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2008 AMC 12B Problems/Problem 13

Revision as of 14:16, 19 January 2012 by Jhggins (talk | contribs) (→Problem)

Problem

Vertex $E$ of equilateral $\triangle{ABE}$ is in the interior of unit square $ABCD$. Let $R$ be the region consisting of all points inside $ABCD$ and outside $\triangle{ABE}$ whose distance from $AD$ is between $\frac{1}{3}$ and $\frac{2}{3}$. What is the area of $R$?

$\textbf{(A)}\ \frac{12-5\sqrt3}{72} \qquad \textbf{(B)}\ \frac{12-5\sqrt3}{36} \qquad \textbf{(C)}\ \frac{\sqrt3}{18} \qquad \textbf{(D)}\ \frac{3-\sqrt3}{9} \qquad \textbf{(E)}\ \frac{\sqrt3}{12}$

Solution

The region is the shaded area:

[asy] pair A,B,C,D,E; A=(0,1); B=(1,1); C=(1,0); D=(0,0); E=(1/2,1-sqrt(3)/2); draw(A--B--C--D--cycle); label("A",A,NW); dot(A); label("B",B,NE); dot(B); label("C",C,SE); dot(C); label("D",D,SW); dot(D); draw(A--E--B--cycle); label("E",E,S); dot(E); draw((1/3,0)--(1/3,1)); draw((2/3,0)--(2/3,1)); fill((1/3,0)--(1/3,1-sqrt(3)/3)--E--(2/3,1-sqrt(3)/3)--(2/3,0)--cycle,Black);[/asy]

We can find the area of the shaded region by subtracting the pentagon from the middle third of the square. The area of the middle third of the square is $\left(\frac13\right)(1)=\frac13$. The pentagon can be split into a rectangle and an equilateral triangle.

The base of the equilateral triangle is $\frac13$ and the height is $\left(\frac13\right)\left(\frac12\right)(\sqrt{3})=\frac{\sqrt{3}}{6}$. Thus, the area is $\left(\frac{\sqrt3}{6}\right)\left(\frac13\right)\left(\frac12\right)=\frac{\sqrt3}{36}$.

The base of the rectangle is $\frac13$ and the height is the height of the equilateral triangle minus the height of the smaller equilateral triangle. This is: $\frac{\sqrt3}{2}-\frac{\sqrt3}{6}=\frac{\sqrt3}{3}$ Therefore, the area of the shaded region is

$\frac13-\frac{\sqrt3}{9}-\frac{\sqrt3}{36}=\boxed{\text{(B) }\frac{12-5\sqrt3}{36}}.$

See Also

2008 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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 12 Problems and Solutions
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