HMMT 十一月 2025 · 冲刺赛 · 第 18 题
HMMT November 2025 — Guts Round — Problem 18
题目详情
- [10] Compute the number of ways to divide a 6 × 6 square into 36 triangles, each of which has side lengths √ √ 2, 2, and 2. (Rotations and reflections of a division are considered distinct divisions.) © 2025 HMMT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT November 2025, November 08, 2025 — GUTS ROUND Organization Team Team ID#
解析
- [10] Compute the number of ways to divide a 6 × 6 square into 36 triangles, each of which has side √ √ lengths 2, 2, and 2. (Rotations and reflections of a division are considered distinct divisions.) Proposed by: Derek Liu Answer: 4096 Solution: √ The square’s edges must be lined with 12 triangles’ hypotenuses because sides of 2 cannot add to an ◦ integer length. After placing those 12 triangles, rotate the diagram 45 and observe that the remaining √ area can be divided into 12 squares, each of side length 2. Each triangle must cover half of one of √ these squares by the lemma below (scaled up by a factor of 2). Each square can be covered in 2 ways, 12 so there are 2 ways to cover the entire figure. Lemma 1. Consider any finite polyomino in a unit-length grid. If the polygon is divided into triangles √ of side lengths 1, 1, and 2, then each triangle covers half of one cell of the grid. Proof. Align the grid so that the gridlines are horizontal and vertical. We use the term “edge” to refer ◦ to any edge of any triangle. Clearly, every edge is either horizontal, vertical, or diagonal (45 angle to horizontal). We first prove the following: for any vertical line ℓ which is not a grid line, if some diagonal edge e 1 has left endpoint on ℓ , then there also exists a diagonal edge with right endpoint on ℓ . Without loss of generality, assume e is in the up-right direction, and let P be the left endpoint of e . 1 1 Consider the triangle T that has P on its perimeter and borders (some part of) edge e from above. 1 We consider three cases for T . ◦ • T has a 45 angle at P . Thus, it has a vertical edge at P . Let Q be any point on the interior of this edge. Since ℓ is not a gridline, some triangle on the left of ℓ has Q on its perimeter; this triangle has at least one diagonal edge with right endpoint on ℓ . ◦ • T has a 90 angle at P . Then, T has an up-left edge from P , which is a diagonal edge with right endpoint on ℓ . ◦ • T has a 180 angle at P (i.e., P is not a vertex of T ). Since P is a vertex of some triangle, there must exist a triangle with P as a vertex and a down-left edge from P (bordering T from below); this edge is a diagonal edge with right endpoint on ℓ . © 2025 HMMT In every case, the desired diagonal edge exists. Some examples of each case are shown below (each column corresponds to one case). In each, ℓ is dashed, e is drawn in blue, T is shaded green, and the desired edge is drawn in red. 1 Q P P P Q P P Now assume for sake of contradiction that there exists a diagonal edge e of length 1. Let ℓ be the 0 0 vertical line containing the left endpoint of e . If ℓ is not a gridline, then there exists a diagonal edge 0 0 e with right endpoint on ℓ . Let ℓ be the vertical line containing the left endpoint of e ; if it is not a 1 0 1 1 gridline, there exists a diagonal edge e with right endpoint on ℓ . We continue this process; it cannot 2 1 go on forever because the polyomino is finite, so eventually we reach some edge e with left endpoint n on a vertical gridline. We similarly extend rightwards, giving us a sequence of edges e , e , . . . , e , . . . , e such that the n n − 1 0 − m right endpoint of e and the left endpoint of e lie on the same vertical line, and the left endpoint of i i − 1 e and the right endpoint of e both lie on vertical gridlines. Let d be the distance between these n − m √ two gridlines; the total length of the edges in the sequence must be d 2. Every edge has length 1 or √ √ 2, so the total length can also be written in the form a + b 2; since e has length 1, we know a > 0. 0 √ √ Then a = ( d − b ) 2, contradicting the irrationality of 2. √ Thus, every edge of length 1 is horizontal or vertical, so every edge of length 2 is diagonal. Hence, for any diagonal edge, the vertical lines through its two endpoints are separated by a distance of 1. If any diagonal edge had endpoints not on vertical gridlines, then constructing the same sequence of edges as before would not terminate, as for all i , ℓ would be exactly one unit left of ℓ . Thus, every i i − 1 diagonal edge has both endpoints on vertical gridlines. A similar argument shows both endpoints are on horizontal gridlines as well. Thus, for any triangle in √ the division, its edge of length 2 is the diagonal of some cell of the grid. Thus, the triangle is half of this cell.