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

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~EricShi1685
 
~EricShi1685
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== Solution 4 ==
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Extend line <math>BI</math> to <math>D</math>. Where <math>D</math> is on line <math>AC</math>. Now lets say <math>AD = x</math> and <math>DC = y</math>. Because of the Angle Bisector Theorem, we can say <math>\frac{AB}{BC} = \frac{x}{y}</math>. Knowing that <math>x + y = 26</math> we can solve for <math>x</math> and <math>y</math>. We would get <math>x</math> as <math>\frac{27}{2}</math> and <math>y</math> <math>\frac{25}{2}</math>. By using the Angle Bisector Theorem again, in <math>\triangle ABD</math>, we find that the <math>2 \cdot ID = BD</math>. This is useful because we just need to find <math>BD</math> and multiply by <math>\frac{2}{3}</math> in order to find <math>BI</math>.
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Now we can use the Law of Cosines to find <math>BD</math>. First, we gotta find <math>Cos(a)</math>, and we can use the Law of Cosines for that. Using <math>\traingle ABC</math> we can write the equation, <math>BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot Cos(A)</math>. Plugging in the values we get, <math>25^2 = 27^2 + 26^2 - 2 \cdot 27 \cdot 26 \cdot Cos(a)</math>. Solving for <math>cos(a)</math> we get <math>\frac{5}{9}</math>. We can use <math>cos(a)</math> to find <math>BD</math> by using the Law of Cosines. Using it would give us <math>BD^2 = x^2 + AB^2 - 2 \cdot AB \cdot x \cdot cos(a)</math>. Plugging in the variables would give us, <math>BD^2 = \frac{27^2}{4} + 27^2 - 2 \cdot 27 \cdot \frac{27}{2} \cdot \frac{5}{9}</math>. Solving for <math>BD</math> would give us <math>\frac{135}{6}</math> and multipling by <math>\frac{2}{3}</math> would give us <math>BI</math>, which is <math>\boxed{\textbf{(A) } 15}</math>
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-Andyyang01
  
 
== See Also ==
 
== See Also ==

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

Solution 4

Extend line $BI$ to $D$. Where $D$ is on line $AC$. Now lets say $AD = x$ and $DC = y$. Because of the Angle Bisector Theorem, we can say $\frac{AB}{BC} = \frac{x}{y}$. Knowing that $x + y = 26$ we can solve for $x$ and $y$. We would get $x$ as $\frac{27}{2}$ and $y$ $\frac{25}{2}$. By using the Angle Bisector Theorem again, in $\triangle ABD$, we find that the $2 \cdot ID = BD$. This is useful because we just need to find $BD$ and multiply by $\frac{2}{3}$ in order to find $BI$.

Now we can use the Law of Cosines to find $BD$. First, we gotta find $Cos(a)$, and we can use the Law of Cosines for that. Using $\traingle ABC$ (Error compiling LaTeX. Unknown error_msg) we can write the equation, $BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot Cos(A)$. Plugging in the values we get, $25^2 = 27^2 + 26^2 - 2 \cdot 27 \cdot 26 \cdot Cos(a)$. Solving for $cos(a)$ we get $\frac{5}{9}$. We can use $cos(a)$ to find $BD$ by using the Law of Cosines. Using it would give us $BD^2 = x^2 + AB^2 - 2 \cdot AB \cdot x \cdot cos(a)$. Plugging in the variables would give us, $BD^2 = \frac{27^2}{4} + 27^2 - 2 \cdot 27 \cdot \frac{27}{2} \cdot \frac{5}{9}$. Solving for $BD$ would give us $\frac{135}{6}$ and multipling by $\frac{2}{3}$ would give us $BI$, which is $\boxed{\textbf{(A) } 15}$

-Andyyang01

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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