Problem:
Let β³ABC have side lengths AB=5,BC=9, and CA=10. The tangents to the circumcircle of β³ABC at B and C intersect at point D, and AD intersects the circumcircle at Pξ =A. The length of AP is equal to nmβ, where m and n are relatively prime positive integers. Find m+n.
Solution:
Note that β DBP=β DAB and β DCP=β DAC because the two pairs of angles subtend arcs BP and CP, respectively. It follows that β³DBPβΌβ³DAB and β³DCPβΌβ³DAC.
Then
BPABβ=BDADβ=CDADβ=CPACβ.β
Because AB=5 and AC=10, it follows that CP=2β BP. Let y=BP; then CP=2y. Applying Ptolemy's Theorem to cyclic quadrilateral ABPC yields
Quadrilateral ABPC is cyclic, implying that β BPC=180βββ BAC, so cos(β BPC)=βcos(β BAC). Applying the Law of Cosines to β³BAC yields
cos(β BAC)=2β 5β 1052+102β92β=2511β.β
Substituting this into (2) yields
5y2+2544βy2=81βΉy=1345β.β
Substituting this value into (1) yields
AP=920ββ 1345β=13100β.β
The requested sum is 100+13=113β.
OR
Let T be the transformation of the plane that consists of an inversion of the plane with respect to the circle with center A and radius ABβ ACβ followed by a reflection of the plane across the angle bisector of β BAC. Note that T is an involution, that is, it is its own inverse. Then T(B)=C and T(C)=B. Line BC goes to a circle Ο passing through A, T(B), and T(C), so it is the circumcircle of β³ABC. Thus T(BC)=Ο and T(Ο)=BC. Line BD is tangent to Ο, so T(BD) is a circle passing through A and C and is tangent to line BC. Similarly, T(CD) is a circle passing through A and B and is tangent to line BC. Hence T(D)=Dβ² is the second intersection of the above circles. Line ADβ² is the radical axis of the two circles, and line BC is tangent to both circles. Therefore ADβ² intersects BC at M, where
Therefore cot(β CAM)=cot(β PAB) and β CAM=β PAB. Because β ACB=β APB, it follows that β³CAMβΌβ³PAB, and APβ AM=ABβ AC. The remainder follows as in the second solution.
