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

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==Problem==
 
==Problem==
  
Triangle <math>ABC</math> has <math>AB = 13, BC = 14</math>, and <math>AC = 15</math>. The points <math>D, E</math>, and <math>F</math> are the midpoints of <math>\overline{AB}, \overline{BC}</math>, and <math>\overline{AC}</math> respectively. Let <math>X \neq E</math> be the intersection of the circumcircles of <math>\triangle BDE</math> and <math>\triangle CEF</math>. What is <math>XA + XB + XC</math>?
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Triangle <math>ABC</math> has <math>AB = 13, BC = 14</math>, and <math>AC = 15</math>. The points <math>D, E</math>, and <math>F</math> are the midpoints of <math>\overline{AB}, \overline{BC}</math>, and <math>\overline{AC}</math> respectively. Let <math>X \neq F</math> be the intersection of the circumcircles of <math>\triangle BDE</math> and <math>\triangle CEF</math>. What is <math>XA + XB + XC</math>?
  
 
<math>\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}</math>
 
<math>\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}</math>
  
==Solution 1==
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==Solutions==
Answer: (C)
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===Solution 1 (Coordinates)===
 
 
 
Let us also consider the circumcircle of <math>\triangle ADF</math>.
 
Let us also consider the circumcircle of <math>\triangle ADF</math>.
  
Note that if we draw the perpendicular bisector of each side, we will have the circumcenter of <math>\triangle ABC</math> which is <math>P</math>, Also, since <math>m\angle ADP = m\angle AFP = 90^\circ</math>. <math>ADPF</math> is cyclic, similarly, <math>BDPE</math> and <math>CEPF</math> are also cyclic. With this, we know that the circumcircles of <math>\triangle ADF</math>, <math>\triangle BDE</math> and <math>\triangle CEF</math> all intercept at <math>P</math>, so <math>P</math> is <math>X</math>.
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Note that if we draw the perpendicular bisector of each side, we will have the circumcenter of <math>\triangle ABC</math> which is <math>P</math>, Also, since <math>m\angle ADP = m\angle AFP = 90^\circ</math>. <math>ADPF</math> is cyclic, similarly, <math>BDPE</math> and <math>CEPF</math> are also cyclic. With this, we know that the circumcircles of <math>\triangle ADF</math>, <math>\triangle BDE</math> and <math>\triangle CEF</math> all intersect at <math>P</math>, so <math>P</math> is <math>X</math>.
  
 
The question now becomes calculate the sum of distance from each vertices to the circumcenter.  
 
The question now becomes calculate the sum of distance from each vertices to the circumcenter.  
  
We can do it will coordinate geometry, note that <math>XA = XB = XC</math> because of <math>X</math> being circumcenter.
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We can calculate the distances with coordinate geometry. (Note that <math>XA = XB = XC</math> because <math>X</math> is the circumcenter.)
  
 
Let <math>A = (5,12)</math>, <math>B = (0,0)</math>, <math>C = (14, 0)</math>, <math>X= (x_0, y_0)</math>
 
Let <math>A = (5,12)</math>, <math>B = (0,0)</math>, <math>C = (14, 0)</math>, <math>X= (x_0, y_0)</math>
  
Then <math>X</math> is on the line <math>x = 7</math> and also the line with slope <math>-\frac{5}{12}</math> and passes through <math>(2.5, 6)</math>.  
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Then <math>X</math> is on the line <math>x = 7</math> and also the line with slope <math>-\frac{5}{12}</math> that passes through <math>(2.5, 6)</math>.  
  
 
<math>y_0 = 6-\frac{45}{24} = \frac{33}{8}</math>
 
<math>y_0 = 6-\frac{45}{24} = \frac{33}{8}</math>
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and <math>XA +XB+XC = 3XB = 3\sqrt{7^2 + \left(\frac{33}{8}\right)^2} = 3\times\frac{65}{8}=\frac{195}{8}</math>
 
and <math>XA +XB+XC = 3XB = 3\sqrt{7^2 + \left(\frac{33}{8}\right)^2} = 3\times\frac{65}{8}=\frac{195}{8}</math>
  
==Solution 2==
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===Solution 2 (Algebra)===
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>:
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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>, such that <math>s=\frac{a+b+c}{2}</math> and <math>R</math> 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>
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After a few algebraic manipulations:
 
After a few algebraic manipulations:
  
<math>\Rightarrow R = \frac{65}{16} \Rightarrow D=2R=\frac{65}{8} \Rightarrow 3D = \boxed{\frac{195}{8}}</math>.
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<math>\Rightarrow R=\frac{65}{16} \Rightarrow XA = XB = XC = \frac{65}{8} \Rightarrow XA + XB + XC = \boxed{\frac{195}{8}}</math>.
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===Solution 3 (Homothety)===
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Let <math>O</math> be the circumcenter of <math>\triangle ABC,</math> and <math>h_A</math> denote the length of the altitude from <math>A.</math> Note that a homothety centered at <math>B</math> with ratio <math>\frac{1}{2}</math> takes the circumcircle of <math>\triangle BAC</math> to the circumcircle of <math>\triangle BDE</math>. It also takes the point diametrically opposite <math>B</math> on the circumcircle of <math>\triangle BAC</math> to <math>O.</math> Therefore, <math>O</math> lies on the circumcircle of <math>\triangle BDE.</math> Similarly, it lies on the circumcircle of <math>\triangle CEF.</math> By Pythagorean triples, <math>h_A=12.</math> Finally, our answer is <cmath>3R=3\cdot \frac{abc}{4\{ABC\}}=3\cdot \frac{abc}{2ah_A}=3\cdot \frac{bc}{2h_A}=\boxed{\frac{195}{8}.}</cmath>
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 +
 
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===Solution 4 (basically Solution 1 but without coordinates)===
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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>
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==Solution 5==
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<center>
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[[File:Screen Shot 2021-08-06 at 7.30.10 PM.png|300px]]
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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>
  
 
== 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}}
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{{MAA Notice}}

Revision as of 22:33, 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

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 $\triangle CED$ 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 (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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All AMC 12 Problems and Solutions

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