HMMT 十一月 2017 · 冲刺赛 · 第 24 题

HMMT November 2017 — Guts Round — Problem 24

[ 12 ] Triangle ABC has side lengths AB = 15 , BC = 18 , CA = 20. Extend CA and CB to points D and E respectively such that DA = AB = BE . Line AB intersects the circumcircle of CDE at P and Q . Find the length of P Q . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT November 2017, November 11, 2017 — GUTS ROUND Organization Team Team ID# D´ ej` a vu?

解析

[ 12 ] Triangle ABC has side lengths AB = 15 , BC = 18 , CA = 20. Extend CA and CB to points D and E respectively such that DA = AB = BE . Line AB intersects the circumcircle of CDE at P and Q . Find the length of P Q . Proposed by: Yuan Yao Answer: 37 WLOG suppose that P is closer to A than to B . Let DA = AB = BE = c = 15 , BC = a = 18 , CA = b = 20 , P A = x , and QB = y . By Power of a Point on B and A , we get ac = ( x + c ) y and bc = ( y + c ) x , respectively. Subtracting the two equations gives cy − cx = ac − bc ⇒ y − x = a − b . Substituting 2 y = x + a − b into the first equation gives ac = ( x + c )( x + a − b ) = x + ( a − b + c ) x + ac − bc , which √ 2 ( b − a − c )+ ( a − b + c ) +4 bc is a quadratic with unique positive solution x = . Thus, 2 √ √ 2 2 P Q = x + y + c = ( y − x )+2 x + c = ( a − b + c )+( b − a − c )+ ( a − b + c ) + 4 bc = 13 + 4 · 20 · 15 = 37 .