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Difference between revisions of "2012 AMC 12A Problems/Problem 18"

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<math> \textbf{(A)}\ 15\qquad\textbf{(B)}\ 5+\sqrt{26}+3\sqrt{3}\qquad\textbf{(C)}\ 3\sqrt{26}\qquad\textbf{(D)}\ \frac{2}{3}\sqrt{546}\qquad\textbf{(E)}\ 9\sqrt{3} </math>
 
<math> \textbf{(A)}\ 15\qquad\textbf{(B)}\ 5+\sqrt{26}+3\sqrt{3}\qquad\textbf{(C)}\ 3\sqrt{26}\qquad\textbf{(D)}\ \frac{2}{3}\sqrt{546}\qquad\textbf{(E)}\ 9\sqrt{3} </math>
  
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== Solution ==
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== Solution 1 ==
  
 
Inscribe circle <math>C</math> of radius <math>r</math> inside triangle <math>ABC</math> so that it meets <math>AB</math> at <math>Q</math>, <math>BC</math> at <math>R</math>, and <math>AC</math> at <math>S</math>. Note that angle bisectors of triangle <math>ABC</math> are concurrent at the center <math>O</math>(also <math>I</math>) of circle <math>C</math>. Let <math>x=QB</math>, <math>y=RC</math> and <math>z=AS</math>. Note that <math>BR=x</math>, <math>SC=y</math> and <math>AQ=z</math>. Hence <math>x+z=27</math>, <math>x+y=25</math>, and <math>z+y=26</math>. Subtracting the last 2 equations we have <math>x-z=-1</math> and adding this to the first equation we have <math>x=13</math>.  
 
Inscribe circle <math>C</math> of radius <math>r</math> inside triangle <math>ABC</math> so that it meets <math>AB</math> at <math>Q</math>, <math>BC</math> at <math>R</math>, and <math>AC</math> at <math>S</math>. Note that angle bisectors of triangle <math>ABC</math> are concurrent at the center <math>O</math>(also <math>I</math>) of circle <math>C</math>. Let <math>x=QB</math>, <math>y=RC</math> and <math>z=AS</math>. Note that <math>BR=x</math>, <math>SC=y</math> and <math>AQ=z</math>. Hence <math>x+z=27</math>, <math>x+y=25</math>, and <math>z+y=26</math>. Subtracting the last 2 equations we have <math>x-z=-1</math> and adding this to the first equation we have <math>x=13</math>.  
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Since the radius of circle <math>O</math> is perpendicular to <math>BC</math> at <math>R</math>, we have by the pythagorean theorem <math>BO^2=BI^2=r^2+x^2=56+169=225</math> so that <math>BI=15</math>.
 
Since the radius of circle <math>O</math> is perpendicular to <math>BC</math> at <math>R</math>, we have by the pythagorean theorem <math>BO^2=BI^2=r^2+x^2=56+169=225</math> so that <math>BI=15</math>.
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== Solution 2 ==
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We can use mass points and Stewart's to solve this problem. Because we are looking at the Incenter we then label <math>A</math> with a mass of 25, <math>B</math> with <math>26</math>, and <math>C</math> with <math>27</math>. We also label where the angle bisectors intersect the opposite side <math>A'</math>, <math>B'</math>, and <math>C'</math> correspondingly. It follows then that point <math>B'</math> has mass 52. Which means that <math>\overline{BB'}</math> is split into a <math>2:1</math> ratio. We can then use Stewart's to find <math>\overline{BB'}</math>. So we have <math>25^2\frac{27}{2} + 27^2\frac{25}{2} = \frac{25 \cdot 26 \cdot 27}{4} + 26\overline{BB'}^2</math>. Solving we get <math>\overline{BB'} = \frac{45}{2}</math>. Plugging it in we get <math>\overline{BI} = 15</math>. Therefore the answer is <math>\boxed{(E)\:6.}</math>
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-Solution by '''arowaaron'''
  
 
== See Also ==
 
== See Also ==

Revision as of 18:57, 16 July 2017

Problem

Triangle $ABC$ has $AB=27$, $AC=26$, and $BC=25$. Let $I$ denote the intersection of the internal angle bisectors of $\triangle ABC$. What is $BI$?

$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 5+\sqrt{26}+3\sqrt{3}\qquad\textbf{(C)}\ 3\sqrt{26}\qquad\textbf{(D)}\ \frac{2}{3}\sqrt{546}\qquad\textbf{(E)}\ 9\sqrt{3}$

Solution 1

Inscribe circle $C$ of radius $r$ inside triangle $ABC$ so that it meets $AB$ at $Q$, $BC$ at $R$, and $AC$ at $S$. Note that angle bisectors of triangle $ABC$ are concurrent at the center $O$(also $I$) of circle $C$. Let $x=QB$, $y=RC$ and $z=AS$. Note that $BR=x$, $SC=y$ and $AQ=z$. Hence $x+z=27$, $x+y=25$, and $z+y=26$. Subtracting the last 2 equations we have $x-z=-1$ and adding this to the first equation we have $x=13$.

By Heron's formula for the area of a triangle we have that the area of triangle $ABC$ is $\sqrt{39(14)(13)(12)}$. On the other hand the area is given by $(1/2)25r+(1/2)26r+(1/2)27r$. Then $39r=\sqrt{39(14)(13)(12)}$ so that $r^2=56$.

Since the radius of circle $O$ is perpendicular to $BC$ at $R$, we have by the pythagorean theorem $BO^2=BI^2=r^2+x^2=56+169=225$ so that $BI=15$.

Solution 2

We can use mass points and Stewart's to solve this problem. Because we are looking at the Incenter we then label $A$ with a mass of 25, $B$ with $26$, and $C$ with $27$. We also label where the angle bisectors intersect the opposite side $A'$, $B'$, and $C'$ correspondingly. It follows then that point $B'$ has mass 52. Which means that $\overline{BB'}$ is split into a $2:1$ ratio. We can then use Stewart's to find $\overline{BB'}$. So we have $25^2\frac{27}{2} + 27^2\frac{25}{2} = \frac{25 \cdot 26 \cdot 27}{4} + 26\overline{BB'}^2$. Solving we get $\overline{BB'} = \frac{45}{2}$. Plugging it in we get $\overline{BI} = 15$. Therefore the answer is $\boxed{(E)\:6.}$

-Solution by arowaaron

See Also

2012 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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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