Problem:
In △ABC with side lengths AB=13,BC=14, and CA=15, let M be the midpoint of BC. Let P be the point on the circumcircle of △ABC such that M is on AP. There exists a unique point Q on segment AM such that ∠PBQ=∠PCQ. Then AQ can be written as nm, where m and n are relatively prime positive integers. Find m+n.
Solution:
Answer (247):
Let Q′ be the reflection of P over M. Then MP=MQ′, and the diagonals of quadrilateral BPCQ′ bisect each other, so the quadrilateral is a parallelogram. It follows that ∠PCQ′=∠PBQ′, and Q′ is the required point Q. By the Length of the Median Formula, AM=212AB2+2AC2−BC2=237. The power of point M in the circumcircle is 7⋅7=AM⋅PM=237⋅PM, from which PM=744937. Because M is the midpoint of PQ, it follows that AQ=AM−PM=14899. The requested sum is 99+148=247. Note that the Length of the Median Formula follows immediately from applications of the Law of Cosines to △AMB and △AMC.
OR
As above, AM=237. Note that ∠APB=∠C and ∠APC=∠B. Applying the Law of Sines to △PBQ and △PCQ gives
As above, CQBQ=ACAB, so Q lies on the Apollonius circle of △ABC corresponding to vertex A.
Let D and D′ be the respective intersections with line BC of the interior and exterior angle bisectors from A. Then CDBD=CD′BD′=1513. Thus BD=213 and BD′=91. Recall that DD′ is the diameter of the Apollonius circle. It follows that the radius of the Apollonius circle is 4195.
Denote by O the center of this Apollonius circle, and let S be the projection of O onto the median AM. By the above work, OD=4195 and OM=4197.
Let E be the foot of the altitude from A to BC. Because AE2=AB2−BE2=AC2−CE2, and BE+CE=14, it follows that BE=5,CE=9, and EM=2. As in the first solution, AM=237. Because △AEM and △OSM are similar, it follows that
4197SM=MOSM=MAEM=2372,
which implies SM=437197. Finally, because S is the midpoint of AQ, it follows that