February, 2017

Geometry Difficulty 4.8 AIME Prove it United States

Problem:
Let $ABCD$ be a quadrilateral with side lengths $AB=2$ , $BC=3$ , $CD=5$ , and $DA=4$ . What is the maximum possible radius of a circle inscribed in quadrilateral $ABCD$ ?

Solution

Solution:
Let the tangent lengths be $a, b, c, d$ so that
$\begin{aligned} & a+b=2 \\ & b+c=3 \\ & c+d=5 \\ & d+a=4 \end{aligned}$
Then $b=2-a$ and $c=1+a$ and $d=4-a$ . The radius of the inscribed circle of quadrilateral $ABCD$ is given by
$\sqrt{\frac{abc+abd+acd+bcd}{a+b+c+d}} = \sqrt{\frac{-7a^2+16a+8}{7}}$
This is clearly maximized when $a=\frac{8}{7}$ which leads to a radius of $\sqrt{\frac{120}{49}}=\frac{2\sqrt{30}}{7}$ .

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