Problem #1628


Triangle $ABC$ has $AB = 13, BC = 14$, and $AC = 15$. The points $D, E$, and $F$ are the midpoints of $\overline{AB}, \overline{BC}$, and $\overline{AC}$ respectively. Let $X \not= E$ be the intersection of the circumcircles of $\Delta BDE$ and $\Delta CEF$. What is $XA + XB + XC$?

$\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 14\sqrt{3} \qquad \textbf{(C)}\ \frac{195}{8} \qquad \textbf{(D)}\ \frac{129\sqrt{7}}{14} \qquad \textbf{(E)}\ \frac{69\sqrt{2}}{4}$

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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.
  • If the problem is multiple choice, enter the appropriate (capital) letter.
  • Enter points with parentheses, like so: (4,5)
  • Complex numbers should be entered in rectangular form unless otherwise specified, like so: 3+4i. If there is no real component, enter only the imaginary component (i.e. 2i, NOT 0+2i).

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