2011 AMC 12B Problems/Problem 20


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 (Coordinates)

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 intersect at $P$, so $P$ is $X$.

The question now becomes calculating the sum of the distance from each vertex 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 $-\frac{5}{12}$ that 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 (Algebra)

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.

Let $M$ & $N$ be $\triangle BDE$ & $\triangle CEF$'s circumcircles' respective centers. Since $\triangle BDE$ & $\triangle CEF$ are congruent, the distance $M$ & $N$ each are from $\overline{BC}$ are equal, so $\overline{MN} || \overline{BC}$. The angle between $\overline {MN}$ & $\overline{EX}$ is $90^{\circ}$, and since $\overline{MN} || \overline{BC}$, $\angle XEC$ is also $90^{\circ}$. $\triangle XEC$ is a right triangle inscribed in a circle, so $\overline{XC}$ must be the diameter of $N$. Using the same logic & reasoning, we could deduce that $XA$ & $XB$ are also circumdiameters.

Since the circumcircles are congruent, circumdiameters $XA$, $XB$, and $XC$ 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 $\sqrt{(s)(s-a)(s-b)(s-c)} = \frac{abc}{4R}$, such that $s=\frac{a+b+c}{2}$ 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:

$\Rightarrow R=\frac{65}{16} \Rightarrow XA = XB = XC = \frac{65}{8} \Rightarrow XA + XB + XC = \boxed{\frac{195}{8}}$.

Solution 3 (Homothety)

Let $O$ be the circumcenter of $\triangle ABC,$ and $h_A$ denote the length of the altitude from $A.$ Note that a homothety centered at $B$ with ratio $\frac{1}{2}$ takes the circumcircle of $\triangle BAC$ to the circumcircle of $\triangle BDE$. It also takes the point diametrically opposite $B$ on the circumcircle of $\triangle BAC$ to $O.$ Therefore, $O$ lies on the circumcircle of $\triangle BDE.$ Similarly, it lies on the circumcircle of $\triangle CEF.$ By Pythagorean triples, $h_A=12.$ Finally, our answer is \[3R=3\cdot \frac{abc}{4\{ABC\}}=3\cdot \frac{abc}{2ah_A}=3\cdot \frac{bc}{2h_A}=\boxed{\frac{195}{8}.}\]

Solution 4 (basically Solution 1 but without coordinates)

Since Solution 1 has already proven that the circumcenter of $\triangle ABC$ coincides with $X$, we'll go from there. Note that the radius of the circumcenter of any given triangle is $\frac{a}{2\sin{A}}$, and since $b=15$ and $\sin{B}=\frac{12}{13}$, it can be easily seen that $XA = XB = XC = \frac{65}{8}$ and therefore our answer is \[3\cdot \frac{65}{8}=\boxed{\frac{195}{8}}.\]

Solution 5

Screen Shot 2021-08-06 at 7.30.10 PM.png

Since $ED$ is a midline of $\triangle CAB,$ we have that $\triangle CED \sim \triangle CAB$ with a side length ratio of $1:2.$

Consider a homothety of scale factor $2$ with on $\triangle CED$ concerning point $C$. Note that this sends $(CEDX)$ to $(ABCC')$ with $CX=XC'.$ By properties of homotheties, $C,X,$ and $C'$ are collinear. Similarly, we obtain that $BX=XB',$ with all three points collinear. Let $O$ denote the circumcenter of $\triangle ABC.$ It is well-known that $OX \perp CC'$ and analogously $OX \perp BB'.$ However, there is only one perpendicular line to $OX$ passing through $X,$, therefore, $O$ coincides with $X.$

It follows that $AX=BX=CX=R,$ where $R$ is the circumradius of $\triangle ABC,$ and this can be computed using the formula \[R=\frac{abc}{4[ABC]},\] from which we quickly obtain \[R=\frac{65}{8} \implies AX+BX+CX=\boxed{\frac{195}{8}}.\]

Solution 6 (Trigonometry)


$\angle BXE = \angle BDE$, $\angle CXE = \angle CFE$, as the angles are on the same circle.

$\triangle BDE \sim \triangle ABC$, $\triangle CFE \sim \triangle ABC$

$\angle BDE = \angle A$, $\angle CFE = \angle A$

$\angle BXE = \angle A$, $\angle CXE = \angle A$

Therefore $\angle BXE = \angle CXE$, and $XE$ is the angle bisector of $\triangle XBC$. By the angle bisector theorem $\frac{XB}{XC} = \frac{BE}{CE} = 1$, $XB = XC$. In a similar fashion $XA = XB = XC = R$, where $R$ is the circumcircle of $\triangle ABC$.

By the law of cosine, $\cos A = \frac{13^2 + 15^2 - 14^2}{2 \cdot 13 \cdot 15} = \frac{33}{65}$, $\sin A = \sqrt{1 - \left(\frac{33}{65}\right)^2} = \frac{56}{65}$

By the extended law of sines, $2R = \frac{BC}{\sin A} = \frac{14}{\frac{56}{65}} = \frac{65}{4}$, $R = \frac{65}{8}$

$XA + XB + XC = 3 R = \boxed{\textbf{(C) } \frac{195}{8}}$


See also

2011 AMC 12B (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png