Difference between revisions of "2023 AMC 8 Problems/Problem 24"
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Revision as of 00:44, 25 January 2023
Contents
Problem
Isosceles 
has equal side lengths 
and 
. In the figure below, segments are drawn parallel to 
so that the shaded portions of have the same area. The heights of the two unshaded portions are 11 and 5 units, respectively. What is the height of 
of ?
- Add asymptote diagram*
(note: diagrams are not necessarily drawn to scale)
Solution 1
First, we notice that the smaller isosceles triangles are similar to the larger isosceles triangles. We can find that the area of the grey part in the first triangle is ![$[\text{ABC}]\cdot\left(1-\left(\tfrac{11}{h}\right)^2\right)$](http://latex.artofproblemsolving.com/c/c/d/ccd19fe2d21c98e12c3f7a6e1f715aee46085b3d.png)
. Similarly, we can find that the area of the grey part in the second triangle is ![$[\text{ABC}]\cdot\left(\tfrac{h-5}{h}\right)^2$](http://latex.artofproblemsolving.com/0/1/c/01c770040e946593b2026999898568707d5b5538.png)
. These areas are equal, so 
. Simplifying yields 
so 
.
~MathFun1000
Video Solution 1 by OmegaLearn (Using Similarity)
Video Solution 2 by SpreadTheMathLove(Using Area-Similarity Relaitionship)
https://www.youtube.com/watch?v=GTlkTwxSxgo
See Also
| 2023 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. 
