem in $\triangle CEF$ gives $CF=\sqrt{4-4r}$. Also, $CD=CA-AD=2-3r$, so \[DF=DC+CF=2-3r+\sqrt{4-4r}.\] Noting that $DE=4r$, we can now use the Pythagorean theorem in $\triangle DEF$ to get
\[(2-3r+\sqrt{4-4r})^2+r^2=16r^2.\]
Solving this quadratic is somewhat tedious, but the constant terms cancel, so the computation isn't terrible. Solving gives $3r=\sqrt{240}-14$ for a final answer of $\boxed{254}$.
Notice that C, E and the point of tangency to circle C for circle E will be collinear because C and E intersect the tangent line at a right angle, implying they must be on the same line. | // Block 1
import graph; size(7.99cm);
real labelscalefactor = 0.5;
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);
pen dotstyle = black;
real xmin = 4.087153740193288, xmax = 11.08175859031552, ymin = -4.938019122704778, ymax = 1.194137062512079;
draw(circle((7.780000000000009,-1.320000000000002), 2.000000000000000));
draw(circle((7.271934046987930,-1.319179731427737), 1.491933384829670));
draw(circle((9.198812158392690,-0.8249788848962227), 0.4973111282761416));
draw((5.780002606580324,-1.316771019595571)--(9.779997393419690,-1.323228980404432));
draw((9.198812158392690,-0.8249788848962227)--(9.198009254448635,-1.322289365031666));
draw((7.271934046987930,-1.319179731427737)--(9.198812158392690,-0.8249788848962227));
draw((9.198812158392690,-0.8249788848962227)--(7.780000000000009,-1.320000000000002));
dot((7.780000000000009,-1.320000000000002),dotstyle);
label("$C$", (7.707377218471464,-1.576266740352400), NE * labelscalefactor);
dot((7.271934046987930,-1.319179731427737),dotstyle);
label("$D$", (7.303064016111554,-1.276266740352400), NE * labelscalefactor);
dot((9.198812158392690,-0.8249788848962227),dotstyle);
label("$E$", (9.225301294671791,-0.7792624249832147), NE * labelscalefactor);
dot((9.198009254448635,-1.322289365031666),dotstyle);
label("$F$", (9.225301294671791,-1.276266740352400), NE * labelscalefactor);
dot((9.779997393419690,-1.323228980404432),dotstyle);
label("$B$", (9.810012253929656,-1.276266740352400), NE * labelscalefactor);
dot((5.780002606580324,-1.316771019595571),dotstyle);
label("$A$", (5.812051070003994,-1.276266740352400), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
// Block 2
import graph; size(7.99cm); real labelscalefactor = 0.5; pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); pen dotstyle = black; real xmin = 4.087153740193288, xmax = 11.08175859031552, ymin = -4.938019122704778, ymax = 1.194137062512079; draw(circle((7.780000000000009,-1.320000000000002), 2.000000000000000)); draw(circle((7.271934046987930,-1.319179731427737), 1.491933384829670)); draw(circle((9.198812158392690,-0.8249788848962227), 0.4973111282761416)); draw((5.780002606580324,-1.316771019595571)--(9.779997393419690,-1.323228980404432)); draw((9.198812158392690,-0.8249788848962227)--(9.198009254448635,-1.322289365031666)); draw((7.271934046987930,-1.319179731427737)--(9.198812158392690,-0.8249788848962227)); draw((9.198812158392690,-0.8249788848962227)--(7.780000000000009,-1.320000000000002)); dot((7.780000000000009,-1.320000000000002),dotstyle); label("$C$", (7.707377218471464,-1.576266740352400), NE * labelscalefactor); dot((7.271934046987930,-1.319179731427737),dotstyle); label("$D$", (7.303064016111554,-1.276266740352400), NE * labelscalefactor); dot((9.198812158392690,-0.8249788848962227),dotstyle); label("$E$", (9.225301294671791,-0.7792624249832147), NE * labelscalefactor); dot((9.198009254448635,-1.322289365031666),dotstyle); label("$F$", (9.225301294671791,-1.276266740352400), NE * labelscalefactor); dot((9.779997393419690,-1.323228980404432),dotstyle); label("$B$", (9.810012253929656,-1.276266740352400), NE * labelscalefactor); dot((5.780002606580324,-1.316771019595571),dotstyle); label("$A$", (5.812051070003994,-1.276266740352400), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); | [] |
486 | Let $ABCDE$ be a pentagon inscribed in a circle such that $AB = CD = 3$, $BC = DE = 10$, and $AE= 14$. The sum of the lengths of all diagonals of $ABCDE$ is equal to $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$ ?
$\textbf{(A) }129\qquad \textbf{(B) }247\qquad \textbf{(C) }353\qquad \textbf{(D) }391\qquad \textbf{(E) }421\qquad$ | 2014 AMC 12B Problems/Problem 24 | Let $BE=a$, $AD=b$, and $AC=CE=BD=c$. Let $F$ be on $AE$ such that $CF \perp AE$.
In $\triangle CFE$ we have $\cos\theta = -\cos(\pi-\theta)=-7/c$. We use the Law of Cosines on $\triangle ABC$ to get $60\cos\theta = 109-c^2$. Eliminating $\cos\theta$ we get $c^3-109c-420=0$ which factorizes as
\[(c+7)(c+5)(c-12)=0.\]Discarding the negative roots we have $c=12$. Thus $BD=AC=CE=12$. For $BE=a$, we use Ptolemy's theorem on cyclic quadrilateral $ABCE$ to get $a=44/3$. For $AD=b$, we use Ptolemy's theorem on cyclic quadrilateral $ACDE$ to get $b=27/2$.
The sum of the lengths of the diagonals is $12+12+12+\tfrac{44}{3}+\tfrac{27}{2} = \tfrac{385}{6}$ so the answer is $385 + 6 = \fbox{\textbf{(D) }391}$ | // Block 1
size(200);
defaultpen(linewidth(0.4)+fontsize(10));
pen s = linewidth(0.8)+fontsize(8);
pair O,A,B,C,D,E0,F;
O=origin;
A= dir(198);
path c = CR(O,1);
real r = 0.13535;
B = IP(c, CR(A,3*r));
C = IP(c, CR(B,10*r));
D = IP(c, CR(C,3*r));
E0 = OP(c, CR(D,10*r));
F = foot(C,A,E0);
dot("$A$", A, A-O);
dot("$B$", B, B-O);
dot("$C$", C, C-O);
dot("$D$", D, D-O);
dot("$E$", E0, E0-O);
dot("$F$", F, F-C);
label("$c$",A--C,S);
label("$c$",E0--C,W);
label("$7$",F--E0,S);
label("$7$",F--A,S);
label("$3$",A--B,2*W);
label("$10$",B--C,2*N);
label("$3$",C--D,2*NE);
label("$10$",D--E0,E);
draw(A--B--C--D--E0--A, black+0.8);
draw(CR(O,1), s);
draw(A--C--E0, royalblue);
draw(C--F, royalblue+dashed);
draw(rightanglemark(E0,F,C,2));
MA("\theta",A,B,C,0.075);
MA("\pi-\theta",C,E0,A,0.1);
// Block 2
size(200); defaultpen(linewidth(0.4)+fontsize(10)); pen s = linewidth(0.8)+fontsize(8); pair O,A,B,C,D,E0,F; O=origin; A= dir(198); path c = CR(O,1); real r = 0.13535; B = IP(c, CR(A,3*r)); C = IP(c, CR(B,10*r)); D = IP(c, CR(C,3*r)); E0 = OP(c, CR(D,10*r)); F = foot(C,A,E0); dot("$A$", A, A-O); dot("$B$", B, B-O); dot("$C$", C, C-O); dot("$D$", D, D-O); dot("$E$", E0, E0-O); dot("$F$", F, F-C); label("$c$",A--C,S); label("$c$",E0--C,W); label("$7$",F--E0,S); label("$7$",F--A,S); label("$3$",A--B,2*W); label("$10$",B--C,2*N); label("$3$",C--D,2*NE); label("$10$",D--E0,E); draw(A--B--C--D--E0--A, black+0.8); draw(CR(O,1), s); draw(A--C--E0, royalblue); draw(C--F, royalblue+dashed); draw(rightanglemark(E0,F,C,2)); MA("\theta",A,B,C,0.075); MA("\pi-\theta",C,E0,A,0.1); | [] |
487 | Completely describe the set of all right triangles with positive integer-valued legs such that when four
copies of the triangle are arranged in square formation shown below, the incenters of the four triangles
lie on the extensions of the sides of the smaller square. (Note: the incenter of a triangle is the center
of the circle inscribed in that triangle.) | 2014 UMO Problems/Problem 3 | Let $I$ be the incenter of a triangle. Drop $I$ onto the three sides of the triangle, and let the points be $X, Y, Z$
Finally, let $a = AB, b = BC$ so that $a > b$ and let $s = BZ$.
Note that $Z$ is also a corner of the square, so $s = a - b$. But then, $AC = CX + AZ = a + b - 2(a-b) = 3b-a$. From the Pythogorean theorem, we also know that $AC^2 = a^2+b^2$. Therefore,
\[a^2 + b^2 = (3b-a)^2 = 9b^2-6ab + a^2\]
\[\Longleftrightarrow\]
\[8b^2 = 6ab\]
\[\Longleftrightarrow\]
\[b = \frac34 a\]
So, the only solutions are of the form $(3k, 4k, 5k).$ | // Block 1
size(200);
path T=((0,0)--(4,0)--(4,.3)--(3.7,.3)--(3.7,0)--(4,0)--(4,3)--(0,0));
D(shift(10,0)*rotate(53)*T,black);
D(shift(15,5)*rotate(233)*T,black);
D(shift(15,0)*rotate(143)*T,black);
D(shift(10,5)*rotate(323)*T,black);
pair A, B, C, X, Y, Z, I;
A = shift(10,0)*rotate(53)*(0,0);
B = shift(10,0)*rotate(53)*(4, 0);
C = shift(10,0)*rotate(53)*(4, 3);
I = incenter(A, B, C);
X = foot(I, B, C);
Y = foot(I, A, C);
Z = foot(I, A, B);
D(incircle(A, B, C), black);
D(I -- X, black);
D(I -- Y, black);
D(I -- Z, black);
label("X", X, NE);
label("Y", Y, W);
label("Z", Z, S);
label("A", A, S);
label("B", B, NE);
label("C", C, NW);
label("I", I, N);
D(A--Y, green);
D(A--Z, green);
D(B--Z, red);
D(B--X, red);
D(C--Y, blue);
D(C--X, blue);
// Block 2
size(200); path T=((0,0)--(4,0)--(4,.3)--(3.7,.3)--(3.7,0)--(4,0)--(4,3)--(0,0)); D(shift(10,0)*rotate(53)*T,black); D(shift(15,5)*rotate(233)*T,black); D(shift(15,0)*rotate(143)*T,black); D(shift(10,5)*rotate(323)*T,black); pair A, B, C, X, Y, Z, I; A = shift(10,0)*rotate(53)*(0,0); B = shift(10,0)*rotate(53)*(4, 0); C = shift(10,0)*rotate(53)*(4, 3); I = incenter(A, B, C); X = foot(I, B, C); Y = foot(I, A, C); Z = foot(I, A, B); D(incircle(A, B, C), black); D(I -- X, black); D(I -- Y, black); D(I -- Z, black); label("X", X, NE); label("Y", Y, W); label("Z", Z, S); label("A", A, S); label("B", B, NE); label("C", C, NW); label("I", I, N); D(A--Y, green); D(A--Z, green); D(B--Z, red); D(B--X, red); D(C--Y, blue); D(C--X, blue); | [] |
488 | Find the smallest and largest possible distances between the centers of two circles of radius
$1$ such that there is an equilateral triangle of side length $1$ with two vertices on one of
the circles and the third vertex on the second circle. | 2014 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 4 | The smallest distance would be found if the two circles were externally tangent, so testing that and messing around with it yields:
Where $P$ is the point of tangency. This clearly works, so the smallest distance would be $2\cdot1=\boxed{2}$
The largest distance would be found by first finding the closest place to the edge of a circle to place a line segment with side length $1$ (a side of the triangle), then adding the other two sides outwards like shown:
The largest distance would be found by first finding the closest place to the edge of a circle to place a line segment with side length $1$ (a side of the triangle), then adding the other two sides outwards like shown:
The segments $AB=BC=\frac {\sqrt3}{2}$ because of $30-60-90$ triangles. From this diagram, we can see that the distance between the centers is $\boxed{1+\sqrt3}$. | // Block 1
size(300 pt);
draw(circle((0,0),1));
draw(circle((2,0),1));
draw((1,0)--(0.5,0.86602),linewidth(.5));
draw((0,0)--(0.5,0.86602),linewidth(.5));
draw((1,0)--(1.5,0.86602),linewidth(.5));
draw((2,0)--(1.5,0.86602),linewidth(.5));
draw((0.5,0.86602)--(1.5,0.86602),linewidth(.5));
dot((0,0));
dot((2,0));
dot((1,0));
dot((0.5,0.86602));
dot((1.5,0.86602));
label("$P$",(1,0),SW);
label("$1$",(0.25,0.43301),NW);
label("$1$",(1.75,0.43301),NE);
label("$1$",(0.75,0.43301),SW);
label("$1$",(1.25,0.43301),SE);
label("$1$",(1,0.86602),N);
// Block 2
size(300pt);
draw(circle((0,0),3));
draw(circle((8.19615,0),3));
draw((2.59807,1.5)--(2.59807,-1.5),linewidth(.5));
draw((2.59807,1.5)--(5.19615,0),linewidth(.5));
draw((5.19615,0)--(2.59807,-1.5),linewidth(.5));
draw((0,0)--(2.59807,-1.5),linewidth(.5));
draw((0,0)--(2.59807,1.5),linewidth(.5));
draw((0,0)--(8.19615,0),linewidth(.5));
dot((2.59807,1.5));
dot((2.59807,-1.5));
dot((5.19615,0));
dot((8.19615,0));
dot((0,0));
label("$1$",(4.09807,0.5),N);
label("$\frac {\sqrt3}{2}$",(4.09807,0),SW);
label("$1$",(6.69615,0),S);
label("$A$",(0,0),SW);
label("$B$",(2.59807,0),SW);
label("$C$",(5.19615,0),SSW);
// Block 3
size(300 pt); draw(circle((0,0),1)); draw(circle((2,0),1)); draw((1,0)--(0.5,0.86602),linewidth(.5)); draw((0,0)--(0.5,0.86602),linewidth(.5)); draw((1,0)--(1.5,0.86602),linewidth(.5)); draw((2,0)--(1.5,0.86602),linewidth(.5)); draw((0.5,0.86602)--(1.5,0.86602),linewidth(.5)); dot((0,0)); dot((2,0)); dot((1,0)); dot((0.5,0.86602)); dot((1.5,0.86602)); label("$P$",(1,0),SW); label("$1$",(0.25,0.43301),NW); label("$1$",(1.75,0.43301),NE); label("$1$",(0.75,0.43301),SW); label("$1$",(1.25,0.43301),SE); label("$1$",(1,0.86602),N);
// Block 4
size(300pt); draw(circle((0,0),3)); draw(circle((8.19615,0),3)); draw((2.59807,1.5)--(2.59807,-1.5),linewidth(.5)); draw((2.59807,1.5)--(5.19615,0),linewidth(.5)); draw((5.19615,0)--(2.59807,-1.5),linewidth(.5)); draw((0,0)--(2.59807,-1.5),linewidth(.5)); draw((0,0)--(2.59807,1.5),linewidth(.5)); draw((0,0)--(8.19615,0),linewidth(.5)); dot((2.59807,1.5)); dot((2.59807,-1.5)); dot((5.19615,0)); dot((8.19615,0)); dot((0,0)); label("$1$",(4.09807,0.5),N); label("$\frac {\sqrt3}{2}$",(4.09807,0),SW); label("$1$",(6.69615,0),S); label("$A$",(0,0),SW); label("$B$",(2.59807,0),SW); label("$C$",(5.19615,0),SSW); | [] |
489 | Triangle $ABC$ has side lengths $AB = 12$, $BC = 25$, and $CA = 17$. Rectangle $PQRS$ has vertex $P$ on $\overline{AB}$, vertex $Q$ on $\overline{AC}$, and vertices $R$ and $S$ on $\overline{BC}$. In terms of the side length $PQ = \omega$, the area of $PQRS$ can be expressed as the quadratic polynomial \[Area(PQRS) = \alpha \omega - \beta \omega^2.\]
Then the coefficient $\beta = \frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2015 AIME II Problems/Problem 7 | Similar triangles can also solve the problem.
First, solve for the area of the triangle. $[ABC] = 90$. This can be done by Heron's Formula or placing an $8-15-17$ right triangle on $AC$ and solving. (The $8$ side would be collinear with line $AB$)
After finding the area, solve for the altitude to $BC$. Let $E$ be the intersection of the altitude from $A$ and side $BC$. Then $AE = \frac{36}{5}$.
Solving for $BE$ using the Pythagorean Formula, we get $BE = \frac{48}{5}$. We then know that $CE = \frac{77}{5}$.
Now consider the rectangle $PQRS$. Since $SR$ is collinear with $BC$ and parallel to $PQ$, $PQ$ is parallel to $BC$ meaning $\Delta APQ$ is similar to $\Delta ABC$.
Let $F$ be the intersection between $AE$ and $PQ$. By the similar triangles, we know that $\frac{PF}{FQ}=\frac{BE}{EC} = \frac{48}{77}$. Since $PF+FQ=PQ=\omega$. We can solve for $PF$ and $FQ$ in terms of $\omega$. We get that $PF=\frac{48}{125} \omega$ and $FQ=\frac{77}{125} \omega$.
Let's work with $PF$. We know that $PQ$ is parallel to $BC$ so $\Delta APF$ is similar to $\Delta ABE$. We can set up the proportion:
$\frac{AF}{PF}=\frac{AE}{BE}=\frac{3}{4}$. Solving for $AF$, $AF = \frac{3}{4} PF = \frac{3}{4} \cdot \frac{48}{125} \omega = \frac{36}{125} \omega$.
We can solve for $PS$ then since we know that $PS=FE$ and $FE= AE - AF = \frac{36}{5} - \frac{36}{125} \omega$.
Therefore, $[PQRS] = PQ \cdot PS = \omega (\frac{36}{5} - \frac{36}{125} \omega) = \frac{36}{5}\omega - \frac{36}{125} \omega^2$.
This means that $\beta = \frac{36}{125} \Rightarrow (m,n) = (36,125) \Rightarrow m+n = \boxed{161}$. | // Block 1
unitsize(20);
pair A,B,C,E,F,P,Q,R,S;
A=(48/5,36/5);
B=(0,0);
C=(25,0);
E=(48/5,0);
F=(48/5,18/5);
P=(24/5,18/5);
Q=(173/10,18/5);
S=(24/5,0);
R=(173/10,0);
draw(A--B--C--cycle);
draw(P--Q);
draw(Q--R);
draw(R--S);
draw(S--P);
draw(A--E,dashed);
label("$A$",A,N);
label("$B$",B,SW);
label("$C$",C,SE);
label("$E$",E,SE);
label("$F$",F,NE);
label("$P$",P,NW);
label("$Q$",Q,NE);
label("$R$",R,SE);
label("$S$",S,SW);
draw(rightanglemark(B,E,A,12));
dot(E);
dot(F);
// Block 2
unitsize(20); pair A,B,C,E,F,P,Q,R,S; A=(48/5,36/5); B=(0,0); C=(25,0); E=(48/5,0); F=(48/5,18/5); P=(24/5,18/5); Q=(173/10,18/5); S=(24/5,0); R=(173/10,0); draw(A--B--C--cycle); draw(P--Q); draw(Q--R); draw(R--S); draw(S--P); draw(A--E,dashed); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,SE); label("$E$",E,SE); label("$F$",F,NE); label("$P$",P,NW); label("$Q$",Q,NE); label("$R$",R,SE); label("$S$",S,SW); draw(rightanglemark(B,E,A,12)); dot(E); dot(F); | [] |
490 | A cylindrical barrel with radius $4$ feet and height $10$ feet is full of water. A solid cube with side length $8$ feet is set into the barrel so that the diagonal of the cube is vertical. The volume of water thus displaced is $v$ cubic feet. Find $v^2$. | 2015 AIME II Problems/Problem 9 | We can use the same method as in Solution 2 to find the side length of the equilateral triangle, which is $4\sqrt3$. From here, its area is
\[\dfrac{\bigl(4\sqrt3\bigr)^2\sqrt3}4=12\sqrt3.\]
The leg of the isosceles right triangle is $\dfrac{4\sqrt3}{\sqrt2}=2\sqrt6$, and the horizontal distance from the vertex to the base of the tetrahedron is $4$ (the radius of the cylinder), so we can find the height, as shown in the diagram.
The height from the tip to the base is $2\sqrt2$, so the volume is $\dfrac{12\sqrt3\cdot2\sqrt2}3=8\sqrt6$, and thus the answer is $\boxed{384}$.
-integralarefun | // Block 1
import olympiad;
pair V, T, B;
V = (-4, 0);
B = origin;
T = (0, 2*sqrt(2));
draw(V--B--T--cycle);
draw(rightanglemark(V, B, T));
label("Vertex", V, W);
label("Tip", T, N);
label("Base", B, SE);
label("$4$", V--B, S);
label("$2\sqrt6$", V--T, NW);
// Block 2
import olympiad; pair V, T, B; V = (-4, 0); B = origin; T = (0, 2*sqrt(2)); draw(V--B--T--cycle); draw(rightanglemark(V, B, T)); label("Vertex", V, W); label("Tip", T, N); label("Base", B, SE); label("$4$", V--B, S); label("$2\sqrt6$", V--T, NW); | [] |
491 | Centered at each lattice point in the coordinate plane are a circle radius $\frac{1}{10}$ and a square with sides of length $\frac{1}{5}$ whose sides are parallel to the coordinate axes. The line segment from $(0,0)$ to $(1001, 429)$ intersects $m$ of the squares and $n$ of the circles. Find $m + n$. | 2016 AIME I Problems/Problem 14 | See if you can solve the problem with the following.
Solution to Solution 2
This is mostly a clarification to Solution 1, but let's take the diagram for the origin to $(7,3)$. We have the origin circle and square intersected, then two squares, then the circle and square at $(7,3)$. If we take the circle and square at the origin out of the diagram, we will be able to repeat the resulting segment (with its circles and squares) end to end from $(0,0)$ to $(1001,429)$, which forms the line we need without overlapping. Since $143$ of these segments are needed to do this, and $3$ squares and $1$ circle are intersected with each, there are $143 \cdot (3+1) = 572$ squares and circles intersected. Adding the circle and square that are intersected at the origin back into the picture, we get that there are $572+2=\boxed{574}$ squares and circles intersected in total.
Solution to Solution 2 without a diagram
This solution is a more systematic approach for finding when the line intersects the squares and circles. Because $1001 = 7*11*13$ and $429=3*11*13$, the slope of our line is $\frac{3}{7}$, and we only need to consider the line in the rectangle from the origin to $(7,3)$, and we can iterate the line $11*13=143$ times. First, we consider how to figure out if the line intersects a square. Given a lattice point $(x_1, y_1)$, we can think of representing a square centered at that lattice point as all points equal to $(x_1 \pm a, y_1 \pm b)$ s.t. $0 \leq a,b \leq \frac{1}{10}$. If the line $y = \frac{3}{7}x$ intersects the square, then we must have $\frac{y_1 + b}{x_1 + a} = \frac{3}{7}$. The line with the least slope that intersects the square intersects at the bottom right corner and the line with the greatest slope that intersects the square intersects at the top left corner; thus we must have that $\frac{3}{7}$ lies in between these slopes, or that $\frac{y_1-\frac{1}{10}}{x_1+\frac{1}{10}} \leq \frac{3}{7} \leq \frac{y_1+\frac{1}{10}}{x_1-\frac{1}{10}}$. Simplifying, $3x_1 - 1 \leq 7y_1 \leq 3x_1 + 1$. Because $y$ can only equal $0, 1, 2, 3$, we just do casework based on the values of $y$ and find that the points $(2, 1)$ and $(5, 2)$ are intersected just at the corner of the square and $(0, 0), (7, 3)$ are intersected through the center of the square. However, we disregard one of $(0, 0)$ and $(7, 3)$, WLOG $(0, 0)$, since we just use it in our count for the next of the 143 segments. Therefore, in one of our "segments", 3 squares are intersected and 1 circle is intersected giving 4 total. Thus our answer is $143*4 = 572$. HOWEVER, we cannot forget that we ignored $(0, 0)$, which contributes another square and circle to our count, making the final answer $572 + 2 = \boxed{574}$.
-Patrick4President | // Block 1
size(12cm);draw((0,0)--(7,3));draw(box((0,0),(7,3)),dotted);
for(int i=0;i<8;++i)for(int j=0; j<4; ++j){dot((i,j),linewidth(1));draw(box((i-.1,j-.1),(i+.1,j+.1)),linewidth(.5));draw(circle((i,j),.1),linewidth(.5));}
// Block 2
size(12cm);draw((0,0)--(7,3));draw(box((0,0),(7,3)),dotted); for(int i=0;i<8;++i)for(int j=0; j<4; ++j){dot((i,j),linewidth(1));draw(box((i-.1,j-.1),(i+.1,j+.1)),linewidth(.5));draw(circle((i,j),.1),linewidth(.5));} | [] |
492 | Equilateral $\triangle ABC$ has side length $600$. Points $P$ and $Q$ lie outside the plane of $\triangle ABC$ and are on opposite sides of the plane. Furthermore, $PA=PB=PC$, and $QA=QB=QC$, and the planes of $\triangle PAB$ and $\triangle QAB$ form a $120^{\circ}$ dihedral angle (the angle between the two planes). There is a point $O$ whose distance from each of $A,B,C,P,$ and $Q$ is $d$. Find $d$. | 2016 AIME II Problems/Problem 14 | To make numbers more feasible, we'll scale everything down by a factor of $100$ so that $\overline{AB}=\overline{BC}=\overline{AC}=6$. We should also note that $P$ and $Q$ must lie on the line that is perpendicular to the plane of $ABC$ and also passes through the circumcenter of $ABC$ (due to $P$ and $Q$ being equidistant from $A$, $B$, $C$), let $D$ be the altitude from $C$ to $AB$. We can draw a vertical cross-section of the figure then: We let $\angle PDI=\alpha$ so $\angle QDI=120^{\circ}-\alpha$, also note that $\overline{PO}=\overline{QO}=\overline{CO}=d$. Because $I$ is the centroid of $ABC$, we know that ratio of $\overline{CI}$ to $\overline{DI}$ is $2:1$. Since we've scaled the figure down, the length of $CD$ is $3\sqrt{3}$, from this it's easy to know that $\overline{CI}=2\sqrt{3}$ and $\overline{DI}=\sqrt{3}$. The following two equations arise: \begin{align} \sqrt{3}\tan{\left(\alpha\right)}+\sqrt{3}\tan{\left(120^{\circ}-\alpha\right)}&=2d \\ \sqrt{3}\tan{\left(\alpha\right)} - d &= \sqrt{d^{2}-12} \end{align} Using trig identities for the tangent, we find that \begin{align*} \sqrt{3}\tan{\left(120^{\circ}-\alpha\right)}&=\sqrt{3}\left(\frac{\tan{\left(120^{\circ}\right)}+\tan{\left(\text{-}\alpha\right)}}{1-\tan{\left(120^{\circ}\right)}\tan{\left(\text{-}\alpha\right)}}\right) \\ &= \sqrt{3}\left(\frac{\text{-}\sqrt{3}+\tan{\left(\text{-}\alpha\right)}}{1+\sqrt{3}\tan{\left(\text{-}\alpha\right)}}\right) \\ &= \sqrt{3}\left(\frac{\text{-}\sqrt{3}-\tan{\left(\alpha\right)}}{1-\sqrt{3}\tan{\left(\alpha\right)}}\right) \\ &= \frac{\sqrt{3}\tan{\left(\alpha\right)}+3}{\sqrt{3}\tan{\left(\alpha\right)}-1}.\end{align*} Okay, now we can plug this into $\left(1\right)$ to get: \begin{align}\sqrt{3}\tan{\left(\alpha\right)}+\frac{\sqrt{3}\tan{\left(\alpha\right)}+3}{\sqrt{3}\tan{\left(\alpha\right)}-1}&=2d \\ \sqrt{3}\tan{\left(\alpha\right)} - d &= \sqrt{d^{2}-12} \end{align} Notice that $\alpha$ only appears in the above system of equations in the form of $\sqrt{3}\tan{\left(\alpha\right)}$, we can set $\sqrt{3}\tan{\left(\alpha\right)}=a$ for convenience since we really only care about $d$. Now we have \begin{align}a+\frac{a+3}{a-1}&=2d \\ a - d &= \sqrt{d^{2}-12} \end{align} Looking at $\left(2\right)$, it's tempting to square it to get rid of the square-root so now we have: \begin{align*}a^{2}-2ad+d^{2}&=d^{2}-12 \\ a - 2ad &= \text{-}12 \end{align*} See the sneaky $2d$ in the above equation? That we means we can substitute it for $a+\frac{a+3}{a-1}$: \begin{align*}a^{2}-2ad+d^{2}&=d^{2}-12 \\ a^{2} - a\left(a+\frac{a+3}{a-1}\right) &= \text{-}12 \\ a^{2}-a^{2}-\frac{a^{2}+3a}{a-1} &=\text{-}12 \\ -\frac{a^{2}+3a}{a-1}&=\text{-}12 \\ \text{-}a^{2}-3a&=\text{-}12a+12 \\ 0 &= a^{2}-9a+12 \end{align*} Use the quadratic formula, we find that $a=\frac{9\pm\sqrt{9^{2}-4\left(1\right)\left(12\right)}}{2\left(1\right)}=\frac{9\pm\sqrt{33}}{2}$ - the two solutions were expected because $a$ can be $\angle PDI$ or $\angle QDI$. We can plug this into $\left(1\right)$: \begin{align*}a+\frac{a+3}{a-1}&=2d \\ \frac{9\pm\sqrt{33}}{2}+\frac{\frac{9\pm\sqrt{33}}{2}+3}{\frac{9\pm\sqrt{33}}{2}-1}=2d \\ \frac{9\pm\sqrt{33}}{2}+\frac{15\pm\sqrt{33}}{7\pm\sqrt{33}} &= 2d\end{align*} I'll use $a=\frac{9+\sqrt{33}}{2}$ because both values should give the same answer for $d$. \begin{align*} \frac{9+\sqrt{33}}{2}+\frac{15+\sqrt{33}}{7+\sqrt{33}} &= 2d \\ \frac{\left(9+\sqrt{33}\right)\left(7+\sqrt{33}\right)+\left(2\right)\left(15+\sqrt{33}\right)}{\left(2\right)\left(7+\sqrt{33}\right)} &= 2d \\ \frac{63+33+16\sqrt{33}+30+2\sqrt{33}}{14+2\sqrt{33}} &= 2d \\ \frac{126+18\sqrt{33}}{14+2\sqrt{33}} &= 2d \\ 9 &= 2d \\ \frac{9}{2} &= d\end{align*} Wait! Before you get excited, remember that we scaled the entire figure by $100$?? That means that the answer is $d=100\times\frac{9}{2}=\boxed{450}$.
-fatant | pair C, D, I, P, Q, O; D=(0,0); C=(5.196152,0); P=(1.732051,7.37228); I=(1.732051,0); Q=(1.732051,-1.62772); O=(1.732051,2.87228); draw(C--Q--D--P--cycle); draw(C--D, dashed); draw(P--Q, dotted); draw(O--C, dotted); label("$C$", C, E); label("$D$", D, W); label("$I$", I, NW); label("$P$", P, N); label("$Q$", Q, S); label("$O$", O, SW); dot(O); dot(I); | [] |
492 | Equilateral $\triangle ABC$ has side length $600$. Points $P$ and $Q$ lie outside the plane of $\triangle ABC$ and are on opposite sides of the plane. Furthermore, $PA=PB=PC$, and $QA=QB=QC$, and the planes of $\triangle PAB$ and $\triangle QAB$ form a $120^{\circ}$ dihedral angle (the angle between the two planes). There is a point $O$ whose distance from each of $A,B,C,P,$ and $Q$ is $d$. Find $d$. | 2016 AIME II Problems/Problem 14 | Let $M$ be the midpoint of $\overline{AB}$ and $X$ the center of $\triangle ABC$. Then \[P, O, Q, M, X, C\] all lie in the same vertical plane. We can make the following observations:
The equilateral triangle has side length $600$, so $MC=300\sqrt{3}$ and $X$ divides $MC$ so that $MX=100\sqrt{3}$ and $XC=200\sqrt{3}$;
$O$ is the midpoint of $PQ$ since $O$ is equidistant from $A, B, C, P, Q$ – it is also the circumcenter of $\triangle PCQ$;
$\angle PMQ=120^{\circ}$, the dihedral angle.
To make calculations easier, we will denote $100\sqrt{3}=m$, so that $MX=m$ and $XC=2m$.
Denote $PX=p$ and $QX=q$, where the tangent addition formula on $\triangle PMQ$ yields \[\frac{\tan\measuredangle PMX+\tan\measuredangle QMX}{1-\tan\measuredangle PMX\tan\measuredangle QMX}=\tan(120^{\circ})=-\sqrt{3}.\] Using $\tan\measuredangle PMX=\frac{p}{m}$ and $\tan\measuredangle QMX=\frac{q}{m}$, we have \[\frac{\frac{p}{m}+\frac{q}{m}}{1-\frac{p}{m}\cdot\frac{q}{m}}=-\sqrt{3}.\] After multiplying both numerator and denominator by $m^{2}$ we have \[\frac{(p+q)m}{m^{2}-pq}=-\sqrt{3}.\] But note that $pq=(2m)(2m)=4m^{2}$ by power of a point at $X$, where we deduce by symmetry that $MM^{\prime}=MX=m$ on the diagram below:
Thus \begin{align*} \frac{(p+q)m}{m^{2}-4m^{2}}=-\sqrt{3} \\ \frac{p+q}{-3m}=-\sqrt{3} \\ p+q=\left(-\sqrt{3}\right)\left(-3m\right) \\ p+q=3\sqrt{3}\cdot m.\end{align*} Earlier we assigned the variable $m$ to the length $100\sqrt{3}$ which implies $PQ=\left(3\sqrt{3}\right)\left(100\sqrt{3}\right)=900$. Thus the distance $d$ is equal to $\frac{PQ}{2}=\boxed{450}$. | // Block 1
unitsize(20);
pair P = (0, 12);
pair Q = (0, -3);
pair O = (P+Q)/2;
pair M = (-3, 0);
pair X = (0, 0);
pair C = (6, 0);
draw(P--O--Q);
draw(M--X--C);
draw(P--M--Q, blue);
draw(Q--C--P);
draw(circle((0, 4.5), 7.5));
label("$P$", P, N);
label("$Q$", Q, S);
label("$O$", O, E);
dot(O);
label("$M$", M, W);
label("$X$", X, NE);
label("$C$", C, E);
label("$m$", (M+X)/2, N);
label("$2m$", (X+C)/2, N);
// Block 2
unitsize(20);
pair P = (0, 12);
pair Q = (0, -3);
pair O = (P+Q)/2;
pair M = (-3, 0);
pair Mprime = (-6, 0);
pair X = (0, 0);
pair C = (6, 0);
draw(P--O--Q);
draw(Mprime--M--X--C);
draw(P--M--Q, blue);
draw(Q--C--P);
draw(circle((0, 4.5), 7.5));
label("$P$", P, N);
label("$Q$", Q, S);
label("$O$", O, E);
dot(O);
label("$M$", M, S);
label("$M^{\prime}$", Mprime, W);
label("$X$", X, SE);
label("$C$", C, E);
label("$m$", (Mprime+M)/2, N);
label("$m$", (M+X)/2, N);
label("$2m$", (X+C)/2, N);
// Block 3
unitsize(20); pair P = (0, 12); pair Q = (0, -3); pair O = (P+Q)/2; pair M = (-3, 0); pair X = (0, 0); pair C = (6, 0); draw(P--O--Q); draw(M--X--C); draw(P--M--Q, blue); draw(Q--C--P); draw(circle((0, 4.5), 7.5)); label("$P$", P, N); label("$Q$", Q, S); label("$O$", O, E); dot(O); label("$M$", M, W); label("$X$", X, NE); label("$C$", C, E); label("$m$", (M+X)/2, N); label("$2m$", (X+C)/2, N);
// Block 4
unitsize(20); pair P = (0, 12); pair Q = (0, -3); pair O = (P+Q)/2; pair M = (-3, 0); pair Mprime = (-6, 0); pair X = (0, 0); pair C = (6, 0); draw(P--O--Q); draw(Mprime--M--X--C); draw(P--M--Q, blue); draw(Q--C--P); draw(circle((0, 4.5), 7.5)); label("$P$", P, N); label("$Q$", Q, S); label("$O$", O, E); dot(O); label("$M$", M, S); label("$M^{\prime}$", Mprime, W); label("$X$", X, SE); label("$C$", C, E); label("$m$", (Mprime+M)/2, N); label("$m$", (M+X)/2, N); label("$2m$", (X+C)/2, N); | [] |
493 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 10A Problems/Problem 24 | Let $AD$ intersect $OB$ at $E$ and $OC$ at $F.$
$\overarc{AB}= \overarc{BC}= \overarc{CD}=\theta$
$\angle{BAD}=\frac{1}{2} \cdot \overarc{BCD}=\theta=\angle{AOB}$
From there, $\triangle{OAB} \sim \triangle{ABE}$, thus:
$\frac{OA}{AB} = \frac{AB}{BE} = \frac{OB}{AE}$
$OA = OB$ because they are both radii of $\odot{O}$. Since $\frac{OA}{AB} = \frac{OB}{AE}$, we have that $200 = AB = AE$. Similarly, $CD = DF$.
$OE = 100\sqrt{2} = \frac{OB}{2}$ and $EF=\frac{BC}{2}=100$ , so $AD=AE + EF + FD = 200 + 100 + 200 = \boxed{\textbf{(E) } 500}$
Note
The first angle chase is done by the central angle theorem. Note that $\angle BAD$ is an inscribed angle from $BD$ to $A$, which is always $\frac{1}{2}$ of the angle of $\overarc{BD}$. Hence follows.
Furthermore, we get $OE=100\sqrt{2}$ from the similar triangles mentioned. We have: $\frac{200}{200\sqrt{2}}=\frac{BE}{200}$. Hence, $BE=100\sqrt{2}$ and so $OE=OB-BE=100\sqrt{2}$.
~mathboy282 | // Block 1
size(250);
defaultpen(linewidth(0.4));
//Variable Declarations
real RADIUS;
pair A, B, C, D, E, F, O;
RADIUS=3;
//Variable Definitions
A=RADIUS*dir(148.414);
B=RADIUS*dir(109.471);
C=RADIUS*dir(70.529);
D=RADIUS*dir(31.586);
O=(0,0);
//Path Definitions
path quad= A -- B -- C -- D -- cycle;
//Initial Diagram
draw(Circle(O, RADIUS), linewidth(0.8));
draw(quad, linewidth(0.8));
label("$A$",A,W);
label("$B$",B,NW);
label("$C$",C,NE);
label("$D$",D,ENE);
label("$O$",O,S);
label("$\theta$",O,3N);
//Radii
draw(O--A);
draw(O--B);
draw(O--C);
draw(O--D);
//Construction
E=extension(B,O,A,D);
label("$E$",E,NE);
F=extension(C,O,A,D);
label("$F$",F,NE);
//Angle marks
draw(anglemark(C,O,B));
// Block 2
size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,ENE); label("$O$",O,S); label("$\theta$",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction E=extension(B,O,A,D); label("$E$",E,NE); F=extension(C,O,A,D); label("$F$",F,NE); //Angle marks draw(anglemark(C,O,B)); | [] |
493 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 10A Problems/Problem 24 | To save us from getting big numbers with lots of zeros behind them, let's divide all side lengths by $200$ for now, then multiply it back at the end of our solution.
Construct quadrilateral $ABCD$ on the circle with $AD$ being the missing side (Notice that since the side length is less than the radius, it will be very small on the top of the circle). Now, draw the radii from center $O$ to $A,B,C,$ and $D$. Let the intersection of $BD$ and $OC$ be point $E$. Notice that $BD$ and $OC$ are perpendicular because $BCDO$ is a kite.
We set lengths $BE=ED$ equal to $x$ (Solution 1.1 begins from here). By the Pythagorean Theorem,
\[\sqrt{1^2-x^2}+\sqrt{(\sqrt{2})^2-x^2}=\sqrt{2}\]
We solve for $x$:
\[1-x^2+2-x^2+2\sqrt{(1-x^2)(2-x^2)}=2\]
\[2\sqrt{(1-x^2)(2-x^2)}=2x^2-1\]
\[4(1-x^2)(2-x^2)=(2x^2-1)^2\]
\[8-12x^2+4x^4=4x^4-4x^2+1\]
\[8x^2=7\]
\[x=\frac{\sqrt{14}}{4}\]
By Ptolemy's Theorem,
\[AB \cdot CD + BC \cdot AD = AC \cdot BD = BD^2 = (2 \cdot BE)^2\]
Substituting values,
\[1^2+1 \cdot AD = 4{\left( \frac{\sqrt{14}}{4} \right)}^2\]
\[1+AD=\frac{7}{2}\]
\[AD=\frac{5}{2}\]
Finally, we multiply back the $200$ that we divided by at the beginning of the problem to get $AD=\boxed{\textbf{(E) } 500}$. | // Block 1
size(250);
defaultpen(linewidth(0.4));
//Variable Declarations
real RADIUS;
pair A, B, C, D, E, O;
RADIUS=3;
//Variable Definitions
A=RADIUS*dir(148.414);
B=RADIUS*dir(109.471);
C=RADIUS*dir(70.529);
D=RADIUS*dir(31.586);
E=extension(B,D,O,C);
O=(0,0);
//Path Definitions
path quad= A -- B -- C -- D -- cycle;
//Initial Diagram
draw(Circle(O, RADIUS), linewidth(0.8));
draw(quad, linewidth(0.8));
label("$A$",A,W);
label("$B$",B,NW);
label("$C$",C,NE);
label("$D$",D,E);
label("$E$",E,WSW);
label("$O$",O,S);
//Radii
draw(O--A);
draw(O--B);
draw(O--C);
draw(O--D);
//Construction
draw(B--D);
draw(rightanglemark(C,E,D));
// Block 2
size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); E=extension(B,D,O,C); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,E); label("$E$",E,WSW); label("$O$",O,S); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction draw(B--D); draw(rightanglemark(C,E,D)); | [] |
493 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 10A Problems/Problem 24 | size(250);
defaultpen(linewidth(0.4));
//Variable Declarations
real RADIUS;
pair A, B, C, D, O;
RADIUS=3;
//Variable Definitions
A=RADIUS*dir(148.414);
B=RADIUS*dir(109.471);
C=RADIUS*dir(70.529);
D=RADIUS*dir(31.586);
P=RADIUS*dir(270);
O=(0,0);
//Path Definitions
path quad= A -- B -- C -- D -- cycle;
//Initial Diagram
draw(circle(O, RADIUS), linewidth(0.8));
draw(quad, linewidth(0.8));
label("<imath>A</imath>",A,W);
label("<imath>B</imath>",B,NW);
label("<imath>C</imath>",C,NE);
label("<imath>D</imath>",D,E);
label("<imath>O</imath>",O,S);
//Radii
draw(O--A);
draw(O--B);
draw(O--C);
draw(O--D);
draw(P--A);
draw(P--B);
draw(P--C);
draw(P--D);
Draw some point <imath>P</imath> somewhere on major arc <imath>\overarc{AD}</imath>. Let <imath>\angle AOB \cong \angle BOC \cong \angle COD = 2\theta</imath>. Then, by inscribed angle <imath>\angle APB \cong \angle BPC \cong \angle CPD = \theta</imath>. By extended law of sines, we have <imath>\frac{200}{\text{sin}(\theta)} = 2(200\sqrt{2}) \implies \text{sin}(\theta) = \frac{\sqrt{2}}{4}</imath>. Using the triple angle sine formula, we have <imath>\text{sin}(3\theta)=3\text{sin}(\theta) - 4\text{sin}^3(\theta) = \frac{5\sqrt{2}}{8}</imath>.
Then, again by extended law of sines, we have <imath>\frac{AD}{\text{sin}(3\theta)}=400\sqrt{2}</imath>. Thus, <imath>AD = \text{sin}(3\theta) 400\sqrt{2} = \boxed{500}</imath>.
==Solution 3 (HARD Algebra)== | // Block 1
size(250);
defaultpen(linewidth(0.4));
//Variable Declarations
real RADIUS;
pair A, B, C, D, O;
RADIUS=3;
//Variable Definitions
A=RADIUS*dir(148.414);
B=RADIUS*dir(109.471);
C=RADIUS*dir(70.529);
D=RADIUS*dir(31.586);
P=RADIUS*dir(270);
O=(0,0);
//Path Definitions
path quad= A -- B -- C -- D -- cycle;
//Initial Diagram
draw(circle(O, RADIUS), linewidth(0.8));
draw(quad, linewidth(0.8));
label("<imath>A</imath>",A,W);
label("<imath>B</imath>",B,NW);
label("<imath>C</imath>",C,NE);
label("<imath>D</imath>",D,E);
label("<imath>O</imath>",O,S);
//Radii
draw(O--A);
draw(O--B);
draw(O--C);
draw(O--D);
draw(P--A);
draw(P--B);
draw(P--C);
draw(P--D);
Draw some point <imath>P</imath> somewhere on major arc <imath>\overarc{AD}</imath>. Let <imath>\angle AOB \cong \angle BOC \cong \angle COD = 2\theta</imath>. Then, by inscribed angle <imath>\angle APB \cong \angle BPC \cong \angle CPD = \theta</imath>. By extended law of sines, we have <imath>\frac{200}{\text{sin}(\theta)} = 2(200\sqrt{2}) \implies \text{sin}(\theta) = \frac{\sqrt{2}}{4}</imath>. Using the triple angle sine formula, we have <imath>\text{sin}(3\theta)=3\text{sin}(\theta) - 4\text{sin}^3(\theta) = \frac{5\sqrt{2}}{8}</imath>.
Then, again by extended law of sines, we have <imath>\frac{AD}{\text{sin}(3\theta)}=400\sqrt{2}</imath>. Thus, <imath>AD = \text{sin}(3\theta) 400\sqrt{2} = \boxed{500}</imath>.
==Solution 3 (HARD Algebra)==
<asy>
size(250);
defaultpen(linewidth(0.4));
//Variable Declarations
real RADIUS;
pair A, B, C, D, O;
RADIUS=3;
//Variable Definitions
A=RADIUS*dir(148.414);
B=RADIUS*dir(109.471);
C=RADIUS*dir(70.529);
D=RADIUS*dir(31.586);
O=(0,0);
//Path Definitions
path quad= A -- B -- C -- D -- cycle;
//Initial Diagram
draw(Circle(O, RADIUS), linewidth(0.8));
draw(quad, linewidth(0.8));
label("$A$",A,W);
label("$B$",B,NW);
label("$C$",C,NE);
label("$D$",D,E);
label("$O$",O,S);
//Radii
draw(O--A);
draw(O--B);
draw(O--C);
draw(O--D);
// Block 2
size(250);
defaultpen(linewidth(0.4));
//Variable Declarations
real RADIUS;
pair A, B, C, D, O;
RADIUS=3;
//Variable Definitions
A=RADIUS*dir(148.414);
B=RADIUS*dir(109.471);
C=RADIUS*dir(70.529);
D=RADIUS*dir(31.586);
O=(0,0);
//Path Definitions
path quad= A -- B -- C -- D -- cycle;
//Initial Diagram
draw(Circle(O, RADIUS), linewidth(0.8));
draw(quad, linewidth(0.8));
label("$A$",A,W);
label("$B$",B,NW);
label("$C$",C,NE);
label("$D$",D,E);
label("$O$",O,S);
//Radii
draw(O--A);
draw(O--B);
draw(O--C);
draw(O--D);
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Let quadrilateral $ABCD$ be inscribed in circle $O$, where $AD$ is the side of unknown length. Draw the radii from center $O$ to all four vertices of the quadrilateral, and draw the altitude of $\triangle BOC$ such that it passes through side $AD$ at the point $G$ and meets side $BC$ at point $H$.
By the Pythagorean Theorem, the length of $OH$ is
\begin{align*} \sqrt{CO^2 - HC^2} &= \sqrt{(200\sqrt{2})^2 - \left(\frac{200}{2}\right)^2} \\ &= \sqrt{80000 - 10000} \\ &= \sqrt{70000} \\ &= 100\sqrt{7}. \end{align*}
Note that $[ABCDO] = [AOB] + [BOC] + [COD] = [AOD] + [ABCD].$ Let the length of $OG$ be $h$ and the length of $AD$ be $x$; then we have that
$[AOB] + [BOC] + [COD] = \frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} = \frac{x \times h}{2} + \frac{(100\sqrt{7} - h)(200 + x)}{2} = [AOD] + [ABCD].$
Furthermore,
\begin{align*} h &= \sqrt{OD^2 - GD^2} \\ &= \sqrt{(200\sqrt{2})^2 - \left(\frac{x}{2}\right)^2} \\ &= \sqrt{80000 - \frac{x^2}{4}} \end{align*}
Substituting this value of $h$ into the previous equation and evaluating for $x$, we get:
\[\frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} = \frac{x \times h}{2} + \frac{(100\sqrt{7} - h)(200 + x)}{2}\]
\[\frac{3 \times 200 \times 100\sqrt{7}}{2} = \frac{x\sqrt{80000 - \frac{x^2}{4}}}{2} + \frac{\left(100\sqrt{7} - \sqrt{80000 - \frac{x^2}{4}}\right)(200 + x)}{2}\]
\[60000\sqrt{7} = \left(x\sqrt{80000 - \frac{x^2}{4}}\right) + \left(20000\sqrt{7}\right) + \left(100x\sqrt{7}\right) - \left(200\sqrt{80000 - \frac{x^2}{4}}\right) - \left(x\sqrt{80000 - \frac{x^2}{4}}\right)\]
\[40000\sqrt{7} = 100x\sqrt{7} - 200\sqrt{80000 - \frac{x^2}{4}}\]
\[400\sqrt{7} = x\sqrt{7} - 2\sqrt{80000 - \frac{x^2}{4}}\]
\[(x - 400)\sqrt{7} = 2\sqrt{80000 - \frac{x^2}{4}}\]
\[7(x-400)^2 = 4\left(80000 - \frac{x^2}{4}\right)\]
\[7x^2 - 5600x + 1120000 = 320000 - x^2\]
\[8x^2 - 5600x + 800000 = 0\]
\[x^2 - 700x + 100000 = 0\]
The roots of this quadratic are found by using the quadratic formula:
\begin{align*} x &= \frac{-(-700) \pm \sqrt{(-700)^2 - 4 \times 1 \times 100000}}{2 \times 1} \\ &= \frac{700 \pm \sqrt{490000 - 400000}}{2} \\ &= \frac{700}{2} \pm \frac{\sqrt{90000}}{2} \\ &= 350 \pm \frac{300}{2} \\ &= 200, 500 \end{align*}
If the length of $AD$ is $200$, then $ABCD$ would be a square. Thus, the radius of the circle would be
\[\frac{\sqrt{200^2 + 200^2}}{2} = \frac{\sqrt{80000}}{2} = \frac{200\sqrt{2}}{2} = 100\sqrt{2}\]
Which is a contradiction. Therefore, our answer is $\boxed{500}.$ | [] |
493 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 10A Problems/Problem 24 | Construct quadrilateral $ABCD$ on the circle with $AD$ being the missing side (Notice that since the side length is less than the radius, it will be very small on the top of the circle). Now, draw the radii from center $O$ to $A,B,C,$ and $D$. Apply the law of cosines on $\Delta BOC$; let $\theta = \angle BOC$. We get the following equation: \[(BC)^{2}=(OB)^{2}+(OC)^{2}-2\cdot OB \cdot OC\cdot \cos\theta\] Substituting the values in, we get \[(200)^{2}=2\cdot (200)^{2}+ 2\cdot (200)^{2}- 2\cdot 2\cdot (200)^{2}\cdot \cos\theta\] Canceling out, we get \[\cos\theta=\frac{3}{4}\]
Because $\angle AOB$, $\angle BOC$, and $\angle COD$ are congruent, $\angle AOD = 3\theta$. To find the remaining side ($AD$), we simply have to apply the law of cosines to $\Delta AOD$. Now, to find $\cos 3\theta$, we can derive a formula that only uses $\cos\theta$: \[\cos 3\theta=\cos (2\theta+\theta)= \cos 2\theta \cos\theta- (2\sin\theta \cos\theta) \cdot \sin \theta\] \[\cos 3\theta= \cos\theta (\cos 2\theta-2\sin^{2}\theta)=\cos\theta (2\cos^{2}\theta-3+2\cos^{2}\theta)\] \[\Rightarrow \cos 3\theta=4\cos^{3}\theta-3\cos\theta\] It is useful to memorize the triple angle formulas ($\cos 3\theta=4\cos^{3}\theta-3\cos\theta, \sin 3\theta=3\sin\theta-4\sin^{3}\theta$). Plugging in $\cos\theta=\frac{3}{4}$, we get $\cos 3\theta= -\frac{9}{16}$. Now, applying law of cosines on triangle $OAD$, we get \[(AD)^{2}= 2\cdot (200)^{2}+ 2\cdot (200)^{2}+2\cdot 200\sqrt2 \cdot 200\sqrt2 \cdot \frac{9}{16}\] \[\Rightarrow 2\cdot (200)^{2} \cdot (1+1+ \frac{9}{8})=(200)^{2}\cdot \frac{25}{4}\] \[AD=200 \cdot \frac{5}{2}=\boxed{500}\] | // Block 1
size(250);
defaultpen(linewidth(0.4));
//Variable Declarations
real RADIUS;
pair A, B, C, D, O;
RADIUS=3;
//Variable Definitions
A=RADIUS*dir(148.414);
B=RADIUS*dir(109.471);
C=RADIUS*dir(70.529);
D=RADIUS*dir(31.586);
O=(0,0);
//Path Definitions
path quad= A -- B -- C -- D -- cycle;
//Initial Diagram
draw(Circle(O, RADIUS), linewidth(0.8));
draw(quad, linewidth(0.8));
label("$A$",A,W);
label("$B$",B,NW);
label("$C$",C,NE);
label("$D$",D,E);
label("$O$",O,S);
label("$\theta$",O,3N);
//Radii
draw(O--A);
draw(O--B);
draw(O--C);
draw(O--D);
//Angle mark for BOC
draw(anglemark(C,O,B));
// Block 2
size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,E); label("$O$",O,S); label("$\theta$",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Angle mark for BOC draw(anglemark(C,O,B)); | [] |
493 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 10A Problems/Problem 24 | Construct quadrilateral $ABCD$ on the circle $O$ with $AD$ being the desired side. Then, drop perpendiculars from $A$ and $D$ to the extended line of $\overline{BC}$ and let these points be $E$ and $F$, respectively. Also, let $\theta = \angle BOC$. From the Law of Cosines on $\triangle BOC$, we have $\cos \theta = \frac{3}{4}$.
Now, since $\triangle BOC$ is isosceles with $\overline{OB} \cong \overline{OC}$, we have that $\angle BCO = \angle CBO = 90 - \frac{\theta}{2}$. In addition, we know that $\overline{BC} \cong \overline{CD}$ as they are both equal to $200$ and $\overline{OB} \cong \overline{OC} \cong \overline{OD}$ as they are both radii of the same circle. By SSS Congruence, we have that $\triangle OBC \cong \triangle OCD$, so we have that $\angle OCD = \angle BCO = 90 - \frac{\theta}{2}$, so $\angle DCF = \theta$.
Thus, we have $\frac{FC}{DC} = \cos \theta = \frac{3}{4}$, so $FC = 150$. Similarly, $BE = 150$, and $AD = 150 + 200 + 150 = \boxed{500}$. | // Block 1
size(250);
defaultpen(linewidth(0.4));
//Variable Declarations
real RADIUS;
pair A, B, C, D, E, F, O;
RADIUS=3;
//Variable Definitions
A=RADIUS*dir(148.414);
B=RADIUS*dir(109.471);
C=RADIUS*dir(70.529);
D=RADIUS*dir(31.586);
E=foot(A,B,C);
F=foot(D,B,C);
O=(0,0);
//Path Definitions
path quad= A -- B -- C -- D -- cycle;
//Initial Diagram
draw(Circle(O, RADIUS), linewidth(0.8));
draw(quad, linewidth(0.8));
label("$A$",A,W);
label("$B$",B,NW);
label("$C$",C,NE);
label("$D$",D,ENE);
label("$O$",O,S);
label("$\theta$",O,3N);
//Radii
draw(O--A);
draw(O--B);
draw(O--C);
draw(O--D);
//Construction
draw(A--E);
draw(E--B);
draw(C--F);
draw(F--D);
label("$E$",E,NW);
label("$F$",F,NE);
//Angle marks
draw(anglemark(C,O,B));
draw(rightanglemark(A,E,B));
draw(rightanglemark(C,F,D));
// Block 2
size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); E=foot(A,B,C); F=foot(D,B,C); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,ENE); label("$O$",O,S); label("$\theta$",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction draw(A--E); draw(E--B); draw(C--F); draw(F--D); label("$E$",E,NW); label("$F$",F,NE); //Angle marks draw(anglemark(C,O,B)); draw(rightanglemark(A,E,B)); draw(rightanglemark(C,F,D)); | [] |
493 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 10A Problems/Problem 24 | Let $s = 200$. Let $O$ be the center of the circle. Then $AC$ is twice the altitude of $\triangle OBC$ to $\overline{OB}$. Since $\triangle OBC$ is isosceles we can compute its area to be $\frac{s^2 \sqrt{7}}{4}$, hence $CA = 2 \cdot \tfrac{2 \cdot s^2\sqrt7/4}{s\sqrt2} = s\sqrt{\frac{7}{2}}$.
Now by Ptolemy's Theorem we have $CA^2 = s^2 + AD \cdot s \implies AD = \left(\frac{7}{2}-1\right)s.$ This gives us: \[\boxed{\textbf{(E) } 500.}\] | // Block 1
pathpen = black; pointpen = black;
size(6cm);
draw(unitcircle);
pair A = D("A", dir(50), dir(50));
pair B = D("B", dir(90), dir(90));
pair C = D("C", dir(130), dir(130));
pair D = D("D", dir(170), dir(170));
pair O = D("O", (0,0), dir(-90));
draw(A--C, red);
draw(B--D, blue+dashed);
draw(A--B--C--D--cycle);
draw(A--O--C);
draw(O--B);
// Block 2
pathpen = black; pointpen = black; size(6cm); draw(unitcircle); pair A = D("A", dir(50), dir(50)); pair B = D("B", dir(90), dir(90)); pair C = D("C", dir(130), dir(130)); pair D = D("D", dir(170), dir(170)); pair O = D("O", (0,0), dir(-90)); draw(A--C, red); draw(B--D, blue+dashed); draw(A--B--C--D--cycle); draw(A--O--C); draw(O--B); | [] |
493 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 10A Problems/Problem 24 | Label the points as shown, and let $\angle{EOF} = \theta$. Since $\overline{OB} = \overline{OC}$, and $\triangle{OFE} \sim \triangle{OCB}$, we get that $\angle{EFO} = 90-\frac{\theta}{2}$. We assign $\alpha$ to $90-\frac{\theta}{2}$ for simplicity.
From here, by vertical angles $\angle{CFD} = \alpha$. Also, since $\triangle{OCB} \cong \triangle{ODC}$, $\angle{OCD} = \alpha$. This means that $\angle{CDF} = 180-2\alpha = \theta$, which leads to $\triangle{OCB} \sim \triangle{DCF}$.
Since we know that $\overline{CD} = 200$, $\overline{DF} = 200$, and by similar reasoning $\overline{AE} = 200$.
Finally, again using similar triangles, we get that $\overline{CF} = 100\sqrt{2}$, which means that $\overline{OF} = \overline{OC} - \overline{CF} = 200\sqrt{2} - 100\sqrt{2} = 100\sqrt{2}$. We can again apply similar triangles (or use Power of a Point) to get $\overline{EF} = 100$, and finally $\overline{AD} = \overline{AE}+\overline{EF}+\overline{FD} = 200+100+200=\boxed{\textbf{(E)}500}$ - ColtsFan10 | // Block 1
size(250);
defaultpen(linewidth(0.4));
//Variable Declarations, L is used to write alpha= statement
real RADIUS;
pair A, B, C, D, E, F, O, L;
RADIUS=3;
//Variable Definitions
A=RADIUS*dir(148.414);
B=RADIUS*dir(109.471);
C=RADIUS*dir(70.529);
D=RADIUS*dir(31.586);
E=extension(A,D,O,B);
F=extension(A,D,O,C);
L=midpoint(C--D);
O=(0,0);
//Path Definitions
path quad = A -- B -- C -- D -- cycle;
//Initial Diagram
draw(circle(O, RADIUS), linewidth(0.8));
draw(quad, linewidth(0.8));
label("$A$",A,NW);
label("$B$",B,NW);
label("$C$",C,NE);
label("$D$",D,NE);
label("$E$",E,SW);
label("$F$",F,SE);
label("$O$",O,SE);
dot(O,linewidth(5));
//Radii
draw(O--A);
draw(O--B);
draw(O--C);
draw(O--D);
//Construction
label("$\alpha = 90-\frac{\theta}{2}$",L,5NE,rgb(128, 0, 0));
draw(anglemark(C,O,B));
label("$\theta$",O,3N);
draw(anglemark(E,F,O));
label("$\alpha$",F,3SW);
draw(anglemark(D,F,C));
label("$\alpha$",F,3NE);
draw(anglemark(F,C,D));
label("$\alpha$",C,3SSE);
draw(anglemark(C,D,F));
label("$\theta$",(RADIUS-0.04)*dir(31.586),3WNW);
// Block 2
size(250); defaultpen(linewidth(0.4)); //Variable Declarations, L is used to write alpha= statement real RADIUS; pair A, B, C, D, E, F, O, L; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); E=extension(A,D,O,B); F=extension(A,D,O,C); L=midpoint(C--D); O=(0,0); //Path Definitions path quad = A -- B -- C -- D -- cycle; //Initial Diagram draw(circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("$A$",A,NW); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,NE); label("$E$",E,SW); label("$F$",F,SE); label("$O$",O,SE); dot(O,linewidth(5)); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction label("$\alpha = 90-\frac{\theta}{2}$",L,5NE,rgb(128, 0, 0)); draw(anglemark(C,O,B)); label("$\theta$",O,3N); draw(anglemark(E,F,O)); label("$\alpha$",F,3SW); draw(anglemark(D,F,C)); label("$\alpha$",F,3NE); draw(anglemark(F,C,D)); label("$\alpha$",C,3SSE); draw(anglemark(C,D,F)); label("$\theta$",(RADIUS-0.04)*dir(31.586),3WNW); | [] |
493 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 10A Problems/Problem 24 | Claim: $[ABCD]$ is an isosceles trapezoid.
Proof: Notice that $[ABCD]$ is cyclic, triangle $BOC$ is isosceles, and triangle $AOB$ is congruent to $DOC$ by SSS congruence. Therefore, $\angle BAD = 180 - \angle BCD = 180-(\angle BCO + \angle DCO)=180-(\angle CBO+\angle ABO) = 180 - \angle ABC = \angle CDA$. Hence, $[ABCD]$ is an isosceles trapezoid.
Let $\angle CDA=\alpha$. Notice that the length of the altitude from $C$ to $AD$ is $200sin(\alpha)$. Furthermore, the length of the altitude from $O$ to $BC$ is $100\sqrt{7}$ by the Pythagorean theorem. Therefore, the length of the altitude from $O$ to $AD$ is $100\sqrt{7}-200sin\alpha$. Let $F$ the feet of the altitude from $O$ to $AD$. Then, $FD=(200+400cos(\alpha))/2=100+200cos(\alpha)$, because $AOD$ is isosceles.
Therefore, by the Pythagorean theorem, $(100+200cos(\alpha))^2+(100\sqrt{7}-200sin(\alpha))^2=80000$. Simplifying, we have $1+cos(\alpha)=sin(\alpha) \cdot sqrt{7} \implies cos^2(\alpha)+2cos(\alpha)+1=sin^2(\alpha) \cdot 7 = 7-7cos^2(\alpha) \implies 8cos^2(\alpha)+2cos(\alpha) - 6 =0$. Solving this quadratic, we have $cos(\alpha)=\frac{3}{4}, -1$, but $0<\alpha<180 \implies cos(\alpha)=3/4$. Therefore, $AD=200cos(\alpha)+200cos(\alpha)+200=\boxed{500}$
- [mathMagicOPS] | // Block 1
size(250);
defaultpen(linewidth(0.4));
//Variable Declarations
real RADIUS;
pair A, B, C, D, O;
RADIUS=3;
//Variable Definitions
A=RADIUS*dir(148.414);
B=RADIUS*dir(109.471);
C=RADIUS*dir(70.529);
D=RADIUS*dir(31.586);
O=(0,0);
//Path Definitions
path quad= A -- B -- C -- D -- cycle;
//Initial Diagram
draw(Circle(O, RADIUS), linewidth(0.8));
draw(quad, linewidth(0.8));
label("$A$",A,W);
label("$B$",B,NW);
label("$C$",C,NE);
label("$D$",D,E);
label("$O$",O,S);
//Radii
draw(O--A);
draw(O--B);
draw(O--C);
draw(O--D);
// Block 2
size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,E); label("$O$",O,S); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); | [] |
493 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 10A Problems/Problem 24 | Since $AB$, $BC$, and $CD$ are congruent chords, the triangles formed by connecting these sides to the circumcenter $O$ (i.e., $\triangle AOB$, $\triangle BOC$, $\triangle COD$) are congruent isosceles triangles.
Let $\alpha = \angle OCB$. In $\triangle BOC$, the side length $BC=200$ and $OC = OB = 200\sqrt{2}$. By dropping an altitude from $O$ to $BC$, we bisect $BC$. Half the base is $\frac{200}{2} = 100$.
Using the right triangle formed, we find $\cos(\alpha)$:
\[\cos(\alpha) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{100}{200\sqrt{2}} = \frac{\sqrt{2}}{4}\]
Since $\triangle BOC \cong \triangle COD$, we have $m\angle OCB = m\angle OCD = \alpha$.
The interior angle measure $m\angle DCB$ is given by:
\[m\angle DCB = m\angle OCB + m\angle OCD = 2\alpha\]
We calculate $\cos(2\alpha)$ using the double-angle identity:
\[\cos(2\alpha) = 2\cos^2(\alpha) - 1\]
\[\cos(2\alpha) = 2\left(\frac{\sqrt{2}}{4}\right)^2 - 1 = \frac{1}{4} - 1 = -\frac{3}{4}\]
Now, apply the Law of Cosines to $\triangle BCD$ to find $BD^2$:
\[BD^2 = BC^2 + CD^2 - 2(BC)(CD)\cos(\angle DCB)\]
\[BD^2 = 200^2 + 200^2 - 2(200)(200)\left(-\frac{3}{4}\right)\]
\[BD^2 = 80000 + 60000 = 140000\]
Since $ABCD$ is a cyclic quadrilateral, the opposite angles are supplementary:
\[m\angle DAB = 180^\circ - m\angle DCB = 180^\circ - 2\alpha\]
Thus, we find $\cos(\angle DAB)$:
\[\cos(\angle DAB) = \cos(180^\circ - 2\alpha) = -\cos(2\alpha)\]
\[\cos(\angle DAB) = -\left(-\frac{3}{4}\right) = \frac{3}{4}\]
Finally, we apply the Law of Cosines to $\triangle DAB$ to find the side length $AD$. Let $AD = a$:
\[140000 = 200^2 + a^2 - 2(200)(a)\left(\frac{3}{4}\right)\]
\[140000 = 40000 + a^2 - 300a\]
Rearranging the terms yields the quadratic equation:
\[a^2 - 300a - 100000 = 0\]
We solve this equation by factoring:
\[(a - 500)(a + 200) = 0\]
The two possible solutions for $a$ are $a=500$ and $a=-200$. Since $a$ represents a side length, it must be positive, so $a = 500$, which means $AD = 500$, or $\boxed{\textbf{(E) } 500}$.
~Voidling | // Block 1
size(250);
defaultpen(linewidth(0.4));
//Variable Declarations
real RADIUS;
pair A, B, C, D, E, F, O;
RADIUS=3;
//Variable Definitions
A=RADIUS*dir(148.414);
B=RADIUS*dir(109.471);
C=RADIUS*dir(70.529);
D=RADIUS*dir(31.586);
O=(0,0);
//Path Definitions
path quad= A -- B -- C -- D -- cycle;
//Initial Diagram
draw(Circle(O, RADIUS), linewidth(0.8));
draw(quad, linewidth(0.8));
label("$A$",A,W);
label("$B$",B,NW);
label("$C$",C,NE);
label("$D$",D,ENE);
label("$O$",O,S);
label("$\theta$",O,3N);
//Radii
draw(O--A);
draw(O--B);
draw(O--C);
draw(O--D);
//Construction
E=extension(B,O,A,D);
label("$E$",E,NE);
F=extension(C,O,A,D);
label("$F$",F,NE);
//Angle marks
draw(anglemark(C,O,B));
// Block 2
size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,ENE); label("$O$",O,S); label("$\theta$",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction E=extension(B,O,A,D); label("$E$",E,NE); F=extension(C,O,A,D); label("$F$",F,NE); //Angle marks draw(anglemark(C,O,B)); | [] |
494 | Circles with centers $P, Q$ and $R$, having radii $1, 2$ and $3$, respectively, lie on the same side of line $l$ and are tangent to $l$ at $P', Q'$ and $R'$, respectively, with $Q'$ between $P'$ and $R'$. The circle with center $Q$ is externally tangent to each of the other two circles. What is the area of triangle $PQR$?
$\textbf{(A) } 0\qquad \textbf{(B) } \sqrt{\frac{2}{3}}\qquad\textbf{(C) } 1\qquad\textbf{(D) } \sqrt{6}-\sqrt{2}\qquad\textbf{(E) }\sqrt{\frac{3}{2}}$ | 2016 AMC 12A Problems/Problem 15 | Notice that we can find $[P'PQRR']$ in two different ways: $[P'PQQ']+[Q'QRR']$ and $[PQR]+[P'PRR'].$ Since we want $[PQR],$ we use the latter method, so we have $[P'PQQ']+[Q'QRR']=[PQR]+[P'PRR'].$
$\break$
$P'Q'=\sqrt{PQ^2-(QQ'-PP')^2}=\sqrt{3^2-1^2}=\sqrt{8}=2\sqrt{2}$. Additionally, $Q'R'=\sqrt{QR^2-(RR'-QQ')^2}=\sqrt{5^2-1^2}=\sqrt{24}=2\sqrt{6}$. Therefore, $[P'PQQ']=\frac{P'P+Q'Q}{2}*2\sqrt{2}=\frac{1+2}{2}*2\sqrt{2}=3\sqrt{2}$. Similarly, $[Q'QRR']=5\sqrt6$. We can calculate $[P'PRR']$ easily because $P'R'=P'Q'+Q'R'=2\sqrt{2}+2\sqrt{6}$. $[P'PRR']=4\sqrt{2}+4\sqrt{6}$. $\newline$
Plugging into first equation, the two sums of areas, $3\sqrt{2}+5\sqrt{6}=4\sqrt{2}+4\sqrt{6}+[PQR],$ so therefore $[PQR]=\boxed{\textbf{(D) }\sqrt{6}-\sqrt{2}.}$ | // Block 1
size(250);
defaultpen(linewidth(0.4));
//Variable Declarations
pair P,Q,R,Pp,Qp,Rp;
pair A,B;
//Variable Definitions
A=(-5, 0);
B=(8, 0);
P=(-2.828,1);
Q=(0,2);
R=(4.899,3);
Pp=foot(P,A,B);
Qp=foot(Q,A,B);
Rp=foot(R,A,B);
path PQR = P--Q--R--cycle;
//Initial Diagram
dot(P);
dot(Q);
dot(R);
dot(Pp);
dot(Qp);
dot(Rp);
draw(Circle(P, 1), linewidth(0.8));
draw(Circle(Q, 2), linewidth(0.8));
draw(Circle(R, 3), linewidth(0.8));
draw(A--B,Arrows);
label("$P$",P,N);
label("$Q$",Q,N);
label("$R$",R,N);
label("$P'$",Pp,S);
label("$Q'$",Qp,S);
label("$R'$",Rp,S);
label("$l$",B,E);
//Added lines
draw(PQR);
draw(P--Pp);
draw(Q--Qp);
draw(R--Rp);
//Angle marks
draw(rightanglemark(P,Pp,B));
draw(rightanglemark(Q,Qp,B));
draw(rightanglemark(R,Rp,B));
// Block 2
size(250); defaultpen(linewidth(0.4)); //Variable Declarations pair P,Q,R,Pp,Qp,Rp; pair A,B; //Variable Definitions A=(-5, 0); B=(8, 0); P=(-2.828,1); Q=(0,2); R=(4.899,3); Pp=foot(P,A,B); Qp=foot(Q,A,B); Rp=foot(R,A,B); path PQR = P--Q--R--cycle; //Initial Diagram dot(P); dot(Q); dot(R); dot(Pp); dot(Qp); dot(Rp); draw(Circle(P, 1), linewidth(0.8)); draw(Circle(Q, 2), linewidth(0.8)); draw(Circle(R, 3), linewidth(0.8)); draw(A--B,Arrows); label("$P$",P,N); label("$Q$",Q,N); label("$R$",R,N); label("$P'$",Pp,S); label("$Q'$",Qp,S); label("$R'$",Rp,S); label("$l$",B,E); //Added lines draw(PQR); draw(P--Pp); draw(Q--Qp); draw(R--Rp); //Angle marks draw(rightanglemark(P,Pp,B)); draw(rightanglemark(Q,Qp,B)); draw(rightanglemark(R,Rp,B)); | [] |
494 | Circles with centers $P, Q$ and $R$, having radii $1, 2$ and $3$, respectively, lie on the same side of line $l$ and are tangent to $l$ at $P', Q'$ and $R'$, respectively, with $Q'$ between $P'$ and $R'$. The circle with center $Q$ is externally tangent to each of the other two circles. What is the area of triangle $PQR$?
$\textbf{(A) } 0\qquad \textbf{(B) } \sqrt{\frac{2}{3}}\qquad\textbf{(C) } 1\qquad\textbf{(D) } \sqrt{6}-\sqrt{2}\qquad\textbf{(E) }\sqrt{\frac{3}{2}}$ | 2016 AMC 12A Problems/Problem 15 | The above diagram can be achieved relatively simply using basic knowledge of the Pythagorean theorem and the fact that the radius from the center to the point of tangency is perpendicular to the tangent line. From there, observe that $[PQRY]$ can be calculated in two ways: $[\triangle PQX] + [QZYX] + [\triangle QRZ]$ and $[\triangle PRY] + [\triangle PQR]$. Solving, we get:
\begin{align*} [\triangle PQR] &= [PQRY] - [\triangle PRY] \\ &= [\triangle PQX] + [QZYX] + [\triangle QRZ] - [\triangle PRY] \\ &= \sqrt{2} + 2\sqrt{6} + \sqrt{6}- 2\sqrt{2}-2\sqrt{6} \\ &= \boxed{\textbf{(D)} \sqrt{6}-\sqrt{2}} \end{align*}
- ColtsFan10, diagram partially borrowed from Solution 1 | // Block 1
// Initial Pen Sizing
size(250);
defaultpen(linewidth(0.4));
defaultpen(fontsize(10pt));
// Variable Declarations
pair P,Q,R,Pp,Qp,Rp,X,Y,Z,A,B;
// Variable Definitions
A=(-5, 0);
B=(8, 0);
P=(-2.828,1);
Q=(0,2);
R=(4.899,3);
X=(0,1);
Y=(4.899,1);
Z=(4.899,2);
Pp=foot(P,A,B);
Qp=foot(Q,A,B);
Rp=foot(R,A,B);
path PQR = P--Q--R--cycle;
//Initial Diagram
dot(P);
dot(Q);
dot(R);
dot(Pp);
dot(Qp);
dot(Rp);
dot(X);
dot(Y);
dot(Z);
draw(Circle(P, 1), linewidth(0.3));
draw(Circle(Q, 2), linewidth(0.3));
draw(Circle(R, 3), linewidth(0.3));
draw(A--B,Arrows);
label("$P$",P,N);
label("$Q$",Q,N);
label("$R$",R,N);
label("$P'$",Pp,S);
label("$Q'$",Qp,S);
label("$R'$",Rp,S);
label("$l$",B,E);
label("$X$",X,NE);
label("$Y$",Y,E);
label("$Z$",Z,E);
//Added lines
filldraw(PQR,gray(0.8));
draw(P--Pp,linetype("8 8"));
draw(Q--Qp,linetype("8 8"));
draw(R--Rp,linetype("8 8"));
draw(P--Y,linetype("8 8"));
draw(Q--Z,linetype("8 8"));
//Angle marks
draw(rightanglemark(R,Y,P));
draw(rightanglemark(Q,X,P));
draw(rightanglemark(R,Z,Q));
//Length labeling
label("$2\sqrt{2}$",P--X,fontsize(8pt));
label("$2\sqrt{6}$",X--Y,fontsize(8pt));
label("$2\sqrt{6}$",Q--Z,fontsize(8pt));
label("$1$",R--Z,E,fontsize(8pt));
label("$1$",Z--Y,E,fontsize(8pt));
label("$3$",(-2.828,1.3)--Q,W,fontsize(8pt));
label("$5$",Q--R,N,fontsize(8pt));
// Block 2
// Initial Pen Sizing size(250); defaultpen(linewidth(0.4)); defaultpen(fontsize(10pt)); // Variable Declarations pair P,Q,R,Pp,Qp,Rp,X,Y,Z,A,B; // Variable Definitions A=(-5, 0); B=(8, 0); P=(-2.828,1); Q=(0,2); R=(4.899,3); X=(0,1); Y=(4.899,1); Z=(4.899,2); Pp=foot(P,A,B); Qp=foot(Q,A,B); Rp=foot(R,A,B); path PQR = P--Q--R--cycle; //Initial Diagram dot(P); dot(Q); dot(R); dot(Pp); dot(Qp); dot(Rp); dot(X); dot(Y); dot(Z); draw(Circle(P, 1), linewidth(0.3)); draw(Circle(Q, 2), linewidth(0.3)); draw(Circle(R, 3), linewidth(0.3)); draw(A--B,Arrows); label("$P$",P,N); label("$Q$",Q,N); label("$R$",R,N); label("$P'$",Pp,S); label("$Q'$",Qp,S); label("$R'$",Rp,S); label("$l$",B,E); label("$X$",X,NE); label("$Y$",Y,E); label("$Z$",Z,E); //Added lines filldraw(PQR,gray(0.8)); draw(P--Pp,linetype("8 8")); draw(Q--Qp,linetype("8 8")); draw(R--Rp,linetype("8 8")); draw(P--Y,linetype("8 8")); draw(Q--Z,linetype("8 8")); //Angle marks draw(rightanglemark(R,Y,P)); draw(rightanglemark(Q,X,P)); draw(rightanglemark(R,Z,Q)); //Length labeling label("$2\sqrt{2}$",P--X,fontsize(8pt)); label("$2\sqrt{6}$",X--Y,fontsize(8pt)); label("$2\sqrt{6}$",Q--Z,fontsize(8pt)); label("$1$",R--Z,E,fontsize(8pt)); label("$1$",Z--Y,E,fontsize(8pt)); label("$3$",(-2.828,1.3)--Q,W,fontsize(8pt)); label("$5$",Q--R,N,fontsize(8pt)); | [] |
495 | Circle $C_0$ has radius $1$, and the point $A_0$ is a point on the circle. Circle $C_1$ has radius $r<1$ and is internally tangent to $C_0$ at point $A_0$. Point $A_1$ lies on circle $C_1$ so that $A_1$ is located $90^{\circ}$ counterclockwise from $A_0$ on $C_1$. Circle $C_2$ has radius $r^2$ and is internally tangent to $C_1$ at point $A_1$. In this way a sequence of circles $C_1,C_2,C_3,\ldots$ and a sequence of points on the circles $A_1,A_2,A_3,\ldots$ are constructed, where circle $C_n$ has radius $r^n$ and is internally tangent to circle $C_{n-1}$ at point $A_{n-1}$, and point $A_n$ lies on $C_n$ $90^{\circ}$ counterclockwise from point $A_{n-1}$, as shown in the figure below. There is one point $B$ inside all of these circles. When $r = \frac{11}{60}$, the distance from the center $C_0$ to $B$ is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2017 AIME II Problems/Problem 12 | Using the invariance again as in Solution 3, assume $B$ is $d$ away from the origin. The locus of possible points is a circle with radius $d$. Consider the following diagram.
Let the distance from $B$ to $(1,0)$ be $x$. As $B$ is invariant, $x = r(BB' + x) \implies x = r\frac{d\sqrt{2}}{1-r}$. Then by Power of a Point, $x(BB' + x) = (1-d)(1+d) \implies xr(BB' + x) = r(1-d)(1+d) \implies x^2 = r(1-d^2) \implies d^2 = \left(1 + \frac{2r}{(1-r)^2}\right)$. Solving, $d = \frac{49}{61}$. Our answer is $49+61=\boxed{110}$
-Isogonal | // Block 1
size(7cm);
draw(circle((0,0), 49/61));
draw((0,0)--(0.790110185, 0.144853534));
draw((0,0)--(-0.144853534, 0.790110185));
draw((-0.144853534, 0.790110185)--(1,0));
draw((0,0)--(1,0));
draw(rightanglemark((-0.144853534, 0.790110185), (0,0), (0.790110185, 0.144853534), 3));
label("$O$", (0,0), SW);
label("$(1,0)$", (1,0), E);
label("$B$", (0.790110185, 0.144853534), NE);
label("$B'$", (-0.144853534, 0.790110185), N);
label("$d$", (0.5 * 49/61, 0), S);
// Block 2
size(7cm); draw(circle((0,0), 49/61)); draw((0,0)--(0.790110185, 0.144853534)); draw((0,0)--(-0.144853534, 0.790110185)); draw((-0.144853534, 0.790110185)--(1,0)); draw((0,0)--(1,0)); draw(rightanglemark((-0.144853534, 0.790110185), (0,0), (0.790110185, 0.144853534), 3)); label("$O$", (0,0), SW); label("$(1,0)$", (1,0), E); label("$B$", (0.790110185, 0.144853534), NE); label("$B'$", (-0.144853534, 0.790110185), N); label("$d$", (0.5 * 49/61, 0), S); | [] |
496 | The diameter $AB$ of a circle of radius $2$ is extended to a point $D$ outside the circle so that $BD=3$. Point $E$ is chosen so that $ED=5$ and line $ED$ is perpendicular to line $AD$. Segment $AE$ intersects the circle at a point $C$ between $A$ and $E$. What is the area of $\triangle ABC$?
$\textbf{(A)}\ \frac{120}{37}\qquad\textbf{(B)}\ \frac{140}{39}\qquad\textbf{(C)}\ \frac{145}{39}\qquad\textbf{(D)}\ \frac{140}{37}\qquad\textbf{(E)}\ \frac{120}{31}$ | 2017 AMC 12B Problems/Problem 18 | Let $O$ be the center of the circle. Note that $EC + CA = EA = \sqrt{AD^2 + DE^2} = \sqrt{(2+2+3)^2 + 5^2} = \sqrt{74}$. However, by Power of a Point, $(EC)(EC + CA) = EO^2 - R^2 = (2+3)^2 + 5^2 - 2^2 = 25 + 25 - 4 = 46 \implies EC = \frac{46}{\sqrt{74}}$, so $AC = \sqrt{74} - \frac{46}{\sqrt{74}} = \frac{28}{\sqrt{74}}$. Now $BC = \sqrt{AB^2 - AC^2} = \sqrt{4^2 - \frac{28^2}{74}} = \sqrt{\frac{16 \cdot 74 - 28^2}{74}} = \sqrt{\frac{1184 - 784}{74}} = \frac{20}{\sqrt{74}}$. Since $\angle ACB = 90^{\circ}, [ABC] = \frac{1}{2} \cdot BC \cdot AC = \frac{1}{2} \cdot \frac{20}{\sqrt{74}} \cdot \frac{28}{\sqrt{74}} = \boxed{\textbf{(D)}\ \frac{140}{37}}$. | // Block 1
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(8.865514650638614cm);
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 = -6.36927122464312, xmax = 11.361758076634109, ymin = -3.789601803155515, ymax = 7.420015026296013; /* image dimensions */
draw((-2.,0.)--(0.6486486486486486,1.8918918918918919)--(2.,0.)--cycle);
/* draw figures */
draw(circle((0.,0.), 2.));
draw((-2.,0.)--(5.,5.));
draw((5.,5.)--(5.,0.));
draw((5.,0.)--(-2.,0.));
draw((-2.,0.)--(0.6486486486486486,1.8918918918918919));
draw((0.6486486486486486,1.8918918918918919)--(2.,0.));
draw((2.,0.)--(-2.,0.));
draw((2.,0.)--(5.,5.));
draw((0.,0.)--(5.,5.));
/* dots and labels */
dot((0.,0.),dotstyle);
label("$O$", (-0.10330578512396349,-0.39365890308038826), NE * labelscalefactor);
dot((-2.,0.),dotstyle);
label("$A$", (-2.2370398196844437,-0.42371149511645134), NE * labelscalefactor);
dot((2.,0.),dotstyle);
label("$B$", (2.045454545454548,-0.36360631104432517), NE * labelscalefactor);
dot((5.,0.),dotstyle);
label("$D$", (4.900450788880542,-0.42371149511645134), NE * labelscalefactor);
dot((5.,5.),dotstyle);
label("$E$", (5.06574004507889,5.15104432757325), NE * labelscalefactor);
dot((0.6486486486486486,1.8918918918918919),linewidth(3.pt) + dotstyle);
label("$C$", (0.48271975957926694,2.100706235912847), 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(8.865514650638614cm); 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 = -6.36927122464312, xmax = 11.361758076634109, ymin = -3.789601803155515, ymax = 7.420015026296013; /* image dimensions */ draw((-2.,0.)--(0.6486486486486486,1.8918918918918919)--(2.,0.)--cycle); /* draw figures */ draw(circle((0.,0.), 2.)); draw((-2.,0.)--(5.,5.)); draw((5.,5.)--(5.,0.)); draw((5.,0.)--(-2.,0.)); draw((-2.,0.)--(0.6486486486486486,1.8918918918918919)); draw((0.6486486486486486,1.8918918918918919)--(2.,0.)); draw((2.,0.)--(-2.,0.)); draw((2.,0.)--(5.,5.)); draw((0.,0.)--(5.,5.)); /* dots and labels */ dot((0.,0.),dotstyle); label("$O$", (-0.10330578512396349,-0.39365890308038826), NE * labelscalefactor); dot((-2.,0.),dotstyle); label("$A$", (-2.2370398196844437,-0.42371149511645134), NE * labelscalefactor); dot((2.,0.),dotstyle); label("$B$", (2.045454545454548,-0.36360631104432517), NE * labelscalefactor); dot((5.,0.),dotstyle); label("$D$", (4.900450788880542,-0.42371149511645134), NE * labelscalefactor); dot((5.,5.),dotstyle); label("$E$", (5.06574004507889,5.15104432757325), NE * labelscalefactor); dot((0.6486486486486486,1.8918918918918919),linewidth(3.pt) + dotstyle); label("$C$", (0.48271975957926694,2.100706235912847), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ | [] |
497 | $ABCD$ is a parallelogram. $g$ is a line passing $A$. Prove that the distance from $C$ to $g$ is either the sum or the difference of the distance from $B$ to $g$, and the distance from $D$ to $g$. | 2017 Indonesia MO Problems/Problem 1 | In order to prove that the distance from $C$ to $g$ is either the sum or the difference of the distance from $B$ to $g$, and the distance from $D$ to $g$, we will use casework to show that the statement is true for all scenarios.
Case 1: $g$ passes through $D$ or $B$
If $g$ passes through $D$, then the distance from $g$ to $D$ is zero. The distance from $g$ to $B$ is the same as the distance from $g$ to $C$ because $BC$ is parallel to $g$. That means the distance from $C$ to $g$ is the sum of the distance from from $B$ to $g$ and from $D$ to $g$. By using similar steps, if $g$ passes through $B$, the distance from $C$ to $g$ is the sum of the distance from from $B$ to $g$ and from $D$ to $g$.
Case 2: $g$ passes through $C$
By SSS Congruency, $\triangle ACD \cong \triangle CAB$. Since the area of the two triangles is the same, the distance from $D$ to $g$ equals the distance from $B$ to $g$. Because the distance from $C$ to $g$ is zero, the distance from $C$ to $g$ is the difference of the distance from from $B$ to $g$ and from $D$ to $g$.
Case 3: $g$ passes through $CD$ or $BC$
Let $E$ be the intersection of lines $DC$ and $g$, and let $X_1, X_2, X_3$ be points on $g$ such that $CX_1, BX_2, DX_3 \perp g$. Also, let $a = AB$, $x = DE$, and $b = BX_2$, making $EC = a-x$.
By Alternate Interior Angles Theorem and Vertical Angle Theorem, $\angle EAB = \angle AED = \angle CEX_1$. Thus, by AA Similarity, $\triangle X_2BA \sim \triangle X_3DE \sim \triangle X_1CE$.
Using similar triangle ratios, we have $DX_3 = \tfrac{bx}{a}$ and $X_1 = \tfrac{ba-bx}{a}$. Thus, we have $BX_2 - DX_3 = \frac{ba-bx}{a} = CX_1$, so the distance from $C$ to $g$ is the difference between the distance from $D$ to $g$ and the distance from $B$ to $g$ if $g$ passes through $CD$. By symmetry, we can also show that the distance from $C$ to $g$ is the difference between the distance from $D$ to $g$ and the distance from $B$ to $g$ if $g$ passes through $BC$.
Case 4: $g$ does not intersect the parallelogram at any other points
Let $E$ be the intersection of lines $DC$ and $g$, and let $X_1, X_2, X_3$ be points on $g$ such that $DX_1, CX_2, BX_3 \perp g$. Also, let $a = AB$, $x = DE$, and $b = BX_3$, making $EC = a+x$.
By Alternate Interior Angles Theorem, $\angle CEA = \angle BAX_3$. Thus, by AA Similarity, $\triangle EDX_1 \sim \triangle ECX_2 \sim \triangle ABX_3$.
Using similar triangle ratios, we have $DX_1 = \tfrac{bx}{a}$ and $CX_2 = \tfrac{ba+bx}{a}$. Thus, we have $BX_3 + DX_1 = \frac{ba+bx}{a} = CX_2$, so the distance from $C$ to $g$ is the sum of the distance from $D$ to $g$ and the distance from $B$ to $g$ if $g$ passes through $CD$.
In all of the cases, the sum or difference of the distance from $B$ to $g$ and the distance from $D$ to $g$ is equal to the distance from $C$ to $g$. | // Block 1
pair A=(0,0),B=(50,0),C=(60,40),D=(10,40);
draw(A--B--C--D--A);
dot(A);
label("A",A,SW);
dot(B);
label("B",B,SE);
dot(C);
label("C",C,NE);
dot(D);
label("D",D,NW);
pair e=(48,40),XA=(55.082,45.902),XB=(29.508,24.59),XC=(25.574,21.311);
draw(A--XA--C);
draw(D--XC);
draw(B--XB);
dot(e);
label("E",e,N);
dot(XA);
label("$X_1$",XA,N);
dot(XB);
label("$X_2$",XB,E);
dot(XC);
label("$X_3$",XC,S);
// Block 2
pair A=(0,0),B=(50,0),C=(60,40),D=(10,40);
draw(A--B--C--D--A);
dot(A);
label("A",A,SW);
dot(B);
label("B",B,SE);
dot(C);
label("C",C,NE);
dot(D);
label("D",D,NW);
pair e=(-40,40),XA=(-15,15),XB=(10,-10),XC=(25,-25);
draw(D--e--XC--B);
draw(D--XA);
draw(C--XB);
draw(B--XC);
dot(e);
label("E",e,N);
dot(XA);
label("$X_1$",XA,S);
dot(XB);
label("$X_2$",XB,S);
dot(XC);
label("$X_3$",XC,S);
// Block 3
pair A=(0,0),B=(50,0),C=(60,40),D=(10,40); draw(A--B--C--D--A); dot(A); label("A",A,SW); dot(B); label("B",B,SE); dot(C); label("C",C,NE); dot(D); label("D",D,NW); pair e=(48,40),XA=(55.082,45.902),XB=(29.508,24.59),XC=(25.574,21.311); draw(A--XA--C); draw(D--XC); draw(B--XB); dot(e); label("E",e,N); dot(XA); label("$X_1$",XA,N); dot(XB); label("$X_2$",XB,E); dot(XC); label("$X_3$",XC,S);
// Block 4
pair A=(0,0),B=(50,0),C=(60,40),D=(10,40); draw(A--B--C--D--A); dot(A); label("A",A,SW); dot(B); label("B",B,SE); dot(C); label("C",C,NE); dot(D); label("D",D,NW); pair e=(-40,40),XA=(-15,15),XB=(10,-10),XC=(25,-25); draw(D--e--XC--B); draw(D--XA); draw(C--XB); draw(B--XC); dot(e); label("E",e,N); dot(XA); label("$X_1$",XA,S); dot(XB); label("$X_2$",XB,S); dot(XC); label("$X_3$",XC,S); | [] |
498 | Suppose n points on the circumference of a circle
are joined by straight line segments in all possible ways
and that no point that is not one of the original n points
is contained in more than two of the segments. How
many triangles are formed by the segments? Count all
triangles whose sides lie along the segments, including
triangles that overlap with other triangles. For example,
for n = 3, there is one triangle and for n = 4 (shown
in the diagram), there are 8 triangles. | 2017 UNCO Math Contest II Problems/Problem 9 | Since any $3$ pairwise intersecting lines make a triangle, so we can split the possibilities into $4$ cases:
Case 1: All vertices of the triangle are points on the circle
Any $3$ points will work, so it's ${n}\choose{3}$
Case 2: 2 vertices of the triangle are points on the circle
There will be $4$ points to choose from, but you also have to multiply by the number of orientations ($4$), so $4$${n}\choose{4}$
Case 3: 1 vertex of the triangle is a point on the circle
There will be $5$ points to choose from, but you also have to multiply by the number of orientations ($5$), so $5$${n}\choose{5}$
Case 4: All vertices of the triangle are points on the circle
Any $6$ points will work, so it's ${n}\choose{6}$
Thus there are a total of $\boxed{\binom{n}{6}+5\binom{n}{5}+4\binom{n}{4}+\binom{n}{3}}$ different triangles. | // Block 1
pair A=dir(0),B=dir(80),C=dir(160);
draw(unitcircle);
draw(A--B--C--A);
// Block 2
pair A=dir(0),B=dir(80),C=dir(160),D=dir(310);
draw(unitcircle);
draw(A--C--B--D);
// Block 3
pair A=dir(0),B=dir(80),C=dir(160),D=dir(210),E=dir(300);
draw(unitcircle);
draw(A--C--E);
draw(B--D);
// Block 4
pair A=dir(0),B=dir(130),C=dir(160),D=dir(210),E=dir(270),F=dir(320);
draw(unitcircle);
draw(A--D);
draw(B--E);
draw(C--F);
// Block 5
pair A=dir(0),B=dir(80),C=dir(160); draw(unitcircle); draw(A--B--C--A);
// Block 6
pair A=dir(0),B=dir(80),C=dir(160),D=dir(310); draw(unitcircle); draw(A--C--B--D);
// Block 7
pair A=dir(0),B=dir(80),C=dir(160),D=dir(210),E=dir(300); draw(unitcircle); draw(A--C--E); draw(B--D);
// Block 8
pair A=dir(0),B=dir(130),C=dir(160),D=dir(210),E=dir(270),F=dir(320); draw(unitcircle); draw(A--D); draw(B--E); draw(C--F); | [] |
499 | Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths $3$ and $4$ units. In the corner where those sides meet at a right angle, he leaves a small unplanted square $S$ so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from $S$ to the hypotenuse is $2$ units. What fraction of the field is planted?
$\textbf{(A) } \frac{25}{27} \qquad \textbf{(B) } \frac{26}{27} \qquad \textbf{(C) } \frac{73}{75} \qquad \textbf{(D) } \frac{145}{147} \qquad \textbf{(E) } \frac{74}{75}$ | 2018 AMC 10A Problems/Problem 23 | Note that the hypotenuse of the field is $5,$ and the area of the field is $6.$ Let $x$ be the side-length of square $S.$
We partition the field into a red triangle, a yellow triangle, and a green triangle, as shown below:
Let the brackets denote areas. By area addition, we set up an equation for $x:$
\begin{align*} [\text{Red Triangle}]+[\text{Yellow Triangle}]+[\text{Green Triangle}]&=[\text{Field}] \\ \frac{3x}{2}+\frac{4x}{2}+\frac{5\cdot2}{2}&=6, \end{align*}
from which $x=\frac27.$ Therefore, the answer is \[\frac{[\text{Field}]-[S]}{[\text{Field}]}=\frac{6-x^2}{6}=\boxed{\textbf{(D) } \frac{145}{147}}.\]
~MRENTHUSIASM | // Block 1
/* Edited by MRENTHUSIASM */
size(180);
pair A, B, C, D, F;
A = origin;
B = (4,0);
C = (0,3);
D = (2/7,2/7);
F = foot(D,B,C);
fill(A--D--C--cycle, red);
fill(A--D--B--cycle, yellow);
fill(B--D--C--cycle, green);
draw(A--B--C--cycle);
label("$5$", midpoint(B--C), NE);
label("$4$", midpoint(A--B), S);
label("$3$", midpoint(A--C), W);
label("$2$", midpoint(D--F), SE);
label("$S$", midpoint(A--D));
label("$x$", midpoint((0,2/7)--D), N);
label("$x$", midpoint((2/7,0)--D), E);
draw((2/7,0)--D--(0,2/7));
draw(A--D^^B--D^^C--D, dashed);
draw(D--F, dashed);
// Block 2
/* Edited by MRENTHUSIASM */ size(180); pair A, B, C, D, F; A = origin; B = (4,0); C = (0,3); D = (2/7,2/7); F = foot(D,B,C); fill(A--D--C--cycle, red); fill(A--D--B--cycle, yellow); fill(B--D--C--cycle, green); draw(A--B--C--cycle); label("$5$", midpoint(B--C), NE); label("$4$", midpoint(A--B), S); label("$3$", midpoint(A--C), W); label("$2$", midpoint(D--F), SE); label("$S$", midpoint(A--D)); label("$x$", midpoint((0,2/7)--D), N); label("$x$", midpoint((2/7,0)--D), E); draw((2/7,0)--D--(0,2/7)); draw(A--D^^B--D^^C--D, dashed); draw(D--F, dashed); | [] |
499 | Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths $3$ and $4$ units. In the corner where those sides meet at a right angle, he leaves a small unplanted square $S$ so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from $S$ to the hypotenuse is $2$ units. What fraction of the field is planted?
$\textbf{(A) } \frac{25}{27} \qquad \textbf{(B) } \frac{26}{27} \qquad \textbf{(C) } \frac{73}{75} \qquad \textbf{(D) } \frac{145}{147} \qquad \textbf{(E) } \frac{74}{75}$ | 2018 AMC 10A Problems/Problem 23 | Let the square have side length $x$. Connect the upper-right vertex of square $S$ with the two vertices of the triangle's hypotenuse. This divides the triangle in several regions whose areas must add up to the area of the whole triangle, which is $6$.
Square $S$ has area $x^2$, and the two thin triangle regions have area $\dfrac{x(3-x)}{2}$ and $\dfrac{x(4-x)}{2}$. The final triangular region with the hypotenuse as its base and height $2$ has area $5$. Thus, we have \[x^2+\dfrac{x(3-x)}{2}+\dfrac{x(4-x)}{2}+5=6.\]
Solving gives $x=\dfrac{2}{7}$. The area of $S$ is $\dfrac{4}{49}$ and the desired ratio is $\dfrac{6-\tfrac{4}{49}}{6}=\boxed{\textbf{(D) } \frac{145}{147}}$.
Alternatively, once you get $x=\frac{2}{7}$, you can avoid computation by noticing that there is a denominator of $7$, so the answer must have a factor of $7$ in the denominator, which only $\frac{145}{147}$ does. | // Block 1
/* Edited by MRENTHUSIASM */
size(180);
pair A, B, C, D, F;
A = origin;
B = (4,0);
C = (0,3);
D = (2/7,2/7);
F = foot(D,B,C);
fill(A--(2/7,0)--D--(0,2/7)--cycle, lightgray);
draw(A--B--C--cycle);
label("$5$", midpoint(B--C), NE);
label("$4$", midpoint(A--B), S);
label("$3$", midpoint(A--C), W);
label("$2$", midpoint(D--F), SE);
label("$S$", midpoint(A--D));
label("$x$", midpoint((0,2/7)--D), N);
label("$x$", midpoint((2/7,0)--D), E);
draw((2/7,0)--D--(0,2/7));
draw(B--D^^C--D, dashed);
draw(D--F, dashed);
// Block 2
/* Edited by MRENTHUSIASM */ size(180); pair A, B, C, D, F; A = origin; B = (4,0); C = (0,3); D = (2/7,2/7); F = foot(D,B,C); fill(A--(2/7,0)--D--(0,2/7)--cycle, lightgray); draw(A--B--C--cycle); label("$5$", midpoint(B--C), NE); label("$4$", midpoint(A--B), S); label("$3$", midpoint(A--C), W); label("$2$", midpoint(D--F), SE); label("$S$", midpoint(A--D)); label("$x$", midpoint((0,2/7)--D), N); label("$x$", midpoint((2/7,0)--D), E); draw((2/7,0)--D--(0,2/7)); draw(B--D^^C--D, dashed); draw(D--F, dashed); | [] |
499 | Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths $3$ and $4$ units. In the corner where those sides meet at a right angle, he leaves a small unplanted square $S$ so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from $S$ to the hypotenuse is $2$ units. What fraction of the field is planted?
$\textbf{(A) } \frac{25}{27} \qquad \textbf{(B) } \frac{26}{27} \qquad \textbf{(C) } \frac{73}{75} \qquad \textbf{(D) } \frac{145}{147} \qquad \textbf{(E) } \frac{74}{75}$ | 2018 AMC 10A Problems/Problem 23 | On the diagram above, find two smaller triangles similar to the large one with side lengths $3$, $4$, and $5$; consequently, the segments with length $\frac{5}{2}$ and $\frac{10}{3}$.
With $\ell$ being the side length of the square, we need to find an expression for $\ell$. Using the hypotenuse, we can see that $\frac{3}{2}+\frac{8}{3}+\frac{5}{4}\ell+\frac{5}{3}\ell=5$. Simplifying, $\frac{35}{12}\ell=\frac{5}{6}$, or $\ell=\frac27$.
A different calculation would yield $\ell+\frac{3}{4}\ell+\frac{5}{2}=3$, so $\frac{7}{4}\ell=\frac{1}{2}$. In other words, $\ell=\frac{2}{7}$, while to check, $\ell+\frac{4}{3}\ell+\frac{10}{3}=4$. As such, $\frac{7}{3}\ell=\frac{2}{3}$, and $\ell=\frac{2}{7}$.
Finally, we get $A(\Square S)=\ell^2=\frac{4}{49}$, to finish. As a proportion of the triangle with area $6$, the answer would be $1-\frac{4}{49\cdot6}=1-\frac{2}{147}=\frac{145}{147}$, so $\boxed{\textbf{(D) } \frac{145}{147}}$ is correct. | // Block 1
/* Edited by MRENTHUSIASM */
size(180);
pair A, B, C, D, F;
A = origin;
B = (4,0);
C = (0,3);
D = (2/7,2/7);
F = foot(D,B,C);
fill(A--(2/7,0)--D--(0,2/7)--cycle, lightgray);
draw(A--B--C--cycle);
label("$4$", midpoint(A--B), S);
label("$3$", midpoint(A--C), W);
label("$2$", midpoint(D--F), SE);
label("$S$", midpoint(A--D));
label("$\ell$", midpoint((0,2/7)--D), N);
label("$\ell$", midpoint((2/7,0)--D), E);
label("$\ell$", midpoint((2/7,2/7+5/2)--(0,2/7+5/2)), S);
label("$\ell$", midpoint((2/7+10/3,2/7)--(2/7+10/3,0)), W);
label("$\frac{5}{2}$", midpoint((2/7,2/7+5/2)--D), E);
label("$\frac{10}{3}$", midpoint((2/7+10/3,2/7)--D), N);
draw((2/7,0)--D--(0,2/7));
draw((2/7,2/7+5/2)--D^^(2/7+10/3,2/7)--D, dashed);
draw((2/7,2/7+5/2)--(0,2/7+5/2)^^(2/7+10/3,2/7)--(2/7+10/3,0), dashed);
draw(D--F, dashed);
// Block 2
/* Edited by MRENTHUSIASM */ size(180); pair A, B, C, D, F; A = origin; B = (4,0); C = (0,3); D = (2/7,2/7); F = foot(D,B,C); fill(A--(2/7,0)--D--(0,2/7)--cycle, lightgray); draw(A--B--C--cycle); label("$4$", midpoint(A--B), S); label("$3$", midpoint(A--C), W); label("$2$", midpoint(D--F), SE); label("$S$", midpoint(A--D)); label("$\ell$", midpoint((0,2/7)--D), N); label("$\ell$", midpoint((2/7,0)--D), E); label("$\ell$", midpoint((2/7,2/7+5/2)--(0,2/7+5/2)), S); label("$\ell$", midpoint((2/7+10/3,2/7)--(2/7+10/3,0)), W); label("$\frac{5}{2}$", midpoint((2/7,2/7+5/2)--D), E); label("$\frac{10}{3}$", midpoint((2/7+10/3,2/7)--D), N); draw((2/7,0)--D--(0,2/7)); draw((2/7,2/7+5/2)--D^^(2/7+10/3,2/7)--D, dashed); draw((2/7,2/7+5/2)--(0,2/7+5/2)^^(2/7+10/3,2/7)--(2/7+10/3,0), dashed); draw(D--F, dashed); | [] |
499 | Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths $3$ and $4$ units. In the corner where those sides meet at a right angle, he leaves a small unplanted square $S$ so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from $S$ to the hypotenuse is $2$ units. What fraction of the field is planted?
$\textbf{(A) } \frac{25}{27} \qquad \textbf{(B) } \frac{26}{27} \qquad \textbf{(C) } \frac{73}{75} \qquad \textbf{(D) } \frac{145}{147} \qquad \textbf{(E) } \frac{74}{75}$ | 2018 AMC 10A Problems/Problem 23 | Let $AD=x$. Note that $\triangle DEF$ is a $3{-}4{-}5$ triangle, so $EF=\frac{5}{4}x$ and $FD=\frac{3}{4}x$. $BF=BD+FD=4-x+\frac{3}{4}x=4-\frac{1}{4}x$. We know that $GE$ is $2$ from the problem so $GF=2+\frac{5}{4}x$. $\triangle FGB$ is also a $3{-}4{-}5$ triangle with $GF:BF=3:5$. We now have $3\left(4-\frac{1}{4}x\right)=5\left(2+\frac{5}{4}x\right)$. Solving this equation, we get that $x=\frac{2}{7}$ so the area of $S$ is $\frac{4}{49}$. The area of the triangle is $\frac{3\cdot 4}{2}=6$ so the fraction of field that is unplanted is $\frac{\frac{4}{49}}{6}=\frac{2}{147}$. Thus, the fraction of the field that is planted is $1-\frac{2}{147}=\boxed{\textbf{(D) } \frac{145}{147}}$.
~Heavytoothpaste | // Block 1
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(180);
real labelscalefactor = 1.5; /* changes label-to-point distance */
// pen dps = linewidth(0.5) + fontsize(10);
// defaultpen(dps); /* default pen style */
// pen dotstyle = black; /* point style */
real xmin = -1.6030465381283199, xmax = 7.095084767820557, ymin = -1.3624649422453508, ymax = 4.065350676871526; /* image dimensions */
/* draw figures */
draw((0,0)--(0,3));
draw((0,0)--(4,0));
draw((4,0)--(0,3));
draw((0,0.2857142857142857)--(0.2857142857142857,0.2857142857142857));
draw((0.2857142857142857,0.2857142857142857)--(0.2857142857142857,0));
draw((0.07142857142857142,0)--(1.4857142857142858,1.885714285714286));
label("$A$",(0, 0),SW*labelscalefactor);
label("$B$",(4,0),SE*labelscalefactor);
label("$C$",(0, 3),N*labelscalefactor);
label("$D$",(0.2857142857142857,0),S*labelscalefactor);
label("$E$",(0.2857142857142857, 0.2857142857142857),E*labelscalefactor);
label("$F$",(0.0714285714, 0),S*labelscalefactor);
label("$G$", (1.49, 1.89), NE*labelscalefactor);
/* dots and labels */
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(180); real labelscalefactor = 1.5; /* changes label-to-point distance */ // pen dps = linewidth(0.5) + fontsize(10); // defaultpen(dps); /* default pen style */ // pen dotstyle = black; /* point style */ real xmin = -1.6030465381283199, xmax = 7.095084767820557, ymin = -1.3624649422453508, ymax = 4.065350676871526; /* image dimensions */ /* draw figures */ draw((0,0)--(0,3)); draw((0,0)--(4,0)); draw((4,0)--(0,3)); draw((0,0.2857142857142857)--(0.2857142857142857,0.2857142857142857)); draw((0.2857142857142857,0.2857142857142857)--(0.2857142857142857,0)); draw((0.07142857142857142,0)--(1.4857142857142858,1.885714285714286)); label("$A$",(0, 0),SW*labelscalefactor); label("$B$",(4,0),SE*labelscalefactor); label("$C$",(0, 3),N*labelscalefactor); label("$D$",(0.2857142857142857,0),S*labelscalefactor); label("$E$",(0.2857142857142857, 0.2857142857142857),E*labelscalefactor); label("$F$",(0.0714285714, 0),S*labelscalefactor); label("$G$", (1.49, 1.89), NE*labelscalefactor); /* dots and labels */ clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ | [] |
499 | Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths $3$ and $4$ units. In the corner where those sides meet at a right angle, he leaves a small unplanted square $S$ so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from $S$ to the hypotenuse is $2$ units. What fraction of the field is planted?
$\textbf{(A) } \frac{25}{27} \qquad \textbf{(B) } \frac{26}{27} \qquad \textbf{(C) } \frac{73}{75} \qquad \textbf{(D) } \frac{145}{147} \qquad \textbf{(E) } \frac{74}{75}$ | 2018 AMC 10A Problems/Problem 23 | Denote $A,B,C$ to be the three vertices of the triangular field. Also denote $A,M,D,N$ to be the vertices of the square $S$. Let $X$ be on $BC$ such that $AC\parallel DX$ and $Y$ be on $BC$ such that $AB\parallel DY$. Let $P$ and $Q$ be the foot of the altitudes from $X$ to $AC$ and from $Y$ to $AB$ respectively.
Note that $\triangle ABC \sim \triangle DYX \sim \triangle PXC \sim \triangle QBY$. Thus, $PC = x \cdot \frac34$ and $QB = x \cdot \frac43$, making
\begin{align*} DX &= 3-x-\dfrac{3}{4}x = 3-\dfrac{7}{4}x, \\ MQ &= 4-x-\dfrac{4}{3}x = 4-\dfrac{7}{3}x. \end{align*}
Also from the similarity ratio is the fact that $CX = \frac54 x$ and $BY = \frac53 x$, making
\[XY = 5 - \dfrac{5}{4}x - \dfrac{5}{3}x = 5 - \dfrac{35}{12}x.\]
Computing the area of $\triangle XDY$ in two ways gives an equation for $x$:
\begin{align*} \left(3-\dfrac{7}{4}x\right)\left(4-\dfrac{7}{3}x\right) &= 2\cdot \left(5 - \dfrac{35}{12}x\right) \\ 10-\dfrac{35}{6}x &= \dfrac{49}{12}x^2 - 14x + 12 \\ \dfrac{49}{12}x^2 - \dfrac{49}{6}x + 2 &= 0 \\ 49x^2 - 98x + 24 &= 0 \\ x&=\dfrac{2}{7} \text{ or } \dfrac{12}{7}. \end{align*}
But $x=\dfrac{12}{7}$ is extraneous. Thus, the area of square $S = x^2 = \dfrac{4}{49}$, making the portion of the field that is planted being \[1 - \dfrac{\tfrac{4}{49}}{6} = 1 - \dfrac{2}{147} = \boxed{\textbf{(D) } \frac{145}{147}}.\]
-Solution by sml1809 | // Block 1
size(240);
pair A, B, C, D, F, X, Y, P, Q, M, N;
A = origin; label(A, "$A$", SW);
B = (4,0); label(B, "$B$", S);
C = (0,3); label(C, "$C$", W);
D = (2/7,2/7); label(D, "$D$", NE);
F = foot(D,B,C); label(F, "$F$", NE);
X = (2/7,39/14); label(X, "$X$", NE, red);
Y = (76/21,2/7); label(Y, "$Y$", NE, red);
P = foot(X,A,C); label(P, "$P$", W, red);
Q = foot(Y,A,B); label(Q, "$Q$", S, red);
M = (2/7,0); label(M, "$M$", S);
N = (0,2/7); label(N, "$N$", W);
fill(A--(2/7,0)--D--(0,2/7)--cycle, lightgray);
draw(A--B--C--cycle);
draw((2/7,0)--D--(0,2/7));
label("$x$", midpoint(A--M), S);
label("$x$", midpoint(A--N), W);
label("$2$", midpoint(D--F), SE);
draw(D--F);
draw(D--X, red);
draw(D--Y, red);
draw(X--P, red);
draw(Y--Q, red);
// Block 2
size(240); pair A, B, C, D, F, X, Y, P, Q, M, N; A = origin; label(A, "$A$", SW); B = (4,0); label(B, "$B$", S); C = (0,3); label(C, "$C$", W); D = (2/7,2/7); label(D, "$D$", NE); F = foot(D,B,C); label(F, "$F$", NE); X = (2/7,39/14); label(X, "$X$", NE, red); Y = (76/21,2/7); label(Y, "$Y$", NE, red); P = foot(X,A,C); label(P, "$P$", W, red); Q = foot(Y,A,B); label(Q, "$Q$", S, red); M = (2/7,0); label(M, "$M$", S); N = (0,2/7); label(N, "$N$", W); fill(A--(2/7,0)--D--(0,2/7)--cycle, lightgray); draw(A--B--C--cycle); draw((2/7,0)--D--(0,2/7)); label("$x$", midpoint(A--M), S); label("$x$", midpoint(A--N), W); label("$2$", midpoint(D--F), SE); draw(D--F); draw(D--X, red); draw(D--Y, red); draw(X--P, red); draw(Y--Q, red); | [] |
500 | Line segment $\overline{AB}$ is a diameter of a circle with $AB = 24$. Point $C$, not equal to $A$ or $B$, lies on the circle. As point $C$ moves around the circle, the centroid (center of mass) of $\triangle ABC$ traces out a closed curve missing two points. To the nearest positive integer, what is the area of the region bounded by this curve?
$\textbf{(A) } 25 \qquad \textbf{(B) } 38 \qquad \textbf{(C) } 50 \qquad \textbf{(D) } 63 \qquad \textbf{(E) } 75$ | 2018 AMC 10B Problems/Problem 12 | By the Inscribed Angle Theorem, $\triangle ABC$ is a right triangle with $\angle C=90^{\circ}.$ So, its circumcenter is the midpoint of $\overline{AB},$ and its median from $C$ is half as long as $\overline{AB}.$ For each $\triangle ABC,$ let $O$ and $G$ be its circumcenter and centroid, respectively. It follows that $OA=OB=OC=12.$ In any triangle, since the centroid divides each median into parts in the ratio $2:1,$ with the centroid being twice as close to the midpoint of a side as it is to the opposite vertex, we have $OG=\frac13 OC=4.$
As shown below, $\triangle ABC_1$ and $\triangle ABC_2$ are two shapes of $\triangle ABC$ with centroids $G_1$ and $G_2,$ respectively:
Therefore, point $G$ traces out a circle (missing two points) with the center $O$ and the radius $\overline{OG},$ as indicated in red. To the nearest positive integer, the area of the region bounded by the red curve is $\pi\cdot OG^2=16\pi\approx\boxed{\textbf{(C) } 50}.$
~MRENTHUSIASM ~megacleverstarfish15 | // Block 1
/* Made by MRENTHUSIASM */
size(200);
pair O, A, B, C1, C2, G1, G2, M1, M2;
O = (0,0);
A = (-12,0);
B = (12,0);
C1 = (36/5,48/5);
C2 = (-96/17,-180/17);
G1 = O + 1/3 * C1;
G2 = O + 1/3 * C2;
M1 = (4,0);
M2 = (-4,0);
draw(Circle(O,12));
draw(Circle(O,4),red);
dot("$O$", O, (3/5,-4/5), linewidth(4.5));
dot("$A$", A, W, linewidth(4.5));
dot("$B$", B, E, linewidth(4.5));
dot("$C_1$", C1, dir(C1), linewidth(4.5));
dot("$C_2$", C2, dir(C2), linewidth(4.5));
dot("$G_1$", G1, 1.5*E, linewidth(4.5));
dot("$G_2$", G2, 1.5*W, linewidth(4.5));
draw(A--B^^A--C1--B^^A--C2--B);
draw(O--C1^^O--C2);
dot(M1,red+linewidth(0.8),UnFill);
dot(M2,red+linewidth(0.8),UnFill);
// Block 2
/* Made by MRENTHUSIASM */ size(200); pair O, A, B, C1, C2, G1, G2, M1, M2; O = (0,0); A = (-12,0); B = (12,0); C1 = (36/5,48/5); C2 = (-96/17,-180/17); G1 = O + 1/3 * C1; G2 = O + 1/3 * C2; M1 = (4,0); M2 = (-4,0); draw(Circle(O,12)); draw(Circle(O,4),red); dot("$O$", O, (3/5,-4/5), linewidth(4.5)); dot("$A$", A, W, linewidth(4.5)); dot("$B$", B, E, linewidth(4.5)); dot("$C_1$", C1, dir(C1), linewidth(4.5)); dot("$C_2$", C2, dir(C2), linewidth(4.5)); dot("$G_1$", G1, 1.5*E, linewidth(4.5)); dot("$G_2$", G2, 1.5*W, linewidth(4.5)); draw(A--B^^A--C1--B^^A--C2--B); draw(O--C1^^O--C2); dot(M1,red+linewidth(0.8),UnFill); dot(M2,red+linewidth(0.8),UnFill); | [] |
501 | Let $ABCDEF$ be a regular hexagon with side length $1$. Denote by $X$, $Y$, and $Z$ the midpoints of sides $\overline {AB}$, $\overline{CD}$, and $\overline{EF}$, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of $\triangle ACE$ and $\triangle XYZ$?
$\textbf{(A)}\ \frac {3}{8}\sqrt{3} \qquad \textbf{(B)}\ \frac {7}{16}\sqrt{3} \qquad \textbf{(C)}\ \frac {15}{32}\sqrt{3} \qquad \textbf{(D)}\ \frac {1}{2}\sqrt{3} \qquad \textbf{(E)}\ \frac {9}{16}\sqrt{3}$ | 2018 AMC 10B Problems/Problem 24 | The desired area (hexagon $MPNQOR$) consists of an equilateral triangle ($\triangle MNO$) and three right triangles ($\triangle MPN,\triangle NQO,$ and $\triangle ORM$).
Notice that $\overline {AD}$ (not shown) and $\overline {BC}$ are parallel. $\overline {XY}$ divides transversals $\overline {AB}$ and $\overline {CD}$ into a $1:1$ ratio (This can be shown by similar triangles.). Thus, it must also divide transversal $\overline {AC}$ and transversal $\overline {CO}$ into a $1:1$ ratio. By symmetry, the same applies for $\overline {CE}$ and $\overline {EA}$ as well as $\overline {EM}$ and $\overline {AN}.$
In $\triangle ACE,$ we see that $\frac{[MNO]}{[ACE]} = \frac{1}{4}$ and $\frac{[MPN]}{[ACE]} = \frac{1}{8}.$ Our desired area becomes \[\left(\frac{1}{4}+3 \cdot \frac{1}{8}\right) \cdot \frac{(\sqrt{3})^2 \cdot \sqrt{3}}{4} = \boxed{\textbf{(C)}\ \frac {15}{32}\sqrt{3}}.\] | // Block 1
/* Made by MRENTHUSIASM */
size(200);
draw(polygon(6));
pair A, B, C, D, E, F, X, Y, Z, M, N, O, P, Q, R;
A = dir(120);
B = dir(60);
C = dir(0);
D = dir(300);
E = dir(240);
F = dir(180);
X = midpoint(A--B);
Y = midpoint(C--D);
Z = midpoint(E--F);
M = intersectionpoint(A--E,X--Z);
N = intersectionpoint(A--C,X--Y);
O = intersectionpoint(C--E,Y--Z);
P = intersectionpoint(A--C,X--Z);
Q = intersectionpoint(C--E,X--Y);
R = intersectionpoint(A--E,Y--Z);
fill(M--P--N--Q--O--R--cycle,mediumgray);
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("$E$",E,1.5*dir(E),linewidth(4));
dot("$F$",F,1.5*dir(F),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("$M$",M,1.5*dir(165),linewidth(4));
dot("$N$",N,1.5*dir(45),linewidth(4));
dot("$O$",O,1.5*dir(-75),linewidth(4));
dot("$P$",P,1.5*dir(105),linewidth(4));
dot("$Q$",Q,1.5*dir(-15),linewidth(4));
dot("$R$",R,1.5*dir(-135),linewidth(4));
draw(A--C--E--cycle^^X--Y--Z--cycle);
draw(M--N--O--cycle,dashed);
// Block 2
/* Made by MRENTHUSIASM */ size(200); draw(polygon(6)); pair A, B, C, D, E, F, X, Y, Z, M, N, O, P, Q, R; A = dir(120); B = dir(60); C = dir(0); D = dir(300); E = dir(240); F = dir(180); X = midpoint(A--B); Y = midpoint(C--D); Z = midpoint(E--F); M = intersectionpoint(A--E,X--Z); N = intersectionpoint(A--C,X--Y); O = intersectionpoint(C--E,Y--Z); P = intersectionpoint(A--C,X--Z); Q = intersectionpoint(C--E,X--Y); R = intersectionpoint(A--E,Y--Z); fill(M--P--N--Q--O--R--cycle,mediumgray); 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("$E$",E,1.5*dir(E),linewidth(4)); dot("$F$",F,1.5*dir(F),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("$M$",M,1.5*dir(165),linewidth(4)); dot("$N$",N,1.5*dir(45),linewidth(4)); dot("$O$",O,1.5*dir(-75),linewidth(4)); dot("$P$",P,1.5*dir(105),linewidth(4)); dot("$Q$",Q,1.5*dir(-15),linewidth(4)); dot("$R$",R,1.5*dir(-135),linewidth(4)); draw(A--C--E--cycle^^X--Y--Z--cycle); draw(M--N--O--cycle,dashed); | [] |
501 | Let $ABCDEF$ be a regular hexagon with side length $1$. Denote by $X$, $Y$, and $Z$ the midpoints of sides $\overline {AB}$, $\overline{CD}$, and $\overline{EF}$, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of $\triangle ACE$ and $\triangle XYZ$?
$\textbf{(A)}\ \frac {3}{8}\sqrt{3} \qquad \textbf{(B)}\ \frac {7}{16}\sqrt{3} \qquad \textbf{(C)}\ \frac {15}{32}\sqrt{3} \qquad \textbf{(D)}\ \frac {1}{2}\sqrt{3} \qquad \textbf{(E)}\ \frac {9}{16}\sqrt{3}$ | 2018 AMC 10B Problems/Problem 24 | Now, if we look at the figure, we can see that the complement of the hexagon we are trying to find is composed of $3$ isosceles trapezoids (namely $AXZF,CYXB,$ and $EZYD$) and $3$ right triangles (namely $\triangle XPN,\triangle YQO,$ and $\triangle ZRM$).
Finding the trapezoid's area, we know that one base of each trapezoid is just the side length of the hexagon, which is $1,$ and the other base is $\frac{3}{2}$ (it is halfway in between the side and the longest diagonal, which has length $2$) with a height of $\frac{\sqrt{3}}{4}$ (by using the Pythagorean Theorem and the fact that it is an isosceles trapezoid) to give each trapezoid having an area of $\frac{5\sqrt{3}}{16}$ for a total area of $\frac{15\sqrt{3}}{16}.$ (Alternatively, we could have calculated the area of hexagon $ABCDEF$ and subtracted the area of $\triangle XYZ,$ which, as we showed before, had a side length of $\frac{3}{2}$).
Now, we need to find the area of each of the small triangles, which, if we look at the triangle that has a vertex on $X,$ is similar to the triangle with a base of $YC = \frac12.$ Using similar triangles, we calculate the base to be $\frac{1}{4}$ and the height to be $\frac{\sqrt{3}}{4}$ giving us an area of $\frac{\sqrt{3}}{32}$ per triangle, and a total area of $\frac{3\sqrt{3}}{32}.$ Adding the two areas together, we get $\frac{15\sqrt{3}}{16} + \frac{3\sqrt{3}}{32} = \frac{33\sqrt{3}}{32}.$ Finding the total area, we get $6 \cdot 1^2 \cdot \frac{\sqrt{3}}{4}=\frac{3\sqrt{3}}{2}.$ Taking the complement, we get $\frac{3\sqrt{3}}{2} - \frac{33\sqrt{3}}{32} = \boxed{\textbf{(C)}\ \frac {15}{32}\sqrt{3}}.$ | // Block 1
/* Made by MRENTHUSIASM */
size(200);
draw(polygon(6));
pair A, B, C, D, E, F, X, Y, Z, M, N, O, P, Q, R;
A = dir(120);
B = dir(60);
C = dir(0);
D = dir(300);
E = dir(240);
F = dir(180);
X = midpoint(A--B);
Y = midpoint(C--D);
Z = midpoint(E--F);
M = intersectionpoint(A--E,X--Z);
N = intersectionpoint(A--C,X--Y);
O = intersectionpoint(C--E,Y--Z);
P = intersectionpoint(A--C,X--Z);
Q = intersectionpoint(C--E,X--Y);
R = intersectionpoint(A--E,Y--Z);
fill(M--P--N--Q--O--R--cycle,mediumgray);
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("$E$",E,1.5*dir(E),linewidth(4));
dot("$F$",F,1.5*dir(F),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("$M$",M,1.5*dir(165),linewidth(4));
dot("$N$",N,1.5*dir(45),linewidth(4));
dot("$O$",O,1.5*dir(-75),linewidth(4));
dot("$P$",P,1.5*dir(105),linewidth(4));
dot("$Q$",Q,1.5*dir(-15),linewidth(4));
dot("$R$",R,1.5*dir(-135),linewidth(4));
draw(A--C--E--cycle^^X--Y--Z--cycle);
draw(M--N--O--cycle,dashed);
// Block 2
/* Made by MRENTHUSIASM */ size(200); draw(polygon(6)); pair A, B, C, D, E, F, X, Y, Z, M, N, O, P, Q, R; A = dir(120); B = dir(60); C = dir(0); D = dir(300); E = dir(240); F = dir(180); X = midpoint(A--B); Y = midpoint(C--D); Z = midpoint(E--F); M = intersectionpoint(A--E,X--Z); N = intersectionpoint(A--C,X--Y); O = intersectionpoint(C--E,Y--Z); P = intersectionpoint(A--C,X--Z); Q = intersectionpoint(C--E,X--Y); R = intersectionpoint(A--E,Y--Z); fill(M--P--N--Q--O--R--cycle,mediumgray); 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("$E$",E,1.5*dir(E),linewidth(4)); dot("$F$",F,1.5*dir(F),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("$M$",M,1.5*dir(165),linewidth(4)); dot("$N$",N,1.5*dir(45),linewidth(4)); dot("$O$",O,1.5*dir(-75),linewidth(4)); dot("$P$",P,1.5*dir(105),linewidth(4)); dot("$Q$",Q,1.5*dir(-15),linewidth(4)); dot("$R$",R,1.5*dir(-135),linewidth(4)); draw(A--C--E--cycle^^X--Y--Z--cycle); draw(M--N--O--cycle,dashed); | [] |
501 | Let $ABCDEF$ be a regular hexagon with side length $1$. Denote by $X$, $Y$, and $Z$ the midpoints of sides $\overline {AB}$, $\overline{CD}$, and $\overline{EF}$, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of $\triangle ACE$ and $\triangle XYZ$?
$\textbf{(A)}\ \frac {3}{8}\sqrt{3} \qquad \textbf{(B)}\ \frac {7}{16}\sqrt{3} \qquad \textbf{(C)}\ \frac {15}{32}\sqrt{3} \qquad \textbf{(D)}\ \frac {1}{2}\sqrt{3} \qquad \textbf{(E)}\ \frac {9}{16}\sqrt{3}$ | 2018 AMC 10B Problems/Problem 24 | We could also subtract $\triangle APM,\triangle CQN,$ and $\triangle ERO$ from $\triangle ACE.$
Since $\angle BAF = 120^{\circ}$ and $\angle BAC = \angle FAE = 30^{\circ},$ we have $\angle CAE = \angle BAF-\angle BAC-\angle FAE=60^{\circ}.$
Since $AX=BX$ and $FZ=EZ,$ we have $AF \parallel XZ,$ from which $\angle AMX= \angle FAM = 30^{\circ}.$
We can show that $\triangle APM$ is $30$-$60$-$90$ using a similar method, $\triangle CQN$ and $\triangle ERO$ are also $30$-$60$-$90.$
Since $AC=CE=AE=\sqrt{3},$ we have $[ACE]=AC^2 \cdot \frac{\sqrt{3}}{4}=3 \cdot \frac{\sqrt{3}}{4} = \frac{3 \sqrt{3}}{4}.$
Since $AX= \frac{1}{2}$ and $AP = AX \cdot \frac{\sqrt{3}}{2}= \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4},$ we have $PM = AP \cdot \sqrt{3} = \frac{\sqrt{3}}{4} \cdot \sqrt{3} = \frac{3}{4}.$
Note that \[[APM]=[CQN]=[ERO]=\frac{1}{2} \cdot AP \cdot PM = \frac{1}{2} \cdot \frac{\sqrt{3}}{4} \cdot \frac{3}{4} = \frac{3 \sqrt{3}}{32}.\]
Therefore, we get
\[[PNQORM]=[ACE]-[APM]-[CQN]-[ERO]=\frac{3 \sqrt{3}}{4} - 3 \cdot \frac{3 \sqrt{3}}{32} = \frac{24 \sqrt{3}}{32} - \frac{9 \sqrt{3}}{32} = \boxed{\textbf{(C)}\ \frac {15}{32}\sqrt{3}}.\]
~isabelchen | // Block 1
/* Made by MRENTHUSIASM */
size(200);
draw(polygon(6));
pair A, B, C, D, E, F, X, Y, Z, M, N, O, P, Q, R;
A = dir(120);
B = dir(60);
C = dir(0);
D = dir(300);
E = dir(240);
F = dir(180);
X = midpoint(A--B);
Y = midpoint(C--D);
Z = midpoint(E--F);
M = intersectionpoint(A--E,X--Z);
N = intersectionpoint(A--C,X--Y);
O = intersectionpoint(C--E,Y--Z);
P = intersectionpoint(A--C,X--Z);
Q = intersectionpoint(C--E,X--Y);
R = intersectionpoint(A--E,Y--Z);
fill(M--P--N--Q--O--R--cycle,mediumgray);
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("$E$",E,1.5*dir(E),linewidth(4));
dot("$F$",F,1.5*dir(F),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("$M$",M,1.5*dir(165),linewidth(4));
dot("$N$",N,1.5*dir(45),linewidth(4));
dot("$O$",O,1.5*dir(-75),linewidth(4));
dot("$P$",P,1.5*dir(105),linewidth(4));
dot("$Q$",Q,1.5*dir(-15),linewidth(4));
dot("$R$",R,1.5*dir(-135),linewidth(4));
draw(A--C--E--cycle^^X--Y--Z--cycle);
// Block 2
/* Made by MRENTHUSIASM */ size(200); draw(polygon(6)); pair A, B, C, D, E, F, X, Y, Z, M, N, O, P, Q, R; A = dir(120); B = dir(60); C = dir(0); D = dir(300); E = dir(240); F = dir(180); X = midpoint(A--B); Y = midpoint(C--D); Z = midpoint(E--F); M = intersectionpoint(A--E,X--Z); N = intersectionpoint(A--C,X--Y); O = intersectionpoint(C--E,Y--Z); P = intersectionpoint(A--C,X--Z); Q = intersectionpoint(C--E,X--Y); R = intersectionpoint(A--E,Y--Z); fill(M--P--N--Q--O--R--cycle,mediumgray); 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("$E$",E,1.5*dir(E),linewidth(4)); dot("$F$",F,1.5*dir(F),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("$M$",M,1.5*dir(165),linewidth(4)); dot("$N$",N,1.5*dir(45),linewidth(4)); dot("$O$",O,1.5*dir(-75),linewidth(4)); dot("$P$",P,1.5*dir(105),linewidth(4)); dot("$Q$",Q,1.5*dir(-15),linewidth(4)); dot("$R$",R,1.5*dir(-135),linewidth(4)); draw(A--C--E--cycle^^X--Y--Z--cycle); | [] |
501 | Let $ABCDEF$ be a regular hexagon with side length $1$. Denote by $X$, $Y$, and $Z$ the midpoints of sides $\overline {AB}$, $\overline{CD}$, and $\overline{EF}$, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of $\triangle ACE$ and $\triangle XYZ$?
$\textbf{(A)}\ \frac {3}{8}\sqrt{3} \qquad \textbf{(B)}\ \frac {7}{16}\sqrt{3} \qquad \textbf{(C)}\ \frac {15}{32}\sqrt{3} \qquad \textbf{(D)}\ \frac {1}{2}\sqrt{3} \qquad \textbf{(E)}\ \frac {9}{16}\sqrt{3}$ | 2018 AMC 10B Problems/Problem 24 | We partition hexagon $ABCDEF$ into $48$ congruent $30^\circ\text{-}60^\circ\text{-}90^\circ$ triangles, as shown below:
Let the brackets denote areas. Note that the desired region contains $15$ of the $48$ small triangles, so the answer is \[\frac{15}{48}[ABCDEF]=\frac{15}{48}\cdot\frac{3\sqrt3}{2}=\boxed{\textbf{(C)}\ \frac {15}{32}\sqrt{3}}.\]
~AlexLikeMath ~MRENTHUSIASM | // Block 1
/* Made by MRENTHUSIASM */
size(200);
pair A, B, C, D, E, F, X, Y, Z, M, N, O, P, Q, R;
A = dir(120);
B = dir(60);
C = dir(0);
D = dir(300);
E = dir(240);
F = dir(180);
X = midpoint(A--B);
Y = midpoint(C--D);
Z = midpoint(E--F);
M = intersectionpoint(A--E,X--Z);
N = intersectionpoint(A--C,X--Y);
O = intersectionpoint(C--E,Y--Z);
P = intersectionpoint(A--C,X--Z);
Q = intersectionpoint(C--E,X--Y);
R = intersectionpoint(A--E,Y--Z);
fill(M--P--N--Q--O--R--cycle,mediumgray);
draw(A--D^^B--E^^C--F^^X--Y--Z--cycle^^midpoint(A--F)--midpoint(B--C)--midpoint(D--E)--cycle,red);
draw(A--C--E--cycle^^M--N--O--cycle^^M--midpoint(F--Z)^^M--F+1/4*(A-F)^^N--midpoint(X--B)^^N--B+1/4*(C-B)^^O--midpoint(Y--D)^^O--D+1/4*(E-D),blue);
draw(polygon(6));
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("$E$",E,1.5*dir(E),linewidth(4));
dot("$F$",F,1.5*dir(F),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(M^^N^^O^^P^^Q^^R,linewidth(4));
// Block 2
/* Made by MRENTHUSIASM */ size(200); pair A, B, C, D, E, F, X, Y, Z, M, N, O, P, Q, R; A = dir(120); B = dir(60); C = dir(0); D = dir(300); E = dir(240); F = dir(180); X = midpoint(A--B); Y = midpoint(C--D); Z = midpoint(E--F); M = intersectionpoint(A--E,X--Z); N = intersectionpoint(A--C,X--Y); O = intersectionpoint(C--E,Y--Z); P = intersectionpoint(A--C,X--Z); Q = intersectionpoint(C--E,X--Y); R = intersectionpoint(A--E,Y--Z); fill(M--P--N--Q--O--R--cycle,mediumgray); draw(A--D^^B--E^^C--F^^X--Y--Z--cycle^^midpoint(A--F)--midpoint(B--C)--midpoint(D--E)--cycle,red); draw(A--C--E--cycle^^M--N--O--cycle^^M--midpoint(F--Z)^^M--F+1/4*(A-F)^^N--midpoint(X--B)^^N--B+1/4*(C-B)^^O--midpoint(Y--D)^^O--D+1/4*(E-D),blue); draw(polygon(6)); 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("$E$",E,1.5*dir(E),linewidth(4)); dot("$F$",F,1.5*dir(F),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(M^^N^^O^^P^^Q^^R,linewidth(4)); | [] |
501 | Let $ABCDEF$ be a regular hexagon with side length $1$. Denote by $X$, $Y$, and $Z$ the midpoints of sides $\overline {AB}$, $\overline{CD}$, and $\overline{EF}$, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of $\triangle ACE$ and $\triangle XYZ$?
$\textbf{(A)}\ \frac {3}{8}\sqrt{3} \qquad \textbf{(B)}\ \frac {7}{16}\sqrt{3} \qquad \textbf{(C)}\ \frac {15}{32}\sqrt{3} \qquad \textbf{(D)}\ \frac {1}{2}\sqrt{3} \qquad \textbf{(E)}\ \frac {9}{16}\sqrt{3}$ | 2018 AMC 10B Problems/Problem 24 | If we try to coordinate bash this problem, it's going to look very ugly with a lot of radicals. However, we can alter and skew the diagram in such a way that all ratios of lengths and areas stay the same while making it a lot easier to work with. Then, we can find the ratio of the area of the wanted region to the area of $ABCDEF$ then apply it to the old diagram.
The isosceles right triangle with a leg length of $3$ in the new diagram is $\triangle XYZ$ in the old diagram. We see that if we want to take the area of the new hexagon, we must subtract $\frac{3}{4}$ from the area of $\triangle XYZ$ (the red triangles), giving us $\frac{15}{4}.$ However, we need to take the ratio of this area to the area of $ABCDEF,$ which is $\frac{\frac{15}{4}}{12}=\frac{5}{16}.$ Now we know that our answer is $\frac{5}{16} \cdot \frac{3\sqrt{3}}{2}=\boxed{\textbf{(C)}\ \frac {15}{32}\sqrt{3}}.$ | // Block 1
unitsize(1cm);
draw((0,0)--(4,0)--(6,3.464)--(2,3.464)--(0,0));
draw((2,0)--(1,1.732));
draw((5,1.732)--(4,3.464));
draw((1.5, 0.866)--(3, 3.464)--(4.5, 0.866)--cycle);
draw((2,0)--(2,3.464)--(5,1.732)--cycle);
// Block 2
unitsize(1cm);
fill((1,4)--(1,3.5)--(2,3)--cycle,red);
fill((1,1)--(1.5,1)--(1,2)--cycle,red);
fill((3,1)--(3.5,1.5)--(4,1)--cycle,red);
draw((1,0)--(1,4),gray(.7));
draw((2,0)--(2,4),gray(.7));
draw((3,0)--(3,4),gray(.7));
draw((0,1)--(4,1),gray(.7));
draw((0,2)--(4,2),gray(.7));
draw((0,3)--(4,3),gray(.7));
draw((0,0)--(4,0)--(4,4)--(0,4)--(0,0));
draw((2,0)--(0,2));
draw((4,2)--(2,4));
draw((1,1)--(1,4)--(4,1)--cycle);
draw((0,4)--(2,0)--(4,2)--cycle);
// Block 3
unitsize(1cm); draw((0,0)--(4,0)--(6,3.464)--(2,3.464)--(0,0)); draw((2,0)--(1,1.732)); draw((5,1.732)--(4,3.464)); draw((1.5, 0.866)--(3, 3.464)--(4.5, 0.866)--cycle); draw((2,0)--(2,3.464)--(5,1.732)--cycle);
// Block 4
unitsize(1cm); fill((1,4)--(1,3.5)--(2,3)--cycle,red); fill((1,1)--(1.5,1)--(1,2)--cycle,red); fill((3,1)--(3.5,1.5)--(4,1)--cycle,red); draw((1,0)--(1,4),gray(.7)); draw((2,0)--(2,4),gray(.7)); draw((3,0)--(3,4),gray(.7)); draw((0,1)--(4,1),gray(.7)); draw((0,2)--(4,2),gray(.7)); draw((0,3)--(4,3),gray(.7)); draw((0,0)--(4,0)--(4,4)--(0,4)--(0,0)); draw((2,0)--(0,2)); draw((4,2)--(2,4)); draw((1,1)--(1,4)--(4,1)--cycle); draw((0,4)--(2,0)--(4,2)--cycle); | [] |
501 | Let $ABCDEF$ be a regular hexagon with side length $1$. Denote by $X$, $Y$, and $Z$ the midpoints of sides $\overline {AB}$, $\overline{CD}$, and $\overline{EF}$, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of $\triangle ACE$ and $\triangle XYZ$?
$\textbf{(A)}\ \frac {3}{8}\sqrt{3} \qquad \textbf{(B)}\ \frac {7}{16}\sqrt{3} \qquad \textbf{(C)}\ \frac {15}{32}\sqrt{3} \qquad \textbf{(D)}\ \frac {1}{2}\sqrt{3} \qquad \textbf{(E)}\ \frac {9}{16}\sqrt{3}$ | 2018 AMC 10B Problems/Problem 24 | (~MRENTHUSIASM's diagram + modifications)
Lemma 1: $MP$ is midsegment of $\triangle ANE$.
As $ZX$ is a midsegment of trapezoid $FEBA$, we know that $FA||ZX||EB$. Furthermore, by subdividing trapezoid $FEBA$, $ZM,MP,PX$ are midsegments of $\triangle EFA, \triangle ANE, \triangle, ANB$ respectively.
Lemma 2: $[ACE] = \frac12[ABCDEF]$
Due to radial symmetry about the center, $G$, we only have to prove $[AGC] = [CBA]$. As $\triangle AGB$ and $\triangle CBG$ are equilateral triangles, $AC$ therefore divides rhombus $ABCG$ into 2 congruent halves. $\triangle AGC$ and $\triangle CBA$ thus each contain two such congruent triangles, which means that their areas are equal.
Lemma 3: $[AMP] = \frac1{16}[ABCDEF]$
From Lemma 1, $\triangle AMP \simeq \triangle AEN$ with a scale factor of $\frac12$, which means $[AMP] = \frac14[AEN]$. Furthermore, due to symmetry of $\triangle AEN$ and $\triangle CEN$ about $EN$, $[AEN] = \frac12[ACE]$. Therefore, $[AMP] = \frac18[ACE]$, or (using lemma 2) $\frac1{16}[ABCDEF]$.
Lemma 4: $[ABCDEF] = \frac{3\sqrt3}2$
Because the hexagon is regular, $[ABCDEF] = 6[ABG] = \frac{6\sqrt3}4 = \frac{3\sqrt3}2$. (For more info on where $\frac{\sqrt3}4$ came from, use the search term "area of equilateral triangle")
Final step:
To find the area of the shaded part, we seek $[ACE] - 3[AMP]$. Using lemmas 2 and 3, the expression resolves to $\frac12[ABCDEF] - 3\frac1{16}[ABCDEF] = \frac5{16} \cdot \frac{3\sqrt3}2 = \boxed{\textbf{(C)}\ \frac{15}{32}\sqrt{3}}$.
~math660 | // Block 1
/* Made by MRENTHUSIASM */
size(200);
draw(polygon(6));
pair A, B, C, D, E, F, X, Y, Z, M, N, O, P, Q, R, G;
A = dir(120);
B = dir(60);
C = dir(0);
D = dir(300);
E = dir(240);
F = dir(180);
X = midpoint(A--B);
Y = midpoint(C--D);
Z = midpoint(E--F);
G = midpoint(A--D);
M = intersectionpoint(A--E,X--Z);
N = intersectionpoint(A--C,X--Y);
O = intersectionpoint(C--E,Y--Z);
P = intersectionpoint(A--C,X--Z);
Q = intersectionpoint(C--E,X--Y);
R = intersectionpoint(A--E,Y--Z);
fill(M--P--N--Q--O--R--cycle,mediumgray);
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("$E$",E,1.5*dir(E),linewidth(4));
dot("$F$",F,1.5*dir(F),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(O^^Q^^R,linewidth(4));
dot("$N$",N,1.5*dir(N),linewidth(4));
dot("$M$",M,1.5*dir(M),linewidth(4));
dot("$P$",P,1.5*dir(P),linewidth(4));
dot("$G$",G,1.5*dir(G),linewidth(4));
draw(A--C--E--cycle^^X--Y--Z--cycle);
draw(A--G^^C--G);
draw(N--B);
draw(N--E--A--cycle^^M--P, red);
// Block 2
/* Made by MRENTHUSIASM */ size(200); draw(polygon(6)); pair A, B, C, D, E, F, X, Y, Z, M, N, O, P, Q, R, G; A = dir(120); B = dir(60); C = dir(0); D = dir(300); E = dir(240); F = dir(180); X = midpoint(A--B); Y = midpoint(C--D); Z = midpoint(E--F); G = midpoint(A--D); M = intersectionpoint(A--E,X--Z); N = intersectionpoint(A--C,X--Y); O = intersectionpoint(C--E,Y--Z); P = intersectionpoint(A--C,X--Z); Q = intersectionpoint(C--E,X--Y); R = intersectionpoint(A--E,Y--Z); fill(M--P--N--Q--O--R--cycle,mediumgray); 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("$E$",E,1.5*dir(E),linewidth(4)); dot("$F$",F,1.5*dir(F),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(O^^Q^^R,linewidth(4)); dot("$N$",N,1.5*dir(N),linewidth(4)); dot("$M$",M,1.5*dir(M),linewidth(4)); dot("$P$",P,1.5*dir(P),linewidth(4)); dot("$G$",G,1.5*dir(G),linewidth(4)); draw(A--C--E--cycle^^X--Y--Z--cycle); draw(A--G^^C--G); draw(N--B); draw(N--E--A--cycle^^M--P, red); | [] |
502 | Square $ABCD$ has side length $30$. Point $P$ lies inside the square so that $AP = 12$ and $BP = 26$. The centroids of $\triangle{ABP}$, $\triangle{BCP}$, $\triangle{CDP}$, and $\triangle{DAP}$ are the vertices of a convex quadrilateral. What is the area of that quadrilateral?
$\textbf{(A) }100\sqrt{2}\qquad\textbf{(B) }100\sqrt{3}\qquad\textbf{(C) }200\qquad\textbf{(D) }200\sqrt{2}\qquad\textbf{(E) }200\sqrt{3}$ | 2018 AMC 12B Problems/Problem 13 | As shown below, let $M_1,M_2,M_3,M_4$ be the midpoints of $\overline{AB},\overline{BC},\overline{CD},\overline{DA},$ respectively, and $G_1,G_2,G_3,G_4$ be the centroids of $\triangle{ABP},\triangle{BCP},\triangle{CDP},\triangle{DAP},$ respectively.
By SAS, we conclude that $\triangle G_1G_2P\sim\triangle M_1M_2P, \triangle G_2G_3P\sim\triangle M_2M_3P, \triangle G_3G_4P\sim\triangle M_3M_4P,$ and $\triangle G_4G_1P\sim\triangle M_4M_1P.$ By the properties of centroids, the ratio of similitude for each pair of triangles is $\frac{2}{3}.$
Note that quadrilateral $M_1M_2M_3M_4$ is a square of side-length $15\sqrt2.$ It follows that:
Since $\overline{G_1G_2}\parallel\overline{M_1M_2},\overline{G_2G_3}\parallel\overline{M_2M_3},\overline{G_3G_4}\parallel\overline{M_3M_4},$ and $\overline{G_4G_1}\parallel\overline{M_4M_1}$ by the Converse of the Corresponding Angles Postulate, we have $\angle G_1G_2G_3=\angle G_2G_3G_4=\angle G_3G_4G_1=\angle G_4G_1G_2=90^\circ.$
Since $G_1G_2=\frac23M_1M_2, G_2G_3=\frac23M_2M_3, G_3G_4=\frac23M_3M_4,$ and $G_4G_1=\frac23M_4M_1$ by the ratio of similitude, we have $G_1G_2=G_2G_3=G_3G_4=G_4G_1=10\sqrt2.$
Together, quadrilateral $G_1G_2G_3G_4$ is a square of side-length $10\sqrt2,$ so its area is $\left(10\sqrt2\right)^2=\boxed{\textbf{(C) }200}.$
Remark
This solution shows that, if point $P$ is within square $ABCD,$ then the shape and the area of quadrilateral $G_1G_2G_3G_4$ are independent of the location of $P.$ Let the brackets denote areas. More generally, $G_1G_2G_3G_4$ is always a square of area \[[G_1G_2G_3G_4]=\left(\frac23\right)^2[M_1M_2M_3M_4]=\frac49[M_1M_2M_3M_4]=\frac29[ABCD].\] On the other hand, the location of $G_1G_2G_3G_4$ is dependent on the location of $P.$
~RandomPieKevin ~Kyriegon ~MRENTHUSIASM | // Block 1
/* Made by MRENTHUSIASM */
unitsize(210);
pair B = (0, 0), A = (0, 1), D = (1, 1), C = (1, 0), P = (1/4, 2/3);
pair M1 = midpoint(A--B);
pair M2 = midpoint(B--C);
pair M3 = midpoint(C--D);
pair M4 = midpoint(D--A);
pair G1 = centroid(A,B,P);
pair G2 = centroid(B,C,P);
pair G3 = centroid(C,D,P);
pair G4 = centroid(D,A,P);
filldraw(M1--M2--P--cycle,red);
filldraw(M2--M3--P--cycle,yellow);
filldraw(M3--M4--P--cycle,green);
filldraw(M4--M1--P--cycle,lightblue);
draw(A--B--C--D--cycle);
draw(M1--M2--M3--M4--cycle);
draw(G1--G2--G3--G4--cycle);
dot(P);
defaultpen(fontsize(10pt));
draw(A--P--B);
draw(C--P--D);
label("$A$", A, W);
label("$B$", B, W);
label("$C$", C, E);
label("$D$", D, E);
label("$P$", P, N);
label("$M_1$", M1, W);
label("$M_2$", M2, S);
label("$M_3$", M3, E);
label("$M_4$", M4, N);
label("$G_1$", G1, 1.5S);
label("$G_2$", G2, 1.5E);
label("$G_3$", G3, 1.5NE);
label("$G_4$", G4, 1.5E);
dot(A);
dot(B);
dot(C);
dot(D);
dot(M1);
dot(M2);
dot(M3);
dot(M4);
dot(G1);
dot(G2);
dot(G3);
dot(G4);
// Block 2
/* Made by MRENTHUSIASM */ unitsize(210); pair B = (0, 0), A = (0, 1), D = (1, 1), C = (1, 0), P = (1/4, 2/3); pair M1 = midpoint(A--B); pair M2 = midpoint(B--C); pair M3 = midpoint(C--D); pair M4 = midpoint(D--A); pair G1 = centroid(A,B,P); pair G2 = centroid(B,C,P); pair G3 = centroid(C,D,P); pair G4 = centroid(D,A,P); filldraw(M1--M2--P--cycle,red); filldraw(M2--M3--P--cycle,yellow); filldraw(M3--M4--P--cycle,green); filldraw(M4--M1--P--cycle,lightblue); draw(A--B--C--D--cycle); draw(M1--M2--M3--M4--cycle); draw(G1--G2--G3--G4--cycle); dot(P); defaultpen(fontsize(10pt)); draw(A--P--B); draw(C--P--D); label("$A$", A, W); label("$B$", B, W); label("$C$", C, E); label("$D$", D, E); label("$P$", P, N); label("$M_1$", M1, W); label("$M_2$", M2, S); label("$M_3$", M3, E); label("$M_4$", M4, N); label("$G_1$", G1, 1.5S); label("$G_2$", G2, 1.5E); label("$G_3$", G3, 1.5NE); label("$G_4$", G4, 1.5E); dot(A); dot(B); dot(C); dot(D); dot(M1); dot(M2); dot(M3); dot(M4); dot(G1); dot(G2); dot(G3); dot(G4); | [] |
503 | The solutions to the equation $(z+6)^8=81$ are connected in the complex plane to form a convex regular polygon, three of whose vertices are labeled $A,B,$ and $C$. What is the least possible area of $\triangle ABC?$
$\textbf{(A) } \frac{1}{6}\sqrt{6} \qquad \textbf{(B) } \frac{3}{2}\sqrt{2}-\frac{3}{2} \qquad \textbf{(C) } 2\sqrt3-3\sqrt2 \qquad \textbf{(D) } \frac{1}{2}\sqrt{2} \qquad \textbf{(E) } \sqrt 3-1$ | 2018 AMC 12B Problems/Problem 16 | Recall that translations preserve the shapes and the sizes for all objects. We translate the solutions to the given equation $6$ units right, so they become the solutions to the equation $z^8=81.$
We rewrite $z$ to the polar form \[z=r(\cos\theta+i\sin\theta)=r\operatorname{cis}\theta,\] where $r$ is the magnitude of $z$ such that $r\geq0,$ and $\theta$ is the argument of $z$ such that $0\leq\theta<2\pi.$
By De Moivre's Theorem, we have \[z^8=r^8\operatorname{cis}(8\theta)={\sqrt3}^8(1),\] from which
$r^8={\sqrt3}^8,$ so $r=\sqrt3.$
$\begin{cases} \begin{aligned} \cos(8\theta) &= 1 \\ \sin(8\theta) &= 0 \end{aligned}, \end{cases}$ so $\theta=0,\frac{\pi}{4},\frac{\pi}{2},\frac{3\pi}{4},\pi,\frac{5\pi}{4},\frac{3\pi}{2},\frac{7\pi}{4}.$
In the complex plane, the solutions to the equation $z^8=81$ are the vertices of a regular octagon with center $0$ and radius $\sqrt3.$
The least possible area of $\triangle ABC$ occurs when $A,B,$ and $C$ are the consecutive vertices of the octagon. For simplicity purposes, let $A=\sqrt3\operatorname{cis}\frac{\pi}{4}=\frac{\sqrt6}{2}+\frac{\sqrt6}{2}i, B=\sqrt3\operatorname{cis}\frac{\pi}{2}=\sqrt3i,$ and $C=\sqrt3\operatorname{cis}\frac{3\pi}{4}=-\frac{\sqrt6}{2}+\frac{\sqrt6}{2}i,$ as shown below.
Note that $\triangle ABC$ has base $AC=\sqrt6$ and height $\sqrt3-\frac{\sqrt6}{2},$ so its area is \[\frac12\cdot\sqrt6\cdot\left(\sqrt3-\frac{\sqrt6}{2}\right)=\boxed{\textbf{(B) } \frac{3}{2}\sqrt{2}-\frac{3}{2}}.\]
~MRENTHUSIASM | // Block 1
/* Made by MRENTHUSIASM */
size(200);
int xMin = -2;
int xMax = 2;
int yMin = -2;
int yMax = 2;
int numRays = 24;
//Draws a polar grid that goes out to a number of circles
//equal to big, with numRays specifying the number of rays:
void polarGrid(int big, int numRays)
{
for (int i = 1; i < big+1; ++i)
{
draw(Circle((0,0),i), gray+linewidth(0.4));
}
for (int i=0;i<numRays;++i)
draw(rotate(i*360/numRays)*((-big,0)--(big,0)), gray+linewidth(0.4));
}
//Draws the horizontal gridlines
void horizontalLines()
{
for (int i = yMin+1; i < yMax; ++i)
{
draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4));
}
}
//Draws the vertical gridlines
void verticalLines()
{
for (int i = xMin+1; i < xMax; ++i)
{
draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4));
}
}
horizontalLines();
verticalLines();
polarGrid(xMax,numRays);
draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5));
draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5));
label("Re",(xMax,0),2*E);
label("Im",(0,yMax),2*N);
//The n such that we're taking the nth roots of unity multiplied by 2.
int n = 8;
pair A[];
for(int i = 0; i <= n-1; i+=1) {
A[i] = rotate(360*i/n)*(sqrt(3),0);
}
label("$A$",A[1],1.5*NE,UnFill);
label("$B$",A[2],1.5*NE,UnFill);
label("$C$",A[3],1.5*NW,UnFill);
fill(A[1]--A[2]--A[3]--cycle,green);
draw(A[1]--A[3]^^A[0]--A[1]--A[2]--A[3]--A[4]--A[5]--A[6]--A[7]--cycle,red);
for(int i = 0; i< n; ++i) dot(A[i],red+linewidth(4.5));
// Block 2
/* Made by MRENTHUSIASM */ size(200); int xMin = -2; int xMax = 2; int yMin = -2; int yMax = 2; int numRays = 24; //Draws a polar grid that goes out to a number of circles //equal to big, with numRays specifying the number of rays: void polarGrid(int big, int numRays) { for (int i = 1; i < big+1; ++i) { draw(Circle((0,0),i), gray+linewidth(0.4)); } for (int i=0;i<numRays;++i) draw(rotate(i*360/numRays)*((-big,0)--(big,0)), gray+linewidth(0.4)); } //Draws the horizontal gridlines void horizontalLines() { for (int i = yMin+1; i < yMax; ++i) { draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4)); } } //Draws the vertical gridlines void verticalLines() { for (int i = xMin+1; i < xMax; ++i) { draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4)); } } horizontalLines(); verticalLines(); polarGrid(xMax,numRays); draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("Re",(xMax,0),2*E); label("Im",(0,yMax),2*N); //The n such that we're taking the nth roots of unity multiplied by 2. int n = 8; pair A[]; for(int i = 0; i <= n-1; i+=1) { A[i] = rotate(360*i/n)*(sqrt(3),0); } label("$A$",A[1],1.5*NE,UnFill); label("$B$",A[2],1.5*NE,UnFill); label("$C$",A[3],1.5*NW,UnFill); fill(A[1]--A[2]--A[3]--cycle,green); draw(A[1]--A[3]^^A[0]--A[1]--A[2]--A[3]--A[4]--A[5]--A[6]--A[7]--cycle,red); for(int i = 0; i< n; ++i) dot(A[i],red+linewidth(4.5)); | [] |
504 | Ajay is standing at point $A$ near Pontianak, Indonesia, $0^\circ$ latitude and $110^\circ \text{ E}$ longitude. Billy is standing at point $B$ near Big Baldy Mountain, Idaho, USA, $45^\circ \text{ N}$ latitude and $115^\circ \text{ W}$ longitude. Assume that Earth is a perfect sphere with center $C.$ What is the degree measure of $\angle ACB?$
$\textbf{(A) }105 \qquad \textbf{(B) }112\frac{1}{2} \qquad \textbf{(C) }120 \qquad \textbf{(D) }135 \qquad \textbf{(E) }150 \qquad$ | 2018 AMC 12B Problems/Problem 23 | This solution refers to the Diagram section.
Let $D$ be the orthogonal projection of $B$ onto the equator. Note that $\angle BDA = \angle BDC = 90^\circ$ and $\angle BCD = 45^\circ.$ Recall that $115^\circ \text{ W}$ longitude is the same as $245^\circ \text{ E}$ longitude, so $\angle ACD=135^\circ.$
We obtain the following diagram:
Without loss of generality, let $AC=BC=1.$ For tetrahedron $ABCD:$
Since $\triangle BCD$ is an isosceles right triangle, we have $BD=CD=\frac{\sqrt2}{2}.$
In $\triangle ACD,$ we apply the Law of Cosines to get $AD=\sqrt{AC^2+CD^2-2\cdot AC\cdot CD\cdot\cos\angle ACD}=\frac{\sqrt{10}}{2}.$
In right $\triangle ABD,$ we apply the Pythagorean Theorem to get $AB=\sqrt{AD^2+BD^2}=\sqrt{3}.$
In $\triangle ABC,$ we apply the Law of Cosines to get $\cos\angle ACB=\frac{AC^2+BC^2-AB^2}{2\cdot AC\cdot BC}=-\frac12,$ so $\angle ACB=\boxed{\textbf{(C) }120}$ degrees.
~MRENTHUSIASM | // Block 1
/* Made by MRENTHUSIASM */
size(250);
import graph3;
import solids;
currentprojection=orthographic((0.2,-0.5,0.2));
triple A, B, C, D;
A = (1,0,0);
B = (-1/2,1/2,sqrt(2)/2);
C = (0,0,0);
D = (-1/2,1/2,0);
draw(unitsphere,white,light=White);
draw(surface(A--B--C--cycle),yellow);
dot(A^^B^^C^^D,linewidth(4.5));
draw(Circle(C,1,(0,0,1))^^A--B--D--C--B^^C--A--D);
label("$A$",A,3*dir(A));
label("$B$",B,3*dir(B));
label("$C$",C,3*(0,0,-1));
label("$D$",D,3*(-1/2,-1/2,0));
// Block 2
/* Made by MRENTHUSIASM */ size(250); import graph3; import solids; currentprojection=orthographic((0.2,-0.5,0.2)); triple A, B, C, D; A = (1,0,0); B = (-1/2,1/2,sqrt(2)/2); C = (0,0,0); D = (-1/2,1/2,0); draw(unitsphere,white,light=White); draw(surface(A--B--C--cycle),yellow); dot(A^^B^^C^^D,linewidth(4.5)); draw(Circle(C,1,(0,0,1))^^A--B--D--C--B^^C--A--D); label("$A$",A,3*dir(A)); label("$B$",B,3*dir(B)); label("$C$",C,3*(0,0,-1)); label("$D$",D,3*(-1/2,-1/2,0)); | [] |
504 | Ajay is standing at point $A$ near Pontianak, Indonesia, $0^\circ$ latitude and $110^\circ \text{ E}$ longitude. Billy is standing at point $B$ near Big Baldy Mountain, Idaho, USA, $45^\circ \text{ N}$ latitude and $115^\circ \text{ W}$ longitude. Assume that Earth is a perfect sphere with center $C.$ What is the degree measure of $\angle ACB?$
$\textbf{(A) }105 \qquad \textbf{(B) }112\frac{1}{2} \qquad \textbf{(C) }120 \qquad \textbf{(D) }135 \qquad \textbf{(E) }150 \qquad$ | 2018 AMC 12B Problems/Problem 23 | This solution refers to the Diagram section.
Let $D$ be the orthogonal projection of $B$ onto the equator. Note that $\angle BDA = \angle BDC = 90^\circ$ and $\angle BCD = 45^\circ.$ Recall that $115^\circ \text{ W}$ longitude is the same as $245^\circ \text{ E}$ longitude, so $\angle ACD=135^\circ.$
Without loss of generality, let $AC=BC=1.$ As shown below, we place Earth in the $xyz$-plane with $C=(0,0,0)$ such that the positive $x$-axis runs through $A,$ the positive $y$-axis runs through $0^\circ$ latitude and $160^\circ \text{ W}$ longitude, and the positive $z$-axis runs through the North Pole.
It follows that $A=(1,0,0)$ and $D=(-t,t,0)$ for some positive number $t.$ Since $\triangle BCD$ is an isosceles right triangle, we have $B=\left(-t,t,\sqrt{2}t\right).$ By the Distance Formula, we get $(-t)^2+t^2+\left(\sqrt{2}t\right)^2=1,$ from which $t=\frac12.$
As $\vec{A} = \begin{pmatrix}1 \\ 0 \\ 0 \end{pmatrix}$ and $\vec{B} = \begin{pmatrix}-1/2 \\ 1/2 \\ \sqrt2/2 \end{pmatrix},$ we obtain \[\cos\angle ACB=\frac{\vec{A}\bullet\vec{B}}{\left\lVert\vec{A}\right\rVert\left\lVert\vec{B}\right\rVert}=-\frac12\] by the dot product, so $\angle ACB=\boxed{\textbf{(C) }120}$ degrees.
~MRENTHUSIASM | // Block 1
/* Made by MRENTHUSIASM */
size(300);
import graph3;
import solids;
currentprojection=orthographic((0.2,-0.5,0.2));
triple A, B, C, D;
A = (1,0,0);
B = (-1/2,1/2,sqrt(2)/2);
C = (0,0,0);
D = (-1/2,1/2,0);
draw(unitsphere,white,light=White);
dot(A^^B^^C^^D,linewidth(4.5));
draw(Circle(C,1,(0,0,1))^^B--C--D--cycle);
label("$A$",A,5*dir((2.5,-3,0)));
label("$B$",B,3*dir(B));
label("$C$",C,1.5*(1,0,-1));
label("$D$",D,3*(-1/2,-1/2,0));
draw((-1.5,0,0)--(1.5,0,0),linewidth(1.25),EndArrow3(10));
draw((0,-1.5,0)--(0,1.5,0),linewidth(1.25),EndArrow3(10));
draw((0,0,-1.5)--(0,0,1.5),linewidth(1.25),EndArrow3(10));
label("$x$",(1.5,0,0),2*dir((1.5,0,0)));
label("$y$",(0,1.5,0),3*dir((0,1.5,0)));
label("$z$",(0,0,1.5),2*dir((0,0,1.5)));
// Block 2
/* Made by MRENTHUSIASM */ size(300); import graph3; import solids; currentprojection=orthographic((0.2,-0.5,0.2)); triple A, B, C, D; A = (1,0,0); B = (-1/2,1/2,sqrt(2)/2); C = (0,0,0); D = (-1/2,1/2,0); draw(unitsphere,white,light=White); dot(A^^B^^C^^D,linewidth(4.5)); draw(Circle(C,1,(0,0,1))^^B--C--D--cycle); label("$A$",A,5*dir((2.5,-3,0))); label("$B$",B,3*dir(B)); label("$C$",C,1.5*(1,0,-1)); label("$D$",D,3*(-1/2,-1/2,0)); draw((-1.5,0,0)--(1.5,0,0),linewidth(1.25),EndArrow3(10)); draw((0,-1.5,0)--(0,1.5,0),linewidth(1.25),EndArrow3(10)); draw((0,0,-1.5)--(0,0,1.5),linewidth(1.25),EndArrow3(10)); label("$x$",(1.5,0,0),2*dir((1.5,0,0))); label("$y$",(0,1.5,0),3*dir((0,1.5,0))); label("$z$",(0,0,1.5),2*dir((0,0,1.5))); | [] |
505 | Circles $\omega_1$, $\omega_2$, and $\omega_3$ each have radius $4$ and are placed in the plane so that each circle is externally tangent to the other two. Points $P_1$, $P_2$, and $P_3$ lie on $\omega_1$, $\omega_2$, and $\omega_3$ respectively such that $P_1P_2=P_2P_3=P_3P_1$ and line $P_iP_{i+1}$ is tangent to $\omega_i$ for each $i=1,2,3$, where $P_4 = P_1$. See the figure below. The area of $\triangle P_1P_2P_3$ can be written in the form $\sqrt{a}+\sqrt{b}$ for positive integers $a$ and $b$. What is $a+b$?
$\textbf{(A) }546\qquad\textbf{(B) }548\qquad\textbf{(C) }550\qquad\textbf{(D) }552\qquad\textbf{(E) }554$ | 2018 AMC 12B Problems/Problem 25 | First, note that because the $\angle P_1=\angle P_2=\angle P_3=\pi/3$, the arcs inside the shaded equilateral triangle are each $2\pi/3$. Also, the distances between the centers of any two of the $3$ given circles are each $8$.
Draw the circle $\Gamma$ concentric with $\omega_1$ with radius $2$. Because the arc of $\omega_1$ inside said triangle is $2\pi/3$, $\Gamma$ touches $P_1P_3$, say at a point $X$. Thus, $P_1P_3$ is a common tangent of $\omega_3$ and $\Gamma$, and it can be seen from inspection of the given diagram that the line is an common internal tangent.
The length of the common internal tangent segment $XP_3$ of $\Gamma$ and $\omega_3$ is then $\sqrt{8^2-(2+4)^2}=2\sqrt{7}$, and it is easily seen that $XP_1=4\sin \pi/3=2\sqrt{3}$.
Because $P_1P_3=2(\sqrt{3}+\sqrt{7})$, the area of the shaded equilateral triangle is $\sqrt{3}(\sqrt{3}+\sqrt{7})^2=10\sqrt{3}+6\sqrt{7}$. We get $\sqrt{300}+\sqrt{252}\Rightarrow\boxed{\textbf{(D) }552}.$
~crazyeyemoody907 | // Block 1
unitsize(12);
pair A = (0, 8/sqrt(3)), B = rotate(-120)*A, C = rotate(120)*A;
real theta = 41.5;
pair P1 = rotate(theta)*(2+2*sqrt(7/3), 0), P2 = rotate(-120)*P1, P3 = rotate(120)*P1;
filldraw(P1--P2--P3--cycle, gray(0.9));
draw(Circle(A, 4));
draw(Circle(B, 4));
draw(Circle(C, 4));
dot(P1);
dot(P2);
dot(P3);
defaultpen(fontsize(10pt));
label("$P_1$", P1, E*1.5);
label("$P_2$", P2, SW*1.5);
label("$P_3$", P3, N);
label("$\omega_1$", A, W*30);
label("$\omega_2$", B, E*17);
label("$\omega_3$", C, W*17);
label("$\Gamma$", A, W*15);
draw(Circle(A,2),red);
pair X=foot(A,P1,P3);
dot(X,blue);
draw(A--X,blue);
label("$2\sqrt{7}$", X--P3);
label("$2\sqrt{3}$",X--P1);
label("$X$",X,dir(-80),blue);
// Block 2
unitsize(12); pair A = (0, 8/sqrt(3)), B = rotate(-120)*A, C = rotate(120)*A; real theta = 41.5; pair P1 = rotate(theta)*(2+2*sqrt(7/3), 0), P2 = rotate(-120)*P1, P3 = rotate(120)*P1; filldraw(P1--P2--P3--cycle, gray(0.9)); draw(Circle(A, 4)); draw(Circle(B, 4)); draw(Circle(C, 4)); dot(P1); dot(P2); dot(P3); defaultpen(fontsize(10pt)); label("$P_1$", P1, E*1.5); label("$P_2$", P2, SW*1.5); label("$P_3$", P3, N); label("$\omega_1$", A, W*30); label("$\omega_2$", B, E*17); label("$\omega_3$", C, W*17); label("$\Gamma$", A, W*15); draw(Circle(A,2),red); pair X=foot(A,P1,P3); dot(X,blue); draw(A--X,blue); label("$2\sqrt{7}$", X--P3); label("$2\sqrt{3}$",X--P1); label("$X$",X,dir(-80),blue); | [] |
505 | Circles $\omega_1$, $\omega_2$, and $\omega_3$ each have radius $4$ and are placed in the plane so that each circle is externally tangent to the other two. Points $P_1$, $P_2$, and $P_3$ lie on $\omega_1$, $\omega_2$, and $\omega_3$ respectively such that $P_1P_2=P_2P_3=P_3P_1$ and line $P_iP_{i+1}$ is tangent to $\omega_i$ for each $i=1,2,3$, where $P_4 = P_1$. See the figure below. The area of $\triangle P_1P_2P_3$ can be written in the form $\sqrt{a}+\sqrt{b}$ for positive integers $a$ and $b$. What is $a+b$?
$\textbf{(A) }546\qquad\textbf{(B) }548\qquad\textbf{(C) }550\qquad\textbf{(D) }552\qquad\textbf{(E) }554$ | 2018 AMC 12B Problems/Problem 25 | Suppose $P_1P_2 = x$, noticed that $OMP_1H$ is rectangle.
\[OO_1 = \frac{O_1O_2}{\sqrt{3}} = \frac83\sqrt{3}\]
\[OH = P_1M = \frac{1}{2}P_1P_2 = \frac12 x\]
\[O_1H = O_1P_1 - HP_1 = O_1P_1 - OM = O_1P_1 - \frac{P_1P_2}{2\sqrt{3}} = 4 - \frac{\sqrt{3}}{6}x\]
$\triangle O_1HO$ is right triangle, we can use Pythagorean theorem to establish an equation,
\[OH^2 + O_1H^2 = OO_1^2\]
\[\left(\frac12 x\right)^2 + \left(4 - \frac{\sqrt{3}}{6}x\right)^2 = \left(\frac83\sqrt{3}\right)^2\]
\[x^2 - 4\sqrt{3}x - 16 = 0\]
\[x = 2\sqrt{3} + 2\sqrt{7}\]
The side length of the equilateral triangle is $2(\sqrt{7} + \sqrt{3})$, so its area is $(\sqrt{7} + \sqrt{3})^2\sqrt{3} = 10\sqrt{3} + 6\sqrt{7} = \sqrt{300} + \sqrt{252} \implies 300 + 252 = \boxed{\textbf{(D) }552}$.
~reda_mandymath | // Block 1
import geometry;
unitsize(12);
pair A = (0, 8/sqrt(3)), B = rotate(-120)*A, C = rotate(120)*A;
real theta = 41.5;
pair P1 = rotate(theta)*(2+2*sqrt(7/3), 0), P2 = rotate(-120)*P1, P3 = rotate(120)*P1;
filldraw(P1--P2--P3--cycle, gray(0.9));
draw(Circle(A, 4));
draw(Circle(B, 4));
draw(Circle(C, 4));
dot(P1);
dot(P2);
dot(P3);
pair M = extension(P3, origin, P1, P2);
pair H = extension(A, P1, origin, rotate(90) * M);
dot(M ^^ H ^^ origin ^^ A);
draw(H -- P1 ^^ P3 -- M, dashed);
draw(A -- H -- origin -- cycle, red);
markrightangle(A, H, origin, 0.2 * markangleradius(), red);
defaultpen(fontsize(10pt));
label("$P_1$", P1, E*1.5);
label("$P_2$", P2, SW*1.5);
label("$P_3$", P3, N);
label("$\omega_1$", A, W*17);
label("$\omega_2$", B, E*17);
label("$\omega_3$", C, W*17);
label("$M$", M, SE);
label("$H$", H, NE);
label("$O_1$", A, N);
label("$O$", origin, NE);
// Block 2
import geometry; unitsize(12); pair A = (0, 8/sqrt(3)), B = rotate(-120)*A, C = rotate(120)*A; real theta = 41.5; pair P1 = rotate(theta)*(2+2*sqrt(7/3), 0), P2 = rotate(-120)*P1, P3 = rotate(120)*P1; filldraw(P1--P2--P3--cycle, gray(0.9)); draw(Circle(A, 4)); draw(Circle(B, 4)); draw(Circle(C, 4)); dot(P1); dot(P2); dot(P3); pair M = extension(P3, origin, P1, P2); pair H = extension(A, P1, origin, rotate(90) * M); dot(M ^^ H ^^ origin ^^ A); draw(H -- P1 ^^ P3 -- M, dashed); draw(A -- H -- origin -- cycle, red); markrightangle(A, H, origin, 0.2 * markangleradius(), red); defaultpen(fontsize(10pt)); label("$P_1$", P1, E*1.5); label("$P_2$", P2, SW*1.5); label("$P_3$", P3, N); label("$\omega_1$", A, W*17); label("$\omega_2$", B, E*17); label("$\omega_3$", C, W*17); label("$M$", M, SE); label("$H$", H, NE); label("$O_1$", A, N); label("$O$", origin, NE); | [] |
506 | In $\triangle ABC$, the sides have integer lengths and $AB=AC$. Circle $\omega$ has its center at the incenter of $\triangle ABC$. An excircle of $\triangle ABC$ is a circle in the exterior of $\triangle ABC$ that is tangent to one side of the triangle and tangent to the extensions of the other two sides. Suppose that the excircle tangent to $\overline{BC}$ is internally tangent to $\omega$, and the other two excircles are both externally tangent to $\omega$. Find the minimum possible value of the perimeter of $\triangle ABC$. | 2019 AIME I Problems/Problem 11 | First, assume $BC=2$ and $AB=AC=x$. The triangle can be scaled later if necessary. Let $I$ be the incenter and let $r$ be the inradius. Let the points at which the incircle intersects $AB$, $BC$, and $CA$ be denoted $M$, $N$, and $O$, respectively.
Next, we calculate $r$ in terms of $x$. Note the right triangle formed by $A$, $I$, and $M$. The length $IM$ is equal to $r$. Using the Pythagorean Theorem, the length $AN$ is $\sqrt{x^2-1}$, so the length $AI$ is $\sqrt{x^2-1}-r$. Note that $BN$ is half of $BC=2$, and by symmetry caused by the incircle, $BN=BM$ and $BM=1$, so $MA=x-1$. Applying the Pythagorean Theorem to $AIM$, we get
\[r^2+(x-1)^2=\left(\sqrt{x^2-1}-r\right)^2.\]
Expanding yields
\[r^2+x^2-2x+1=x^2-1-2r\sqrt{x^2-1}+r^2,\]
which can be simplified to
\[2r\sqrt{x^2-1}=2x-2.\]
Dividing by $2$ and then squaring results in
\[r^2(x^2-1)=(x-1)^2,\]
and isolating $r^2$ gets us
\[r^2=\frac{(x-1)^2}{x^2-1}=\frac{(x-1)^2}{(x+1)(x-1)}=\frac{x-1}{x+1},\]
so $r=\sqrt{\frac{x-1}{x+1}}$.
We then calculate the radius of the excircle tangent to $BC$. We denote the center of the excircle $E_N$ and the radius $r_N$.
Consider the quadrilateral formed by $M$, $I$, $E_N$, and the point at which the excircle intersects the extension of $AB$, which we denote $H$. By symmetry caused by the excircle, $BN=BH$, so $BH=1$.
Note that triangles $MBI$ and $NBI$ are congruent, and $HBE$ and $NBE$ are also congruent. Denoting the measure of angles $MBI$ and $NBI$ measure $\alpha$ and the measure of angles $HBE$ and $NBE$ measure $\beta$, straight angle $MBH=2\alpha+2\beta$, so $\alpha + \beta=90^\circ$. This means that angle $IBE$ is a right angle, so it forms a right triangle.
Setting the base of the right triangle to $IE$, the height is $BN=1$ and the base consists of $IN=r$ and $EN=r_N$. Triangles $INB$ and $BNE$ are similar to $IBE$, so $\frac{IN}{BN}=\frac{BN}{EN}$, or $\frac{r}{1}=\frac{1}{r_N}$. This makes $r_N$ the reciprocal of $r$, so $r_N=\sqrt{\frac{x+1}{x-1}}$.
Circle $\omega$'s radius can be expressed by the distance from the incenter $I$ to the bottom of the excircle with center $E_N$. This length is equal to $r+2r_N$, or $\sqrt{\frac{x-1}{x+1}}+2\sqrt{\frac{x+1}{x-1}}$. Denote this value $r_\omega$.
Finally, we calculate the distance from the incenter $I$ to the closest point on the excircle tangent to $AB$, which forms another radius of circle $\omega$ and is equal to $r_\omega$. We denote the center of the excircle $E_M$ and the radius $r_M$. We also denote the points where the excircle intersects $AB$ and the extension of $BC$ using $J$ and $K$, respectively. In order to calculate the distance, we must find the distance between $I$ and $E_M$ and subtract off the radius $r_M$.
We first must calculate the radius of the excircle. Because the excircle is tangent to both $AB$ and the extension of $AC$, its center must lie on the angle bisector formed by the two lines, which is parallel to $BC$. This means that the distance from $E_M$ to $K$ is equal to the length of $AN$, so the radius is also $\sqrt{x^2-1}$.
Next, we find the length of $IE_M$. We can do this by forming the right triangle $IAE_M$. The length of leg $AI$ is equal to $AN$ minus $r$, or $\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}$. In order to calculate the length of leg $AE_M$, note that right triangles $AJE_M$ and $BNA$ are congruent, as $JE_M$ and $NA$ share a length of $\sqrt{x^2-1}$, and angles $E_MAJ$ and $NAB$ add up to the right angle $NAE_M$. This means that $AE_M=BA=x$.
Using Pythagorean Theorem, we get
\[IE_M=\sqrt{\left(\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}\right)^2+x^2}.\]
Bringing back
\[r_\omega=IE_M-r_M\]
and substituting in some values, the equation becomes
\[r_\omega=\sqrt{\left(\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}\right)^2+x^2}-\sqrt{x^2-1}.\]
Rearranging and squaring both sides gets
\[\left(r_\omega+\sqrt{x^2-1}\right)^2=\left(\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}\right)^2+x^2.\]
Distributing both sides yields
\[r_\omega^2+2r_\omega\sqrt{x^2-1}+x^2-1=x^2-1-2\sqrt{x^2-1}\sqrt{\frac{x-1}{x+1}}+\frac{x-1}{x+1}+x^2.\]
Canceling terms results in
\[r_\omega^2+2r_\omega\sqrt{x^2-1}=-2\sqrt{x^2-1}\sqrt{\frac{x-1}{x+1}}+\frac{x-1}{x+1}+x^2.\]
Since
\[-2\sqrt{x^2-1}\sqrt{\frac{x-1}{x+1}}=-2\sqrt{(x+1)(x-1)\frac{x-1}{x+1}}=-2(x-1),\]
We can further simplify to
\[r_\omega^2+2r_\omega\sqrt{x^2-1}=-2(x-1)+\frac{x-1}{x+1}+x^2.\]
Substituting out $r_\omega$ gets
\[\left(\sqrt{\frac{x-1}{x+1}}+2\sqrt{\frac{x+1}{x-1}}\right)^2+2\left(\sqrt{\frac{x-1}{x+1}}+2\sqrt{\frac{x+1}{x-1}}\right)\sqrt{x^2-1}=-2(x-1)+\frac{x-1}{x+1}+x^2\]
which when distributed yields
\[\frac{x-1}{x+1}+4+4\left(\frac{x+1}{x-1}\right)+2(x-1+2(x+1))=-2(x-1)+\frac{x-1}{x+1}+x^2.\]
After some canceling, distributing, and rearranging, we obtain
\[4\left(\frac{x+1}{x-1}\right)=x^2-8x-4.\]
Multiplying both sides by $x-1$ results in
\[4x+4=x^3-x^2-8x^2+8x-4x+4,\]
which can be rearranged into
\[x^3-9x^2=0\]
and factored into
\[x^2(x-9)=0.\]
This means that $x$ equals $0$ or $9$, and since a side length of $0$ cannot exist, $x=9$.
As a result, the triangle must have sides in the ratio of $9:2:9$. Since the triangle must have integer side lengths, and these values share no common factors greater than $1$, the triangle with the smallest possible perimeter under these restrictions has a perimeter of $9+2+9=\boxed{020}$. ~emerald_block | // Block 1
unitsize(1cm);
var x = 9;
pair A = (0,sqrt(x^2-1));
pair B = (-1,0);
pair C = (1,0);
dot(Label("$A$",A,NE),A);
dot(Label("$B$",B,SW),B);
dot(Label("$C$",C,SE),C);
draw(A--B--C--cycle);
var r = sqrt((x-1)/(x+1));
pair I = (0,r);
dot(Label("$I$",I,SE),I);
draw(circle(I,r));
draw(Label("$r$"),I--I+r*SSW,dashed);
pair M = intersectionpoint(A--B,circle(I,r));
pair N = (0,0);
pair O = intersectionpoint(A--C,circle(I,r));
dot(Label("$M$",M,W),M);
dot(Label("$N$",N,S),N);
dot(Label("$O$",O,E),O);
var rN = sqrt((x+1)/(x-1));
pair EN = (0,-rN);
dot(Label("$E_N$",EN,SE),EN);
draw(circle(EN,rN));
draw(Label("$r_N$"),EN--EN+rN*SSW,dashed);
pair AB = (-1-2/(x-1),-2rN);
pair AC = (1+2/(x-1),-2rN);
draw(B--AB,EndArrow);
draw(C--AC,EndArrow);
pair H = intersectionpoint(B--AB,circle(EN,rN));
dot(Label("$H$",H,W),H);
var rM = sqrt(x^2-1);
pair EM = (-x,rM);
dot(Label("$E_M$",EM,SW),EM);
draw(Label("$r_M$"),EM--EM+rM*SSE,dashed);
pair CB = (-x-1,0);
pair CA = (-2/x,sqrt(x^2-1)+2(sqrt(x^2-1)/x));
draw(B--CB,EndArrow);
draw(A--CA,EndArrow);
pair J = intersectionpoint(A--B,circle(EM,rM));
pair K = intersectionpoint(B--CB,circle(EM,rM));
dot(Label("$J$",J,W),J);
dot(Label("$K$",K,S),K);
draw(arc(EM,rM,-100,15),Arrows);
// Block 2
unitsize(1cm); var x = 9; pair A = (0,sqrt(x^2-1)); pair B = (-1,0); pair C = (1,0); dot(Label("$A$",A,NE),A); dot(Label("$B$",B,SW),B); dot(Label("$C$",C,SE),C); draw(A--B--C--cycle); var r = sqrt((x-1)/(x+1)); pair I = (0,r); dot(Label("$I$",I,SE),I); draw(circle(I,r)); draw(Label("$r$"),I--I+r*SSW,dashed); pair M = intersectionpoint(A--B,circle(I,r)); pair N = (0,0); pair O = intersectionpoint(A--C,circle(I,r)); dot(Label("$M$",M,W),M); dot(Label("$N$",N,S),N); dot(Label("$O$",O,E),O); var rN = sqrt((x+1)/(x-1)); pair EN = (0,-rN); dot(Label("$E_N$",EN,SE),EN); draw(circle(EN,rN)); draw(Label("$r_N$"),EN--EN+rN*SSW,dashed); pair AB = (-1-2/(x-1),-2rN); pair AC = (1+2/(x-1),-2rN); draw(B--AB,EndArrow); draw(C--AC,EndArrow); pair H = intersectionpoint(B--AB,circle(EN,rN)); dot(Label("$H$",H,W),H); var rM = sqrt(x^2-1); pair EM = (-x,rM); dot(Label("$E_M$",EM,SW),EM); draw(Label("$r_M$"),EM--EM+rM*SSE,dashed); pair CB = (-x-1,0); pair CA = (-2/x,sqrt(x^2-1)+2(sqrt(x^2-1)/x)); draw(B--CB,EndArrow); draw(A--CA,EndArrow); pair J = intersectionpoint(A--B,circle(EM,rM)); pair K = intersectionpoint(B--CB,circle(EM,rM)); dot(Label("$J$",J,W),J); dot(Label("$K$",K,S),K); draw(arc(EM,rM,-100,15),Arrows); | [] |
506 | In $\triangle ABC$, the sides have integer lengths and $AB=AC$. Circle $\omega$ has its center at the incenter of $\triangle ABC$. An excircle of $\triangle ABC$ is a circle in the exterior of $\triangle ABC$ that is tangent to one side of the triangle and tangent to the extensions of the other two sides. Suppose that the excircle tangent to $\overline{BC}$ is internally tangent to $\omega$, and the other two excircles are both externally tangent to $\omega$. Find the minimum possible value of the perimeter of $\triangle ABC$. | 2019 AIME I Problems/Problem 11 | Before we start thinking about the problem, let’s draw it out;
For the sake of space, I've drawn only 2 of the 3 excircles because the third one looks the same as the second large one because the triangle is isosceles. By the incenter-excenter lemma, $AII_A$ and $BII_B$ are collinear, $E$ is the tangent of circle $I_B$ to $AC$, $F$ is the tangent of that circle to the extension of $BC$, and $J$ is the tangent of the circle to the extension of $BA$. The interesting part of the diagram is circle $\omega$, which is internally tangent to circle $I_A$ yet externally tangent to circle $I_B$. Therefore, perhaps we can relate the radius of this circle to the semiperimeter of triangle $ABC$.
We can see that the radius of circle $\omega$ is $2r_{I_A}+r$ using the incenter and A-excenter of our main triangle. This radius is also equal to $BI_B - BI - r_{I_B}$ from the incenter and B-excenter of our triangle. Thus, we can solve for each of these separately in terms of the lengths of the triangle and set them equal to each other to form an equation.
To find the left hand side of the equation, we have to first find $r$ and $r_{I_A}$. Let $a = AB = AC, b = BD = DC,$ and $h = AD$. Then since the perimeter of the triangle is $2a+2b$, the semiperimeter is $a+b$.
Now let's take a look at triangle $BDI$. Because $BI$ is the angle bisector of $\angle B$, by the angle bisector theorem, $\frac{AI}{ID} = \frac{BA}{BD} \implies \frac{h-r}{r} = \frac{a}{b}$. Rearranging, we get $r = \frac{hb}{a+b}$.
Take a look at triangle $AGI$. $AG = a - GB = a - BD = a-b$, $AI = h-r = \frac{ha}{a+b}$ (angle bisector theorem), and $GI = r = \frac{hb}{a+b}$. Now let's analyze triangle $AHI_A$. $AH = AB + BH = AB+ BD = a+b$, $AI_A = h+r_{I_A}$, and $HI_A = r_{I_A}$. Since $\angle GAI = \angle HAI_A$ and $\angle IGA = \angle I_AHA = 90^{\circ}$, triangle $AGI$ and $AHI_A$ are similar by AA. Then $\frac{r_{I_A}}{r} = \frac{h+r_{I_A}}{h-r} \implies r_{I_A} = r \cdot \frac{h+r_{I_A}}{h-r} = \frac{hb}{a+b} \cdot \frac{h+r_{I_A}}{\frac{ha}{a+b}} = \frac{b(h+r_{I_A})}{a}$. Now, solving yields $r_{I_A} = \frac{hb}{a-b}$.
Finally, the left hand side of our equation is \[\frac{2hb}{a-b} + \frac{hb}{a+b}\]
Now let's look at triangle $BFI_B$. How will we find $BI_B$? Let's first try to find $BF$ and $I_BF$ in terms of the lengths of the triangle. We recognize:
$BF = BC + CF = BC + CK$. We really want to have $CA$ instead of $CK$, and $AK$ looks very similar in length to $DC$, so let's try to prove that they are equal.
$BJ = BF$, so we can try to add these two and see if we get anything interesting. We have:
$BJ + BF = BA + AJ + BC + CF = BA + AE + BC + CE = BA + BC + CA$, which is our perimeter. Thus, $BF = a+b$.
Triangle $BDI$ is similar to triangle $BFI_B$ by AA, and we know that $BD = b$, and $ID = r = \frac{hb}{a+b}$, so thus $I_BF = BF \cdot \frac{ID}{BD} = (a+b) \cdot \frac{\frac{hb}{a+b}}{b} = \frac{hb}{b} = h$. Thus, the height of this triangle is $h$ by similarity ratios, the same height as vertex $A$. By the Pythagorean Theorem, $BI_B = \sqrt{(a+b)^2 + h^2}$ and by similarity ratios, $BI = \frac{b}{a+b} \cdot \sqrt{(a+b)^2 + h^2}$. Finally, $r_{I_B} = I_BF = h$, and thus the right hand side of our equation is \[\sqrt{(a+b)^2 + h^2} - \frac{b}{a+b} \cdot \sqrt{(a+b)^2 + h^2} - h = \sqrt{(a+b)^2 + h^2}(1 - \frac{b}{a+b}) - h = \sqrt{(a+b)^2 + h^2} \cdot \frac{a}{a+b} - h\].
Setting the two equal, we have \[\frac{2hb}{a-b} + \frac{hb}{a+b} = \sqrt{(a+b)^2 + h^2} \cdot \frac{a}{a+b} - h\]
Multiplying both sides by $(a+b)(a-b)$ we have $2hb(a+b) + hb(a-b) = \sqrt{(a+b)^2 + h^2} \cdot a(a-b) - h(a^2 - b^2)$
From here, let $b = 1$ arbitrarily; note that we can always scale this value to fit the requirements later. Thus our equation is $2h(a+1) + h(a-1) = \sqrt{(a+1)^2 + h^2} \cdot a(a-1) - h(a^2 - 1)$. Now since $h = \sqrt{a^2 - b^2}$, we can plug into our equation:
$2h(a+1) + h(a-1) = \sqrt{a² + 2a + 1 + a^2 - b^2 } \cdot a(a-1) - h(a^2 - 1)$. Remembering $b = 1$;
$\implies 2h(a+1) + h(a-1) = \sqrt{2a^2 + 2a} \cdot a(a-1) - h(a^2 - 1)$
$\implies 3ha + h + ha^2 - h = \sqrt{2a^2 + 2a} \cdot a(a-1)$
$\implies ah(3+a) = a(a-1) \cdot \sqrt{2a^2 + 2a}$
$\implies h^2(3+a)^2 = (a-1)^2 \cdot 2a(a+1)$
$\implies (a^2 - 1) (3 + a)^2 = 2a(a+1)(a - 1)^2$
$\implies (3+ a)^2 = 2a(a-1)$
$\implies 9 + 6a + a^2 = 2a^2 - 2a$
$\implies a^2 - 8a - 9 = 0$
$\implies (a-9)(a+1) = 0$
$\implies a = 9$ because the side lengths have to be positive numbers. Furthermore, because our values for $a$ and $b$ are relatively prime, we don't have to scale down our triangle further, and we are done. Therefore, our answer is $2a + 2b = 18 + 2 = \boxed{020}$
~KingRavi | // Block 1
unitsize(1cm);
var x = 9;
pair A = (0,sqrt(x^2-1));
pair B = (-1,0);
pair C = (1,0);
dot(Label("$A$",A,NE),A);
dot(Label("$B$",B,SW),B);
dot(Label("$C$",C,SE),C);
draw(A--B--C--cycle);
var r = sqrt((x-1)/(x+1));
pair I = (0,r);
dot(Label("$I$",I,SE),I);
draw(circle(I,r));
pair G = intersectionpoint(A--B,circle(I,r));
pair D = (0,0);
dot(Label("$G$",G,W),G);
dot(Label("$D$",D,SSE),D);
draw(Label("$r$"),I--G,dashed);
var rA = sqrt((x+1)/(x-1));
pair IA = (0,-rA);
dot(Label("$I_A$",IA,SE),IA);
draw(circle(IA,rA));
pair AB = (-1-2/(x-1),-2rA);
pair AC = (1+2/(x-1),-2rA);
draw(B--AB,EndArrow);
draw(C--AC,EndArrow);
pair H = intersectionpoint(B--AB,circle(IA,rA));
dot(Label("$H$",H,W),H);
draw(Label("$r_{I_A}$"),IA--H,dashed);
var rB = sqrt(x^2-1);
pair IB = (x,rB);
dot(Label("$I_B$",IB,SE),IB);
pair BC = (x+1,0);
pair BA = (2/x,sqrt(x^2-1)+2(sqrt(x^2-1)/x));
draw(C--BC,EndArrow);
draw(A--BA,EndArrow);
pair E = intersectionpoint(A--C,circle(IB,rB));
pair F = intersectionpoint(C--BC,circle(IB,rB));
dot(Label("$E$",E,SE),E);
dot(Label("$F$",F,S),F);
draw(Label("$r_{I_B}$"),IB--F,dashed);
draw(circle(IB,rB));
draw(A--IA);
draw(B--IB);
pair J = intersectionpoint(B--BA,circle(IB,rB));
dot(Label("$J$",J,W),J);
draw(circle(I,r+2rA));
pair W = intersectionpoint(B--F,circle(I,r+2rA));
dot(Label("$\omega$",W,SSE),W);
// Block 2
unitsize(1cm); var x = 9; pair A = (0,sqrt(x^2-1)); pair B = (-1,0); pair C = (1,0); dot(Label("$A$",A,NE),A); dot(Label("$B$",B,SW),B); dot(Label("$C$",C,SE),C); draw(A--B--C--cycle); var r = sqrt((x-1)/(x+1)); pair I = (0,r); dot(Label("$I$",I,SE),I); draw(circle(I,r)); pair G = intersectionpoint(A--B,circle(I,r)); pair D = (0,0); dot(Label("$G$",G,W),G); dot(Label("$D$",D,SSE),D); draw(Label("$r$"),I--G,dashed); var rA = sqrt((x+1)/(x-1)); pair IA = (0,-rA); dot(Label("$I_A$",IA,SE),IA); draw(circle(IA,rA)); pair AB = (-1-2/(x-1),-2rA); pair AC = (1+2/(x-1),-2rA); draw(B--AB,EndArrow); draw(C--AC,EndArrow); pair H = intersectionpoint(B--AB,circle(IA,rA)); dot(Label("$H$",H,W),H); draw(Label("$r_{I_A}$"),IA--H,dashed); var rB = sqrt(x^2-1); pair IB = (x,rB); dot(Label("$I_B$",IB,SE),IB); pair BC = (x+1,0); pair BA = (2/x,sqrt(x^2-1)+2(sqrt(x^2-1)/x)); draw(C--BC,EndArrow); draw(A--BA,EndArrow); pair E = intersectionpoint(A--C,circle(IB,rB)); pair F = intersectionpoint(C--BC,circle(IB,rB)); dot(Label("$E$",E,SE),E); dot(Label("$F$",F,S),F); draw(Label("$r_{I_B}$"),IB--F,dashed); draw(circle(IB,rB)); draw(A--IA); draw(B--IB); pair J = intersectionpoint(B--BA,circle(IB,rB)); dot(Label("$J$",J,W),J); draw(circle(I,r+2rA)); pair W = intersectionpoint(B--F,circle(I,r+2rA)); dot(Label("$\omega$",W,SSE),W); | [] |
507 | Triangle $ABC$ has side lengths $AB=4$, $BC=5$, and $CA=6$. Points $D$ and $E$ are on ray $AB$ with $AB<AD<AE$. The point $F \neq C$ is a point of intersection of the circumcircles of $\triangle ACD$ and $\triangle EBC$ satisfying $DF=2$ and $EF=7$. Then $BE$ can be expressed as $\tfrac{a+b\sqrt{c}}{d}$, where $a$, $b$, $c$, and $d$ are positive integers such that $a$ and $d$ are relatively prime, and $c$ is not divisible by the square of any prime. Find $a+b+c+d$. | 2019 AIME I Problems/Problem 13 | Notice that \[\angle DFE=\angle CFE-\angle CFD=\angle CBE-\angle CAD=180-B-A=C.\]By the Law of Cosines, \[\cos C=\frac{AC^2+BC^2-AB^2}{2\cdot AC\cdot BC}=\frac34.\]Then, \[DE^2=DF^2+EF^2-2\cdot DF\cdot EF\cos C=32\implies DE=4\sqrt2.\]Let $X=\overline{AB}\cap\overline{CF}$, $a=XB$, and $b=XD$. Then, \[XA\cdot XD=XC\cdot XF=XB\cdot XE\implies b(a+4)=a(b+4\sqrt2)\implies b=a\sqrt2.\]However, since $\triangle XFD\sim\triangle XAC$, $XF=\tfrac{4+a}3$, but since $\triangle XFE\sim\triangle XBC$, \[\frac75=\frac{4+a}{3a}\implies a=\frac54\implies BE=a+a\sqrt2+4\sqrt2=\frac{5+21\sqrt2}4,\]and the requested sum is $5+21+2+4=\boxed{032}$.
(Solution by TheUltimate123) | // Block 1
unitsize(20);
pair A, B, C, D, E, F, X, O1, O2;
A = (0, 0); B = (4, 0);
C = intersectionpoints(circle(A, 6), circle(B, 5))[0];
D = B + (5/4 * (1 + sqrt(2)), 0); E = D + (4 * sqrt(2), 0);
F = intersectionpoints(circle(D, 2), circle(E, 7))[1];
X = extension(A, E, C, F);
O1 = circumcenter(C, A, D);
O2 = circumcenter(C, B, E);
filldraw(A--B--C--cycle, lightcyan, deepcyan);
filldraw(D--E--F--cycle, lightmagenta, deepmagenta);
draw(B--D, gray(0.6));
draw(C--F, gray(0.6));
draw(circumcircle(C, A, D), dashed);
draw(circumcircle(C, B, E), dashed);
dot("$A$", A, dir(A-O1));
dot("$B$", B, dir(240));
dot("$C$", C, dir(120));
dot("$D$", D, dir(40));
dot("$E$", E, dir(E-O2));
dot("$F$", F, dir(270));
dot("$X$", X, dir(140));
label("$6$", (C+A)/2, dir(C-A)*I, deepcyan);
label("$5$", (C+B)/2, dir(B-C)*I, deepcyan);
label("$4$", (A+B)/2, dir(A-B)*I, deepcyan);
label("$7$", (F+E)/2, dir(F-E)*I, deepmagenta);
label("$2$", (F+D)/2, dir(D-F)*I, deepmagenta);
label("$4\sqrt{2}$", (D+E)/2, dir(E-D)*I, deepmagenta);
label("$a$", (B+X)/2, dir(B-X)*I, gray(0.3));
label("$a\sqrt{2}$", (D+X)/2, dir(D-X)*I, gray(0.3));
// Block 2
unitsize(20); pair A, B, C, D, E, F, X, O1, O2; A = (0, 0); B = (4, 0); C = intersectionpoints(circle(A, 6), circle(B, 5))[0]; D = B + (5/4 * (1 + sqrt(2)), 0); E = D + (4 * sqrt(2), 0); F = intersectionpoints(circle(D, 2), circle(E, 7))[1]; X = extension(A, E, C, F); O1 = circumcenter(C, A, D); O2 = circumcenter(C, B, E); filldraw(A--B--C--cycle, lightcyan, deepcyan); filldraw(D--E--F--cycle, lightmagenta, deepmagenta); draw(B--D, gray(0.6)); draw(C--F, gray(0.6)); draw(circumcircle(C, A, D), dashed); draw(circumcircle(C, B, E), dashed); dot("$A$", A, dir(A-O1)); dot("$B$", B, dir(240)); dot("$C$", C, dir(120)); dot("$D$", D, dir(40)); dot("$E$", E, dir(E-O2)); dot("$F$", F, dir(270)); dot("$X$", X, dir(140)); label("$6$", (C+A)/2, dir(C-A)*I, deepcyan); label("$5$", (C+B)/2, dir(B-C)*I, deepcyan); label("$4$", (A+B)/2, dir(A-B)*I, deepcyan); label("$7$", (F+E)/2, dir(F-E)*I, deepmagenta); label("$2$", (F+D)/2, dir(D-F)*I, deepmagenta); label("$4\sqrt{2}$", (D+E)/2, dir(E-D)*I, deepmagenta); label("$a$", (B+X)/2, dir(B-X)*I, gray(0.3)); label("$a\sqrt{2}$", (D+X)/2, dir(D-X)*I, gray(0.3)); | [] |
508 | Let $\overline{AB}$ be a chord of a circle $\omega$, and let $P$ be a point on the chord $\overline{AB}$. Circle $\omega_1$ passes through $A$ and $P$ and is internally tangent to $\omega$. Circle $\omega_2$ passes through $B$ and $P$ and is internally tangent to $\omega$. Circles $\omega_1$ and $\omega_2$ intersect at points $P$ and $Q$. Line $PQ$ intersects $\omega$ at $X$ and $Y$. Assume that $AP=5$, $PB=3$, $XY=11$, and $PQ^2 = \frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2019 AIME I Problems/Problem 15 | Let $O_1$ and $O_2$ be the centers of $\omega_1$ and $\omega_2$, respectively. There is a homothety at $A$ sending $\omega$ to $\omega_1$ that sends $B$ to $P$ and $O$ to $O_1$, so $\overline{OO_2}\parallel\overline{O_1P}$. Similarly, $\overline{OO_1}\parallel\overline{O_2P}$, so $OO_1PO_2$ is a parallelogram. Moreover, \[\angle O_1QO_2=\angle O_1PO_2=\angle O_1OO_2,\]hence $OO_1O_2Q$ is cyclic. However, \[OO_1=O_2P=O_2Q,\]so $OO_1O_2Q$ is an isosceles trapezoid. Since $\overline{O_1O_2}\perp\overline{XY}$, $\overline{OQ}\perp\overline{XY}$, so $Q$ is the midpoint of $\overline{XY}$.
By Power of a Point, $PX\cdot PY=PA\cdot PB=15$. Since $PX+PY=XY=11$ and $XQ=11/2$, \[XP=\frac{11-\sqrt{61}}2\implies PQ=XQ-XP=\frac{\sqrt{61}}2\implies PQ^2=\frac{61}4,\]
and the requested sum is $61+4=\boxed{065}$.
(Solution by TheUltimate123)
Note
One may solve for $PX$ first using PoAP, $PX = \frac{11}{2} - \frac{\sqrt{61}}{2}$. Then, notice that $PQ^2$ is rational but $PX^2$ is not, also $PX = \frac{XY}{2} - \frac{\sqrt{61}}{2}$. The most likely explanation for this is that $Q$ is the midpoint of $XY$, so that $XQ = \frac{11}{2}$ and $PQ=\frac{\sqrt{61}}{2}$. Then our answer is $m+n=61+4=\boxed{065}$. One can rigorously prove this using the methods above | // Block 1
size(8cm);
pair O, A, B, P, O1, O2, Q, X, Y;
O=(0, 0);
A=dir(140); B=dir(40);
P=(3A+5B)/8;
O1=extension((A+P)/2, (A+P)/2+(0, 1), A, O);
O2=extension((B+P)/2, (B+P)/2+(0, 1), B, O);
Q=intersectionpoints(circle(O1, length(A-O1)), circle(O2, length(B-O2)))[1];
X=intersectionpoint(P -- (P+(P-Q)*100), circle(O, 1));
Y=intersectionpoint(Q -- (Q+(Q-P)*100), circle(O, 1));
draw(circle(O, 1));
draw(circle(O1, length(A-O1)));
draw(circle(O2, length(B-O2)));
draw(A -- B); draw(X -- Y); draw(A -- O -- B); draw(O1 -- P -- O2);
dot("$O$", O, S);
dot("$A$", A, A);
dot("$B$", B, B);
dot("$P$", P, dir(70));
dot("$Q$", Q, dir(200));
dot("$O_1$", O1, SW);
dot("$O_2$", O2, SE);
dot("$X$", X, X);
dot("$Y$", Y, Y);
// Block 2
size(8cm); pair O, A, B, P, O1, O2, Q, X, Y; O=(0, 0); A=dir(140); B=dir(40); P=(3A+5B)/8; O1=extension((A+P)/2, (A+P)/2+(0, 1), A, O); O2=extension((B+P)/2, (B+P)/2+(0, 1), B, O); Q=intersectionpoints(circle(O1, length(A-O1)), circle(O2, length(B-O2)))[1]; X=intersectionpoint(P -- (P+(P-Q)*100), circle(O, 1)); Y=intersectionpoint(Q -- (Q+(Q-P)*100), circle(O, 1)); draw(circle(O, 1)); draw(circle(O1, length(A-O1))); draw(circle(O2, length(B-O2))); draw(A -- B); draw(X -- Y); draw(A -- O -- B); draw(O1 -- P -- O2); dot("$O$", O, S); dot("$A$", A, A); dot("$B$", B, B); dot("$P$", P, dir(70)); dot("$Q$", Q, dir(200)); dot("$O_1$", O1, SW); dot("$O_2$", O2, SE); dot("$X$", X, X); dot("$Y$", Y, Y); | [] |
508 | Let $\overline{AB}$ be a chord of a circle $\omega$, and let $P$ be a point on the chord $\overline{AB}$. Circle $\omega_1$ passes through $A$ and $P$ and is internally tangent to $\omega$. Circle $\omega_2$ passes through $B$ and $P$ and is internally tangent to $\omega$. Circles $\omega_1$ and $\omega_2$ intersect at points $P$ and $Q$. Line $PQ$ intersects $\omega$ at $X$ and $Y$. Assume that $AP=5$, $PB=3$, $XY=11$, and $PQ^2 = \frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2019 AIME I Problems/Problem 15 | $PX \cdot PY=AP \cdot PB=5 \cdot 3=15$ by power of a point. Also, $PX+PY=XY=11$, so $PX$ and $PY$ are solutions to the quadratic $x^2-11x+15=0$ so $PX$ and $PY$ is $\frac{11\pm\sqrt{61}}{2}$ in some order. Now, because we want $PQ^2$ and it is known to be rational, we can guess that $PQ$ is irrational or the problem would simply ask for $PQ$. We can also figure out that since $PQ^2$ is rational, $PQ$ is $\sqrt{\text{[something]}}$. $PQ=QX-PX$, and chances are low that $QX$ is some number with a square root plus or minus $\frac{\sqrt{61}}{2}$ to cancel out the $\frac{\sqrt{61}}{2}$ in $PX$, so one can see that $PQ^2$ is most likely to be $\left(\frac{\sqrt{61}}{2}\right)^2=\frac{61}{4}$, and our answer is $61+4=\boxed{065}$
Note : If our answer is correct, then $QX=\frac{11}{2}$, which made $Q$ the midpoint of $XY$, a feature that occurs often in AIME problems, so that again made our answer probable. Midpoints have many properties and there is a lot of ways to show if a point is the midpoint of a segment. Even if the answer is wrong, it's still the same as leaving it blank and 065 is a good guess. ~Ddk001 | // Block 1
size(8cm);
pair O, A, B, P, O1, O2, Q, X, Y;
O=(0, 0);
A=dir(140); B=dir(40);
P=(3A+5B)/8;
O1=extension((A+P)/2, (A+P)/2+(0, 1), A, O);
O2=extension((B+P)/2, (B+P)/2+(0, 1), B, O);
Q=intersectionpoints(circle(O1, length(A-O1)), circle(O2, length(B-O2)))[1];
X=intersectionpoint(P -- (P+(P-Q)*100), circle(O, 1));
Y=intersectionpoint(Q -- (Q+(Q-P)*100), circle(O, 1));
draw(circle(O, 1));
draw(circle(O1, length(A-O1)));
draw(circle(O2, length(B-O2)));
draw(A -- B,red); draw(X -- Y,green);
dot("$A$", A, A);
dot("$B$", B, B);
dot("$P$", P, dir(70),blue);
dot("$Q$", Q, dir(200));
dot("$X$", X, X);
dot("$Y$", Y, Y);
label("$3$", (A+P)/2, N, red);
label("$5$", (B+P)/2, N, red);
draw(brace(X,Y));
label("$11$",brace(X,Y),dir(20));
// Block 2
size(8cm); pair O, A, B, P, O1, O2, Q, X, Y; O=(0, 0); A=dir(140); B=dir(40); P=(3A+5B)/8; O1=extension((A+P)/2, (A+P)/2+(0, 1), A, O); O2=extension((B+P)/2, (B+P)/2+(0, 1), B, O); Q=intersectionpoints(circle(O1, length(A-O1)), circle(O2, length(B-O2)))[1]; X=intersectionpoint(P -- (P+(P-Q)*100), circle(O, 1)); Y=intersectionpoint(Q -- (Q+(Q-P)*100), circle(O, 1)); draw(circle(O, 1)); draw(circle(O1, length(A-O1))); draw(circle(O2, length(B-O2))); draw(A -- B,red); draw(X -- Y,green); dot("$A$", A, A); dot("$B$", B, B); dot("$P$", P, dir(70),blue); dot("$Q$", Q, dir(200)); dot("$X$", X, X); dot("$Y$", Y, Y); label("$3$", (A+P)/2, N, red); label("$5$", (B+P)/2, N, red); draw(brace(X,Y)); label("$11$",brace(X,Y),dir(20)); | [] |
509 | In convex quadrilateral $KLMN$ side $\overline{MN}$ is perpendicular to diagonal $\overline{KM}$, side $\overline{KL}$ is perpendicular to diagonal $\overline{LN}$, $MN = 65$, and $KL = 28$. The line through $L$ perpendicular to side $\overline{KN}$ intersects diagonal $\overline{KM}$ at $O$ with $KO = 8$. Find $MO$. | 2019 AIME I Problems/Problem 6 | NOTE: this solution is wrong. The equation is correct due to similar triangles as described in solution 8, not PoP.
Because $\angle KLN = \angle KMN = 90^{\circ}$, $KLMN$ is a cyclic quadrilateral. (THE FOLLOWING SENTENCE IS WRONG) Hence, by Power of Point, \[KO\cdot KM = KL^2 \implies KM=\dfrac{28^2}{8}=98 \implies MO=98-8=\boxed{090}\] as desired.
~Mathkiddie | // Block 1
size(250);
real h = sqrt(98^2+65^2);
real l = sqrt(h^2-28^2);
pair K = (0,0);
pair N = (h, 0);
pair M = ((98^2)/h, (98*65)/h);
pair L = ((28^2)/h, (28*l)/h);
pair P = ((28^2)/h, 0);
pair O = ((28^2)/h, (8*65)/h);
draw(K--L--N);
draw(K--M--N--cycle);
draw(L--M);
label("K", K, SW);
label("L", L, NW);
label("M", M, NE);
label("N", N, SE);
draw(L--P);
label("P", P, S);
dot(O);
label("O", shift((1,1))*O, NNE);
label("28", scale(1/2)*L, W);
label("65", ((M.x+N.x)/2, (M.y+N.y)/2), NE);
// Block 2
size(250); real h = sqrt(98^2+65^2); real l = sqrt(h^2-28^2); pair K = (0,0); pair N = (h, 0); pair M = ((98^2)/h, (98*65)/h); pair L = ((28^2)/h, (28*l)/h); pair P = ((28^2)/h, 0); pair O = ((28^2)/h, (8*65)/h); draw(K--L--N); draw(K--M--N--cycle); draw(L--M); label("K", K, SW); label("L", L, NW); label("M", M, NE); label("N", N, SE); draw(L--P); label("P", P, S); dot(O); label("O", shift((1,1))*O, NNE); label("28", scale(1/2)*L, W); label("65", ((M.x+N.x)/2, (M.y+N.y)/2), NE); | [] |
509 | In convex quadrilateral $KLMN$ side $\overline{MN}$ is perpendicular to diagonal $\overline{KM}$, side $\overline{KL}$ is perpendicular to diagonal $\overline{LN}$, $MN = 65$, and $KL = 28$. The line through $L$ perpendicular to side $\overline{KN}$ intersects diagonal $\overline{KM}$ at $O$ with $KO = 8$. Find $MO$. | 2019 AIME I Problems/Problem 6 | First, let $P$ be the intersection of $LO$ and $KN$ as shown above. Note that $m\angle KPL = 90^{\circ}$ as given in the problem. Since $\angle KPL \cong \angle KLN$ and $\angle PKL \cong \angle LKN$, $\triangle PKL \sim \triangle LKN$ by AA similarity. Similarly, $\triangle KMN \sim \triangle KPO$. Using these similarities we see that
\[\frac{KP}{KL} = \frac{KL}{KN}\]
\[KP = \frac{KL^2}{KN} = \frac{28^2}{KN} = \frac{784}{KN}\]
and
\[\frac{KP}{KO} = \frac{KM}{KN}\]
\[KP = \frac{KO \cdot KM}{KN} = \frac{8\cdot KM}{KN}\]
Combining the two equations, we get
\[\frac{8\cdot KM}{KN} = \frac{784}{KN}\]
\[8 \cdot KM = 28^2\]
\[KM = 98\]
Since $KM = KO + MO$, we get $MO = 98 -8 = \boxed{090}$.
Solution by vedadehhc | // Block 1
size(250);
real h = sqrt(98^2+65^2);
real l = sqrt(h^2-28^2);
pair K = (0,0);
pair N = (h, 0);
pair M = ((98^2)/h, (98*65)/h);
pair L = ((28^2)/h, (28*l)/h);
pair P = ((28^2)/h, 0);
pair O = ((28^2)/h, (8*65)/h);
draw(K--L--N);
draw(K--M--N--cycle);
draw(L--M);
label("K", K, SW);
label("L", L, NW);
label("M", M, NE);
label("N", N, SE);
draw(L--P);
label("P", P, S);
dot(O);
label("O", shift((1,1))*O, NNE);
label("28", scale(1/2)*L, W);
label("65", ((M.x+N.x)/2, (M.y+N.y)/2), NE);
// Block 2
size(250); real h = sqrt(98^2+65^2); real l = sqrt(h^2-28^2); pair K = (0,0); pair N = (h, 0); pair M = ((98^2)/h, (98*65)/h); pair L = ((28^2)/h, (28*l)/h); pair P = ((28^2)/h, 0); pair O = ((28^2)/h, (8*65)/h); draw(K--L--N); draw(K--M--N--cycle); draw(L--M); label("K", K, SW); label("L", L, NW); label("M", M, NE); label("N", N, SE); draw(L--P); label("P", P, S); dot(O); label("O", shift((1,1))*O, NNE); label("28", scale(1/2)*L, W); label("65", ((M.x+N.x)/2, (M.y+N.y)/2), NE); | [] |
509 | In convex quadrilateral $KLMN$ side $\overline{MN}$ is perpendicular to diagonal $\overline{KM}$, side $\overline{KL}$ is perpendicular to diagonal $\overline{LN}$, $MN = 65$, and $KL = 28$. The line through $L$ perpendicular to side $\overline{KN}$ intersects diagonal $\overline{KM}$ at $O$ with $KO = 8$. Find $MO$. | 2019 AIME I Problems/Problem 6 | Notice that $KLMN$ is inscribed in the circle with diameter $\overline{KN}$ and $XOMN$ is inscribed in the circle with diameter $\overline{ON}$. Furthermore, $(XLN)$ is tangent to $\overline{KL}$. Then, \[KO\cdot KM=KX\cdot KN=KL^2\implies KM=\frac{28^2}{8}=98,\]and $MO=KM-KO=\boxed{090}$.
(Solution by TheUltimate123)
If you're wondering why $KX \cdot KN=KL^2,$ it's because PoP on $(XLN)$ or by $KX \cdot KN=KX \cdot (KX+XN)=KX^2+KX \cdot XN=KX^2+LX^2=KL^2$ (last part by geometric mean theorem / similarity).
Note: the "semicircle" circumscribing points XOMN is not a semicircle. That is just there to tell you that X, O, M, N are indeed concyclic, so ignore the subtlety of the diagram that makes O seems slightly off the marks than it should be. | // Block 1
size(8cm);
pair K, L, M, NN, X, O;
K=(-sqrt(98^2+65^2)/2, 0);
NN=(sqrt(98^2+65^2)/2, 0);
L=sqrt(98^2+65^2)/2*dir(180-2*aSin(28/sqrt(98^2+65^2)));
M=sqrt(98^2+65^2)/2*dir(2*aSin(65/sqrt(98^2+65^2)));
X=foot(L, K, NN);
O=extension(L, X, K, M);
draw(K -- L -- M -- NN -- K -- M); draw(L -- NN); draw(arc((K+NN)/2, NN, K));
draw(L -- X, dashed); draw(arc((O+NN)/2, NN, X), dashed);
draw(rightanglemark(K, L, NN, 100));
draw(rightanglemark(K, M, NN, 100));
draw(rightanglemark(L, X, NN, 100));
dot("$K$", K, SW);
dot("$L$", L, unit(L));
dot("$M$", M, unit(M));
dot("$N$", NN, SE);
dot("$X$", X, S);
// Block 2
size(8cm); pair K, L, M, NN, X, O; K=(-sqrt(98^2+65^2)/2, 0); NN=(sqrt(98^2+65^2)/2, 0); L=sqrt(98^2+65^2)/2*dir(180-2*aSin(28/sqrt(98^2+65^2))); M=sqrt(98^2+65^2)/2*dir(2*aSin(65/sqrt(98^2+65^2))); X=foot(L, K, NN); O=extension(L, X, K, M); draw(K -- L -- M -- NN -- K -- M); draw(L -- NN); draw(arc((K+NN)/2, NN, K)); draw(L -- X, dashed); draw(arc((O+NN)/2, NN, X), dashed); draw(rightanglemark(K, L, NN, 100)); draw(rightanglemark(K, M, NN, 100)); draw(rightanglemark(L, X, NN, 100)); dot("$K$", K, SW); dot("$L$", L, unit(L)); dot("$M$", M, unit(M)); dot("$N$", NN, SE); dot("$X$", X, S); | [] |
509 | In convex quadrilateral $KLMN$ side $\overline{MN}$ is perpendicular to diagonal $\overline{KM}$, side $\overline{KL}$ is perpendicular to diagonal $\overline{LN}$, $MN = 65$, and $KL = 28$. The line through $L$ perpendicular to side $\overline{KN}$ intersects diagonal $\overline{KM}$ at $O$ with $KO = 8$. Find $MO$. | 2019 AIME I Problems/Problem 6 | (Diagram by vedadehhc)
Call the base of the altitude from $L$ to $NK$ point $P$. Let $PO=x$. Now, we have that $KP=\sqrt{64-x^2}$ by the Pythagorean Theorem. Once again by Pythagorean, $LO=\sqrt{720+x^2}-x$. Using Power of a Point, we have
\[(KO)(OM)=(LO)(OQ)\] ($Q$ is the intersection of $OL$ with the circle $\neq L$)
\[8(MO)=(\sqrt{720+x^2}-x)(\sqrt{720+x^2}+x)\]
\[8(MO)=720\]
\[MO=\boxed{090}\].
(Solution by RootThreeOverTwo)
\[\]
Remark: Length of OQ
Since $P$ is on the circle’s diameter, $QP = LP = \sqrt{720+x^2}$. So, $OQ = PQ + PO = x + \sqrt{720+x^2}$.
~diyarv | // Block 1
size(250);
real h = sqrt(98^2+65^2);
real l = sqrt(h^2-28^2);
pair K = (0,0);
pair N = (h, 0);
pair M = ((98^2)/h, (98*65)/h);
pair L = ((28^2)/h, (28*l)/h);
pair P = ((28^2)/h, 0);
pair O = ((28^2)/h, (8*65)/h);
draw(K--L--N);
draw(K--M--N--cycle);
draw(L--M);
label("K", K, SW);
label("L", L, NW);
label("M", M, NE);
label("N", N, SE);
draw(L--P);
label("P", P, S);
dot(O);
label("O", shift((1,1))*O, NNE);
label("28", scale(1/2)*L, W);
label("65", ((M.x+N.x)/2, (M.y+N.y)/2), NE);
// Block 2
size(250); real h = sqrt(98^2+65^2); real l = sqrt(h^2-28^2); pair K = (0,0); pair N = (h, 0); pair M = ((98^2)/h, (98*65)/h); pair L = ((28^2)/h, (28*l)/h); pair P = ((28^2)/h, 0); pair O = ((28^2)/h, (8*65)/h); draw(K--L--N); draw(K--M--N--cycle); draw(L--M); label("K", K, SW); label("L", L, NW); label("M", M, NE); label("N", N, SE); draw(L--P); label("P", P, S); dot(O); label("O", shift((1,1))*O, NNE); label("28", scale(1/2)*L, W); label("65", ((M.x+N.x)/2, (M.y+N.y)/2), NE); | [] |
509 | In convex quadrilateral $KLMN$ side $\overline{MN}$ is perpendicular to diagonal $\overline{KM}$, side $\overline{KL}$ is perpendicular to diagonal $\overline{LN}$, $MN = 65$, and $KL = 28$. The line through $L$ perpendicular to side $\overline{KN}$ intersects diagonal $\overline{KM}$ at $O$ with $KO = 8$. Find $MO$. | 2019 AIME I Problems/Problem 6 | Note that since $\angle KLN = \angle KMN$, quadrilateral $KLMN$ is cyclic. Therefore, we have \[\angle LMK = \angle LNK = 90^{\circ} - \angle LKN = \angle KLP,\]so $\triangle KLO \sim \triangle KML$, giving \[\frac{KM}{28} = \frac{28}{8} \implies KM = 98.\] Therefore, $OM = 98-8 = \boxed{90}$. | // Block 1
size(250);
real h = sqrt(98^2+65^2);
real l = sqrt(h^2-28^2);
pair K = (0,0);
pair N = (h, 0);
pair M = ((98^2)/h, (98*65)/h);
pair L = ((28^2)/h, (28*l)/h);
pair P = ((28^2)/h, 0);
pair O = ((28^2)/h, (8*65)/h);
draw(K--L--N);
draw(K--M--N--cycle);
draw(L--M);
label("K", K, SW);
label("L", L, NW);
label("M", M, NE);
label("N", N, SE);
draw(L--P);
label("P", P, S);
dot(O);
label("O", shift((1,1))*O, NNE);
label("28", scale(1/2)*L, W);
label("65", ((M.x+N.x)/2, (M.y+N.y)/2), NE);
// Block 2
size(250); real h = sqrt(98^2+65^2); real l = sqrt(h^2-28^2); pair K = (0,0); pair N = (h, 0); pair M = ((98^2)/h, (98*65)/h); pair L = ((28^2)/h, (28*l)/h); pair P = ((28^2)/h, 0); pair O = ((28^2)/h, (8*65)/h); draw(K--L--N); draw(K--M--N--cycle); draw(L--M); label("K", K, SW); label("L", L, NW); label("M", M, NE); label("N", N, SE); draw(L--P); label("P", P, S); dot(O); label("O", shift((1,1))*O, NNE); label("28", scale(1/2)*L, W); label("65", ((M.x+N.x)/2, (M.y+N.y)/2), NE); | [] |
509 | In convex quadrilateral $KLMN$ side $\overline{MN}$ is perpendicular to diagonal $\overline{KM}$, side $\overline{KL}$ is perpendicular to diagonal $\overline{LN}$, $MN = 65$, and $KL = 28$. The line through $L$ perpendicular to side $\overline{KN}$ intersects diagonal $\overline{KM}$ at $O$ with $KO = 8$. Find $MO$. | 2019 AIME I Problems/Problem 6 | By Pythagorean Theorem, $KM^2+65^2 = KN^2 = 28^2 + LN^2$. Thus, $LN^2 = KM^2 + 65^2 - 28^2$.
By Pythagorean Theorem, $KP^2 + LP^2 = 28^2$, and $PN^2 + LP^2 = LN^2$.
\[PN^2 = (KN - KP)^2 = (\sqrt{KM^2 + 65^2} - KP)^2\]
It follows that
\[(\sqrt{KM^2 + 65^2} - KP)^2 + LP^2 = KM^2 + 65^2 - 28^2\]
\[KM^2 + 65^2 - 2\sqrt{KM^2 + 65^2}(KP) + KP^2 + LP^2 = KM^2 + 65^2 - 28^2\]
Since $KP^2 + LP^2 = 28^2$,
\[KM^2 + 65^2 - 2\sqrt{KM^2 + 65^2}(KP) + 28^2 = KM^2 + 65^2 - 28^2\]
\[-2\sqrt{KM^2 + 65^2}(KP) = -2 \times 28^2\]
\[KP = \frac{28^2}{\sqrt{KM^2 + 65^2}}\]
$\angle OKP = \angle NKM$ (it's the same angle) and $\angle OPK = \angle KMN = 90^{\circ}$. Thus, $\triangle KOP \sim \triangle KNM$.
Thus,
\[\frac{KO}{KN} = \frac{KP}{KM}\]
\[\frac{8}{\sqrt{KM^2 + 65^2}} = \frac{\frac{28^2}{\sqrt{KM^2 + 65^2}}}{KM}\]
Multiplying both sides by $\sqrt{KM^2 + 65^2}$:
\[8 = \frac{28^2}{KM}\]
\[KM = 98\]
Therefore, $OM = 98-8 = \boxed{90}$
~ Solution by adam_zheng | // Block 1
size(250);
real h = sqrt(98^2+65^2);
real l = sqrt(h^2-28^2);
pair K = (0,0);
pair N = (h, 0);
pair M = ((98^2)/h, (98*65)/h);
pair L = ((28^2)/h, (28*l)/h);
pair P = ((28^2)/h, 0);
pair O = ((28^2)/h, (8*65)/h);
draw(K--L--N);
draw(K--M--N--cycle);
draw(L--M);
label("K", K, SW);
label("L", L, NW);
label("M", M, NE);
label("N", N, SE);
draw(L--P);
label("P", P, S);
dot(O);
label("O", shift((1,1))*O, NNE);
label("28", scale(1/2)*L, W);
label("65", ((M.x+N.x)/2, (M.y+N.y)/2), NE);
// Block 2
size(250); real h = sqrt(98^2+65^2); real l = sqrt(h^2-28^2); pair K = (0,0); pair N = (h, 0); pair M = ((98^2)/h, (98*65)/h); pair L = ((28^2)/h, (28*l)/h); pair P = ((28^2)/h, 0); pair O = ((28^2)/h, (8*65)/h); draw(K--L--N); draw(K--M--N--cycle); draw(L--M); label("K", K, SW); label("L", L, NW); label("M", M, NE); label("N", N, SE); draw(L--P); label("P", P, S); dot(O); label("O", shift((1,1))*O, NNE); label("28", scale(1/2)*L, W); label("65", ((M.x+N.x)/2, (M.y+N.y)/2), NE); | [] |
510 | Two different points, $C$ and $D$, lie on the same side of line $AB$ so that $\triangle ABC$ and $\triangle BAD$ are congruent with $AB = 9$, $BC=AD=10$, and $CA=DB=17$. The intersection of these two triangular regions has area $\tfrac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2019 AIME II Problems/Problem 1 | - Diagram by Brendanb4321
Extend $AB$ to form a right triangle with legs $6$ and $8$ such that $AD$ is the hypotenuse and connect the points $CD$ so
that you have a rectangle. (We know that $\triangle ADE$ is a $6-8-10$, since $\triangle DEB$ is an $8-15-17$.) The base $CD$ of the rectangle will be $9+6+6=21$. Now, let $O$ be the intersection of $BD$ and $AC$. This means that $\triangle ABO$ and $\triangle DCO$ are similar with ratio $\frac{21}{9}=\frac73$. Set up a proportion, knowing that the two heights add up to 8. We will let $y$ be the height from $O$ to $DC$, and $x$ be the height of $\triangle ABO$.
\[\frac{7}{3}=\frac{y}{x}\]
\[\frac{7}{3}=\frac{8-x}{x}\]
\[7x=24-3x\]
\[10x=24\]
\[x=\frac{12}{5}\]
This means that the area is $A=\tfrac{1}{2}(9)(\tfrac{12}{5})=\tfrac{54}{5}$. This gets us $54+5=\boxed{059}.$
-Solution by the Math Wizard, Number Magician of the Second Order, Head of the Council of the Geometers | // Block 1
unitsize(10);
pair A = (0,0);
pair B = (9,0);
pair C = (15,8);
pair D = (-6,8);
pair E = (-6,0);
draw(A--B--C--cycle);
draw(B--D--A);
label("$A$",A,dir(-120));
label("$B$",B,dir(-60));
label("$C$",C,dir(60));
label("$D$",D,dir(120));
label("$E$",E,dir(-135));
label("$9$",(A+B)/2,dir(-90));
label("$10$",(D+A)/2,dir(-150));
label("$10$",(C+B)/2,dir(-30));
label("$17$",(D+B)/2,dir(60));
label("$17$",(A+C)/2,dir(120));
draw(D--E--A,dotted);
label("$8$",(D+E)/2,dir(180));
label("$6$",(A+E)/2,dir(-90));
// Block 2
unitsize(10); pair A = (0,0); pair B = (9,0); pair C = (15,8); pair D = (-6,8); pair E = (-6,0); draw(A--B--C--cycle); draw(B--D--A); label("$A$",A,dir(-120)); label("$B$",B,dir(-60)); label("$C$",C,dir(60)); label("$D$",D,dir(120)); label("$E$",E,dir(-135)); label("$9$",(A+B)/2,dir(-90)); label("$10$",(D+A)/2,dir(-150)); label("$10$",(C+B)/2,dir(-30)); label("$17$",(D+B)/2,dir(60)); label("$17$",(A+C)/2,dir(120)); draw(D--E--A,dotted); label("$8$",(D+E)/2,dir(180)); label("$6$",(A+E)/2,dir(-90)); | [] |
510 | Two different points, $C$ and $D$, lie on the same side of line $AB$ so that $\triangle ABC$ and $\triangle BAD$ are congruent with $AB = 9$, $BC=AD=10$, and $CA=DB=17$. The intersection of these two triangular regions has area $\tfrac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2019 AIME II Problems/Problem 1 | - Diagram by Brendanb4321 extended by Duoquinquagintillion
Begin with the first step of solution 1, seeing $AD$ is the hypotenuse of a $6-8-10$ triangle and calling the intersection of $DB$ and $AC$ point $E$. Next, notice $DB$ is the hypotenuse of an $8-15-17$ triangle. Drop an altitude from $E$ with length $h$, so the other leg of the new triangle formed has length $4.5$. Notice we have formed similar triangles, and we can solve for $h$.
\[\frac{h}{4.5} = \frac{8}{15}\]
\[h = \frac{36}{15} = \frac{12}{5}\]
So $\triangle ABE$ has area \[\frac{ \frac{12}{5} \cdot 9}{2} = \frac{54}{5}\]
And $54+5=\boxed{059}.$
- Solution by Duoquinquagintillion | // Block 1
unitsize(10);
pair A = (0,0);
pair B = (9,0);
pair C = (15,8);
pair D = (-6,8);
draw(A--B--C--cycle);
draw(B--D--A);
label("$A$",A,dir(-120));
label("$B$",B,dir(-60));
label("$C$",C,dir(60));
label("$D$",D,dir(120));
label("$9$",(A+B)/2,dir(-90));
label("$10$",(D+A)/2,dir(-150));
label("$10$",(C+B)/2,dir(-30));
label("$17$",(D+B)/2,dir(60));
label("$17$",(A+C)/2,dir(120));
draw(D--(-6,0)--A,dotted);
label("$8$",(D+(-6,0))/2,dir(180));
label("$6$",(A+(-6,0))/2,dir(-90));
draw((4.5,0)--(4.5,2.4),dotted);
label("$h$", (4.5,1.2), dir(180));
label("$4.5$", (6,0), dir(90));
// Block 2
unitsize(10); pair A = (0,0); pair B = (9,0); pair C = (15,8); pair D = (-6,8); draw(A--B--C--cycle); draw(B--D--A); label("$A$",A,dir(-120)); label("$B$",B,dir(-60)); label("$C$",C,dir(60)); label("$D$",D,dir(120)); label("$9$",(A+B)/2,dir(-90)); label("$10$",(D+A)/2,dir(-150)); label("$10$",(C+B)/2,dir(-30)); label("$17$",(D+B)/2,dir(60)); label("$17$",(A+C)/2,dir(120)); draw(D--(-6,0)--A,dotted); label("$8$",(D+(-6,0))/2,dir(180)); label("$6$",(A+(-6,0))/2,dir(-90)); draw((4.5,0)--(4.5,2.4),dotted); label("$h$", (4.5,1.2), dir(180)); label("$4.5$", (6,0), dir(90)); | [] |
510 | Two different points, $C$ and $D$, lie on the same side of line $AB$ so that $\triangle ABC$ and $\triangle BAD$ are congruent with $AB = 9$, $BC=AD=10$, and $CA=DB=17$. The intersection of these two triangular regions has area $\tfrac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2019 AIME II Problems/Problem 1 | First, let's define H as the intersection of CB and DA, and define G as the midpoint of AB. Next, let E and F be the feet of the perpendicular lines from C and D to line AB respectively. We get a pleasing line of symmetry HG where A maps to B, C maps to D, and E maps to F. We notice that 8-15-17 and 8-6-10 are both pythagorean triples and we test that theory. Then, we find AB = 15 - 6 = 9 and GB = 4.5. Since △CEB~△HGB, we find that HG = 2.4. Then, the area of △HGB is 5.4 and the total overlap is 10.8 = 54/5. Noting that GCD(54,5)=1, we add them to get 59.
Note: I omitted some computation
~ Afly (talk) | //Made by Afly. I used some resources. //Took me 10 min to get everything right. import olympiad; unitsize(18); pair A = (0,0); pair B = (0,8); pair C = (6,0); pair D = (15,0); pair E = (21,0); pair F = (21,8); pair G = (21/2,0); pair H = intersectionpoints(B--D,C--F)[0]; pen dash1 = linetype(new real [] {9,9})+linewidth(1); pen solid1 = linetype(new real [] {9,0})+linewidth(1); pen dash2 = linetype(new real [] {3,3})+linewidth(1); fill(C--G--H--cycle,rgb(3/4,1/4,1/4)); fill(D--G--H--cycle,rgb(3/4,3/4,1/4)); draw(C--A--B,dash1); draw(C--B--D--C,solid1); draw(F--E--D,dash1); draw(F--D--C--F,solid1); draw(G--H,dash2); draw(brace(D+dir(270),A+dir(270)),solid1); draw(brace(D,C),solid1); draw(A--A+2*dir(180),dash1,EndArrow); draw(E--E+2*dir(0),dash1,EndArrow); pair L1 = (15/2,-7/2); pair L2 = (21/2,-13/8); label("15",L1); label("8",A--B,W); label("6",A--C,S); label("10",B--C,SW); label("17",B--D,NE); label("9",L2); label("4.5",G--D,S); label("2.4",G--H,W); markscalefactor = 1/16; draw(rightanglemark(H,G,D)); draw(rightanglemark(B,A,C)); draw(rightanglemark(D,E,F)); label("A",C,SW); label("B",D,SE); label("C",B,NW); label("D",F,NE); label("E",A,SW); label("F",E,SE); label("G",G,NW); label("H",H,N); | [] |
511 | Triangle $ABC$ has side lengths $AB=7, BC=8,$ and $CA=9.$ Circle $\omega_1$ passes through $B$ and is tangent to line $AC$ at $A.$ Circle $\omega_2$ passes through $C$ and is tangent to line $AB$ at $A.$ Let $K$ be the intersection of circles $\omega_1$ and $\omega_2$ not equal to $A.$ Then $AK=\tfrac mn,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | 2019 AIME II Problems/Problem 11 | -Diagram by Brendanb4321
Note that from the tangency condition that the supplement of $\angle CAB$ with respects to lines $AB$ and $AC$ are equal to $\angle AKB$ and $\angle AKC$, respectively, so from tangent-chord, \[\angle AKC=\angle AKB=180^{\circ}-\angle BAC\] Also note that $\angle ABK=\angle KAC$$^{(*)}$, so $\triangle AKB\sim \triangle CKA$. Using similarity ratios, we can easily find \[AK^2=BK*KC\] However, since $AB=7$ and $CA=9$, we can use similarity ratios to get \[BK=\frac{7}{9}AK, CK=\frac{9}{7}AK\]
Now we use Law of Cosines on $\triangle AKB$: From reverse Law of Cosines, $\cos{\angle BAC}=\frac{11}{21}\implies \cos{(180^{\circ}-\angle BAC)}=\angle AKB=-\frac{11}{21}$
Giving us \[AK^2+\frac{49}{81}AK^2+\frac{22}{27}AK^2=49\] \[\implies \frac{196}{81}AK^2=49\] \[AK=\frac{9}{2}\] so our answer is $9+2=\boxed{011}$.
$^{(*)}$ Let $O$ be the center of $\omega_1$. Then $\angle KAC = 90 - \angle OAK = 90 - \frac{1}{2}(180 - \angle AOK) = \frac{\angle AOK}{2} = \angle ABK$. Thus, $\angle ABK = \angle KAC$
-franchester; $^{(*)}$ by firebolt360
Supplement
In order to get to the Law of Cosines first, we first apply the LOC to $\triangle{ABC},$ obtaining $\angle{BAC}.$
We angle chase before applying the law of cosines to $\angle{AKB}.$
Note that $\angle{ABK}=\angle{KAC}$ and $\angle{KCA}=\angle{KAB}$ from tangent-chord.
Thus, $\angle{AKC}=\angle{AKB}=180^{\circ}-(\angle{ABK}+\angle{KAB}).$
However from our tangent chord, note that:
\[\angle{ABK}+\angle{KAB}=\angle{KAC}+\angle{KAB}=\angle{BAC}.\]
Thus, $\angle{AKB}=180^\circ-\angle{BAC}.$
As an alternative approach, note that the sum of the angles in quadrilateral $ABKC$ is $360^{\circ}$ and we can find $\angle{AKB}=\frac12$ of convex $\angle{BKC},$ which is just:
\[\frac12 \left(360^{\circ}-2(\angle{KAB}+\angle{KBA}\right) = 180^\circ - \angle{BAC}.\]
~mathboy282 | // Block 1
unitsize(20);
pair B = (0,0);
pair A = (2,sqrt(45));
pair C = (8,0);
draw(circumcircle(A,B,(-17/8,0)),rgb(.7,.7,.7));
draw(circumcircle(A,C,(49/8,0)),rgb(.7,.7,.7));
draw(B--A--C--cycle);
label("$A$",A,dir(105));
label("$B$",B,dir(-135));
label("$C$",C,dir(-75));
dot((2.68,2.25));
label("$K$",(2.68,2.25),2*down);
label("$\omega_1$",(-4.5,1));
label("$\omega_2$",(12.75,6));
label("$7$",(A+B)/2,dir(140));
label("$8$",(B+C)/2,dir(-90));
label("$9$",(A+C)/2,dir(60));
// Block 2
unitsize(20); pair B = (0,0); pair A = (2,sqrt(45)); pair C = (8,0); draw(circumcircle(A,B,(-17/8,0)),rgb(.7,.7,.7)); draw(circumcircle(A,C,(49/8,0)),rgb(.7,.7,.7)); draw(B--A--C--cycle); label("$A$",A,dir(105)); label("$B$",B,dir(-135)); label("$C$",C,dir(-75)); dot((2.68,2.25)); label("$K$",(2.68,2.25),2*down); label("$\omega_1$",(-4.5,1)); label("$\omega_2$",(12.75,6)); label("$7$",(A+B)/2,dir(140)); label("$8$",(B+C)/2,dir(-90)); label("$9$",(A+C)/2,dir(60)); | [] |
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