2023 AMC 10A Problem 17

All three triangles are right triangles (each has a corner of the rectangle), and every side must be a whole number. So we need Pythagorean triples, and we can go around the rectangle one triangle at a time.

1. Triangle $ABP$ has legs $AB = 30$ and $BP$, where $BP$ is at most $28$ because $P$ is on side $BC$. The triple $8$-$15$-$17$ doubled is $16$-$30$-$34$, and that is the only right triangle with whole sides, one leg $30$ and the other leg at most $28$. (A leg of $30$ also appears in $30$-$40$-$50$, but $40$ is too long. If you want to be sure, $30^2 + b^2$ is a perfect square for no other whole number $b$ from $1$ to $28$.) So $BP = 16$ and $AP = 34$.
2. That leaves $PC = 28 - 16 = 12$.
3. Triangle $PCQ$ has legs $PC = 12$ and $CQ$, where $CQ$ is at most $30$. Right triangles with a leg of $12$:
- $5$-$12$-$13$, so $CQ = 5$;
- $9$-$12$-$15$ (that is $3$-$4$-$5$ times $3$), so $CQ = 9$;
- $12$-$16$-$20$ (that is $3$-$4$-$5$ times $4$), so $CQ = 16$.

($12$-$35$-$37$ has $35 > 30$, too long.)
4. Triangle $QDA$ has legs $DA = 28$ and $DQ = 30 - CQ$. Test each option:
- $CQ = 5$: $DQ = 25$, and $28^2 + 25^2 = 1409$, not a perfect square.
- $CQ = 9$: $DQ = 21$, and $28^2 + 21^2 = 784 + 441 = 1225 = 35^2$. ✓ (This is $3$-$4$-$5$ times $7$.)
- $CQ = 16$: $DQ = 14$, and $28^2 + 14^2 = 980$, not a perfect square.
5. So $CQ = 9$, $PQ = 15$ and $AQ = 35$.
6. The perimeter of triangle $APQ$ is $AP + PQ + AQ = 34 + 15 + 35 = 84$.

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

The three triangles are linked: the leg left over from one triangle becomes a leg of the next. So you can search in order, and each step leaves only a short list of options for the next step to test. Knowing the common triples ($3$-$4$-$5$, $5$-$12$-$13$, $8$-$15$-$17$, $7$-$24$-$25$) and their multiples turns each step into a quick lookup.

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 $\boxed{\textbf{(A)}\ 84}$.