Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.2 AIME, harder Prove it Croatia

Let $S$ and $T$ be points inside the triangle $ABC$ . The distance from $S$ to lines $AB$ , $BC$ and $CA$ is $10$ , $7$ and $4$ , respectively. The distance from $T$ to these lines is $4$ , $10$ and $16$ , respectively.
Determine the radius of the incircle of triangle $ABC$ .

Solution

Denote $a = |BC|$ , $b = |CA|$ , $c = |AB|$ . Let $P$ be the area, $s = \frac{1}{2}(a + b + c)$ the semiperimeter, and $r$ the radius of the incircle of triangle $ABC$ .
We can divide triangle $ABC$ into three smaller triangles by connecting point $S$ to vertices $A$ , $B$ and $C$ .

By adding the areas of triangles $ABS$ , $BCS$ and $CAS$ , we get the area of triangle $ABC$ , so we have
$2P = 10c + 7a + 4b.$
Analogously, by observing point $T$ we get
$2P = 4c + 10a + 16b.$
If we multiply the first equality by $2$ and add it to the second equality, we get
$6P = 24(a + b + c) = 48s, \quad \text{i.e. } P = 8s.$
Since $P = rs$ , we have $r = 8$ .

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