2023 AMC 10A Problem 17

Triangle $ABP$ is right-angled at $B$ with legs $30$ and $BP \le 28$. Integer right triangles with a leg of $30$ are $16$-$30$-$34$, $30$-$40$-$50$, $30$-$72$-$78$, and $30$-$224$-$226$ (from $(y - x)(y + x) = 900$ with both factors even). Only $BP = 16$ fits, so $AP = 34$ and $PC = 28 - 16 = 12$.

Triangle $PCQ$ is right-angled at $C$ with legs $12$ and $CQ \le 30$. Triples with leg $12$: $5$-$12$-$13$, $9$-$12$-$15$, $16$-$12$-$20$ (also $35$-$12$-$37$, too long). So $CQ \in \{5, 9, 16\}$, giving $DQ = 30 - CQ \in \{25, 21, 14\}$.

Triangle $QDA$ is right-angled at $D$ with legs $28$ and $DQ$. Check: $28^2 + 25^2 = 1409$ (not a square), $28^2 + 21^2 = 1225 = 35^2$, $28^2 + 14^2 = 980$ (not a square). So $DQ = 21$, $CQ = 9$, $PQ = 15$, $AQ = 35$.

The perimeter of $\triangle APQ$ is $34 + 15 + 35 = 84$.

The answer is $\boxed{\textbf{(A)}\ 84}$.

Three integer right triangles chained around a rectangle share their legs through the fixed side lengths, so the search is sequential: each triangle's leg options are a short list that the next triangle prunes. Knowing the common triples ($3$-$4$-$5$, $5$-$12$-$13$, $8$-$15$-$17$, $7$-$24$-$25$) and their multiples makes each step a lookup instead of a computation.

• Stopping at the first Pythagorean triple for leg 12 (5-12-13) without checking that the third triangle is also integer-sided.
• Confusing which side is $30$ and which is $28$, or forgetting that $PC = 28 - BP$ and $DQ = 30 - CQ$.