2023 AMC 10A Problems/Problem 17

Problem

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 $\triangle{APQ}$ ?

$\text{A) } 84 \qquad \text{B) } 86 \qquad \text{C) } 88 \qquad \text{D) } 90 \qquad \text{E) } 92$

Solution

[insert asy diagram]

Using knowledge of common Pythagorean triples and guess and check, we can find that $\triangle{ABP}$ is a $8$ - $15$ - $17$ triangle with side lengths $16$ - $30$ - $34$ , and $\triangle{PCQ}$ and $\triangle{QDA}$ are $3$ - $4$ - $5$ triangles with side lengths $9$ - $12$ - $15$ and $21$ - $28$ - $35$ , respectively.

Adding up the side lengths of $\triangle{APQ}$ gives $34+15+35=\boxed{\textbf{(A) } 84}.$

~ItsMeNoobieboy

2023 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2023 AMC 10A Problems/Problem 17

Problem

Solution

See Also