2023 AMC 10A Problems/Problem 17
Contents
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 ?
Video Solution
https://www.youtube.com/watch?v=RiUlGz-p-LU&t=71s
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 (small enough) Pythagorean triples can be are where the length of the longer leg is a factor of . Testing these, we get that only is a valid solution. Thus, we know that and .
Next, we move on to . The length of is , and the small enough triples are and . Testing 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 ~oinava
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
If you want to see a video solution on this solution, look at Video Solution 1.
-paixiao (minor edits-HW73)
Video Solution 🚀 2 min solve 🚀
~Education, the Study of Everything
Video Solution
https://www.youtube.com/watch?v=bN7Ly70nw_M
Video Solution by OmegaLearn
https://youtu.be/xlFDMuoOd5Q?si=nCVTriSViqfHA2ju
Video Solution by CosineMethod
https://www.youtube.com/watch?v=r8Wa8OrKiZI
Video Solution 1
https://www.youtube.com/watch?v=eO_axHSmum4
-paixiao
VIdeo Solution 2
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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.