Question

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

$84$

$86$

$88$

$90$

$92$

Solution:

Set $A = (0,0),$ $B = (30,0),$ $C = (30,28),$ $D = (0,28),$ with $P = (30, p)$ on $BC$ and $Q = (30 - q, 28)$ on $CD.$ The three right triangles give $AP = \sqrt{30^2 + p^2},$ $QA = \sqrt{28^2 + (30 - q)^2},$ and $PQ = \sqrt{(28 - p)^2 + q^2},$ and all must be integers. Hunt for Pythagorean triples: $p = 16$ makes $AP = 34,$ and $30 - q = 21$ (so $q = 9$ ) makes $QA = 35.$ Then $PQ = \sqrt{12^2 + 9^2} = 15,$ an integer too. So the perimeter of $\triangle APQ$ is $34 + 15 + 35 = 84.$ Thus, A is the correct answer.

2023 AMC 10A Problem 17

Solution:

Problem 17 in Other Years