2012 AMC 12A Problems/Problem 18

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

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