HMMT 二月 2021 · 团队赛 · 第 5 题

HMMT February 2021 — Team Round — Problem 5

[60] A convex polyhedron has n faces that are all congruent triangles with angles 36 , 72 , and 72 . Determine, with proof, the maximum possible value of n . 2

解析

[60] A convex polyhedron has n faces that are all congruent triangles with angles 36 , 72 , and 72 . Determine, with proof, the maximum possible value of n . Proposed by: Handong Wang Answer: 36 Solution: Consider such a polyhedron with V vertices, E edges, and F = n faces. By Euler’s formula we have V + F = E + 2. Next, note that the number of pairs of incident faces and edges is both 2 E and 3 F , so 2 E = 3 F . Now, since our polyhedron is convex, the sum of the degree measures at each vertex is strictly less than 360 = 36 · 10. As all angle measures of the faces of our polyhedron are divisible by 36, the maximum degree measure at a given vertex is 36 · 9 = 324. On the other hand, the total degree measure at all vertices is the total degree measure over all faces, which is 180 F . Thus we have 180 F ≤ 324 V , or 10 F ≤ 18 V . Putting our three conditions together, we have 10 F ≤ 18 V = 18( E + 2 − F ) = 9(2 E ) + 36 − 18 F = 9(3 F ) + 36 − 18 F = 9 F + 36 . Thus F ≤ 36. F = 36 is attainable by taking a 9-gon antiprism with a 9-gon pyramid attached on the top and the bottom. Thus the answer is 36. 2