2022 SSMO Speed Round Problems/Problem 4

Problem

Consider a quadrilateral $ABCD$ with area $120$ and satisfying $AB+CD=AD+BC=24$ . There exists a point $P$ in 3D space such that the distances from $P$ to $AB$ , $BC$ , $CD$ , and $DA$ are all equal to $13$ . Find the volume of $PABCD$ .

Solution

Let $r$ be the inradius of $ABCD$ . Note that $\frac{1}{2}r(AB+BC+CD+DA)=[ABCD]\implies r = \frac{2\cdot120}{24+24} = 5.$ We have $h^2+5^2 = 13^2\implies h = 12,$ meaning the volume of $PABCD$ is $\frac{1}{3}h[ABCD] = \frac{1}{3}(12)(120) = \boxed{480}.$

~pinkpig

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2022 SSMO Speed Round Problems/Problem 4

Problem

Solution