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2013 AMC 10B Problems/Problem 15

Problem

A wire is cut into two pieces, one of length $a$ and the other of length $b$. The piece of length $a$ is bent to form an equilateral triangle, and the piece of length $b$ is bent to form a regular hexagon. The triangle and the hexagon have equal area. What is $\frac{a}{b}$?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{\sqrt{6}}{2}\qquad\textbf{(C)}\ \sqrt{3} \qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \frac{3\sqrt{2}}{2}$

Solution 1

Using the area formulas for an equilateral triangle $\left(\frac{{s}^{2}\sqrt{3}}{4}\right)$ and regular hexagon $\left(\frac{3{s}^{2}\sqrt{3}}{2}\right)$ with side length $s$, plugging $\frac{a}{3}$ and $\frac{b}{6}$ into each equation, we find that $\frac{{a}^{2}\sqrt{3}}{36}=\frac{{b}^{2}\sqrt{3}}{24}$. Simplifying this, we get $\frac{a}{b}=\boxed{\textbf{(B)} \frac{\sqrt{6}}{2}}$

Solution 2

The regular hexagon can be broken into 6 small equilateral triangles, each of which is similar to the big equilateral triangle. The big triangle's area is 6 times the area of one of the little triangles. Therefore each side of the big triangle is $\sqrt{6}$ times the side of the small triangle. The desired ratio is $\frac{3\sqrt{6}}{6}=\frac{\sqrt{6}}{2}\Rightarrow(B).$

Solution 3

In order to avoid fractions we can let $a = 6x$ and $b = 6y$. Then we only need to find $x/y$. The side length of the triangle is $2x$, and the side length of the hexagon is $y$. Thus we have $\frac{(2x)^2\sqrt{3}}{4} = \frac{3y^2\sqrt{3}}{2}$. Solving yields $\frac{x^2}{y^2} = \frac{3}{2}$, so $\frac{x}{y} = \frac{\sqrt{3}}{\sqrt{2}}$. Rationalizing the denominator, the answer is $\boxed{\frac{\sqrt{6}}{2}\text{ (B)}}$.

See also

2013 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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 AMC 10 Problems and Solutions

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