Difference between revisions of "2023 AMC 10A Problems/Problem 17"
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==Problem== | ==Problem== | ||
− | + | 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>? | |
<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> |
Revision as of 02:26, 10 November 2023
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
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
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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