Difference between revisions of "2023 AMC 10A Problems/Problem 17"
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-paixiao | -paixiao | ||
− | ==VIdeo Solution== | + | ==VIdeo Solution 2== |
https://youtu.be/yxfRjwQ8_KM | https://youtu.be/yxfRjwQ8_KM | ||
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
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==Video Solution by TheBeautyofMath== | ==Video Solution by TheBeautyofMath== | ||
https://youtu.be/54dxZvk_W_I | https://youtu.be/54dxZvk_W_I |
Revision as of 19:36, 11 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 ?
Solution 1
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
Solution 2
Let and . We get . Subtracting on both sides, we get . Factoring, we get . Since and are integers, both and have to be even or both have to be odd. We also have . We can pretty easily see now that and . Thus, and . We now get . We do the same trick again. Let and . Thus, . We can get and . Thus, and . We get and by the Pythagorean Theorem, we have . We get . Our answer is A.
If you want to see a video solution on this solution, look at Video Solution 1.
-paixiao
Video Solution 1
https://www.youtube.com/watch?v=eO_axHSmum4
-paixiao
VIdeo Solution 2
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution by TheBeautyofMath
~IceMatrix
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.