PUMaC 2016 · 团队赛 · 第 13 题

PUMaC 2016 — Team Round — Problem 13

(10) Ayase randomly picks a number x ∈ (0 , 1] with uniform probability. He then draws the six points (0 , 0 , 0) , ( x, 0 , 0) , (2 x, 3 x, 0) , (5 , 5 , 2) , (7 , 3 , 0) , (9 , 1 , 4). If the expected value of the m volume of the convex polyhedron formed by these six points can be written as for relatively n prime positive integers m and n , find m + n . 2016 2015

解析

Note that each of these six points lie on a different edge of the tetrahedron with vertices 1 (0 , 0 , 0) , (10 , 0 , 0) , (4 , 6 , 0) , (8 , 2 , 8), which has area A = · 10 · 6 · 8 = 80. The resulting polygon 6 is essentially this tetrahedron with the vertices cut off, and finding the volume of the cut-off pieces is simple. Label the four vertices of the tetrahedron A , . . . , A and the six points Ayase 1 4 draws B , . . . , B , respectively. Then: 1 6 ( ) A B A B A B 1 2 1 3 1 1 E [ A B B B ] = [ A A A A ] · E · · 1 2 3 1 1 2 3 4 A A A A A A 1 2 1 3 1 4 ( ) x x 0 = 80 · E · · √ = 0 10 2 2 33 ( ) A B A B A B 2 2 2 5 2 6 E [ A B B B ] = [ A A A A ] · E · · 2 2 5 6 2 1 3 4 A A A A A A 2 1 2 3 2 4 ( ) 10 − x 1 1 = 80 · E · · = 19 10 2 2 ( ) A B A B A B 3 3 3 5 3 4 E [ A B B B ] = [ A A A A ] · E · · 3 3 5 4 3 1 2 4 A A A A A A 3 1 3 2 3 4 ( ) √ 13(2 − x ) 1 1 15 = 80 · E √ · · = 2 4 2 2 13 ( ) A B A B A B 4 1 4 6 4 4 E [ A B B B ] = [ A A A A ] · E · · 4 1 6 4 4 1 2 3 A A A A A A 4 1 4 2 4 3 ( ) 1 1 3 = 80 · E · · = 30 1 2 4 Which yields: 15 47 E [ B B B B B B ] = 80 − 0 − 19 − − 30 = . 1 2 3 4 5 6 2 2 Thus, the answer is 47 + 2 = 49 . Problem written by Bill Huang.