2010 AIME II Problem 9

Question

Let

ABCDEF

be a regular hexagon. Let

G,

H,

I,

J,

K,

and

L

be the midpoints of sides

AB,

BC,

CD,

DE,

EF,

and

AF,

respectively. The segments

AH,

BI,

CJ,

DK,

EL,

and

FG

bound a smaller regular hexagon. Let the ratio of the area of the smaller hexagon to the area of

ABCDEF

be expressed as a fraction

\frac{m}{n},

where

m

and

n

are relatively prime positive integers. Find

m + n.

Answer

Answer: 11

Concepts:regular polygon coordinate geometry area ratio symmetry

Difficulty rating: 2430

Small hint:

Place the center at the origin; by symmetry the smaller hexagon is regular and concentric, so it suffices to find the distance from the center to one of its vertices

Big hint:

One vertex of the small hexagon is the intersection of lines

AH

and

FG;

with circumradius

1

it is

\left(\frac{5}{7}, \frac{\sqrt{3}}{7}\right)

Solution:

Center the hexagon at the origin with circumradius

1:

A = (1, 0),

B = \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right),

C = \left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right),

and

F = \left(\frac{1}{2}, -\frac{\sqrt{3}}{2}\right).

Then

H = \left(0, \frac{\sqrt{3}}{2}\right)

and

G = \left(\frac{3}{4}, \frac{\sqrt{3}}{4}\right).

Rotation by

60^\circ

permutes the six segments, so the smaller hexagon is regular and concentric, and the area ratio is the square of the ratio of distances from the center to a vertex.

One vertex is the intersection of

AH

and

FG.

Line

AH

is

x + \frac{2}{\sqrt{3}}\,y = 1,

and line

FG

is

y = 3\sqrt{3}\,x - 2\sqrt{3}.

Substituting gives

x + 6x - 4 = 1,

so

x = \frac{5}{7}

and

y = \frac{\sqrt{3}}{7}.

That vertex has squared distance

\frac{25}{49} + \frac{3}{49} = \frac{4}{7}

from the center, while

A

is at distance

1.

The ratio of areas is

\frac{4}{7},

and

m + n = 4 + 7 = 11.