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

(Created page with "== Problem == Triangle <math>ABC</math> has <math>AB=27</math>, <math>AC=26</math>, and <math>BC=25</math>. Let <math>I</math> denote the intersection of the internal angle bis...")
 
(→‎Solution)
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== Solution ==
 
== Solution ==
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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> 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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By Herons formula for the area of a triangle we have that the area of triangle <math>ABC</math> is <math>\sqrt{39(14)(13)(12)}</math>. On the other hand the area is given by <math>(1/2)25r+(1/2)26r+(1/2)27r</math>.  Then <math>39r=\sqrt{39(14)(13)(12)}</math> so that <math>r^2=56</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>.
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC12 box|year=2012|ab=A|num-b=17|num-a=19}}
 
{{AMC12 box|year=2012|ab=A|num-b=17|num-a=19}}

Revision as of 05:03, 13 February 2012

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

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 $O$ 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 Herons 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 $BI=15$.

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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