HMMT 二月 2026 · 冲刺赛 · 第 22 题
HMMT February 2026 — Guts Round — Problem 22
题目详情
- [12] An equilateral triangle-shaped cake of side length 5 is cut into 25 unit equilateral triangle pieces. Jacob selects two distinct edges of the cake, then picks one point independently and uniformly at random on each of the two selected edges. He cuts along the line through these two points. Compute the expected number of pieces of cake after all cuts. Below is an example of the process, with the dots being the selected points and the dashed line being Jacob’s cut. This cut results in 32 pieces.
解析
- [12] An equilateral triangle-shaped cake of side length 5 is cut into 25 unit equilateral triangle pieces. Jacob selects two distinct edges of the cake, then picks one point independently and uniformly at random on each of the two selected edges. He cuts along the line through these two points. Compute the expected number of pieces of cake after all cuts. Below is an example of the process, with the dots being the selected points and the dashed line being Jacob’s cut. This cut results in 32 pieces. Proposed by: Sebastian Attlan 158 Answer: = 31 . 6 5 Solution: It suffices to consider how many of the original 25 pieces Jacob’s cut intersects, as the total number of pieces will be 25 plus this. Label the vertices of the cake ABC and the cut points P and Q such that P is on edge AB and Q is on edge AC . Let AB = 5 . A Q P B C With probability 1 , the cut does not pass through any vertex of any original piece, so we assume this is the case. Claim 1. The number of pieces Jacob’s cut intersects is 2 max( ⌈ AP ⌉ , ⌈ AQ ⌉ ) − 1 . Proof. Let p = ⌈ AP ⌉ and q = ⌈ AQ ⌉ . Jacob’s cut passes through | p − q | edges parallel to BC , p − 1 edges parallel to CA , and q − 1 edges parallel to AB . Because of our assumption, the number of pieces the cut passes through is 1 + | p − q | + ( p − 1) + ( q − 1) = ( p + q ) + | p − q | − 1 = 2 max( p, q ) − 1 . As an example, in the diagram above, p and q are 4 and 3 , and the cut indeed passes through 2 · 4 − 1 = 7 pieces. Since AP and AQ are independent uniformly random real numbers from 0 to 5 , we know p = ⌈ AP ⌉ and q = ⌈ AQ ⌉ are independently and uniformly selected from { 1 , 2 , 3 , 4 , 5 } . For positive integers m ≤ 5 , 2 m the probability that p and q are both at most m is , so the probability their maximum is m is 25 2 2 ( m − 1) m 2 m − 1 − = . Thus, the answer is 25 25 25 5 2 2 2 2 2 ∑ 2 m − 1 1 + 3 + 5 + 7 + 9 165 158 25 + (2 m − 1) = 25 + = 25 + = . 25 25 25 5 m =1 ©2026 HMMT