HMMT 十一月 2025 · 冲刺赛 · 第 11 题
HMMT November 2025 — Guts Round — Problem 11
题目详情
- [8] Jessica has a non-square rectangular sheet of paper with all 4 corners colored differently. She repeats the following process 8 times: she picks one of the rectangle’s two axes of symmetry, then flips the rectangle over that axis. Compute the number of ways she can do this so that each corner ends up in its original position. ◦
解析
- [8] Jessica has a non-square rectangular sheet of paper with all 4 corners colored differently. She repeats the following process 8 times: she picks one of the rectangle’s two axes of symmetry, then flips the rectangle over that axis. Compute the number of ways she can do this so that each corner ends up in its original position. Proposed by: Kira Lewis Answer: 128 Solution 1: Let the corners be A , B , C and D in that order around the rectangle. Each time the sheet is flipped, the side of the sheet facing upward changes. As there are an even number of flips, the side of the sheet facing upward is the same as the original side. Hence, if A ends up in its original corner, then all the other corners will also end up in their original places. During each flip, A will move to one of the corners adjacent to it previous position. In 7 flips, A will end up in the original position of either B or D . After this, there is always a unique way to flip the rectangle so that A ends up in its original position. Thus, there are 2 choices for each of the first seven 7 flips, and the last flip is uniquely determined. This leads to a total of 2 ways to flip the rectangle so that it ends up in the original position, and so the answer is 128 . Solution 2: Label two corners of the rectangle on the left “ L ” and the other two corners on the right “ R ”. Label the horizontal axis as x , vertical axis as y . When the flipping is over y axis, L -corners change the side from left to right or from right to left. When the flipping is over x axis, the L -corners do not change left-right side. Therefore, for each corner to end up in its original position, the axis y should be chosen even times, so the axis x must be chosen even times too. On the other hand, suppose that each of axis x and y is chosen for even times. Applying the same trick to each axis x , y separately, we can conclude that each corner ends up in the same side with respect to both axes. (that is, if a corner starts above x axis, it ends up above x axis. If it starts on the left to y axis, it ends up to the left of y axis). Consequently, each corner ends up at the start position. As a result, the given condition in the problem is true if and only if axes x , y are each selected for even numbers of time. The number of all such combinations is 8 8 8 8 8
-
-
-
- = 128 . 8 6 4 2 0 ◦
-
-