Maths Olympiad Prep

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, 2025

Geometry Difficulty 5.8 AIME, harder Prove it United States

Problem:

In a two-dimensional cave with a parallel floor and ceiling, two stalactites of lengths 1616 and 3636 hang perpendicularly from the ceiling, while two stalagmites of heights 2525 and 4949 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.

Solution

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 coincide, we know that it is 16+362=26\frac{16 + 36}{2} = 26 from the ceiling due to the stalactites, and 25+492=37\frac{25 + 49}{2} = 37 from the ground due to the stalagmites. Therefore the height of the cave is just the sum of these two values, i.e. 6363.

Figure 1

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.