2015 AIME I Problem 4

Question

Point

B

lies on line segment

\overline{AC}

with

AB = 16

and

BC = 4.

Points

D

and

E

lie on the same side of line

AC

forming equilateral triangles

\triangle ABD

and

\triangle BCE.

Let

M

be the midpoint of

\overline{AE},

and

N

be the midpoint of

\overline{CD}.

The area of

\triangle BMN

is

x.

Find

x^2.

Answer

Answer: 507

Concepts:coordinate geometry equilateral triangle distance formula

Difficulty rating: 2390

Small hint:

Put

B

at the origin with

A = (-16, 0)

and

C = (4, 0),

and write

D

and

E

using equilateral-triangle altitudes

Big hint:

Compute the midpoints

M

and

N

and the three distances

BM,

MN,

NB;

triangle

BMN

turns out equilateral

Solution:

Place

B = (0, 0),

A = (-16, 0),

and

C = (4, 0).

Each equilateral triangle has its apex above the midpoint of its base at height

\frac{\sqrt{3}}{2}

times the side, so

D = (-8, 8\sqrt{3})

and

E = (2, 2\sqrt{3}).

The midpoints are

M = (-7, \sqrt{3})

and

N = (-2, 4\sqrt{3}).

Now

BM^2 = 49 + 3 = 52,

BN^2 = 4 + 48 = 52,

and

MN^2 = 25 + 27 = 52,

so

\triangle BMN

is equilateral with side

\sqrt{52}.

Its area is

x = \frac{\sqrt{3}}{4} \cdot 52 = 13\sqrt{3},

so

x^2 = 169 \cdot 3 = 507.