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

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(→‎Video Solution by Math-X (First fully understand the problem!!!))
 
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A rhombic dodecahedron is a solid with <math>12</math> congruent rhombus faces. At every vertex, <math>3</math> or <math>4</math> edges meet, depending on the vertex. How many vertices have exactly <math>3</math> edges meet?
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==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>?
  
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<math>\textbf{(A) }5\qquad\textbf{(B) }6\qquad\textbf{(C) }7\qquad\textbf{(D) }8\qquad\textbf{(E) }9</math>
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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>
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==Solution==
 
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We see there are <math>\frac{12 \cdot 4}{2}</math> edges. We have by Euler's Polyhedral Formula, <math>V-E+F=2</math> meaning <math>V-24+12=2</math> or <math>V=14</math>. Let there be <math>a</math> vertices that have <math>3</math> edges meeting and <math>b</math> vertices that have <math>4</math> edges meeting. Hence, <cmath>a+b=14</cmath>
 
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<cmath>3a+4b=48</cmath>
 
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We find <math>b=6</math> and <math>a=8</math>, hence the answer is <math>8</math>.
 
  
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~SirAppel
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==Solution 1==
 +
<asy>
 +
/* ~ItsMeNoobieboy */
 +
size(200);
 +
pair A, B, C, D, P, Q;
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A = (0,28/30);
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B = (1,28/30);
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C = (1,0);
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D = (0,0);
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P = (1,12/30);
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Q = (21/30,0);
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draw(A--B--C--D--cycle);
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draw(A--P--Q--cycle);
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dot("$A$",A,NW,linewidth(4));
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dot("$B$",B,NE,linewidth(4));
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dot("$C$",C,SE,linewidth(4));
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dot("$D$",D,SW,linewidth(4));
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dot("$P$",P,E,linewidth(4));
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dot("$Q$",Q,S,linewidth(4));
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label("$30$",midpoint(A--B),N);
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label("$16$",midpoint(B--P),E);
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label("$34$",midpoint(A--P),NE, red);
 +
label("$28$",midpoint(A--D),W);
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label("$21$",midpoint(D--Q),S);
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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 <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 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>.
 +
<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 ~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==
 +
{{AMC10 box|year=2023|ab=A|num-b=16|num-a=18}}
 +
{{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

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