Problem #1791


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 $BN^2$?

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

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Instructions for entering answers:

  • Reduce fractions to lowest terms and enter in the form 7/9.
  • Numbers involving pi should be written as 7pi or 7pi/3 as appropriate.
  • Square roots should be written as sqrt(3), 5sqrt(5), sqrt(3)/2, or 7sqrt(2)/3 as appropriate.
  • Exponents should be entered in the form 10^10.
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