2013 AMC 12B Problems/Problem 24

Problem

Let $ABC$ be a triangle where $M$ is the midpoint of $\overline{AC}$ , and $\overline{CN}$ is the angle bisector of $\angle{ACB}$ with $N$ on $\overline{AB}$ . Let $X$ be the intersection of the median $\overline{BM}$ and the bisector $\overline{CN}$ . In addition $\triangle BXN$ is equilateral with $AC=2$ . What is $BN^2$ ?

$\textbf{(A)}\ \frac{10-6\sqrt{2}}{7} \qquad \textbf{(B)}\ \frac{2}{9} \qquad \textbf{(C)}\ \frac{5\sqrt{2}-3\sqrt{3}}{8} \qquad \textbf{(D)}\ \frac{\sqrt{2}}{6} \qquad \textbf{(E)}\ \frac{3\sqrt{3}-4}{5}$

Solution

Let $BN=x$ and $NA=y$ . From the conditions, let's deduct some convenient conditions that seems sufficient to solve the problem.

$M$ is the midpoint of side $AC$ .

This implies that $[ABX]=[CBX]$ . Given that angle $ABX$ is $60$ degrees and angle $BXC$ is $120$ degrees, we can use the area formula to get

$\frac{1}{2}(x+y)x \frac{\sqrt{3}}{2} = \frac{1}{2} x \cdot CX \frac{\sqrt{3}}{2}$

So, $x+y=CX$ .....(1)

$CN$ is angle bisector.

In the triangle $ABC$ , one has $BC/AC=x/y$ , therefore $BC=2x/y$ .....(2)

Furthermore, triangle $BCN$ is similar to triangle $MCX$ , so $BC/CM=CN/CX$ , therefore $BC = (CX+x)/CX = (2x+y)/(x+y)$ ....(3)

By (2) and (3) and the subtraction law of ratios, we get

$BC=2x/y = (2x+y)/(y+x) = y/x$

Therefore $2x^2=y^2$ , or $y=\sqrt{2}x$ . So $BC = 2x/(\sqrt{2}x) = \sqrt{2}$ .

Finally, using the law of cosine for triangle $BCN$ , we get

$2 = BC^2 = x^2 + (2x+y)^2 - x(2x+y) = 3x^2 + 3xy + y^2 = (5+3\sqrt{2})x^2$

$x^2 = \frac{2}{5+3\sqrt{2}} = \frac{10-6\sqrt{2}}{7}.$

2013 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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2013 AMC 12B Problems/Problem 24

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