PUMaC 2012 · 几何（B 组） · 第 8 题

PUMaC 2012 — Geometry (Division B) — Problem 8

[ 8 ] A cyclic quadrilateral ABCD has side lengths AB = 3, BC = AD = 5, and CD = 8. The √ radius of its circumcircle can be written in the form a b/c , where a, b, c are positive integers, a, c are relatively prime, and b is not divisible by the square of any prime. Find a + b + c . 1

解析

[ 8 ] A cyclic quadrilateral ABCD has side lengths AB = 3, BC = AD = 5, and CD = 8. √ What’s the radius of its circumcircle? Your answer can be written in the form a b/c , where a, b, c are positive integers, a, c are relatively prime, and b is not divisible by the square of any prime. Find a + b + c . Solution: Note that ABCD is an isosceles trapezoid with AB parallel to CD . Draw AE parallel to BC with E on segment CD . Since ABCE is a parallelogram, CE = 3 and AE = 5, so DE = 5. Since AD = AE = DE = 5, ◦ ◦ 4 ADE is an equilateral triangle and so m ∠ D = m ∠ C = 60 , and thus m ∠ A = m ∠ B = 120 . √ Applying the law of cosines on 4 ABD gets BD = 9 + 25 + 15 = 7. The area of triangle 1 ◦ abc ◦ 4 ABD is equal to both ab sin 120 and , so 2 R sin 120 = c where c = 7. Solving for R , 2 4 R √ 7 3 we have R = , so a + b + c = 16 . 3 Problem contributed by Chengyue Sun Thanks to teams for correcting the area formula. 6