PUMaC 2014 · 个人决赛（A 组） · 第 1 题

PUMaC 2014 — Individual Finals (Division A) — Problem 1

Let γ be the incircle of 4 ABC (i.e. the circle inscribed in 4 ABC ) for which AB + AC = 3 BC . Let the point where AC is tangent to γ be D . Let the incenter be I . Let the intersection of the circumcircle of 4 BCI with γ that is closer to B be P . Show that P ID is colinear.

解析

Let be the incircle of 4 ABC (i.e. the circle inscribed in 4 ABC ) for which AB + AC = 3 BC . Let the point where AC is tangent to be D . Let the incenter be I . Let the intersection of the circumcircle of 4 BCI with that is closer to B be P . Show that P ID is colinear. Solution: 0 0 Let point P be the diameter of passing through D . Hence we need to show that P and P 0 are the same point, meaning that BP IC is con-cyclic Since AC + AB = 2 AD + CD + BG = 2 AD + CF + F B = 2 AD + BC = 3 BC . Hence AD = BC . 0 0 0 Let us construct C A //CA passing through P . Let F and G be the tangents of with the 0 0 lines BC and BA respectively. We see that BC A ' BCA and let the ratio of the sides BC 0 = k . Extend BP to meet AC at E . 0 BC 0 0 0 Hence we have 2 AD = AG + DE + EA = AG + DE + kP A = AG + DE + kA G = 0 AG + DE + k ( BG BA ) = AG + DE + kBG BA = DE + kBG BG = DE + ( k 1) BG 0 0 0 Similarly, 2 CE = CD + DE + CE = CF + DE + C P k = CF + DE + kC F = CF + DE + 0 k ( BF BC ) = CF + DE + kBF BC = DE + kBF BF = DE + ( k 1) BF 180 \ C Hence we see that CE = AD = BC and hence 4 CBE is isosceles \ CBE = . SInce 2 180 \ C 0 o P ID is the diameter, we see that \ IDC = 90 and hence \ CID = and therefore 2 0 0 BP IC is concyclic, implying that P = P and P ID is colinear.