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2021 JMPSC Invitationals Problems/Problem 10

Revision as of 18:48, 23 September 2024 by Cyclicislscelestrapezoid (talk | contribs) (→‎Solution)
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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

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, edited by CyclicISLscelesTrapezoid

See also

  1. Other 2021 JMPSC Invitationals Problems
  2. 2021 JMPSC Invitationals Answer Key
  3. All JMPSC Problems and Solutions

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