2011 AMC 12B Problems/Problem 20

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 1

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 calculate the distances with coordinate geometry. (Note that $XA = XB = XC$ because $X$ is the 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 $XA +XB+XC = 3XB = 3\sqrt{7^2 + \left(\frac{33}{8}\right)^2} = 3\times\frac{65}{8}=\frac{195}{8}$

Solution 2

Consider an additional circumcircle on $\triangle ADF$ . After drawing the diagram, it is noticed that each triangle has side values: $7$ , $\frac{15}{2}$ , $\frac{13}{2}$ . Thus they are congruent, and their respective circumcircles are. By inspection, we see that $XA$ , $XB$ , and $XC$ are the circumdiameters, and so they are congruent. Therefore, the solution can be found by calculating one of these circumdiameters and multiplying it by a factor of $3$ . We can find the circumradius quite easily with the formula $\sqrt{(s)(s-a)(s-b)(s-c)} = \frac{abc}{4R}$ , s.t. $s=\frac{a+b+c}{2}$ and R is the circumradius. Since $s = \frac{21}{2}$ :

$\sqrt{(\frac{21}{2})(4)(3)(\frac{7}{2})} = \frac{\frac{15}{2}\cdot\frac{13}{2}\cdot 7}{4R}$

After a few algebraic manipulations:

$\Rightarrow R = \frac{65}{16} \Rightarrow D=2R=\frac{65}{8} \Rightarrow 3D = \boxed{\frac{195}{8}}$ .

2011 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

Page

Toolbox

Search

2011 AMC 12B Problems/Problem 20

Contents

Problem

Solution 1

Solution 2

See also