2021 JMPSC Invitationals Problems/Problem 10

Problem

A point $P$ is chosen in isosceles trapezoid $ABCD$ with $AB=4$ , $BC=20$ , $CD=28$ , and $DA=20$ . If the sum of the areas of $PBC$ and $PDA$ is $144$ , then the area of $PAB$ can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime. Find $m+n.$

Solution

t is implied $P$ lies on the line that bisects $AB$ and $DC$ . We have the area of the trapezoid is $16 \cdot 16=256$ since the height is $16$ . Now, subtracting $144$ we have $224=4x+28(16-x)$ for $x$ is the height of $\triangle PAB$ . This means $x=\frac{28}{3}$ , asserting the area of $\triangle PAB$ is $\frac{56}{3} \implies 56+3=\boxed{59}.$ ~Geometry285

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2021 JMPSC Invitationals Problems/Problem 10

Problem

Solution

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