Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2023 AMC 10A Problems/Problem 17"

m (→‎Video Solution by CosineMethod [🔥Fast and Easy🔥])
(→‎Video Solution by Math-X (First fully understand the problem!!!))
 
(4 intermediate revisions by 4 users not shown)
Line 3: Line 3:
  
 
<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>
−
==Video Solution==
 
−
 
−
https://www.youtube.com/watch?v=RiUlGz-p-LU&t=71s
 
  
 
==Solution 1==
 
==Solution 1==
Line 39: Line 36:
 
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 value of one leg is a factor of <math>30</math>. Testing these cases, we get that only <math>(8, 15, 17)</math> is a valid solution because the other triangles result in another leg that is greater than <math>28</math>, the length of <math>\overline{BC}</math>. Thus, we know that <math>BP = 16</math> and <math>AP = 34</math>.  
+
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 possible triples are <math>(3, 4, 5)</math> and <math>(7, 24, 25)</math>. Testing cases 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>.
+
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 <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.
+
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 Little Fermat==
 +
https://youtu.be/h2Pf2hvF1wE?si=FPZ1D4hdcd1QGUK2&t=3834
 +
~little-fermat
 +
==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 ==
 
==Video Solution ==
Line 60: Line 71:
 
==Video Solution by OmegaLearn==
 
==Video Solution by OmegaLearn==
 
https://youtu.be/xlFDMuoOd5Q?si=nCVTriSViqfHA2ju
 
https://youtu.be/xlFDMuoOd5Q?si=nCVTriSViqfHA2ju
 +
==Video Solution==
 +
 +
https://www.youtube.com/watch?v=RiUlGz-p-LU&t=71s
  
 
== Video Solution by CosineMethod ==
 
== Video Solution by CosineMethod ==

Latest revision as of 06:33, 5 November 2024

Problem

Let $ABCD$ be a rectangle with $AB = 30$ and $BC = 28$. Point $P$ and $Q$ lie on $\overline{BC}$ and $\overline{CD}$ respectively so that all sides of $\triangle{ABP}, \triangle{PCQ},$ and $\triangle{QDA}$ have integer lengths. What is the perimeter of $\triangle{APQ}$?

$\textbf{(A) } 84 \qquad \textbf{(B) } 86 \qquad \textbf{(C) } 88 \qquad \textbf{(D) } 90 \qquad \textbf{(E) } 92$

Solution 1

[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.

First, we focus on $\triangle{ABP}$. The length of $AB$ is $30$, and the possible (small enough) Pythagorean triples $\triangle{ABP}$ can be are $(3, 4, 5), (5, 12, 13), (8, 15, 17),$ where the length of the longer leg is a factor of $30$. Testing these, we get that only $(8, 15, 17)$ is a valid solution. Thus, we know that $BP = 16$ and $AP = 34$.

Next, we move on to $\triangle{QDA}$. The length of $AD$ is $28$, and the small enough triples are $(3, 4, 5)$ and $(7, 24, 25)$. Testing again, we get that $(3, 4, 5)$ is our triple. We get the value of $DQ = 21$, and $AQ = 35$.

We know that $CQ = CD - DQ$ which is $9$, and $CP = BC - BP$ which is $12$. $\triangle{CPQ}$ is therefore a right triangle with side length ratios ${3, 4, 5}$, and the hypotenuse is equal to $15$. $\triangle{APQ}$ has side lengths $34, 35,$ and $15,$ so the perimeter is equal to $34 + 35 + 15 = \boxed{\textbf{(A) } 84}.$

~Gabe Horn ~ItsMeNoobieboy ~oinava

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 $\boxed{\textbf{(A) } 84}.$

If you want to see a video solution on this solution, look at Video Solution 1.

-paixiao (minor edits-HW73)

Video Solution by Little Fermat

https://youtu.be/h2Pf2hvF1wE?si=FPZ1D4hdcd1QGUK2&t=3834 ~little-fermat

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

~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

https://youtu.be/yxfRjwQ8_KM

~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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_10A_Problems/Problem_17&oldid=231693"