Difference between revisions of "2023 AMC 10A Problems/Problem 17"
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==Solution 2== | ==Solution 2== | ||
− | Let BP=y and AP=z. We get 30^2+y^2=z^2. Subtracting y^2 on both sides, we get 30^2=z^2-y^2. Factoring, we get 30^2=(z-y)(z+y). Since y and z are integers, both z-y and z+y have to be even or both have to be odd. We also have y<31. We can pretty easily see now that z-y=18 and z+y=50. Thus, y=16 and z=34. We now get CP=12. We do the same trick again. Let DQ=a and AQ=b. Thus, 28^2=(b+a)(b-a). We can get b+a=56 and b-a=14. Thus, b=35 and a=21. We get CQ=9 and by the Pythagorean Theorem, we have PQ=15. We get AP+PQ+AQ=34+15+35=84. Our answer is A. | + | Let <math>BP=y</math> and <math>AP=z</math>. We get <math>30^{2}+y^{2}=z^{2}</math>. Subtracting <math>y^{2}</math> on both sides, we get <math>30^{2}=z^{2}-y^{2}</math>. Factoring, we get <math>30^{2}=(z-y)(z+y)</math>. Since <math>y</math> and <math>z</math> are integers, both <math>z-y</math> and <math>z+y</math> have to be even or both have to be odd. We also have <math>y<31</math>. We can pretty easily see now that <math>z-y=18</math> and <math>z+y=50</math>. Thus, <math>y=16</math> and <math>z=34</math>. We now get <math>CP=12</math>. We do the same trick again. Let <math>DQ=a</math> and <math>AQ=b</math>. Thus, <math>28^{2}=(b+a)(b-a)</math>. We can get <math>b+a=56</math> and <math>b-a=14</math>. Thus, <math>b=35</math> and <math>a=21</math>. We get <math>CQ=9</math> and by the Pythagorean Theorem, we have <math>PQ=15</math>. We get <math>AP+PQ+AQ=34+15+35=84</math>. Our answer is A. |
If you want to see a video solution on this solution, look at Video Solution 1. | If you want to see a video solution on this solution, look at Video Solution 1. | ||
-paixiao | -paixiao | ||
+ | |||
==Video Solution 1== | ==Video Solution 1== | ||
https://www.youtube.com/watch?v=eO_axHSmum4 | https://www.youtube.com/watch?v=eO_axHSmum4 |
Revision as of 19:41, 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 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
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.