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

(solution)
(solution 2 by joml88, revise)
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The [[inscribed circle]] of [[triangle]] <math>ABC</math> is [[tangent]] to <math>\overline{AB}</math> at <math>P_{},</math>  and its [[radius]] is 21.  Given that <math>AP=23</math> and <math>PB=27,</math> find the [[perimeter]] of the triangle.
 
The [[inscribed circle]] of [[triangle]] <math>ABC</math> is [[tangent]] to <math>\overline{AB}</math> at <math>P_{},</math>  and its [[radius]] is 21.  Given that <math>AP=23</math> and <math>PB=27,</math> find the [[perimeter]] of the triangle.
  
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__TOC__
 
== Solution ==
 
== Solution ==
 +
=== Solution 1 ===
 
[[Image:1999_AIME-12.png]]
 
[[Image:1999_AIME-12.png]]
  
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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,  
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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,  
  
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<math></math>\begin{eqnarray*}  
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<cmath>
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\begin{eqnarray*}  
 
[21(50+x)]^2 &=& (50+x)(x)(621)\\
 
[21(50+x)]^2 &=& (50+x)(x)(621)\\
 
441(50+x) &=& 621x\\
 
441(50+x) &=& 621x\\
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180x = 441 \cdot 50 \Longrightarrow x <math>=</math> \frac{245}{2}   
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180x = 441 \cdot 50 &\Longrightarrow & x = \frac{245}{2}   
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<math></math>
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\end{eqnarray*}
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</cmath>
  
 
We want the perimeter, which is <math>2s = 2\left(50 + \frac{245}{2}\right) = \boxed{345}</math>.
 
We want the perimeter, which is <math>2s = 2\left(50 + \frac{245}{2}\right) = \boxed{345}</math>.
 +
 +
=== Solution 2 ===
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Let the incenter be denoted <math>I</math>.  It is commonly known that the incenter is the intersection of the angle bisectors of a triangle.  So let <math>\angle ABI = \angle CBI = \alpha, \angle BAI = \angle CAI = \beta,</math> and <math>\angle BCI = \angle ACI = \gamma.</math>
 +
 +
We have that
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<cmath>
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\begin{eqnarray*} \tan \alpha & = & \frac {21}{27} \\
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\tan \beta & = & \frac {21}{23} \\
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\tan \gamma & = & \frac {21}x. \end{eqnarray*}
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</cmath>
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So naturally we look at <math>\tan \gamma.</math>  But since <math>\gamma = \frac \pi2 - (\beta + \alpha)</math> we have
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<cmath>
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\begin{eqnarray*} \tan \gamma & = & \tan\left(\frac \pi2 - (\beta + \alpha)\right) \\
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& = & \frac 1{\tan(\alpha + \beta)} \\
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\Rightarrow \frac {21}x & = & \frac {1 - \frac {21\cdot 21}{23\cdot 27}}{\frac {21}{27} + \frac {21}{23}} \end{eqnarray*}
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</cmath>
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Doing the algebra, we get <math>x = \frac {245}2.</math>
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 +
The perimeter is therefore <math>2\cdot\frac {245}2 + 2\cdot 23 + 2\cdot 27 = \boxed{345}.</math>
  
 
== See also ==
 
== See also ==

Revision as of 18:30, 14 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 $2\cdot\frac {245}2 + 2\cdot 23 + 2\cdot 27 = \boxed{345}.$

See also

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
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