Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.9 AIME Prove it United States

Problem:
Let $T$ be a triangle with side lengths $26$ , $51$ , and $73$ . Let $S$ be the set of points inside $T$ which do not lie within a distance of $5$ of any side of $T$ . Find the area of $S$ .

Solution

Solution:
Note that the sides of $S$ are parallel to the sides of $T$ , so $S$ is a triangle similar to $T$ .

The semiperimeter of $T$ is $s = \frac{1}{2}(26 + 51 + 73) = 75$ .

By Heron's formula, the area of $T$ is $\sqrt{75 \cdot 49 \cdot 24 \cdot 2} = 420$ .

If $r$ is the inradius of $T$ , then the area of $T$ is $r s$ , so $r = 420 / 75 = 28 / 5$ .

It follows that the inradius of $S$ is $r - 5 = 3 / 5$ , and the ratio of similitude between $S$ and $T$ is $3 / 28$ .

Therefore, the area of $S$ is $420 \cdot (3 / 28)^2 = 135 / 28$ .

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