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

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==Solution 3 ==
 
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(By mastermind.hk16)
 
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By heron's formula,
 
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<math>sr=\sqrt{s(s-a)(s-b)(s-c)}</math> so we have <math>r^2=\frac{14\cdot 13 \cdot 12}{39}</math>. Hence <math>r=2\sqrt{14}</math>.
 
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By the law of cosines on <math>\triangle ABC</math>,
 
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<math>\cos B = \frac{25^2+27^2-26^2}{2\cdot25\cdot27}=\frac{113}{225}</math>
 
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Also <math>\sin^2 \frac{B}{2} = \frac{1-\cos B}{2}</math>. So, <math>\sin \frac{B}{2}=\frac{2\sqrt{14}}{15}</math>.
 
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The area of <math>\triangle BIC</math> can be calculated in 2 ways,
 
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<math>\frac{1}{2} \cdot 25 r=\frac{1}{2}BI \cdot 25 \cdot \sin \frac{B}{2}</math>. Solving this equation yields <math>BI=15</math>.
 
  
 
== See Also ==
 
== See Also ==

Revision as of 22:09, 11 February 2018

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 $\boxed{(A)\:15.}$

-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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