2012 AMC 12A Problems/Problem 18: Difference between revisions

Revision as of 22:00, 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

Solution 3

By heron's formula, $sr=\sqrt{s(s-a)(s-b)(s-c)}$ so we have $r^2=\frac{14\cdot 13 \cdot 12}{39}$ . Hence $r=\sqrt{56}$ .

By the law of cosines on $\triangle ABC$ , $\cos B = \frac{25^2+27^2-26^2}{2\cdot25\cdot27}=\frac{113}{225}$ Also $\sin^2 \frac{B}{2}=\frac{1-\cosB}{2}$ (Error compiling LaTeX. Unknown error_msg). So, $\sin \frac{B}{2}=\frac{2\sqrt{14}}{\sqrt{15}}$ .

The area of $\triangle BIC$ can be calculated in 2 ways, $\frac{1}{2} \cdot 25 r=\frac{1}{2}BI \cdot 25 \cdot \sin \frac{B}{2}$ . Solving this equation yields $BI=15$

@@ Line 18: / Line 18: @@
 -Solution by '''arowaaron'''
+==Solution 3 ==
+By heron's formula,
+<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=\sqrt{56}</math>.
+By the law of cosines on <math>\triangle ABC</math>,
+<math>\cos B = \frac{25^2+27^2-26^2}{2\cdot25\cdot27}=\frac{113}{225}</math>
+Also <math>\sin^2 \frac{B}{2}=\frac{1-\cosB}{2}</math>. So, <math>\sin \frac{B}{2}=\frac{2\sqrt{14}}{\sqrt{15}}</math>.
+The area of <math>\triangle BIC</math> can be calculated in 2 ways,
+<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 ==

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