Difference between revisions of "2024 AMC 8 Problems/Problem 24"
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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 <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 triangle is <math>h^2</math> (the side length is <math>h\sqrt{2}</math>) Thus, | 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 <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 triangle is <math>h^2</math> (the side length is <math>h\sqrt{2}</math>) Thus, | ||
<cmath>144+64-h^2=183,</cmath> | <cmath>144+64-h^2=183,</cmath> | ||
| − | which means that the answer is <math>\boxed{ | + | which means that the answer is <math>\boxed{B}.</math> |
~BS2012 | ~BS2012 | ||
Revision as of 15:27, 26 January 2024
Contents
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 
feet high while the other peak is 
feet high. Each peak forms a 
angle, and the straight sides form a 
angle with the ground. The artwork has an area of 
square feet. The sides of the mountain meet at an intersection point near the center of the artwork, 
feet above the ground. What is the value of 
![[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]](http://latex.artofproblemsolving.com/0/6/2/062be4008fc995e68ebbe56afa0a90b07160a7d6.png)
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 
which means its area is 
Similarly, the area of the second largest right triangle is 
(the side length is 
), and the area of the overlap triangle is 
(the side length is 
) Thus,
![\[144+64-h^2=183,\]](http://latex.artofproblemsolving.com/d/a/3/da33465ebae23a898b76506cd66992f6ff48089f.png)
which means that the answer is 
~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
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=RiSt6_WLfrM
