Difference between revisions of "2021 AIME I Problems/Problem 13"

m (Video Solution)
(Added in Sols 3 and 4 by the others.)
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Denote by <math>O_1</math> and <math>O_2</math> the centers of <math>\omega_1</math> and <math>\omega_2</math> respectively. Set <math>X</math> as the projection of <math>O</math> onto <math>O_1O_2</math>, and denote by <math>Y</math> the intersection of <math>AB</math> with <math>O_1O_2</math>. Note that <math>\ell = XY</math>. Now recall that<cmath>d(O_2Y-O_1Y) = O_2Y^2 - O_1Y^2 = R_2^2 - R_1^2.</cmath>Furthermore, note that<cmath>\begin{align*}d(O_2X - O_1X) &= O_2X^2 - O_1X^2= O_2O^2 - O_1O^2 \\ &= (R_2 + r)^2 - (R_1+r)^2 = (R_2^2 - R_1^2) + 2r(R_2 - R_1).\end{align*}</cmath>Substituting the first equality into the second one and subtracting yields<cmath>2r(R_2 - R_1) = d(O_2Y - O_2X) - d(O_2X - O_1X) = 2dXY,</cmath>which rearranges to the desired.
 
Denote by <math>O_1</math> and <math>O_2</math> the centers of <math>\omega_1</math> and <math>\omega_2</math> respectively. Set <math>X</math> as the projection of <math>O</math> onto <math>O_1O_2</math>, and denote by <math>Y</math> the intersection of <math>AB</math> with <math>O_1O_2</math>. Note that <math>\ell = XY</math>. Now recall that<cmath>d(O_2Y-O_1Y) = O_2Y^2 - O_1Y^2 = R_2^2 - R_1^2.</cmath>Furthermore, note that<cmath>\begin{align*}d(O_2X - O_1X) &= O_2X^2 - O_1X^2= O_2O^2 - O_1O^2 \\ &= (R_2 + r)^2 - (R_1+r)^2 = (R_2^2 - R_1^2) + 2r(R_2 - R_1).\end{align*}</cmath>Substituting the first equality into the second one and subtracting yields<cmath>2r(R_2 - R_1) = d(O_2Y - O_2X) - d(O_2X - O_1X) = 2dXY,</cmath>which rearranges to the desired.
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==Solution 3==
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WLOG assume <math>\omega</math> is a line. Note the angle condition is equivalent to the angle between <math>AB</math> and <math>\omega</math> being <math>60^\circ</math>. We claim the angle between <math>AB</math> and <math>\omega</math> is fixed as <math>\omega</math> varies.
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Proof: Perform an inversion at <math>A</math>, sending <math>\omega_1</math> and <math>\omega_2</math> to two lines <math>\ell_1</math> and <math>\ell_2</math> intersecting at <math>B'</math>. Then <math>\omega</math> is sent to a circle tangent to lines <math>\ell_1</math> and <math>\ell_2</math>, which clearly intersects <math>AB'</math> at a fixed angle. Therefore the angle between <math>AB</math> and <math>\omega</math> is fixed as <math>\omega</math> varies.
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Now simply take <math>\omega</math> to be a line. If <math>\omega</math> intersects <math>\omega_1</math> and <math>\omega_2</math> and <math>X,Y</math>, respectively, and the circles' centers are <math>O_1</math> and <math>O_2</math>, then the projection of <math>O_2</math> to <math>O_1X</math> at <math>F</math> gives that <math>O_2FO_1</math> is a <math>30\text{-}60\text{-}90</math> triangle. Therefore,<cmath>O_1O_2=2O_1F=2(O_1X-O_2Y)=2(961-625)=\boxed{672}.</cmath>
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~spartacle
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==Solution 4==
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Suppose we label the points as shown below: https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvNC9mLzRiM2JjYThjYmZlY2ViZGI0ODhjYzE4YzMyMmM0M2QyOTZlMmU5LmpwZw==&rn=MTU4ODUxMDg3XzczMDI0ODE4MTAwNjA5N184NDQzMjQxMjM3MDQ2NzQ5NjM4X24uanBn. By radical axis, the tangents to <math>\omega</math> at <math>D</math> and <math>E</math> intersect on <math>AB</math>. Thus <math>PDQE</math> is harmonic, so the tangents to <math>\omega</math> at <math>P</math> and <math>Q</math> intersect at <math>X \in DE</math>. Moreover, <math>OX \parallel O_1O_2</math> because both <math>OX</math> and <math>O_1O_2</math> are perpendicular to <math>AB</math>, and <math>OX = 2OP</math> because <math>\angle POQ = 120^{\circ}</math>. Thus<cmath>O_1O_2 = O_1Y - O_2Y = 2 \cdot 961 - 2\cdot 625 = \boxed{672}</cmath>by similar triangles.
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~mathman3880
  
 
==Video Solution==
 
==Video Solution==

Revision as of 20:33, 22 April 2021

Problem

Circles $\omega_1$ and $\omega_2$ with radii $961$ and $625$, respectively, intersect at distinct points $A$ and $B$. A third circle $\omega$ is externally tangent to both $\omega_1$ and $\omega_2$. Suppose line $AB$ intersects $\omega$ at two points $P$ and $Q$ such that the measure of minor arc $\widehat{PQ}$ is $120^{\circ}$. Find the distance between the centers of $\omega_1$ and $\omega_2$.

Solution

Let $O_i$ and $r_i$ be the center and radius of $\omega_i$, and let $O$ and $r$ be the center and radius of $\omega$.

Since $\overline{AB}$ extends to an arc with arc $120^\circ$, the distance from $O$ to $\overline{AB}$ is $r/2$. Let $X=\overline{AB}\cap \overline{O_1O_2}$. Consider $\triangle OO_1O_2$. The line $\overline{AB}$ is perpendicular to $\overline{O_1O_2}$ and passes through $X$. Let $H$ be the foot from $O$ to $\overline{O_1O_2}$; so $HX=r/2$. We have by tangency $OO_1=r+r_1$ and $OO_2=r+r_2$. Let $O_1O_2=d$. [asy] unitsize(3cm); pointpen=black; pointfontpen=fontsize(9);  pair A=dir(110), B=dir(230), C=dir(310);   DPA(A--B--C--A);    pair H = foot(A, B, C);  draw(A--H); pair X = 0.3*B + 0.7*C; pair Y = A+X-H; draw(X--1.3*Y-0.3*X); draw(A--Y, dotted);  pair R1 = 1.3*X-0.3*Y; pair R2 = 0.7*X+0.3*Y; draw(R1--X);   D("O",A,dir(A)); D("O_1",B,dir(B)); D("O_2",C,dir(C)); D("H",H,dir(270)); D("X",X,dir(225)); D("A",R1,dir(180)); D("B",R2,dir(180));  draw(rightanglemark(Y,X,C,3));   [/asy] Since $X$ is on the radical axis of $\omega_1$ and $\omega_2$, it has equal power with respect to both circles, so \[O_1X^2 - r_1^2 = O_2X^2-r_2^2 \implies O_1X-O_2X = \frac{r_1^2-r_2^2}{d}\]since $O_1X+O_2X=d$. Now we can solve for $O_1X$ and $O_2X$, and in particular, \begin{align*} O_1H &= O_1X - HX = \frac{d+\frac{r_1^2-r_2^2}{d}}{2} - \frac{r}{2} \\ O_2H &= O_2X + HX = \frac{d-\frac{r_1^2-r_2^2}{d}}{2} + \frac{r}{2}.  \end{align*} We want to solve for $d$. By the Pythagorean Theorem (twice): \begin{align*} &\qquad -OH^2 = O_2H^2 - (r+r_2)^2 = O_1H^2 - (r+r_1)^2 \\ &\implies \left(d+r-\tfrac{r_1^2-r_2^2}{d}\right)^2 - 4(r+r_2)^2 = \left(d-r+\tfrac{r_1^2-r_2^2}{d}\right)^2 - 4(r+r_1)^2 \\ &\implies 2dr - 2(r_1^2-r_2)^2-8rr_2-4r_2^2 = -2dr+2(r_1^2-r_2^2)-8rr_1-4r_1^2 \\ &\implies 4dr = 8rr_2-8rr_1 \\ &\implies d=2r_2-2r_1 \end{align*} Therefore, $d=2(r_2-r_1) = 2(961-625)=\boxed{672}$.

Solution 2 (Official MAA, Unedited)

Denote by $O_1$, $O_2$, and $O$ the centers of $\omega_1$, $\omega_2$, and $\omega$, respectively. Let $R_1 = 961$ and $R_2 = 625$ denote the radii of $\omega_1$ and $\omega_2$ respectively, $r$ be the radius of $\omega$, and $\ell$ the distance from $O$ to the line $AB$. We claim that\[\dfrac{\ell}{r} = \dfrac{R_2-R_1}{d},\]where $d = O_1O_2$. This solves the problem, for then the $\widehat{PQ} = 120^\circ$ condition implies $\tfrac{\ell}r = \cos 60^\circ = \tfrac{1}{2}$, and then we can solve to get $d = \boxed{672}$. [asy] import olympiad; size(230pt); defaultpen(linewidth(0.8)+fontsize(10pt)); real r1 = 17, r2 = 27, d = 35, r = 18; pair O1 = origin, O2 = (d,0); path w1 = circle(origin,r1), w2 = circle((d,0),r2), w1p = circle(origin,r1+r), w2p = circle((d,0), r2 + r); pair[] X = intersectionpoints(w1,w2), Y = intersectionpoints(w1p,w2p); pair O = Y[1]; path w = circle(Y[1],r); pair Xp = 5 * X[1] - 4 * X[0]; pair[] P = intersectionpoints(Xp--X[0],w); label("$O_1$",origin,N); label("$O_2$",(d,0),N); label("$O$",Y[1],SW); draw(origin--Y[1]--(d,0)--cycle,gray(0.6)); pair T = foot(O,O1,O2), Tp = foot(O,X[0],X[1]); draw(Tp--O--T^^rightanglemark(O,T,O1,60)^^rightanglemark(O,Tp,X[0],60),gray(0.6)); draw(w^^w1^^w2^^P[0]--X[0]); dot(Y[1]^^origin^^(d,0)); label("$X$",T,N,gray(0.6)); label("$Y$",foot(X[0],O1,O2),NE,gray(0.6)); label("$\ell$",(O+Tp)/2,S,gray(0.6)); [/asy]

Denote by $O_1$ and $O_2$ the centers of $\omega_1$ and $\omega_2$ respectively. Set $X$ as the projection of $O$ onto $O_1O_2$, and denote by $Y$ the intersection of $AB$ with $O_1O_2$. Note that $\ell = XY$. Now recall that\[d(O_2Y-O_1Y) = O_2Y^2 - O_1Y^2 = R_2^2 - R_1^2.\]Furthermore, note that\begin{align*}d(O_2X - O_1X) &= O_2X^2 - O_1X^2= O_2O^2 - O_1O^2 \\ &= (R_2 + r)^2 - (R_1+r)^2 = (R_2^2 - R_1^2) + 2r(R_2 - R_1).\end{align*}Substituting the first equality into the second one and subtracting yields\[2r(R_2 - R_1) = d(O_2Y - O_2X) - d(O_2X - O_1X) = 2dXY,\]which rearranges to the desired.

Solution 3

WLOG assume $\omega$ is a line. Note the angle condition is equivalent to the angle between $AB$ and $\omega$ being $60^\circ$. We claim the angle between $AB$ and $\omega$ is fixed as $\omega$ varies.

Proof: Perform an inversion at $A$, sending $\omega_1$ and $\omega_2$ to two lines $\ell_1$ and $\ell_2$ intersecting at $B'$. Then $\omega$ is sent to a circle tangent to lines $\ell_1$ and $\ell_2$, which clearly intersects $AB'$ at a fixed angle. Therefore the angle between $AB$ and $\omega$ is fixed as $\omega$ varies.

Now simply take $\omega$ to be a line. If $\omega$ intersects $\omega_1$ and $\omega_2$ and $X,Y$, respectively, and the circles' centers are $O_1$ and $O_2$, then the projection of $O_2$ to $O_1X$ at $F$ gives that $O_2FO_1$ is a $30\text{-}60\text{-}90$ triangle. Therefore,\[O_1O_2=2O_1F=2(O_1X-O_2Y)=2(961-625)=\boxed{672}.\]

~spartacle

Solution 4

Suppose we label the points as shown below: https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvNC9mLzRiM2JjYThjYmZlY2ViZGI0ODhjYzE4YzMyMmM0M2QyOTZlMmU5LmpwZw==&rn=MTU4ODUxMDg3XzczMDI0ODE4MTAwNjA5N184NDQzMjQxMjM3MDQ2NzQ5NjM4X24uanBn. By radical axis, the tangents to $\omega$ at $D$ and $E$ intersect on $AB$. Thus $PDQE$ is harmonic, so the tangents to $\omega$ at $P$ and $Q$ intersect at $X \in DE$. Moreover, $OX \parallel O_1O_2$ because both $OX$ and $O_1O_2$ are perpendicular to $AB$, and $OX = 2OP$ because $\angle POQ = 120^{\circ}$. Thus\[O_1O_2 = O_1Y - O_2Y = 2 \cdot 961 - 2\cdot 625 = \boxed{672}\]by similar triangles.

~mathman3880

Video Solution

Who wanted to see animated video solutions can see this. I found this really helpful.

https://youtu.be/YtZ8_7i833E

P.S: This video is not made by me. And solution is same like below solutions.

≈@rounak138

See also

2021 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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