Maths Olympiad Prep

Track / Stage 4 / 317 of 340 #577 of 1964

Problem 577

AMC 12 late, AIME early
Geometry Difficulty 5.0 Find the answer HMMT_2

Let ABCA B C be a triangle with AB=13,BC=14,CA=15A B=13, B C=14, C A=15. Let OO be the circumcenter of ABCA B C. Find the distance between the circumcenters of triangles AOBA O B and AOCA O C.

A number or a short expression. Spacing, $ signs and \frac vs / are all fine.

Official solution

Let S,TS, T be the intersections of the tangents to the circumcircle of ABCA B C at A,CA, C and at A,BA, B respectively. Note that ASCOA S C O is cyclic with diameter SOS O, so the circumcenter of AOCA O C is the midpoint of OSO S, and similarly for the other side. So the length we want is 12ST\frac{1}{2} S T. The circumradius RR of ABCA B C can be computed by Heron's formula and K=abc4RK=\frac{a b c}{4 R}, giving R=658R=\frac{65}{8}. A few applications of the Pythagorean theorem and similar triangles gives AT=656,AS=392A T=\frac{65}{6}, A S=\frac{39}{2}, so the answer is 916\frac{91}{6}

Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.