Difference between revisions of "2021 AIME I Problems/Problem 13"
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&\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 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 4dr = 8rr_2-8rr_1 \\ | ||
− | &\implies d=2r_2-2r_1 | + | &\implies d=2r_2-2r_1 |
\end{align*}</cmath> | \end{align*}</cmath> | ||
Therefore, <math>d=2(r_2-r_1) = 2(961-625)=\boxed{672}</math>. | Therefore, <math>d=2(r_2-r_1) = 2(961-625)=\boxed{672}</math>. |
Revision as of 23:01, 13 March 2021
Problem
Circles and with radii and , respectively, intersect at distinct points and . A third circle is externally tangent to both and . Suppose line intersects at two points and such that the measure of minor arc is . Find the distance between the centers of and .
Solution
Let and be the center and radius of , and let and be the center and radius of .
Since extends to an arc with arc , the distance from to is . Let . Consider . The line is perpendicular to and passes through . Let be the foot from to ; so . We have by tangency and . Let . Since is on the radical axis of and , it has equal power with respect to both circles, so since . Now we can solve for and , and in particular, We want to solve for . By the Pythagorean Theorem (twice): Therefore, .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.