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Difference between revisions of "2013 AMC 10B Problems/Problem 15"
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<math> \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} </math>
 
<math> \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} </math>
  
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==Solution==
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==Solution 1==
  
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We have that <math>\frac{(\frac{a}{3})^2 \sqrt3 }{4} =\frac{3 (\frac{b}{6})^2 \sqrt3){2}</math>. Simplifying, we get that <math>\frac{a^2}{b^2}=\frac{3}{2}</math>. Taking the square root and rationalizing the denominator, we see that <math>\frac{a}{b}=\boxed{\textbf{(B) }\frac{\sqrt6}{2}}</math>.
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Using the area formulas for an equilateral triangle <math>\left(\frac{{s}^{2}\sqrt{3}}{4}\right)</math> and regular hexagon <math>\left(\frac{3{s}^{2}\sqrt{3}}{2}\right)</math> with side length <math>s</math>, plugging <math>\frac{a}{3}</math> and <math>\frac{b}{6}</math> into each equation, we find that <math>\frac{{a}^{2}\sqrt{3}}{36}=\frac{{b}^{2}\sqrt{3}}{24}</math>. Simplifying this, we get <math>\frac{a}{b}=\boxed{\textbf{(B)} \frac{\sqrt{6}}{2}}</math>
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==Solution 2==
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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 <math>\sqrt{6}</math> times the side of the small triangle. The desired ratio is <math>\frac{3\sqrt{6}}{6}=\frac{\sqrt{6}}{2}\Rightarrow(B).</math>
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==Solution 3==
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In order to avoid fractions we can let <math>a = 6x</math> and <math>b = 6y</math>. Then we only need to find <math>x/y</math>. The side length of the triangle is <math>2x</math>, and the side length of the hexagon is <math>y</math>. Thus we have <math>\frac{(2x)^2\sqrt{3}}{4} = \frac{3y^2\sqrt{3}}{2}</math>. Solving yields <math>\frac{x^2}{y^2} = \frac{3}{2}</math>, so <math>\frac{x}{y} = \frac{\sqrt{3}}{\sqrt{2}}</math>. Rationalizing the denominator, the answer is <math>\boxed{\frac{\sqrt{6}}{2}\text{ (B)}}</math>.
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== See also ==
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{{AMC10 box|year=2013|ab=B|num-b=14|num-a=16}}
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[[Category:Introductory Geometry Problems]]
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[[Category:Area Problems]]
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{{MAA Notice}}

Latest revision as of 08:18, 21 October 2025

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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