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Difference between revisions of "2024 AMC 8 Problems/Problem 24"

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==Problem==
 
==Problem==
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==Solution 1==
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Jean has made a piece of stained glass art in the shape of two mountains, as shown in the figure below. One mountain peak is <math>8</math> feet high while the other peak is <math>12</math> feet high. Each peak forms a <math>90^\circ</math> angle, and the straight sides form a <math>45^\circ</math> angle with the ground. The artwork has an area of <math>183</math> square feet. The sides of the mountain meet at an intersection point near the center of the artwork, <math>h</math> feet above the ground. What is the value of <math>h?</math>
 +
 
 +
<asy>
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unitsize(.3cm);
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filldraw((0,0)--(8,8)--(11,5)--(18,12)--(30,0)--cycle,gray(0.7),linewidth(1));
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draw((-1,0)--(-1,8),linewidth(.75));
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draw((-1.4,0)--(-.6,0),linewidth(.75));
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draw((-1.4,8)--(-.6,8),linewidth(.75));
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label("$8$",(-1,4),W);
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label("$12$",(31,6),E);
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draw((-1,8)--(8,8),dashed);
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draw((31,0)--(31,12),linewidth(.75));
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draw((30.6,0)--(31.4,0),linewidth(.75));
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draw((30.6,12)--(31.4,12),linewidth(.75));
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draw((31,12)--(18,12),dashed);
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label("$45^{\circ}$",(.75,0),NE,fontsize(10pt));
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label("$45^{\circ}$",(29.25,0),NW,fontsize(10pt));
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draw((8,8)--(7.5,7.5)--(8,7)--(8.5,7.5)--cycle);
 +
draw((18,12)--(17.5,11.5)--(18,11)--(18.5,11.5)--cycle);
 +
draw((11,5)--(11,0),dashed);
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label("$h$",(11,2.5),E);
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</asy>
 +
 
 +
<math>\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 4\sqrt{2} \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 5\sqrt{2}</math>
 +
 
 +
==Solution1==
 +
Extend the "inner part" of the mountain so that the image is two right triangles that overlap in a third right triangle as shown.
 +
<asy>
 +
unitsize(.2cm);
 +
draw((0,0)--(8,8)--(11,5)--(18,12)--(30,0)--cycle,linewidth(1));
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draw((11,5)--(6, 0)--(16, 0)--cycle,linewidth(0.5));
 +
label("$8\sqrt{2}$",(4,4),NW);
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label("$12\sqrt{2}$",(24,6),NE);
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draw((8,8)--(7.5,7.5)--(8,7)--(8.5,7.5)--cycle);
 +
draw((18,12)--(17.5,11.5)--(18,11)--(18.5,11.5)--cycle);
 +
draw((11,5)--(11,0),dashed);
 +
label("$h$",(11,2.5),E);
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</asy>
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The side length of the largest right triangle is <math>12\sqrt{2},</math> which means its area is <math>144.</math> Similarly, the area of the second largest right triangle is <math>64</math> (the side length is <math>8\sqrt{2}</math>), and the area of the overlap is <math>h^2</math> (the side length is <math>h\sqrt{2}</math>). Thus,
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<cmath>144+64-h^2=183,</cmath>
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which means that the answer is <math>\boxed{\mathbf{(B)}\text{ 5}}.</math>
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 +
~BS201
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 +
==Solution 2 (Using a Ruler)==
 +
You can measure <math>h</math> with a ruler(rulers are allowed on the AMC 8), and see that <math>h</math> is closest to <math>\boxed{\mathbf{(B)}\text{ 5}}.</math>
 +
 
 +
~SID-DARTH-VATER
 +
 
 +
(Please note: This is not the best way, as diagrams are NOT DRAWN TO SCALE) ~Anonymous ~Continuous_Pi
 +
 
 +
 
 +
==Video Solution 1 by Math-X (First understand the problem!!!)==
 +
https://youtu.be/BaE00H2SHQM?si=793ml6R_qT79h-aI&t=7651
 +
 
 +
~Math-X
 +
 
 +
==Video Solution (A Clever Explanation You’ll Get Instantly)==
 +
https://youtu.be/5ZIFnqymdDQ?si=D8pGHQYbzJ_eMsEy&t=3678
 +
~hsnacademy
 +
 
 +
==Video Solution (Quick solution) by Parshwa==
 +
https://youtu.be/Ky-NVrOv4ew
 +
 
 +
==Video Solution by Power Solve (easy to understand)==
 +
https://www.youtube.com/watch?v=PlMGNmWIkBg
 +
 
 +
==Video Solution by OmegaLearn.org==
 +
https://youtu.be/KL_CGQrkXXo
 +
 
 +
==Video Solution by SpreadTheMathLove==
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https://www.youtube.com/watch?v=RiSt6_WLfrM
 +
 
 +
== Video Solution by NiuniuMaths (Easy to understand!) ==
 +
https://www.youtube.com/watch?v=8353ngt05N0
 +
 
 +
~NiuniuMaths
 +
 
 +
== Video Solution by CosineMethod [🔥Fast and Easy🔥]==
 +
 
 +
https://www.youtube.com/watch?v=H5Pq8mf-OVk
 +
==Video Solution by Interstigation==
 +
https://youtu.be/ktzijuZtDas&t=3126
 +
==2 minute solve (fast and elegant) by MegaMath==
 +
https://www.youtube.com/watch?v=hUh0hux3xuU
 +
 
 +
==Video solution by Dr. David==
 +
https://youtu.be/ZurHG58d1JM
 +
 
 +
==Video Solution by WhyMath==
 +
https://youtu.be/LSVPJAu-jxw
 +
 
 +
==See Also==
 +
{{AMC8 box|year=2024|num-b=23|num-a=25}}
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{{MAA Notice}}

Latest revision as of 09:57, 18 November 2024

Problem

Jean has made a piece of stained glass art in the shape of two mountains, as shown in the figure below. One mountain peak is $8$ feet high while the other peak is $12$ feet high. Each peak forms a $90^\circ$ angle, and the straight sides form a $45^\circ$ angle with the ground. The artwork has an area of $183$ square feet. The sides of the mountain meet at an intersection point near the center of the artwork, $h$ feet above the ground. What is the value of $h?$

[asy] unitsize(.3cm); filldraw((0,0)--(8,8)--(11,5)--(18,12)--(30,0)--cycle,gray(0.7),linewidth(1)); draw((-1,0)--(-1,8),linewidth(.75)); draw((-1.4,0)--(-.6,0),linewidth(.75)); draw((-1.4,8)--(-.6,8),linewidth(.75)); label("$8$",(-1,4),W); label("$12$",(31,6),E); draw((-1,8)--(8,8),dashed); draw((31,0)--(31,12),linewidth(.75)); draw((30.6,0)--(31.4,0),linewidth(.75)); draw((30.6,12)--(31.4,12),linewidth(.75)); draw((31,12)--(18,12),dashed); label("$45^{\circ}$",(.75,0),NE,fontsize(10pt)); label("$45^{\circ}$",(29.25,0),NW,fontsize(10pt)); draw((8,8)--(7.5,7.5)--(8,7)--(8.5,7.5)--cycle); draw((18,12)--(17.5,11.5)--(18,11)--(18.5,11.5)--cycle); draw((11,5)--(11,0),dashed); label("$h$",(11,2.5),E); [/asy]

$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 4\sqrt{2} \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 5\sqrt{2}$

Solution1

Extend the "inner part" of the mountain so that the image is two right triangles that overlap in a third right triangle as shown. [asy] unitsize(.2cm); draw((0,0)--(8,8)--(11,5)--(18,12)--(30,0)--cycle,linewidth(1)); draw((11,5)--(6, 0)--(16, 0)--cycle,linewidth(0.5)); label("$8\sqrt{2}$",(4,4),NW); label("$12\sqrt{2}$",(24,6),NE); draw((8,8)--(7.5,7.5)--(8,7)--(8.5,7.5)--cycle); draw((18,12)--(17.5,11.5)--(18,11)--(18.5,11.5)--cycle); draw((11,5)--(11,0),dashed); label("$h$",(11,2.5),E); [/asy] The side length of the largest right triangle is $12\sqrt{2},$ which means its area is $144.$ Similarly, the area of the second largest right triangle is $64$ (the side length is $8\sqrt{2}$), and the area of the overlap is $h^2$ (the side length is $h\sqrt{2}$). Thus, \[144+64-h^2=183,\] which means that the answer is $\boxed{\mathbf{(B)}\text{ 5}}.$

~BS201

Solution 2 (Using a Ruler)

You can measure $h$ with a ruler(rulers are allowed on the AMC 8), and see that $h$ is closest to $\boxed{\mathbf{(B)}\text{ 5}}.$

~SID-DARTH-VATER

(Please note: This is not the best way, as diagrams are NOT DRAWN TO SCALE) ~Anonymous ~Continuous_Pi


Video Solution 1 by Math-X (First understand the problem!!!)

https://youtu.be/BaE00H2SHQM?si=793ml6R_qT79h-aI&t=7651

~Math-X

Video Solution (A Clever Explanation You’ll Get Instantly)

https://youtu.be/5ZIFnqymdDQ?si=D8pGHQYbzJ_eMsEy&t=3678 ~hsnacademy

Video Solution (Quick solution) by Parshwa

https://youtu.be/Ky-NVrOv4ew

Video Solution by Power Solve (easy to understand)

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

Video Solution by OmegaLearn.org

https://youtu.be/KL_CGQrkXXo

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=RiSt6_WLfrM

Video Solution by NiuniuMaths (Easy to understand!)

https://www.youtube.com/watch?v=8353ngt05N0

~NiuniuMaths

Video Solution by CosineMethod [🔥Fast and Easy🔥]

https://www.youtube.com/watch?v=H5Pq8mf-OVk

Video Solution by Interstigation

https://youtu.be/ktzijuZtDas&t=3126

2 minute solve (fast and elegant) by MegaMath

https://www.youtube.com/watch?v=hUh0hux3xuU

Video solution by Dr. David

https://youtu.be/ZurHG58d1JM

Video Solution by WhyMath

https://youtu.be/LSVPJAu-jxw

See Also

2024 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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 AJHSME/AMC 8 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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