Difference between revisions of "2023 AMC 10A Problems/Problem 17"

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==Problem==
 
Let <math>ABCD</math> be a rectangle with <math>AB = 30</math> and <math>BC = 28</math>. Point <math>P</math> and <math>Q</math> lie on <math>\overline{BC}</math> and <math>\overline{CD}</math> respectively so that all sides of <math>\triangle{ABP}, \triangle{PCQ},</math> and <math>\triangle{QDA}</math> have integer lengths. What is the perimeter of <math>\triangle{APQ}</math>?
 
Let <math>ABCD</math> be a rectangle with <math>AB = 30</math> and <math>BC = 28</math>. Point <math>P</math> and <math>Q</math> lie on <math>\overline{BC}</math> and <math>\overline{CD}</math> respectively so that all sides of <math>\triangle{ABP}, \triangle{PCQ},</math> and <math>\triangle{QDA}</math> have integer lengths. What is the perimeter of <math>\triangle{APQ}</math>?
  

Revision as of 21:42, 9 November 2023

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

See Also

2023 AMC 10A (ProblemsAnswer KeyResources)
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

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