2023 AMC 10A Problems/Problem 17

Problem

Let $ABCD$ be a rectangle with $AB = 30$ and $BC = 28$ . Point $P$ and $Q$ lie on $\overline{BC}$ and $\overline{CD}$ respectively so that all sides of $\triangle{ABP}, \triangle{PCQ},$ and $\triangle{QDA}$ have integer lengths. What is the perimeter of $\triangle{APQ}$ ?

$\textbf{(A) } 84 \qquad \textbf{(B) } 86 \qquad \textbf{(C) } 88 \qquad \textbf{(D) } 90 \qquad \textbf{(E) } 92$

Solution

We know that all side lengths are integers, so we can test Pythagorean triples for all triangles.

First, we focus on $\triangle{ABP}$ . The length of $AB$ is $30$ , and the possible Pythagorean triples $\triangle{ABP}$ can be are $(3, 4, 5), (5, 12, 13), (8, 15, 17),$ where the value of one leg is a factor of $30$ . Testing these cases, we get that only $(8, 15, 17)$ is a valid solution because the other triangles result in another leg that is greater than $28$ , the length of $\overline{BC}$ . Thus, we know that $BP = 16$ and $AP = 34$ .

Next, we move on to $\triangle{QDA}$ . The length of $AD$ is $28$ , and the possible triples are $(3, 4, 5)$ and $(7, 24, 25)$ . Testing cases again, we get that $(3, 4, 5)$ is our triple. We get the value of $DQ = 21$ , and $AQ = 35$ .

We know that $CQ = CD - DQ$ which is $9$ , and $CP = BC - BP$ which is $12$ . $\triangle{CPQ}$ is therefore a right triangle with side length ratios ${3, 4, 5}$ , and the hypotenuse is equal to $15$ . $\triangle{APQ}$ has side lengths $34, 35,$ and $15,$ so the perimeter is equal to $34 + 35 + 15 = \boxed{\textbf{(A) } 84}.$

~Gabe Horn ~ItsMeNoobieboy

2023 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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2023 AMC 10A Problems/Problem 17

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