PUMaC 2016 · 几何（B 组） · 第 5 题

PUMaC 2016 — Geometry (Division B) — Problem 5

Let V be the volume of the octahedron ABCDEF with A and F opposite, B and E opposite, and C and D opposite, such that AB = AE = EF = BF = 13, BC = DE = BD = CE = 14, √ and CF = CA = AD = F D = 15. If V = a b for positive integers a and b , where b is not divisible by the square of any prime, find a + b .

解析

Notice that CF DA is a rhombus with side length 15. If we consider the plane of the rhombus, by symmetry we have that B and E are right above and below the center of the rhombus, which we will call O . Let AO = x , CO = y , and BO = h . The Pythagorean theorem gives us 2 2 2 2 x + h = AB = 13 2 2 2 2 y + h = BC = 14 2 2 2 2 x + y = AC = 15 . √ 1 Adding the first two equations and subtracting the third one gives us 2 h = 140, i.e. h = 70. √ √ √ √ Also note that x = 169 − 70 = 3 11 and y = 196 − 70 = 3 14. The area of the rhombus √ is thus 2 xy = 18 11 · 14. Thus, the area of pyramid ABCDF is √ √ 1 · 2 xy · h = 6 70 · 11 · 14 = 84 55 . 3 √ Thus, the area of the octahedron is 168 55, giving the answer 168 + 55 = 223 . Problem written by Zhuo Qun Song.