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

Revision as of 15:59, 2 July 2020

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

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 $AT_2 = 26 - (25 - \sqrt{x^2 - 56}) = 1 + \sqrt{x^2 - 56}$ . However, how we try to find $AT_2$ in another way and relate the two expressions. 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 $x = \boxed{\textbf{(A) } 15}$

@@ Line 18: / Line 18: @@
 -Solution by '''arowaaron'''
+== 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>.
+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>. However, how we try to find <math>AT_2</math> in another way and relate the two expressions. 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>
 == 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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