PUMaC 2017 · 几何（A 组） · 第 5 题

PUMaC 2017 — Geometry (Division A) — Problem 5

Rectangle HOM F has HO = 11 and OM = 5. Triangle ABC has orthocenter H and circumcenter O . M is the midpoint of BC and altitude AF meets BC at F . Find the length of BC . ◦

解析

Let M C = x . Then we have M F = 11 and F B = 11 − x . Consider triangle OM C , we have √ √ 2 2 2 2 2 OC = OM + M C . Since OM = 5, then OC = 25 + x . Hence, OA = OC = 25 + x . √ 2 2 2 2 Consider triangle AHO , we have AH = AO − OH , then we have AH = x − 96. Construct a circumcircle of triangle ABC and draw AO intersect the circle at T . Therefore, √ 2 2 2 2 we have AB = AT − BT . Since AT = 2 AO , we have AT = 2 25 + x . Finally, we consider √ 2 2 2 2 2 2 the triangle ABF and the equation AF + F B = AB . Hence, ( x − 96 + 5) + (11 − x ) + 2 2 (11 + x ) = 4( x + 25). We get x = 14 which makes BC = 28 . Problem written by Natthawut Boonsiriphatthanajaroen