Question

Triangle $ABC$ has $AB=13,$ $BC=14,$ and $AC=15.$ The points $D,$ $E,$ and $F$ are the midpoints of $AB,$ $BC,$ and $AC$ respectively. Let $X\ne E$ be the intersection of the circumcircles of $\triangle BDE$ and $\triangle CEF.$ What is $XA+XB+XC?$

$24$

$14\sqrt{3}$

$\dfrac{195}{8}$

$\dfrac{129\sqrt{7}}{14}$

$\dfrac{69\sqrt{2}}{4}$

Solution:

Since $DE\parallel AC$ and $EF\parallel AB,$ we get $\angle BDE=\angle BAC=\angle EFC.$ By the Inscribed Angle Theorem, $\angle BXE=\angle BDE$ and $\angle EXC=\angle EFC,$ so $\angle BXE=\angle EXC.$ With $BE=EC,$ this forces $XB=XC.$

Also $\angle BXC=2\angle BAC,$ so by the Inscribed Angle Theorem $X$ is the circumcenter of $\triangle ABC.$ Hence $XA=XB=XC=R.$

The area of the $13$ - $14$ - $15$ triangle is $84$ by Heron's formula, so $R=\dfrac{13\cdot14\cdot15}{4\cdot84}=\dfrac{65}{8},$ and $XA+XB+XC=3R=\dfrac{195}{8}.$

Thus, the correct answer is C.

2011 AMC 12B Problem 20

Solution:

Problem 20 in Other Years