Difference between revisions of "1999 AIME Problems/Problem 12"

Revision as of 16:39, 19 October 2007

Problem

The inscribed circle of triangle $ABC$ is tangent to $\overline{AB}$ at $P_{},$ and its radius is 21. Given that $AP=23$ and $PB=27,$ find the perimeter of the triangle.

Solution

Solution 1

File:1999 AIME-12.png

Let $Q$ be the tangency point on $\overline{AC}$ , and $R$ on $\overline{BC}$ . By the Two Tangent Theorem, $AP = AQ = 23$ , $BP = BR = 27$ , and $CQ = CR = x$ . Using $rs = A$ , where $s = \frac{27 \cdot 2 + 23 \cdot 2 + x \cdot 2}{2} = 50 + x$ , we get $(21)(50 + x) = A$ . By Heron's formula, $A = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{(50+x)(x)(23)(27)}$ . Equating and squaring both sides,

$\begin{eqnarray*} [21(50+x)]^2 &=& (50+x)(x)(621)\\ 441(50+x) &=& 621x\\ 180x = 441 \cdot 50 &\Longrightarrow & x = \frac{245}{2} \end{eqnarray*}$

We want the perimeter, which is $2s = 2\left(50 + \frac{245}{2}\right) = \boxed{345}$ .

Solution 2

Let the incenter be denoted $I$ . It is commonly known that the incenter is the intersection of the angle bisectors of a triangle. So let $\angle ABI = \angle CBI = \alpha, \angle BAI = \angle CAI = \beta,$ and $\angle BCI = \angle ACI = \gamma.$

We have that $\begin{eqnarray*} \tan \alpha & = & \frac {21}{27} \\ \tan \beta & = & \frac {21}{23} \\ \tan \gamma & = & \frac {21}x. \end{eqnarray*}$ So naturally we look at $\tan \gamma.$ But since $\gamma = \frac \pi2 - (\beta + \alpha)$ we have $\begin{eqnarray*} \tan \gamma & = & \tan\left(\frac \pi2 - (\beta + \alpha)\right) \\ & = & \frac 1{\tan(\alpha + \beta)} \\ \Rightarrow \frac {21}x & = & \frac {1 - \frac {21\cdot 21}{23\cdot 27}}{\frac {21}{27} + \frac {21}{23}} \end{eqnarray*}$ Doing the algebra, we get $x = \frac {245}2.$

The perimeter is therefore $2\cdot\frac {245}2 + 2\cdot 23 + 2\cdot 27 = \boxed{345}.$

@@ Line 7: / Line 7: @@
 [[Image:1999_AIME-12.png]]
-Let <math>Q</math> be the tangency point on <math>\overline{AC}</math>, and <math>R</math> on <math>\overline{BC}</math>. By the Two Tangent Theorem, <math>AP = AQ = 23</math>, <math>BP = BR = 27</math>, and <math>CQ = CR = x</math>. Using <math>rs = A</math>, where <math>s = \frac{27 \cdot 2 + 23 \cdot 2 + x \cdot 2}{2} = 50 + x</math>, we get <math>(21)(50 + x) = A</math>. By [[Heron's formula]], <math>A = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{(50+x)(x)(23)(27)}</math>. Equating and squaring both sides,
+Let <math>Q</math> be the tangency point on <math>\overline{AC}</math>, and <math>R</math> on <math>\overline{BC}</math>. By the [[Two Tangent Theorem]], <math>AP = AQ = 23</math>, <math>BP = BR = 27</math>, and <math>CQ = CR = x</math>. Using <math>rs = A</math>, where <math>s = \frac{27 \cdot 2 + 23 \cdot 2 + x \cdot 2}{2} = 50 + x</math>, we get <math>(21)(50 + x) = A</math>. By [[Heron's formula]], <math>A = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{(50+x)(x)(23)(27)}</math>. Equating and squaring both sides,
 <cmath>

1999 AIME (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search