2008 AMC 12B Problem 13

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}

Take

\overline{AD}

as a side of the square; “distance from

\overline{AD}

between

\tfrac13

and

\tfrac23

” is a vertical strip of width

\tfrac13

and area

\tfrac13

From that strip’s area

\tfrac13,

subtract the part of the strip that lies inside

\triangle ABE

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

\begin{aligned} &\int_{\frac{1}{3}}^{\frac{1}{2}} \sqrt3\,x\,dx \\ &\quad {}+ \int_{\frac{1}{2}}^{\frac{2}{3}} \sqrt3(1 - x)\,dx \\ &= 2\int_{\frac{1}{3}}^{\frac{1}{2}}\sqrt3\,x\,dx \\ &= \frac{5\sqrt3}{36}. \end{aligned}

Therefore

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

Thus, the correct answer is B.

Problem 13 in other years