1992 AMC 12 Problem 9

Question

Five equilateral triangles, each with side

2\sqrt3,

are arranged so they are all on the same side of a line containing one side of each. Along this line, the midpoint of the base of one triangle is a vertex of the next. The area of the region of the plane that is covered by the union of the five triangular regions is

10

12

15

10\sqrt3

12\sqrt3

Answer: E

Concepts:equilateral triangle area inclusion-exclusion overlapping regions

Difficulty rating: 1780

Small hint:

Add the areas of the five large triangles, then subtract their overlaps

Big hint:

Each adjacent overlap is an equilateral triangle with half the original side length

Solution:

Each large triangle has area

\frac{\sqrt3}{4}(2\sqrt3)^2=3\sqrt3.

Each of the four overlaps is an equilateral triangle of side

\sqrt3,

hence area

\frac{3\sqrt3}{4}.

There are no triple overlaps, so the union has area

5(3\sqrt3)-4\left(\frac{3\sqrt3}{4}\right)=12\sqrt3.

Thus the correct answer is E.