Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 2)
(→‎Solution 2)
Line 10: Line 10:
  
 
==Solution 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 <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 <math>\frac{3\sqrt{6}}{6}=\frac{\sqrt{6}}{2}\Rightarrow\boxed{\textbf{(B)}</math>.
+
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 <math>\frac{3\sqrt{6}}{6}=\frac{\sqrt{6}}{2}\Rightarrow\boxed'''(B)'''</math>.
  
 
== See also ==
 
== See also ==

Revision as of 16:27, 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 $\frac{3\sqrt{6}}{6}=\frac{\sqrt{6}}{2}\Rightarrow\boxed'''(B)'''$.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10B_Problems/Problem_15&oldid=58398"