2023 AMC 8 Problems/Problem 19

Problem

An equilateral triangle is placed inside a larger equilateral triangle so that the region between them can be divided into three congruent trapezoids, as shown below. The side length of the inner triangle is $\frac23$ the side length of the larger triangle. What is the ratio of one trapezoid to the area of the inner triangle?

$[asy] // Diagram by TheMathGuyd pair A,B,C; A=(0,1); B=(sqrt(3)/2,-1/2); C=-conj(B); fill(2B--3B--3C--2C--cycle,grey); dot(3A); dot(3B); dot(3C); dot(2A); dot(2B); dot(2C); draw(2A--2B--2C--cycle); draw(3A--3B--3C--cycle); draw(2A--3A); draw(2B--3B); draw(2C--3C); [/asy]$

$\textbf{(A) } 1 : 3 \qquad \textbf{(B) } 3 : 8 \qquad \textbf{(C) } 5 : 12 \qquad \textbf{(D) } 7 : 16 \qquad \textbf{(E) } 4 : 9$

Solution 1

By AA~ similarity triangle we can find the ratio of the area of big: small —> $\frac{9}{4}$ then there are a relative $5$ for the $3$ trapezoids combines. For $1$ trapezoid it is a relative $53$ so now the ratio is $\frac{5}{\frac{3}{4}}$ which can simplify to $\boxed{\text(C) \frac{5}{12}}$

~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat

Solution 2

Subtracting the larger equilateral triangle from the smaller one yields the sum of the three trapezoids. Since the ratio of the side lengths of the larger to the smaller one is $3:2$ , we can set the side lengths as $3$ and $2$ , respectively. So, the sum of the trapezoids is $\frac{9\sqrt{3}}{4}-\frac{4\sqrt{3}}{4}=\frac{5}{4}\sqrt{3}$ . We are also told that the three trapezoids are congruent, thus the area of each of them is $\frac{1}{3} \cdot \frac{5}{4}\sqrt{3}=\frac{5}{12}\sqrt{3}$ . Hence, the ratio is $\frac{\frac{5}{12}\sqrt{3}}{\sqrt{3}}=\boxed{\textbf{(C)}\ \frac{5}{12}}$ .

~MrThinker

Video Solution by OmegaLearn (Using Similar Triangles)

https://youtu.be/bGN-uBsVm0E

Animated Video Solution

https://youtu.be/Xq4LdJJtbDk

Video Solution by SpraedTheMathLove using Area-Similarity Relationship

https://www.youtube.com/watch?v=92hAg3JjqZI

~Star League (https://starleague.us)

2023 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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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