1979 AMC 12 Problem 10

Question

If

P_1P_2P_3P_4P_5P_6

is a regular hexagon whose apothem (distance from the center to the midpoint of a side) is

2,

and

Q_i

is the midpoint of side

P_iP_{i+1}

for

i=1,

2,

3,

4,

then the area of quadrilateral

Q_1Q_2Q_3Q_4

is

6

2\sqrt6

\frac{8\sqrt3}{3}

3\sqrt3

4\sqrt3

The points

Q_i

lie on a circle of radius equal to the apothem

Join the center to the four consecutive side midpoints and decompose the quadrilateral

Let

O

be the center. Consecutive apothems

OQ_i

and

OQ_{i+1}

have length

2

and form a

60^\circ

angle, so each

\triangle OQ_iQ_{i+1}

is equilateral of side

2.

The quadrilateral is the union of three such triangles, and hence has area

3\left(\frac{\sqrt3}{4}\cdot2^2\right)=3\sqrt3.

Therefore, the correct answer is D.

Problem 10 in other years