2021 AIME I Problems/Problem 13

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 $120^{\circ}$ . What is the distance between the centers of $\omega_1$ and $\omega_2$ ?

Solution by pad

Let $O_i$ and $r_i$ be the center and radius $\omega_i$ , 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, 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, $\begin{aligned} O_{1} H & = O_{1} X - H X = \frac{d + \frac{r_{1}^{2} - r_{2}^{2}}{d}}{2} - \frac{r}{2} \\ O_{2} H & = O_{2} X + H X = \frac{d - \frac{r_{1}^{2} - r_{2}^{2}}{d}}{2} + \frac{r}{2} . \end{aligned}$ We want to solve for $d$ . By the Pythagorean Theorem (twice): $\begin{aligned} O H^{2} = O_{2} H^{2} - (r + r_{2})^{2} = O_{1} H^{2} - (r + r_{1})^{2} \\ ⟹ {(d + r - \frac{r_{1}^{2} - r_{2}^{2}}{d})}^{2} - 4 (r + r_{2})^{2} = {(d - r + \frac{r_{1}^{2} - r_{2}^{2}}{d})}^{2} - 4 (r + r_{1})^{2} \\ ⟹ 2 d r - 2 (r_{1}^{2} - r_{2})^{2} - 8 r r_{2} - 4 r_{2}^{2} = - 2 d r + 2 (r_{1}^{2} - r_{2}^{2}) - 8 r r_{1} - 4 r_{1}^{2} \\ ⟹ 4 d r = 8 r r_{2} - 8 r r_{1} \\ ⟹ d = 2 r_{2} - 2 r_{1} . \end{aligned}$ Therefore, $d=2(r_2-r_1) = 2(961-625)=\boxed{672}$ .

2021 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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2021 AIME I Problems/Problem 13

Problem

Solution by pad

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