PUMaC 2019 · 几何（B 组） · 第 3 题

PUMaC 2019 — Geometry (Division B) — Problem 3

Let ∆ ABC be a triangle with circumcenter O and orthocenter H. Let D be a point on the circumcircle of ABC such that AD ⊥ BC. Suppose that AB = 6 , DB = 2 , and the ratio area(∆ ABC ) 2 = 5. Then, if OA is the length of the circumradius, then OA can be written in area(∆ HBC ) m the form , where m, n are relatively prime nonnegative integers. Compute m + n . n Note: The circumradius is the radius of the circumcircle.

解析

Let ∆ ABC be a triangle with circumcenter O and orthocenter H. Let D be a point on the circumcircle of ABC such that AD ⊥ BC. Suppose that AB = 6 , DB = 2 , and the ratio area(∆ ABC ) 2 = 5. Then, if OA is the length of the circumradius, then OA can be written in area(∆ HBC ) m the form , where m, n are relatively prime nonnegative integers. Compute m + n . n Note: The circumradius is the radius of the circumcircle. Proposed by Oliver Thakar. Answer: 29 Solution: The key observation here is that ∆ BDC and ∆ BHC are in fact congruent. Then, AB × AC AB × AC = = 5 . HB × HC DB × DC 2 Because ABCD is an orthogonal cyclic quadrilateral, we get the following relations: AB + 2 2 2 2 2 DC = 4 OA and AC + DB = 4 OA In combination with AB × AC = 5 DB × DC, we 3 2 2 2 2 can solve for OA. We know that AC = DC. Thus, AB + DC = AC + DB reduces 5 9 27 2 2 2 2 2 to 32 + AC = AC , so that AC = 50 . That makes 4 OA = 50 + 4 , so OA = , or 25 2 √ 3 OA = 3 . 2