PUMaC 2017 · 几何（A 组） · 第 8 题
PUMaC 2017 — Geometry (Division A) — Problem 8
题目详情
- Triangle ABC with AB = 4 , BC = 5 , CA = 6 has circumcircle Ω and incircle ω . Let Γ be the circle tangent to Ω and the sides AB, BC , and let X = Γ ∩ Ω. Let Y, Z be distinct points on 2 Ω such that XY, XZ are tangent to ω . Find Y Z . The following fact may be useful: if 4 ABC has incircle ω with center I and radius r , and 4 DEF is the intouch triangle (i.e. D, E, F are intersections of the incircle with BC, CA, AB , respectively) and H is the orthocenter of 4 DEF , then the inversion of X about ω (i.e. the ′ ′ 2 point X on ray IX such that IX · IX = r ) is the midpoint of DH . 1
解析
- Note: the problem as stated has a typo from what was intended. Γ should be tangent to AB and AC ; this is the configuration corresponding to the official solution below, while the configuration given on the contest corresponds to the answer accepted at PUMaC 2017 of 16 . ′ Let G denote the inversion of a point G about ω . Let I and O denote the incenter and circumcenter, respectively, of 4 ABC . Let 4 DEF be the intouch triangle, H its orthocenter, and M the midpoint of DH . Let XY and XZ intersect BC at P and Q and ω at R and S , respectively (WLOG P is closer to B than Q ). ′ Let γ be circumcircle of 4 XP Q . Using the hint, we see that γ passes through the midpoint ′ of DH , as well as the midpoints of DR and DS . Now, Ω passes through the midpoints of DE , ′ EF , and F D , so it is the nine-point circle of 4 DEF . Since Ω is a dilation of ω with center H ′ and factor 1 / 2, Ω passes through the midpoints of HR , HS , and HD . Now, HM = M D and ′ ◦ ′ RM = M S , so DRHS is a parallelogram. It follows that γ is the 180 rotation of Ω about M , thus the two circles are tangent. Inversion preserves tangency, so γ and Ω are tangent. By the dilation γ → Ω centered at X , we find Y Z ‖ BC . By Poncelet’s porism, Y Z is tangent to ω . Now, we calculate the inradius and circumradius of 4 ABC to be √ 7 8 r = , R = √ . 2 7 Let M and N be the midpoints of BC and Y Z , respectively. Then, 9 2 2 2 √ OM = R − BM −→ OM = , 2 7 √ 33 2 2 2 2 2 XN = R − ON = R − (2 r − OM ) −→ XN = . 2 2 Then, Y Z = 33 . Problem written by Bill Huang If you believe that any of these answers is incorrect, or that a problem had multiple reasonable interpretations or was incorrectly stated, you may appeal at http://tinyurl.com/PUMaCappeal2017. All appeals must be in by 1 PM to be considered. 2