Difference between revisions of "2023 AMC 10A Problems/Problem 17"
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<math>\textbf{(A) } 84 \qquad \textbf{(B) } 86 \qquad \textbf{(C) } 88 \qquad \textbf{(D) } 90 \qquad \textbf{(E) } 92</math> | <math>\textbf{(A) } 84 \qquad \textbf{(B) } 86 \qquad \textbf{(C) } 88 \qquad \textbf{(D) } 90 \qquad \textbf{(E) } 92</math> | ||
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+ | 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>? | ||
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+ | <math>\text{A) } 84 \qquad \text{B) } 86 \qquad \text{C) } 88 \qquad \text{D) } 90 \qquad \text{E) } 92</math> | ||
==Solution== | ==Solution== | ||
− | + | We know that all side lengths are integers, so we can test Pythagorean triples for all triangles. | |
+ | |||
+ | First, we focus on <math>\triangle{ABP}</math>. The length of <math>AB</math> is <math>30</math>, and the possible Pythagorean triples <math>\triangle{ABP}</math> can be are <math>(3, 4, 5), (5, 12, 13), (8, 15, 17)</math>, where the value of one leg is a factor of <math>30</math>. Testing these cases, we get that only <math>(8, 15, 17)</math> is a valid solution because the other triangles result in another leg that is greater than <math>28</math>, the length of <math>\overline{BC}</math>. Thus, we know that <math>BP = 16</math> and <math>AP = 34</math>. | ||
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+ | Next, we move on to <math>\triangle{QDA}</math>. The length of <math>AD</math> is <math>28</math>, and the possible triples are <math>(3, 4, 5)</math> and <math>(7, 24, 25)</math>. Testing cases again, we get that <math>(3, 4, 5)</math> is our triple. We get the value of <math>DQ = 21</math>, and <math>AQ = 35</math>. | ||
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+ | We know that <math>CQ = CD - DQ</math> which is <math>9</math>, and <math>CP = BC - BP</math> which is <math>12</math>. <math>\triangle{CPQ}</math> is therefore a right triangle with side length ratios <math>{3, 4, 5}</math>, and the hypotenuse is equal to <math>15</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 | |
− | + | ==See Also== | |
+ | {{AMC10 box|year=2023|ab=A|num-b=16|num-a=18}} | ||
+ | {{MAA Notice}} | ||
− | |||
==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 20:49, 9 November 2023
Contents
[hide]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 ?
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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.