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

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== Problem ==
 
== Problem ==
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The inscribed circle of triangle <math>\displaystyle ABC</math> is tangent to <math>\displaystyle \overline{AB}</math> at <math>\displaystyle P_{},</math>  and its radius is 21.  Given that <math>\displaystyle AP=23</math> and <math>\displaystyle PB=27,</math> find the perimeter of the triangle.
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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.
  
 
== Solution ==
 
== Solution ==
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[[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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<math></math>\begin{eqnarray*}
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[21(50+x)]^2 &=& (50+x)(x)(621)\\
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441(50+x) &=& 621x\\
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180x = 441 \cdot 50 \Longrightarrow x <math>=</math> \frac{245}{2} 
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<math></math>
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We want the perimeter, which is <math>2s = 2\left(50 + \frac{245}{2}\right) = \boxed{345}</math>.
  
 
== See also ==
 
== See also ==
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* [[1999_AIME_Problems/Problem_11|Previous Problem]]
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{{AIME box|year=1999|num-b=11|num-a=13}}
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* [[1999_AIME_Problems/Problem_13|Next Problem]]
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* [[1999 AIME Problems]]
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[[Category:Intermediate Geometry Problems]]

Revision as of 17:22, 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

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,

$$ (Error compiling LaTeX. Unknown error_msg)\begin{eqnarray*} [21(50+x)]^2 &=& (50+x)(x)(621)\\ 441(50+x) &=& 621x\\ 180x = 441 \cdot 50 \Longrightarrow x $=$ \frac{245}{2} $$ (Error compiling LaTeX. Unknown error_msg)

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

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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