Question

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 $\overline{AD}$ is between $\tfrac{1}{3}$ and $\tfrac{2}{3}.$ What is the area of $R?$

$\dfrac{12 - 5\sqrt{3}}{72}$

$\dfrac{12 - 5\sqrt{3}}{36}$

$\dfrac{\sqrt{3}}{18}$

$\dfrac{3 - \sqrt{3}}{9}$

$\dfrac{\sqrt{3}}{12}$

Solution:

Place $A = (0,0),$ $B = (1,0),$ $C = (1,1),$ $D = (0,1),$ so $\overline{AD}$ lies along the $y$ -axis and distance from $\overline{AD}$ is the $x$ -coordinate. The region lies in the strip $\tfrac13 \le x \le \tfrac23,$ which within the square has area $\tfrac13.$

Equilateral $\triangle ABE$ has $E = \left(\tfrac12, \tfrac{\sqrt3}{2}\right),$ with side $AE$ on $y = \sqrt3\,x$ and side $BE$ on $y = \sqrt3(1 - x).$ The area of the triangle inside the strip is $\int_{1/3}^{1/2} \sqrt3\,x\,dx + \int_{1/2}^{2/3} \sqrt3(1 - x)\,dx = 2\int_{1/3}^{1/2}\sqrt3\,x\,dx = \frac{5\sqrt3}{36}.$

Therefore $[R] = \frac13 - \frac{5\sqrt3}{36} = \frac{12 - 5\sqrt3}{36}.$

Thus, the correct answer is B.

2008 AMC 12B Problem 13

Solution:

Problem 13 in Other Years