Difference between revisions of "2013 AMC 10B Problems/Problem 15"

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(Solution 2)
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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>
  
==Solution==
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==Solution 1==
  
 
Using the formulas for area of a regular triangle <math>(\frac{{s}^{2}\sqrt{3}}{4})</math> and regular hexagon <math>(\frac{3{s}^{2}\sqrt{3}}{2})</math> and plugging <math>\frac{a}{3}</math> and <math>\frac{b}{6}</math> into each equation, you find that <math>\frac{{a}^{2}\sqrt{3}}{36}=\frac{{b}^{2}\sqrt{3}}{24}</math>. Simplifying this, you get <math>\frac{a}{b}=\boxed{\textbf{(B)} \frac{\sqrt{6}}{2}}</math>
 
Using the formulas for area of a regular triangle <math>(\frac{{s}^{2}\sqrt{3}}{4})</math> and regular hexagon <math>(\frac{3{s}^{2}\sqrt{3}}{2})</math> and plugging <math>\frac{a}{3}</math> and <math>\frac{b}{6}</math> into each equation, you find that <math>\frac{{a}^{2}\sqrt{3}}{36}=\frac{{b}^{2}\sqrt{3}}{24}</math>. Simplifying this, you 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. Each of the small equilateral triangle's area is <math>\frac{1}{6}</math>th of the area of the big equilateral triangle. Therefore each side of the big triangle is <math>\sqrt{6}</math> times the side of the small triangle. The desired ratio is $\frac{3\sqrt{6}}{6}=\frac{\sqrt{6}}{2}\Rightarrow\Box{B}.
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== See also ==
 
== See also ==
 
{{AMC10 box|year=2013|ab=B|num-b=14|num-a=16}}
 
{{AMC10 box|year=2013|ab=B|num-b=14|num-a=16}}

Revision as of 17:23, 25 December 2013

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 formulas for area of a regular triangle $(\frac{{s}^{2}\sqrt{3}}{4})$ and regular hexagon $(\frac{3{s}^{2}\sqrt{3}}{2})$ and plugging $\frac{a}{3}$ and $\frac{b}{6}$ into each equation, you find that $\frac{{a}^{2}\sqrt{3}}{36}=\frac{{b}^{2}\sqrt{3}}{24}$. Simplifying this, you 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. Each of the small equilateral triangle's area is $\frac{1}{6}$th of the area of the big equilateral triangle. 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\Box{B}.

See also

2013 AMC 10B (ProblemsAnswer KeyResources)
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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