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

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== Problem ==
 
== Problem ==
  
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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 bisectors of <math>\triangle ABC</math>.  What is <math>BI</math>?
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Triangle <math>ABC</math> has <math>AB=27</math>, <math>AC=26</math>, and <math>BC=25</math>.  Let <math>I</math> be the intersection of the internal angle bisectors of <math>\triangle ABC</math>.  What is <math>BI</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>
 
<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 ==
  
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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>.  
+
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>.  
  
−
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>.
+
By Heron's 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>.
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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=\boxed{\textbf{(A) } 15}</math>.
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 +
== Solution 2 ==
 +
 
 +
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 <math>25</math>, <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 <math>52</math>. 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{\textbf{(A) } 15}</math>
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 +
-Solution by '''arowaaron'''
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 +
== Solution 3 ==
 +
 
 +
We can use POP(Power of a point) to solve this problem. First, notice that the area of <math>\triangle ABC</math> is <math>\sqrt{39(39 - 27)(39 - 26)(39 - 25)} = 78\sqrt{14}</math>. Therefore, using the formula that <math>sr = A</math>, where <math>s</math> is the semi-perimeter and <math>r</math> is the length of the inradius, we find that <math>r = 2\sqrt{14}</math>.
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 +
Draw radii to the three tangents, and let the tangent hitting <math>BC</math> be <math>T_1</math>, the tangent hitting <math>AB</math> be <math>T_2</math>, and the tangent hitting <math>AC</math> be <math>T_3</math>. Let <math>BI = x</math>. By the pythagorean theorem, we know that <math>BT_1 = \sqrt{x^2 - 56}</math>. By POP, we also know that <math>BT_2</math> is also <math>\sqrt{x^2 - 56}</math>. Because we know that <math>BC = 25</math>, we find that <math>CT_1 = 25 - \sqrt{x^2 - 56}</math>. We can rinse and repeat and find that <math>AT_2 = 26 - (25 - \sqrt{x^2 - 56}) = 1 + \sqrt{x^2 - 56}</math>. We can find <math>AT_2</math> by essentially coming in from the other way. Since <math>AB = 27</math>, we also know that <math>AT_3 = 27 - \sqrt{x^2 - 56}</math>. By POP, we know that <math>AT_2 = AT_3</math>, so <math>1 + \sqrt{x^2 - 56} = 27 - \sqrt{x^2 - 56}</math>.
 +
 
 +
Let <math>\sqrt{x^2 - 56} = A</math>, for simplicity. We can change the equation into <math>1 + A = 27 - A</math>, which we find <math>A</math> to be <math>13</math>. Therefore, <math>\sqrt{x^2 - 56} = 13</math>, which further implies that <math>x^2 - 56 = 169</math>. After simplifying, we find <math>x^2 = 225</math>, so <math>x = \boxed{\textbf{(A) } 15}</math>
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 +
~EricShi1685
  
 
== 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}}
 +
 +
[[Category:Introductory Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 18:07, 28 July 2021

Problem

Triangle $ABC$ has $AB=27$, $AC=26$, and $BC=25$. Let $I$ be 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 $BI=\boxed{\textbf{(A) } 15}$.

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 $\boxed{\textbf{(A) } 15}$

-Solution by arowaaron

Solution 3

We can use POP(Power of a point) to solve this problem. First, notice that the area of $\triangle ABC$ is $\sqrt{39(39 - 27)(39 - 26)(39 - 25)} = 78\sqrt{14}$. Therefore, using the formula that $sr = A$, where $s$ is the semi-perimeter and $r$ is the length of the inradius, we find that $r = 2\sqrt{14}$.

Draw radii to the three tangents, and let the tangent hitting $BC$ be $T_1$, the tangent hitting $AB$ be $T_2$, and the tangent hitting $AC$ be $T_3$. Let $BI = x$. By the pythagorean theorem, we know that $BT_1 = \sqrt{x^2 - 56}$. By POP, we also know that $BT_2$ is also $\sqrt{x^2 - 56}$. Because we know that $BC = 25$, we find that $CT_1 = 25 - \sqrt{x^2 - 56}$. We can rinse and repeat and find that $AT_2 = 26 - (25 - \sqrt{x^2 - 56}) = 1 + \sqrt{x^2 - 56}$. We can find $AT_2$ by essentially coming in from the other way. Since $AB = 27$, we also know that $AT_3 = 27 - \sqrt{x^2 - 56}$. By POP, we know that $AT_2 = AT_3$, so $1 + \sqrt{x^2 - 56} = 27 - \sqrt{x^2 - 56}$.

Let $\sqrt{x^2 - 56} = A$, for simplicity. We can change the equation into $1 + A = 27 - A$, which we find $A$ to be $13$. Therefore, $\sqrt{x^2 - 56} = 13$, which further implies that $x^2 - 56 = 169$. After simplifying, we find $x^2 = 225$, so $x = \boxed{\textbf{(A) } 15}$

~EricShi1685

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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