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Harvard-MIT Mathematics Tournament

Geometry Difficulty 4.8 AIME Prove it United States

Problem:

The four sides of quadrilateral $ABCD$ are equal in length. Determine the perimeter of $ABCD$ given that it has area $120$ and $AC = 10$ .

Solution

Solution:

Let $M$ be the midpoint of $AC$ . Then triangles $AMB$ , $BMC$ , $CMD$ , and $DMA$ are all right triangles having legs $5$ and $h$ for some $h$ .

The area of $ABCD$ is $120$ , but also $4 \cdot \left(\frac{1}{2} \cdot 5 \cdot h\right) = 10h$ , so $h = 12$ .

Then $AB = BC = CD = DA = \sqrt{12^2 + 5^2} = 13$ , and the perimeter of $ABCD$ is $52$ .

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