HMMT 十一月 2023 · 冲刺赛 · 第 27 题
HMMT November 2023 — Guts Round — Problem 27
题目详情
- [13] Compute the number of ways to color the vertices of a regular heptagon red, green, or blue (with rotations and reflections distinct) such that no isosceles triangle whose vertices are vertices of the heptagon has all three vertices the same color. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT November 2023, November 11, 2023 — GUTS ROUND Organization Team Team ID#
解析
- [13] Compute the number of ways to color the vertices of a regular heptagon red, green, or blue (with rotations and reflections distinct) such that no isosceles triangle whose vertices are vertices of the heptagon has all three vertices the same color. Proposed by: Ethan Liu Answer: 294 Solution: Number the vertices 1 through 7 in order. Then, the only way to have three vertices of a regular heptagon that do not form an isosceles triangle is if they are vertices 1 , 2 , 4 , rotated or reflected. Thus, it is impossible for have four vertices in the heptagon of one color because it is impossible for all subsets of three vertices to form a valid scalene triangle. We then split into two cases: Case 1: Two colors with three vertices each, one color with one vertex. There is only one way to do this up to permutations of color and rotations and reflections; if vertices 1 , 2 , 4 are the same color, of the remaining 4 vertices, only 3 , 5 , 6 form a scalene triangle. Thus, we have 7 possible locations for the vertex with unique color, 3 ways to pick a color for that vertex, and 2 ways to assign the remaining two colors to the two triangles, for a total of 42 ways. Case 2: Two colors with two vertices each, one color with three vertices. There are 3 choices of color 4 for the set of three vertices, 14 possible orientations of the set of three vertices, and choices of 2 which pair of the remaining four vertices is of a particular remaining color; as there are only two of each color, any such assignment is valid. This is a total of total of 3 · 14 · 6 = 252 ways. Thus, the final total is 42 + 252 = 294.