PUMaC 2023 · 几何（B 组） · 第 8 题

PUMaC 2023 — Geometry (Division B) — Problem 8

Let △ ABC have AB = 14, BC = 30, AC = 40 and △ AB C with AB = 7 6, B C = 15 6, √ 5 π ′ ′ ′ ′ AC = 20 6 such that ∠ BAB = . The lines BB and CC intersect at point D . Let O be 12 ′ ′ ′ the circumcenter of △ BCD , and let O be the circumcenter of △ B C D . Then the length of √ a + b c ′ segment OO can be expressed as , where a, b, c, and d are positive integers such that a d and d are relatively prime, and c is not divisible by the square of any prime. Find a + b + c + d . (Write answers on next page.) 1 Name: Team: Write answers in table below: Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 2

解析

Let △ ABC have AB = 14, BC = 30, AC = 40 and △ AB C with AB = 7 6, B C = 15 6, √ ′ ′ 5 π ′ ′ AC = 20 6 such that ∠ BAB = . The lines BB and CC intersect at point D . Let O be 12 ′ ′ ′ the circumcenter of △ BCD , and let O be the circumcenter of △ B C D . Then the length of √ a + b c ′ segment OO can be expressed as , where a, b, c, and d are positive integers such that a d and d are relatively prime, and c is not divisible by the square of any prime. Find a + b + c + d . Proposed by Adam Huang Answer: 55 ′ ′ Note that △ ABC and △ AB C are spirally similar with center of spiral similarity given by √ 5 π 6 A and angle and dilation factor . By properties of spiral similarity, we have that 12 2 ′ ′ ′ ′ D := BB ∩ CC lies on circumcircles ( ABC ) and ( AB C ). Therefore AO is the circumradius √ 6 5 π ′ ′ of △ ABC , and AO = AO by similarity, with ∠ OAO = . To compute R := AO , note by 2 12 abc abc 14 · 30 · 40 Heron that the area of △ ABC is K = 168, so that = K = ⇒ R = = = 25. By 4 R 4 K 4 · 168 √ √ √ 6 6 5 π 25+25 3 ′ 2 2 2 2 ′ Law of Cosines, we have ( OO ) = 25 · (1 + ( ) − 2 · 1 · · cos ), so that OO = , 2 2 12 2 which yields an answer of a + b + c + d = 25 + 25 + 3 + 2 = 55. 3