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Difference between revisions of "2011 AMC 12B Problems/Problem 20"
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https://www.youtube.com/watch?v=Tbzhw9fYsDI
 
https://www.youtube.com/watch?v=Tbzhw9fYsDI
  
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==Solutions==
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==Solution 1 (Coordinates)==
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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>.
  
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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>
  
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===Solution 2 (Algebra)===
+
==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.  
 
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.  
  
Line 43: Line 42:
 
<math>\Rightarrow R=\frac{65}{16} \Rightarrow XA = XB = XC = \frac{65}{8} \Rightarrow XA + XB + XC = \boxed{\frac{195}{8}}</math>.
 
<math>\Rightarrow R=\frac{65}{16} \Rightarrow XA = XB = XC = \frac{65}{8} \Rightarrow XA + XB + XC = \boxed{\frac{195}{8}}</math>.
  
−
===Solution 3 (Homothety)===
+
==Solution 3 (Homothety)==
 
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>
 
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>
  
  
−
===Solution 4 (basically Solution 1 but without coordinates)===
+
==Solution 4 (basically Solution 1 but without coordinates)==
  
 
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>

Revision as of 03:38, 25 December 2022

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 $XA + XB + XC$?

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

Video Solution by Punxsutawney Phil

https://www.youtube.com/watch?v=Tbzhw9fYsDI

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.


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 $\triangle CED$ with respect to 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 \[R=\frac{65}{8} \implies AX+BX+CX=\boxed{\frac{195}{8}}.\]

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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