2023 AMC 10A Problems/Problem 17
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
~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 | ||
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