2013 AMC 12B Problem 24

Question

Let

ABC

be a triangle where

M

is the midpoint of

AC,

and

CN

is the angle bisector of

\angle ACB

with

N

on

AB.

Let

X

be the intersection of the median

BM

and the bisector

CN.

In addition

\triangle BXN

is equilateral and

AC = 2.

What is

BN^2?

\dfrac{10 - 6\sqrt2}{7}

\dfrac{2}{9}

\dfrac{5\sqrt2 - 3\sqrt3}{8}

\dfrac{\sqrt2}{6}

\dfrac{3\sqrt3 - 4}{5}

Let

\alpha = \angle ACN

and

x = BN;

the equilateral triangle gives

\angle BXC = \angle CNA = 120^\circ

Similar triangles yield

BC = \sqrt2

and

CX = (\sqrt2 + 1)x;

then apply the Law of Cosines in

\triangle BCX

Let

\alpha = \angle ACN = \angle NCB

and

x = BN.

Since

\triangle BXN

is equilateral,

\angle BXC = \angle CNA = 120^\circ,

which gives

\triangle ABC \sim \triangle BMC

and

\triangle ANC \sim \triangle BXC.

From the first, with

MC = \tfrac12 AC = 1,

we get

\dfrac{BC}{2} = \dfrac{MC}{BC},

so

BC = \sqrt2.

From the second,

CX = (\sqrt2 + 1)x.

The Law of Cosines in

\triangle BCX

with

\angle BXC = 120^\circ

gives

2 = x^2

+ (\sqrt2 + 1)^2 x^2

+ (\sqrt2 + 1)x^2

= (5 + 3\sqrt2)x^2.

Hence

BN^2 = x^2

= \dfrac{2}{5 + 3\sqrt2}

= \dfrac{10 - 6\sqrt2}{7}.

Thus, the correct answer is A.

Problem 24 in other years