PUMaC 2012 · 组合（A 组） · 第 6 题
PUMaC 2012 — Combinatorics (Division A) — Problem 6
题目详情
- [ 6 ] Two white pentagonal pyramids, with side lengths all the same, are glued to each other at their regular pentagon bases. Some of the resulting 10 faces are colored black. How many rotationally distinguishable colorings may result?
解析
- We do casework on the number of faces colored black. The number of rotationally distinguish- able colorings when k faces are colored black is the same as that when 10 − k faces are black, so it suffices to compute when k = 0 , 1 , 2 , 3 , 4 , 5. Although tedious, most of the cases turn out to be very similar. Indeed, the dual of the shape is a pentagonal prism, so equivalently, the question asks for the number of 2-vertex-colorings of a pentagonal prism (which is how we’ll describe the shape in the following). Without loss of generality, we can fix the orientation of the prism such that the first black vertex to consider lies on the “top” face. • k = 0: 1 way. • k = 1: 1 way. • k = 2: If both black vertices lie on the same pentagonal face, there are 2 ways. Otherwise, the first vertex fixes the “top” face and the second vertex can be any of the 5 vertices of the bottom face, for 7 ways. 2 • k = 3: 3 vertices on a face can happen in 2 ways. 2 vertices on one face can occur in 2 ways, and each way fixes the top face, so the third vertex can be any of the 5 vertices of the bottom face, for 12 ways. • k = 4: If all four vertices lie on a face, there is 1 way. If three lie on a face, there are 2 ways, and the fourth vertex can be any of the 5 vertices of the other face. If two lie on one face and two on the other, we need to be a bit careful: there are 5 if both pairs are adjacent, 5 if neither pair is adjacent, and 5 if one pair is adjacent and the other is non-adjacent. This gives a total of 1 + 2 × 5 + 5 + 5 + 5 = 26 ways. • k = 5: If all five vertices lie on a face, there is 1 way. If four lie on a face, there is 1 way, and the fourth vertex can be any of the 5 vertices of the other face. If three lie on a face and two on the other, there are 2 ways to arrange the three vertices on the top face and ( ) 5 = 10 ways to position the bottom two, for a total of 1 + 5 + 2 × 10 = 26 ways. 2 Summing, there are 2(1 + 1 + 7 + 12 + 26) + 26 = 120 arrangements. Alternatively, we can count the symmetries of this 3D shape. There is 1 identity rotation, 4 ◦ ◦ ◦ ◦ ◦ (72 , 144 , 216 , 288 ) rotations about the apex vertices, and 5 180 rotations about the non- apex vertices that send the shape to itself, and these form the symmetry group of the figure. Respectively, these fix 10 , 2 , and 5 orbits of faces. By the (not-)Burnside’s lemma, the answer 10 2 5 2 +4 · 2 +5 · 2 is = 120 . 10