2023 AMC 10A Problems/Problem 17
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Video Solution by Little Fermat
- 5 Video Solution by Math-X (First fully understand the problem!!!)
- 6 Video Solution 🚀 2 min solve 🚀
- 7 Video Solution
- 8 Video Solution by OmegaLearn
- 9 Video Solution
- 10 Video Solution by CosineMethod
- 11 Video Solution 1
- 12 VIdeo Solution 2
- 13 See Also
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
![[asy] /* ~ItsMeNoobieboy */ size(200); pair A, B, C, D, P, Q; A = (0,28/30); B = (1,28/30); C = (1,0); D = (0,0); P = (1,12/30); Q = (21/30,0); draw(A--B--C--D--cycle); draw(A--P--Q--cycle); dot("$A$",A,NW,linewidth(4)); dot("$B$",B,NE,linewidth(4)); dot("$C$",C,SE,linewidth(4)); dot("$D$",D,SW,linewidth(4)); dot("$P$",P,E,linewidth(4)); dot("$Q$",Q,S,linewidth(4)); label("$30$",midpoint(A--B),N); label("$16$",midpoint(B--P),E); label("$34$",midpoint(A--P),NE, red); label("$28$",midpoint(A--D),W); label("$21$",midpoint(D--Q),S); label("$35$",midpoint(A--Q),SW, red); label("$9$",midpoint(Q--C),S); label("$12$",midpoint(C--P),E); label("$15$",midpoint(Q--P),SE, red); [/asy]](http://latex.artofproblemsolving.com/4/1/f/41fc6aee278a2183fa354f7d7f71a911b18f7b29.png)
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 by Little Fermat
https://youtu.be/h2Pf2hvF1wE?si=FPZ1D4hdcd1QGUK2&t=3834 ~little-fermat
Video Solution by Math-X (First fully understand the problem!!!)
https://youtu.be/GP-DYudh5qU?si=KfPIbpJjHwZQTyH4&t=4479
~Math-X
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
https://www.youtube.com/watch?v=RiUlGz-p-LU&t=71s
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 | ||
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