Difference between revisions of "2007 AIME II Problems/Problem 15"

Revision as of 18:43, 30 March 2007

Problem

Four circles $\omega,$ $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ with the same radius are drawn in the interior of triangle $ABC$ such that $\omega_{A}$ is tangent to sides $AB$ and $AC$ , $\omega_{B}$ to $BC$ and $BA$ , $\omega_{C}$ to $CA$ and $CB$ , and $\omega$ is externally tangent to $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ . If the sides of triangle $ABC$ are $13,$ $14,$ and $15,$ the radius of $\omega$ can be represented in the form $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution

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Solution 1

First, apply Heron's formula to find that the area is $\sqrt{21 \cdot 8 \cdot 7 \cdot 6} = 84$ . Also the semiperimeter is $21$ . So the inradius is $\frac{A}{s} = \frac{84}{21} = 4$ .

Now consider the incenter I. Let the radius of one of the small circles be $r$ . Let the centers of the three little circles tangent to the sides of $\triangle ABC$ be $X$ , $Y$ , and $Z$ . Let the centre of the circle tangent to those three circles be P. A homothety centered at $I$ takes $XYZ$ to $ABC$ with factor $\frac{4 - r}{4}$ . The same homothety takes $P$ to the circumcentre of $\triangle ABC$ , so $\frac{PX}R = \frac{2r}R = \frac{4 - r}4$ , where $R$ is the circumradius of $\triangle ABC$ . The circumradius of $\triangle ABC$ can be easily computed by $R = \frac a{2\sin A}$ , so doing that reveals $R = \frac{65}8$ . Then $\frac{2r}{\frac{65}{8}} = \frac{(4-r)}4 \Longrightarrow \frac{16r}{65} = \frac{1 - r}4 \Longrightarrow \frac{129r}{260} = 1 \Longrightarrow r = \frac{260}{129}$ , so the answer is $389$ .

Solution 2

Consider a 13-14-15 triangle. $A=84.$ [By Heron's Formula or by 5-12-13 and 9-12-15 right triangles.]

The inradius is $r=\frac{A}{s}=\frac{84}{21}=4$ , where $s$ is the semiperimeter. Scale the triangle with the inradius by a linear scale factor, $u.$

The circumradius is $R=\frac{abc}{4rs}=\frac{13\cdot 14\cdot 15}{4\cdot 4\cdot 21}=\frac{65}{8},$ where $a,$ $b,$ and $c$ are the side-lengths. Scale the triangle with the circumradius by a linear scale factor, $v$ .

Cut and combine the triangles, as shown. Then solve for 4u:

$\frac{65}{8}v=8u$

$v=\frac{64}{65}u$

$\displaystyle u+v=1$

$u+\frac{64}{65}u=1$

$\frac{129}{65}u=1$

$4u=\frac{260}{129}$

The solution is $260+129=389$ .

@@ Line 12: / Line 12: @@
 === Solution 2 ===
 [[Image:2007 AIME II-15b.gif]]
 Consider a 13-14-15 triangle.  <math>A=84.</math>  [By Heron's Formula or by 5-12-13 and 9-12-15 right triangles.]
 The inradius is <math>r=\frac{A}{s}=\frac{84}{21}=4</math>, where <math>s</math> is the semiperimeter.  Scale the triangle with the inradius by a linear scale factor, <math>u.</math>
-The circumradius is <math>R=\frac{abc}{4rs}=\frac{13\cdot 14\cdot 15}{4\cdot 4\cdot 21}=\frac{65}{8},</math> where <math>a,</math> <math>b,</math> and <math>c</math> are the side-lengths.  Scale the triangle with the circumradius by a linear scale factor, <math>v</math>.
+The circumradius is <math>R=\frac{abc}{4rs}=\frac{13\cdot 14\cdot 15}{4\cdot 4\cdot 21}=\frac{65}{8},</math> where <math>a,</math> <math>b,</math> and <math>c</math> are the side-lengths.  Scale the triangle with the circumradius by a [[line]]ar scale factor, <math>v</math>.
 Cut and combine the triangles, as shown.  Then solve for 4u:
-<math>\frac{65}{8}v=8u</math>
+:<math>\frac{65}{8}v=8u</math>
-<math>v=\frac{64}{65}u</math>
+:<math>v=\frac{64}{65}u</math>
-<math>u+v=1</math>
+:<math>\displaystyle u+v=1</math>
-<math>u+\frac{64}{65}u=1</math>
+:<math>u+\frac{64}{65}u=1</math>
-<math>\frac{129}{65}u=1</math>
+:<math>\frac{129}{65}u=1</math>
-<math>4u=\frac{260}{129}</math>
+:<math>4u=\frac{260}{129}</math>
-<math>260+129=389</math>
+The solution is <math>260+129=389</math>.
 == See also ==

2007 AIME II (Problems • Answer Key • Resources)
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