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

(official MAA wording)
m (Solution)
Line 2: Line 2:
 
Circles <math>\omega_1</math> and <math>\omega_2</math> with radii <math>961</math> and <math>625</math>, respectively, intersect at distinct points <math>A</math> and <math>B</math>. A third circle <math>\omega</math> is externally tangent to both <math>\omega_1</math> and <math>\omega_2</math>. Suppose line <math>AB</math> intersects <math>\omega</math> at two points <math>P</math> and <math>Q</math> such that the measure of minor arc <math>\widehat{PQ}</math> is <math>120^{\circ}</math>. Find the distance between the centers of <math>\omega_1</math> and <math>\omega_2</math>.
 
Circles <math>\omega_1</math> and <math>\omega_2</math> with radii <math>961</math> and <math>625</math>, respectively, intersect at distinct points <math>A</math> and <math>B</math>. A third circle <math>\omega</math> is externally tangent to both <math>\omega_1</math> and <math>\omega_2</math>. Suppose line <math>AB</math> intersects <math>\omega</math> at two points <math>P</math> and <math>Q</math> such that the measure of minor arc <math>\widehat{PQ}</math> is <math>120^{\circ}</math>. Find the distance between the centers of <math>\omega_1</math> and <math>\omega_2</math>.
  
==Solution by pad==
+
==Solution==
 
Let <math>O_i</math> and <math>r_i</math> be the center and radius <math>\omega_i</math>, and let <math>O</math> and <math>r</math> be the center and radius of <math>\omega</math>.
 
Let <math>O_i</math> and <math>r_i</math> be the center and radius <math>\omega_i</math>, and let <math>O</math> and <math>r</math> be the center and radius of <math>\omega</math>.
  

Revision as of 22:37, 13 March 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 $\omega_i$, 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{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 \boxed{d=2r_2-2r_1}.  \end{align*} Therefore, $d=2(r_2-r_1) = 2(961-625)=\boxed{672}$.

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