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Difference between revisions of "2011 AMC 12B Problems/Problem 20"

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

Revision as of 01:57, 24 July 2017

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 $XA + XB + XC$?

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

Solutions

Solution 1 (Coordinates)

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 intersect 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 (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. By inspection, we see that $XA$, $XB$, and $XC$ are the circumradii, 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:

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

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 \[3R=3\cdot \frac{abc}{4\{ABC\}}=3\cdot \frac{abc}{2ah_A}=3\cdot \frac{bc}{2h_A}=\boxed{\frac{195}{8}.}\]

See also

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

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