2023 AMC 10A Problem 17

P16-20~5 minPrint

Triangles: Area & Pythagorean Diophantine Equations Organized listing / direct enumeration Split into exhaustive cases and handle each

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 $\triangle{APQ}$ ?

Show solution without answering

Key insight

Go around the rectangle one right triangle at a time. Leg 30 forces 16-30-34; then leg 12 allows 5, 9 or 16, and only 9 makes the last triangle work.

Solution

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

Why this works

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.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2)
Introduction to Number Theory, Chapter 7: Algebra With Integers
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.2)
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.2)