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"

(Added a solution)
(→‎Problem)
(24 intermediate revisions by 15 users not shown)
Line 1: Line 1:
 +
==Problem==
 
Let <math>ABCD</math> be a rectangle with <math>AB = 30</math> and <math>BC = 28</math>. Point <math>P</math> and <math>Q</math> lie on <math>\overline{BC}</math> and <math>\overline{CD}</math> respectively so that all sides of <math>\triangle{ABP}, \triangle{PCQ},</math> and <math>\triangle{QDA}</math> have integer lengths. What is the perimeter of <math>\triangle{APQ}</math>?
 
Let <math>ABCD</math> be a rectangle with <math>AB = 30</math> and <math>BC = 28</math>. Point <math>P</math> and <math>Q</math> lie on <math>\overline{BC}</math> and <math>\overline{CD}</math> respectively so that all sides of <math>\triangle{ABP}, \triangle{PCQ},</math> and <math>\triangle{QDA}</math> have integer lengths. What is the perimeter of <math>\triangle{APQ}</math>?
  
 +
<math>\textbf{(A) } 84 \qquad \textbf{(B) } 86 \qquad \textbf{(C) } 88 \qquad \textbf{(D) } 90 \qquad \textbf{(E) } 92</math>
 +
==Video Solution by MegaMath in 1 minute==
  
−
<math>\text{A) } 84 \qquad \text{B) } 86 \qquad \text{C) } 88  \qquad \text{D) } 90 \qquad  \text{E) } 92</math>
+
https://www.youtube.com/watch?v=RiUlGz-p-LU&t=71s
  
−
==Solution==
+
==Solution 1==
−
[insert asy diagram]
+
<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>
  
−
Using knowledge of common Pythagorean triples and guess and check, we can find that <math>\triangle{ABP}</math> is a <math>8</math>-<math>15</math>-<math>17</math> triangle with side lengths <math>16</math>-<math>30</math>-<math>34</math>, and <math>\triangle{PCQ}</math> and <math>\triangle{QDA}</math> are <math>3</math>-<math>4</math>-<math>5</math> triangles with side lengths <math>9</math>-<math>12</math>-<math>15</math> and <math>21</math>-<math>28</math>-<math>35</math>, respectively.
+
We know that all side lengths are integers, so we can test Pythagorean triples for all triangles.  
  
−
Adding up the side lengths of <math>\triangle{APQ}</math> gives <math>34+15+35=\boxed{\textbf{(A) } 84}.</math>
+
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>.
  
−
~ItsMeNoobieboy
+
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>.
 +
 
 +
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>
 +
 
 +
~Gabe Horn ~ItsMeNoobieboy
 +
 
 +
==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.
 +
 
 +
If you want to see a video solution on this solution, look at Video Solution 1.
 +
 
 +
-paixiao
 +
 
 +
==Video Solution by Power Solve (easy to digest!)==
 +
https://www.youtube.com/watch?v=bN7Ly70nw_M
 +
 
 +
==Video Solution by OmegaLearn==
 +
https://youtu.be/xlFDMuoOd5Q?si=nCVTriSViqfHA2ju
 +
 
 +
== Video Solution by CosineMethod [🔥Fast and Easy🔥]==
 +
 
 +
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}}

Revision as of 13:18, 24 February 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 $\triangle{APQ}$?

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

Video Solution by MegaMath in 1 minute

https://www.youtube.com/watch?v=RiUlGz-p-LU&t=71s

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 Pythagorean triples $\triangle{ABP}$ can be are $(3, 4, 5), (5, 12, 13), (8, 15, 17),$ where the value of one leg is a factor of $30$. Testing these cases, we get that only $(8, 15, 17)$ is a valid solution because the other triangles result in another leg that is greater than $28$, the length of $\overline{BC}$. Thus, we know that $BP = 16$ and $AP = 34$.

Next, we move on to $\triangle{QDA}$. The length of $AD$ is $28$, and the possible triples are $(3, 4, 5)$ and $(7, 24, 25)$. Testing cases 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

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.

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

-paixiao

Video Solution by Power Solve (easy to digest!)

https://www.youtube.com/watch?v=bN7Ly70nw_M

Video Solution by OmegaLearn

https://youtu.be/xlFDMuoOd5Q?si=nCVTriSViqfHA2ju

Video Solution by CosineMethod [🔥Fast and Easy🔥]

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=216180"