Difference between revisions of "2023 AMC 10A Problems/Problem 17"
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<math>\triangle{APQ}</math> has side lengths <math>34, 35,</math> and <math>15,</math> so the perimeter is equal to <math>34 + 35 + 15 = \boxed{\textbf{(A) } 84}.</math> | <math>\triangle{APQ}</math> has side lengths <math>34, 35,</math> and <math>15,</math> so the perimeter is equal to <math>34 + 35 + 15 = \boxed{\textbf{(A) } 84}.</math> | ||
− | ~ Gabe Horn | + | ~ Gabe Horn ~ItsMeNoobieboy |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2023|ab=A|num-b=16|num-a=18}} | {{AMC10 box|year=2023|ab=A|num-b=16|num-a=18}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 21:26, 9 November 2023
Problem
Let be a rectangle with and . Point and lie on and respectively so that all sides of and have integer lengths. What is the perimeter of ?
Solution
We know that all side lengths are integers, so we can test Pythagorean triples for all triangles.
First, we focus on . The length of is , and the possible Pythagorean triples can be are , where the value of one leg is a factor of . Testing these cases, we get that only is a valid solution because the other triangles result in another leg that is greater than , the length of . Thus, we know that and .
Next, we move on to . The length of is , and the possible triples are and . Testing cases again, we get that is our triple. We get the value of , and .
We know that which is , and which is . is therefore a right triangle with side length ratios , and the hypotenuse is equal to . has side lengths and so the perimeter is equal to
~ Gabe Horn ~ItsMeNoobieboy
See Also
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 |
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