2013 AMC 10B Problem 15

Question

A wire is cut into two pieces, one of length

a

and the other of length

b.

The piece of length

a

is bent to form an equilateral triangle, and the piece of length

b

is bent to form a regular hexagon. The triangle and the hexagon have equal area. What is

\frac{a}{b}?

1

\dfrac{\sqrt{6}}{2}

\sqrt{3}

2

\dfrac{3\sqrt{2}}{2}

Compare the side length of the large equilateral triangle with the small equilateral triangles inside the hexagon

Areas scale as the square of side length

Let

s

be the side length of the equilateral triangle and

A

its area. Then

a=3s.

A regular hexagon of side length

s

consists of

6

equilateral triangles of side length

s,

so its area is

6A.

Therefore, a regular hexagon of area

A

has side length

\dfrac{s}{\sqrt 6} .

Hence

b = 6\cdot \dfrac{s}{\sqrt 6} = s\sqrt 6 .

It follows that

\dfrac ab = \dfrac{3s}{s\sqrt 6} = \dfrac{3 \sqrt 6}6 = \dfrac{\sqrt 6} 2.

Thus, the correct answer is B.

Problem 15 in other years