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2024 AMC 8 Problems/Problem 24

Revision as of 18:55, 28 August 2024 by Hsnacademy (talk | contribs) (→‎Solution 2 (Using a Ruler))

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]

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

Solution 1

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 $\boxed{\mathbf{(B)}\text{ 5}}.$

~BS2012

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 (A Clever Explanation You’ll Get Instantly)

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

~hsnacademy

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

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

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

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

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