Difference between revisions of "2023 AMC 10A Problems/Problem 17"
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==Solution== | ==Solution== | ||
+ | <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> | ||
+ | |||
We know that all side lengths are integers, so we can test Pythagorean triples for all triangles. | We know that all side lengths are integers, so we can test Pythagorean triples for all triangles. | ||
Revision as of 10:27, 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.