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 $BN^2?$

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

$\dfrac{2}{9}$

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

$\dfrac{\sqrt2}{6}$

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

Answer: A

Concepts:angle bisector equilateral triangle law of cosines similarity

Difficulty rating: 2600

Solution:

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.

2013 AMC 12B Problem 24

Answer: A

Difficulty rating: 2600

Solution:

Problem 24 in Other Years