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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>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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The inscribed circle of triangle <math>ABC</math> is [[tangent]] to <math>\overline{AB}</math> at <math>P_{},</math>  and its [[radius]] is <math>21</math>.  Given that <math>AP=23</math> and <math>PB=27,</math> find the [[perimeter]] of the triangle.
  
 
__TOC__
 
__TOC__
 
== Solution ==
 
== Solution ==
 +
<center><asy>
 +
pathpen = black + linewidth(0.65); pointpen = black;
 +
pair A=(0,0),B=(50,0),C=IP(circle(A,23+245/2),circle(B,27+245/2)), I=incenter(A,B,C);
 +
path P = incircle(A,B,C);
 +
D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle);D(P);
 +
D(MP("P",IP(A--B,P)));
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pair Q=IP(C--A,P),R=IP(B--C,P);
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D(MP("R",R,NE));D(MP("Q",Q,NW));
 +
MP("23",(A+Q)/2,W);MP("27",(B+R)/2,E);
 +
</asy></center>
 +
 
=== Solution 1 ===
 
=== Solution 1 ===
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[[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,  
  

Revision as of 12:11, 26 April 2008

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

[asy] pathpen = black + linewidth(0.65); pointpen = black; pair A=(0,0),B=(50,0),C=IP(circle(A,23+245/2),circle(B,27+245/2)), I=incenter(A,B,C); path P = incircle(A,B,C); D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle);D(P); D(MP("P",IP(A--B,P))); pair Q=IP(C--A,P),R=IP(B--C,P); D(MP("R",R,NE));D(MP("Q",Q,NW)); MP("23",(A+Q)/2,W);MP("27",(B+R)/2,E); [/asy]

Solution 1

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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