HMMT 二月 2025 · 几何 · 第 2 题

HMMT February 2025 — Geometry — Problem 2

In a two-dimensional cave with a parallel floor and ceiling, two stalactites of lengths 16 and 36 hang perpendicularly from the ceiling, while two stalagmites of heights 25 and 49 grow perpendicularly from the ground. If the tips of these four structures form the vertices of a square in some order, compute the height of the cave.

解析

In a two-dimensional cave with a parallel floor and ceiling, two stalactites of lengths 16 and 36 hang perpendicularly from the ceiling, while two stalagmites of heights 25 and 49 grow perpendicularly from the ground. If the tips of these four structures form the vertices of a square in some order, compute the height of the cave. Proposed by: Albert Wang Answer: 63 Solution: Note that the difference in heights between the two stalactites does not equal the difference in heights between the stalagmites. This tells us that the two stalactites form a pair of opposite vertices of the square, and likewise for the stalagmites. As the midpoint of the pairs of structures must then 16+36 25+49 coincide, we know that it is = 26 from the ceiling due to the stalactites, and = 37 from 2 2 the ground due to the stalagmites. Therefore the height of the cave is just the sum of these two values, i.e. 63 .