HMMT 二月 2021 · 冲刺赛 · 第 24 题
HMMT February 2021 — Guts Round — Problem 24
题目详情
- [14] Let P be a point selected uniformly at random in the cube [0 , 1] . The plane parallel to x + y + z = 0 passing through P intersects the cube in a two-dimensional region R . Let t be the expected value of the a 2 perimeter of R . If t can be written as , where a and b are relatively prime positive integers, compute b 100 a + b . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT Spring 2021, March 06, 2021 — GUTS ROUND Organization Team Team ID#
解析
- [14] Let P be a point selected uniformly at random in the cube [0 , 1] . The plane parallel to x + y + z = 0 passing through P intersects the cube in a two-dimensional region R . Let t be the expected value of a 2 the perimeter of R . If t can be written as , where a and b are relatively prime positive integers, b compute 100 a + b . Proposed by: Michael Ren Answer: 12108 Solution: We can divide the cube into 3 regions based on the value of x + y + z which defines the plane: x + y + z < 1 , 1 ≤ x + y + z ≤ 2 , and x + y + z > 2 . The two regions on the ends create tetrahedra, each of which has volume 1 / 6 . The middle region is a triangular antiprism with volume 2 / 3 . √ If our point P lies in the middle region, we can see that we will always get the same value 3 2 for the perimeter of R . Now let us compute the expected perimeter given that we pick a point P in the first region x + y + z < 1 . √ If x + y + z = a, then the perimeter of R will just be 3 2 a, so it is sufficient to find the expected value of a. a is bounded between 0 and 1, and forms a continuous probability distribution with value 2 proportional to a , so we can see with a bit of calculus that its expected value is 3 / 4 . The region x + y + z > 2 is identical to the region x + y + z < 1 , so we get the same expected perimeter. √ √ 9 Thus we have a 2 / 3 of a guaranteed 3 2 perimeter, and a 1 / 3 of having an expected 2 perimeter, 4 √ √ √ 2 1 9 11 2 121 which gives an expected perimeter of · 3 2 + · 2 = . The square of this is , giving an 3 3 4 4 8 extraction of 12108.