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Difference between revisions of "2013 AMC 12B Problems/Problem 24"
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==Solution==
 
==Solution==
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Let <math>BN=x</math> and <math>NA=y</math>. From the conditions, let's deduct some convenient conditions that seems sufficient to solve the problem.
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'''<math>M</math> is the midpoint of side <math>AC</math>.'''
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 +
This implies that <math>[ABX]=[CBX]</math>. Given that angle <math>ABX</math> is <math>60</math> degrees and angle <math>BXC</math> is <math>120</math> degrees, we can use the area formula to get
 +
 +
<cmath>\frac{1}{2}(x+y)x \frac{\sqrt{3}}{2} = \frac{1}{2} x \cdot CX \frac{\sqrt{3}}{2}</cmath>
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 +
So, <math>x+y=CX</math> .....(1)
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 +
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'''<math>CN</math> is angle bisector.'''
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 +
In the triangle <math>ABC</math>, one has <math>BC/AC=x/y</math>, therefore <math>BC=2x/y</math>.....(2)
 +
 +
Furthermore, triangle <math>BCN</math> is similar to triangle <math>MCX</math>, so <math>BC/CM=CN/CX</math>, therefore <math>BC = (CX+x)/CX = (2x+y)/(x+y)</math>....(3)
 +
 +
By (2) and (3) and the subtraction law of ratios, we get
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<cmath>BC=2x/y = (2x+y)/(y+x) = y/x</cmath>
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Therefore <math>2x^2=y^2</math>, or <math>y=\sqrt{2}x</math>. So <math>BC = 2x/(\sqrt{2}x) = \sqrt{2}</math>.
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Finally, using the law of cosine for triangle <math>BCN</math>, we get
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<cmath>2 = BC^2 = x^2 + (2x+y)^2 - x(2x+y) = 3x^2 + 3xy + y^2 = (5+3\sqrt{2})x^2</cmath>
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<cmath>x^2 = \frac{2}{5+3\sqrt{2}} = \frac{10-6\sqrt{2}}{7}.</cmath>
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2013|ab=B|num-b=23|num-a=25}}
 
{{AMC12 box|year=2013|ab=B|num-b=23|num-a=25}}

Revision as of 13:28, 25 February 2013

Problem

Let $ABC$ be a triangle where $M$ is the midpoint of $\overline{AC}$, and $\overline{CN}$ is the angle bisector of $\angle{ACB}$ with $N$ on $\overline{AB}$. Let $X$ be the intersection of the median $\overline{BM}$ and the bisector $\overline{CN}$. In addition $\triangle BXN$ is equilateral with $AC=2$. What is $BN^2$?

$\textbf{(A)}\ \frac{10-6\sqrt{2}}{7} \qquad \textbf{(B)}\ \frac{2}{9} \qquad \textbf{(C)}\ \frac{5\sqrt{2}-3\sqrt{3}}{8} \qquad \textbf{(D)}\ \frac{\sqrt{2}}{6} \qquad \textbf{(E)}\ \frac{3\sqrt{3}-4}{5}$

Solution

Let $BN=x$ and $NA=y$. From the conditions, let's deduct some convenient conditions that seems sufficient to solve the problem.


$M$ is the midpoint of side $AC$.

This implies that $[ABX]=[CBX]$. Given that angle $ABX$ is $60$ degrees and angle $BXC$ is $120$ degrees, we can use the area formula to get

\[\frac{1}{2}(x+y)x \frac{\sqrt{3}}{2} = \frac{1}{2} x \cdot CX \frac{\sqrt{3}}{2}\]

So, $x+y=CX$ .....(1)


$CN$ is angle bisector.

In the triangle $ABC$, one has $BC/AC=x/y$, therefore $BC=2x/y$.....(2)

Furthermore, triangle $BCN$ is similar to triangle $MCX$, so $BC/CM=CN/CX$, therefore $BC = (CX+x)/CX = (2x+y)/(x+y)$....(3)

By (2) and (3) and the subtraction law of ratios, we get

\[BC=2x/y = (2x+y)/(y+x) = y/x\]

Therefore $2x^2=y^2$, or $y=\sqrt{2}x$. So $BC = 2x/(\sqrt{2}x) = \sqrt{2}$.

Finally, using the law of cosine for triangle $BCN$, we get

\[2 = BC^2 = x^2 + (2x+y)^2 - x(2x+y) = 3x^2 + 3xy + y^2 = (5+3\sqrt{2})x^2\]

\[x^2 = \frac{2}{5+3\sqrt{2}} = \frac{10-6\sqrt{2}}{7}.\]

See also

2013 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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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