2021 AIME I Problems/Problem 13
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 .
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, .
Denote by , , and the centers of , , and , respectively. Let and denote the radii of and respectively, be the radius of , and the distance from to the line . We claim thatwhere . This solves the problem, for then the condition implies , and then we can solve to get .
Denote by and the centers of and respectively. Set as the projection of onto , and denote by the intersection of with . Note that . Now recall thatFurthermore, note thatSubstituting the first equality into the second one and subtracting yieldswhich rearranges to the desired.
Solution 3 (Inversion)
WLOG assume is a line. Note the angle condition is equivalent to the angle between and being . We claim the angle between and is fixed as varies.
Proof: Perform an inversion at , sending and to two lines and intersecting at . Then is sent to a circle tangent to lines and , which clearly intersects at a fixed angle. Therefore the angle between and is fixed as varies.
Now simply take to be a line. If intersects and and , respectively, and the circles' centers are and , then the projection of to at gives that is a triangle. Therefore,
Solution 4 (Radical Axis, Harmonic Quadrilaterals, and Similar Triangles)
Suppose we label the points as shown here. By radical axis, the tangents to at and intersect on . Thus is harmonic, so the tangents to at and intersect at . Moreover, because both and are perpendicular to , and because . Thusby similar triangles.
Solution 5 (Olympiad Geometry)
Let be the center of and the radius of for . Let and have radius . Let be the center of . Then, the distance between and the radical axis of and is . It is well-known that the function is linear (see here) and that to compute it, it suffices to project onto line . Moreover, . Hence, we have Cancel out to yield so the answer is .
Who wanted to see animated video solutions can see this. I found this really helpful.
P.S: This video is not made by me. And solution is same like below solutions.
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