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

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<math>\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \textbf{(C)}\ 4\sqrt{2} \qquad \qquad \textbf{(D)}\ 5\sqrt{2} \qquad \textbf{(E)}\ 6</math>
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<math>\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 4\sqrt{2} \qquad \textbf{(D)}\ 5\sqrt{2} \qquad \textbf{(E)}\ 6</math>
  
 
==Solution 1==
 
==Solution 1==

Revision as of 16:26, 26 January 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]

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

Solution 1

Extend the "inner part" of the mountain so that the image is two right triangles that overlap in a third right triangle. 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 triangle is $h^2$ (the side length is $h\sqrt{2}$) Thus, \[144+64-h^2=183,\] which means that the answer is $\boxed{C}.$

~BS2012

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

https://www.youtube.com/watch?v=j6rzEkESmT4

~Math-X

2 minute solve (fast) by MegaMath

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

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 CosineMethod [🔥Fast and Easy🔥]

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