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Difference between revisions of "2021 AIME I Problems/Problem 13"
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Since <math>\overline{AB}</math> extends to an arc with arc <math>120^\circ</math>, the distance from <math>O</math> to <math>\overline{AB}</math> is <math>r/2</math>. Let <math>X=\overline{AB}\cap \overline{O_1O_2}</math>. Consider <math>\triangle OO_1O_2</math>. The line <math>\overline{AB}</math> is perpendicular to <math>\overline{O_1O_2}</math> and passes through <math>X</math>. Let <math>H</math> be the foot from <math>O</math> to <math>\overline{O_1O_2}</math>; so <math>HX=r/2</math>. We have by tangency <math>OO_1=r+r_1</math> and <math>OO_2=r+r_2</math>. Let <math>O_1O_2=d</math>.
 
Since <math>\overline{AB}</math> extends to an arc with arc <math>120^\circ</math>, the distance from <math>O</math> to <math>\overline{AB}</math> is <math>r/2</math>. Let <math>X=\overline{AB}\cap \overline{O_1O_2}</math>. Consider <math>\triangle OO_1O_2</math>. The line <math>\overline{AB}</math> is perpendicular to <math>\overline{O_1O_2}</math> and passes through <math>X</math>. Let <math>H</math> be the foot from <math>O</math> to <math>\overline{O_1O_2}</math>; so <math>HX=r/2</math>. We have by tangency <math>OO_1=r+r_1</math> and <math>OO_2=r+r_2</math>. Let <math>O_1O_2=d</math>.
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[asy]
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<asy>
 
unitsize(3cm);
 
unitsize(3cm);
 
pointpen=black; pointfontpen=fontsize(9);
 
pointpen=black; pointfontpen=fontsize(9);
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[/asy]
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</asy>
 
Since <math>X</math> is on the radical axis of <math>\omega_1</math> and <math>\omega_2</math>, it has equal power, so
 
Since <math>X</math> is on the radical axis of <math>\omega_1</math> and <math>\omega_2</math>, it has equal power, so
 
<cmath> 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} </cmath>since <math>O_1X+O_2X=d</math>. Now we can solve for <math>O_1X</math> and <math>O_2X</math>, and in particular,
 
<cmath> 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} </cmath>since <math>O_1X+O_2X=d</math>. Now we can solve for <math>O_1X</math> and <math>O_2X</math>, and in particular,
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\begin{align*}
+
<cmath>\begin{align*}
 
O_1H &= O_1X - HX = \frac{d+\frac{r_1^2-r_2^2}{d}}{2} - \frac{r}{2} \\
 
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}.  
 
O_2H &= O_2X + HX = \frac{d-\frac{r_1^2-r_2^2}{d}}{2} + \frac{r}{2}.  
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\end{align*}We want to solve for <math>d</math>. By the Pythagorean Theorem (twice):
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\end{align*}</cmath>
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\begin{align*}
+
We want to solve for <math>d</math>. By the Pythagorean Theorem (twice):
 +
<cmath>\begin{align*}
 
&\qquad OH^2 = O_2H^2 - (r+r_2)^2 = O_1H^2 - (r+r_1)^2 \\
 
&\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 \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 \\
Line 53: Line 54:
 
&\implies 4dr = 8rr_2-8rr_1 \\
 
&\implies 4dr = 8rr_2-8rr_1 \\
 
&\implies \boxed{d=2r_2-2r_1}.  
 
&\implies \boxed{d=2r_2-2r_1}.  
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\end{align*}Therefore, <math>d=2(r_2-r_1) = 2(961-625)=\boxed{672}</math>.
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\end{align*}</cmath>
 +
Therefore, <math>d=2(r_2-r_1) = 2(961-625)=\boxed{672}</math>.
  
 
==See also==
 
==See also==

Revision as of 15:56, 11 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 $120^{\circ}$. 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{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 (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
All AIME Problems and Solutions

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