2023 AMC 10A Problem 17

Each of the three triangles is right-angled at a corner, so $AP^2 = 30^2 + BP^2$ and $QA^2 = 28^2 + DQ^2$ must be perfect squares

Use the triples $30$-$16$-$34$ and $21$-$28$-$35;$ then $PQ$ comes out an integer too

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 $PQ = \sqrt{(28 - p)^2 + q^2}$. For $0 \leq p \leq 28$, the equation $(AP-p)(AP+p)=900$ gives only $p=0$ and $p=16$, with $AP=30$ and $AP=34$. Similarly, setting $x=30-q$, the equation $(QA-x)(QA+x)=784$ with $0 \leq x \leq 30$ gives $x=0$ or $x=21$. Testing these four combinations in the formula for $PQ$, only $p=16$, $x=21$ works. Thus $q=9$, $QA=35$, and $PQ=\sqrt{12^2+9^2}=15$. The perimeter of $\triangle APQ$ is $34+15+35=84$. Thus, A is the correct answer.