Difference between revisions of "2023 AMC 10A Problems/Problem 17"
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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. | ||
− | First, we focus on <math>\triangle{ABP}</math>. The length of <math>AB</math> is <math>30</math>, and the possible Pythagorean triples <math>\triangle{ABP}</math> can be are <math>(3, 4, 5), (5, 12, 13), (8, 15, 17),</math> where the | + | First, we focus on <math>\triangle{ABP}</math>. The length of <math>AB</math> is <math>30</math>, and the possible (small enough) Pythagorean triples <math>\triangle{ABP}</math> can be are <math>(3, 4, 5), (5, 12, 13), (8, 15, 17),</math> where the length of the longer leg is a factor of <math>30</math>. Testing these, we get that only <math>(8, 15, 17)</math> is a valid solution. Thus, we know that <math>BP = 16</math> and <math>AP = 34</math>. |
− | Next, we move on to <math>\triangle{QDA}</math>. The length of <math>AD</math> is <math>28</math>, and the | + | Next, we move on to <math>\triangle{QDA}</math>. The length of <math>AD</math> is <math>28</math>, and the small enough triples are <math>(3, 4, 5)</math> and <math>(7, 24, 25)</math>. Testing again, we get that <math>(3, 4, 5)</math> is our triple. We get the value of <math>DQ = 21</math>, and <math>AQ = 35</math>. |
We know that <math>CQ = CD - DQ</math> which is <math>9</math>, and <math>CP = BC - BP</math> which is <math>12</math>. <math>\triangle{CPQ}</math> is therefore a right triangle with side length ratios <math>{3, 4, 5}</math>, and the hypotenuse is equal to <math>15</math>. | We know that <math>CQ = CD - DQ</math> which is <math>9</math>, and <math>CP = BC - BP</math> which is <math>12</math>. <math>\triangle{CPQ}</math> is therefore a right triangle with side length ratios <math>{3, 4, 5}</math>, and the hypotenuse is equal to <math>15</math>. | ||
<math>\triangle{APQ}</math> has side lengths <math>34, 35,</math> and <math>15,</math> so the perimeter is equal to <math>34 + 35 + 15 = \boxed{\textbf{(A) } 84}.</math> | <math>\triangle{APQ}</math> has side lengths <math>34, 35,</math> and <math>15,</math> so the perimeter is equal to <math>34 + 35 + 15 = \boxed{\textbf{(A) } 84}.</math> | ||
− | ~Gabe Horn ~ItsMeNoobieboy | + | ~Gabe Horn ~ItsMeNoobieboy ~oinava |
==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 <math>\boxed{\textbf{(A) } 84}.</math> |
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 (minor edits-HW73) |
+ | |||
+ | ==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 🚀== | ||
+ | |||
+ | https://youtu.be/HVMlhdSurPw | ||
+ | |||
+ | <i> ~Education, the Study of Everything </i> | ||
+ | |||
+ | ==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== | ==Video Solution 1== | ||
https://www.youtube.com/watch?v=eO_axHSmum4 | https://www.youtube.com/watch?v=eO_axHSmum4 | ||
-paixiao | -paixiao | ||
+ | |||
+ | ==VIdeo Solution 2== | ||
+ | |||
+ | https://youtu.be/yxfRjwQ8_KM | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2023|ab=A|num-b=16|num-a=18}} | {{AMC10 box|year=2023|ab=A|num-b=16|num-a=18}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 06:23, 6 September 2024
Contents
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 (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 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.