Difference between revisions of "2023 AMC 8 Problems/Problem 24"

Revision as of 01:22, 25 January 2023

Problem

Isosceles $\triangle ABC$ has equal side lengths $AB$ and $BC$ . In the figure below, segments are drawn parallel to $\overline{AC}$ so that the shaded portions of $\triangle ABC$ have the same area. The heights of the two unshaded portions are 11 and 5 units, respectively. What is the height of $h$ of $\triangle ABC$ ?

Add asymptote diagram*

(note: diagrams are not necessarily drawn to scale)

$\textbf{(A) } 14.6 \qquad \textbf{(B) } 14.8 \qquad \textbf{(C) } 15 \qquad \textbf{(D) } 15.2 \qquad \textbf{(E) } 15.4$

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 gray area in the first triangle is $[\text{ABC}]\cdot\left(1-\left(\tfrac{11}{h}\right)^2\right)$ . Similarly, we can find that the area of the gray part in the second triangle is $[\text{ABC}]\cdot\left(\tfrac{h-5}{h}\right)^2$ . These areas are equal, so $1-\left(\frac{11}{h}\right)^2=\left(\frac{h-5}{h}\right)^2$ . Simplifying yields $10h=146$ so $h=\boxed{\textbf{(A) }14.6}$ .

~MathFun1000 (~edits apex304)

Video Solution 1 by OmegaLearn (Using Similarity)

https://youtu.be/almtw4n-92A

Video Solution 2 by SpreadTheMathLove(Using Area-Similarity Relaitionship)

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

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.

@@ Line 8: / Line 8: @@
 ==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 gray area in the first triangle is <math>[\text{ABC}]\cdot\left(1-\left(\tfrac{11}{h}\right)^2\right)</math>. Similarly, we can find that the area of the grey part in the second triangle is <math>[\text{ABC}]\cdot\left(\tfrac{h-5}{h}\right)^2</math>. These areas are equal, so <math>1-\left(\frac{11}{h}\right)^2=\left(\frac{h-5}{h}\right)^2</math>. Simplifying yields <math>10h=146</math> so <math>h=\boxed{\textbf{(A) }14.6}</math>.
+First, we notice that the smaller isosceles triangles are similar to the larger isosceles triangles. We can find that the area of the gray area in the first triangle is <math>[\text{ABC}]\cdot\left(1-\left(\tfrac{11}{h}\right)^2\right)</math>. Similarly, we can find that the area of the gray part in the second triangle is <math>[\text{ABC}]\cdot\left(\tfrac{h-5}{h}\right)^2</math>. These areas are equal, so <math>1-\left(\frac{11}{h}\right)^2=\left(\frac{h-5}{h}\right)^2</math>. Simplifying yields <math>10h=146</math> so <math>h=\boxed{\textbf{(A) }14.6}</math>.
 ~MathFun1000 (~edits apex304)

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