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

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<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>
  
−
==Solution==
+
==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.  
 
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==
 +
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.
 +
 
 +
-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 ==
 +
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==
 
==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 06:33, 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

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