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

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Since Solution 1 has already proven that the circumcenter of <math>\triangle ABC</math> coincides with <math>X</math>, we'll go from there. Note that the radius of the circumcenter of any given triangle is <math>\frac{a}{2\sin{A}}</math>, and since <math>b=15</math> and <math>\sin{B}=\frac{12}{13}</math>, it can be easily seen that <math>XA = XB = XC = \frac{65}{8}</math> and therefore our answer is <cmath>3\cdot \frac{65}{8}=\boxed{\frac{195}{8}}.</cmath>
 
Since Solution 1 has already proven that the circumcenter of <math>\triangle ABC</math> coincides with <math>X</math>, we'll go from there. Note that the radius of the circumcenter of any given triangle is <math>\frac{a}{2\sin{A}}</math>, and since <math>b=15</math> and <math>\sin{B}=\frac{12}{13}</math>, it can be easily seen that <math>XA = XB = XC = \frac{65}{8}</math> and therefore our answer is <cmath>3\cdot \frac{65}{8}=\boxed{\frac{195}{8}}.</cmath>
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==Solution 5==
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<center><img>https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvZi84L2E4MzA5YjkzYjhmMmMxYmY0YTVjMTNjOTJmNDVkZjE5MWMwZDZjLnBuZw==&rn=U2NyZWVuIFNob3QgMjAyMS0wOC0wNiBhdCA3LjMwLjEwIFBNLnBuZw==</img>
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</center>
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Since <math>ED</math> is a midline of <math>\triangle CAB,</math> we have that <math>\triangle CED \sim \triangle CAB</math> with a side length ratio of <math>1:2.</math>
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Consider a homothety of scale factor <math>2</math> with on <math>\triangle CED</math> with respect to point <math>C.</math> Note that this sends <math>(CEDX)</math> to <math>(ABCC')</math> with <math>CX=XC'.</math> By properties of homtheties, <math>C,X,</math> and <math>C'</math> are collinear. Similarly, we obtain that <math>BX=XB',</math> with all three points collinear. Let <math>O</math> denote the circumcenter of <math>\triangle ABC.</math> It is well-known that <math>OX \perp CC'</math> and analogously <math>OX \perp BB'.</math> However, there is only one perpendicular line to <math>OX</math> passing through <math>X,</math>, therefore, <math>O</math> coincides with <math>X.</math>
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It follows that <math>AX=BX=CX=R,</math> where <math>R</math> is the circumradius of <math>\triangle ABC,</math> and this can be computed using the formula <cmath>R=\frac{abc}{4[ABC]},</cmath> from which we quickly obtain <cmath>R=\frac{65}{8} \implies AX+BX+CX=\boxed{\frac{195}{8}}.</cmath>
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== See also ==
 
== See also ==
 
{{AMC12 box|year=2011|num-b=19|num-a=21|ab=B}}
 
{{AMC12 box|year=2011|num-b=19|num-a=21|ab=B}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:31, 6 August 2021

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 $X \neq F$ 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)

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 $-\frac{5}{12}$ 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. 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 $\sqrt{(s)(s-a)(s-b)(s-c)} = \frac{abc}{4R}$, 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

<img>https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvZi84L2E4MzA5YjkzYjhmMmMxYmY0YTVjMTNjOTJmNDVkZjE5MWMwZDZjLnBuZw==&rn=U2NyZWVuIFNob3QgMjAyMS0wOC0wNiBhdCA3LjMwLjEwIFBNLnBuZw==</img>

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 $\triangle CED$ with respect to point $C.$ Note that this sends $(CEDX)$ to $(ABCC')$ with $CX=XC'.$ By properties of homtheties, $C,X,$ 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 \[R=\frac{65}{8} \implies AX+BX+CX=\boxed{\frac{195}{8}}.\]


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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