2008 AMC 12B Problem 15

Question

On each side of a unit square, an equilateral triangle of side length

1

is constructed. On each new side of each equilateral triangle, another equilateral triangle of side length

1

is constructed. The interiors of the square and the

12

triangles have no points in common. Let

R

be the region formed by the union of the square and all the triangles, and let

S

be the smallest convex polygon that contains

R.

What is the area of the region that is inside

S

but outside

R?

\dfrac{1}{4}

\dfrac{\sqrt{2}}{4}

1

\sqrt{3}

2\sqrt{3}

The gap between

S

and

R

is four small triangles, one at each corner of the square

Each gap triangle has two sides of length

1

meeting at angle

360^\circ - 90^\circ - 4 \cdot 60^\circ;

use

\tfrac12 ab \sin C

The convex hull

S

differs from

R

only near the four corners of the square, where a small triangular gap forms. Each gap triangle has two sides of length

1

(outer edges of adjacent triangles).

The angle between those two sides is

360^\circ - 90^\circ - 4 \cdot 60^\circ = 30^\circ,

so each gap has area

\tfrac12 \cdot 1 \cdot 1 \cdot \sin 30^\circ = \tfrac14.

The total area is

4 \cdot \tfrac14 = 1.

Thus, the correct answer is C.

Problem 15 in other years