Problem:
Triangle ABC is inscribed in circle Ο. Points P and Q are on side AB with AP<AQ. Rays CP and CQ meet Ο again at S and T (other than C ), respectively. If AP=4,PQ=3,QB=6,BT=5, and AS=7, then ST=nmβ, where m and n are relatively prime positive integers. Find m+n.
Solution:
Extend AB through B to R so that BR=8. Because ACBT is cyclic, it follows by the Power of a Point Theorem that CQβ
QT=AQβ
QB=42. Note that PQβ
QR=42=CQβ
QT. By the converse of the Power of a Point Theorem, it follows that CPTR is cyclic. Because CPTR and ACTS are cyclic,
β BRT=β PRT=β PCT=β SCT=β SAT.
Because ABTS is cyclic, it follows that β AST=β RBT. Hence β³AST is similar to β³RBT, from which STASβ=BTRBβ or ST=ASβ
RBBTβ=835β. The requested sum is 35+8=43β.
The problems on this page are the property of the MAA's American Mathematics Competitions
