HMMT 二月 2025 · 团队赛 · 第 7 题
HMMT February 2025 — Team Round — Problem 7
题目详情
- [45] Determine, with proof, whether a square can be dissected into finitely many (not necessarily ◦ ◦ ◦ congruent) triangles, each of which has interior angles 30 , 75 , and 75 .
解析
- [45] Determine, with proof, whether a square can be dissected into finitely many (not necessarily ◦ ◦ ◦ congruent) triangles, each of which has interior angles 30 , 75 , and 75 . Proposed by: Derek Liu Answer: No Solution 1: Assume for sake of contradiction that such a dissection exists. It has exactly half as many ◦ ◦ 30 angles as 75 angles. ◦ Around any intersection point except the square’s vertices, the only angles that can appear are 30 , ◦ ◦ ◦ ◦ 75 , and 180 . The only combinations of these that sum to 180 or 360 are ◦ ◦ 6 · 30 = 180 , ◦ ◦ ◦ 30 + 2 · 75 = 180 , ◦ ◦ 180 = 180 , ◦ ◦ 12 · 30 = 360 , ◦ ◦ ◦ 7 · 30 + 2 · 75 = 360 , ◦ ◦ ◦ 2 · 30 + 4 · 75 = 360 , ◦ ◦ ◦ 6 · 30 + 180 = 360 , ◦ ◦ ◦ ◦ 30 + 2 · 75 + 180 = 360 , ◦ ◦ ◦ 180 + 180 = 360 . ◦ ◦ In particular, around any such point, there are at least half as many 30 angles as 75 angles. ◦ ◦ However, the square’s vertices must each be surrounded by three 30 angles and zero 75 angles, as ◦ ◦ there is no other way to get a sum of 90 . Thus the total number of 30 angles in the dissection must ◦ be at least 12 more than half the number of 75 angles, contradiction. Thus no such dissection exists. Solution 2: Again assume for sake of contradiction that a dissection exists. Interpret the dissection as a graph G , where the vertices of the graph are the vertices of all the triangles, and edges connect each pair of consecutive vertices along a line segment. Call a vertex flat if it is on either the boundary of the square (including its corners) or the interior of an edge of any triangle. Let X be the number of flat vertices and Y be the number of non-flat vertices in G . Let E and F be the number of edges and faces (triangles) in the dissection, respectively. Then ( X + Y ) − E + F = 1. Observing the angle combinations in the first solution, we see that any non-flat vertex must have at least 6 incident edges, and any flat vertex must have at least 4. Thus 2 E ≥ 6 Y + 4 X , so E ≥ 3 Y + 2 X . The sum of the angles of all F triangles is πF . Around any non-flat vertex, such angles sum to 2 π . Around any flat vertex, the angles sum to π , with the exception of the four corners of the square, where they sum to π/ 2 instead. Thus F π = ( X − 4) π + 4( π/ 2) + Y (2 π ) = ( X + 2 Y − 2) π, so F = X + 2 Y − 2. This means X + Y − E + F ≤ ( X + Y ) − (3 Y + 2 X ) + ( X + 2 Y − 2) = − 2 , contradiction. Thus no dissection exists.