HMMT 十一月 2010 · 冲刺赛 · 第 5 题
HMMT November 2010 — Guts Round — Problem 5
题目详情
- [ 6 ] An icosahedron is a regular polyhedron with twenty faces, all of which are equilateral triangles. If an icosahedron is rotated by θ degrees around an axis that passes through two opposite vertices so that it occupies exactly the same region of space as before, what is the smallest possible positive value of θ ?
解析
- [ 6 ] An icosahedron is a regular polyhedron with twenty faces, all of which are equilateral triangles. If an icosahedron is rotated by θ degrees around an axis that passes through two opposite vertices so that it occupies exactly the same region of space as before, what is the smallest possible positive value of θ ? ◦ Answer: 72 Because this polyhedron is regular, all vertices must look the same. Let’s consider just ◦ one vertex. Each triangle has a vertex angle of 60 , so we must have fewer than 6 triangles; if we had ◦ 6, there would be 360 at each vertex and you wouldn’t be able to “fold” the polyhedron up (that is, it would be a flat plane). It’s easy to see that we need at least 3 triangles at each vertex, and this gives a triangular pyramid with only 4 faces. Having 4 triangles meeting at each vertex gives an octahedron (two square pyramids with the squares glued together) with 8 faces. Therefore, an icosahedron has 5 ◦ 360 ◦ triangles meeting at each vertex, so rotating by = 72 gives another identical icosahedron. 5 Guts Round Alternate solution: Euler’s formula tells us that V − E + F = 2, where an icosahedron has V vertices, E edges, and F faces. We’re told that F = 20. Each triangle has 3 edges, and every edge is common to 3(20) 2 triangles, so E = = 30. Additionally, each triangle has 3 vertices, so if every vertex is common 2 3(20) 60 60 to n triangles, then V = = . Plugging this into the formula, we have − 30 + 20 = 2, so n n n ◦ 60 360 ◦ = 12 and n = 5. Again this shows that the rotation is = 72 n 5