2011 AMC 12B Problems/Problem 20

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Problem

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 $X \neq E$ be the intersection of the circumcircles of $\triangle BDE$ and $\triangle CEF$. What is $XA + XB + XC$?

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

Solution

Answer: (C)

Let us also consider the circumcircle of $\triangle ADF$.

Note that if we draw the perpendicular bisector of each side, we will have the circumcenter of $\triangle ABC$ which is $P$, Also, since $m\angle ADP = m\angle AFP = 90^\circ$. $ADPF$ is cyclic, similarly, $BDPE$ and $CEPF$ are also cyclic. With this, we know that the circumcircles of $\triangle ADF$, $\triangle BDE$ and $\triangle CEF$ all intercept at $P$, so $P$ is $X$.

The question now becomes calculate the sum of distance from each vertices to the circumcenter.

We can do it will coordinate geometry, note that $XA = XB = XC$ because of $X$ being circumcenter.

Let $A = (5,12)$, $B = (0,0)$, $C = (14, 0)$, $X= (x_0, y_0)$

Then $X$ is on the line $x = 7$ and also the line with slope $-\frac{5}{12}$ and passes through $(2.5, 6)$.

$y_0 = 6-\frac{45}{24} = \frac{33}{8}$

So $X = (7, \frac{33}{8})$

and $XA +XB+XC = 3XB = 3\sqrt{7^2 + \left(\frac{33}{8}\right)^2} = 3\times\frac{65}{8}=\frac{195}{8}$

See also

2011 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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All AMC 12 Problems and Solutions
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