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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 <math>21</math>.  Given that <math>AP=23</math> and <math>PB=27,</math> find the [[perimeter]] of the triangle.
  
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__TOC__
 
== Solution ==
 
== Solution ==
 +
<center><asy>
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pathpen = black + linewidth(0.65); pointpen = black;
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pair A=(0,0),B=(50,0),C=IP(circle(A,23+245/2),circle(B,27+245/2)), I=incenter(A,B,C);
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path P = incircle(A,B,C);
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D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle);D(P);
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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));
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MP("23",(A+Q)/2,W);MP("27",(B+R)/2,E);
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</asy></center>
 +
 +
=== Solution 1 ===
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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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<cmath>
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\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 = \frac{245}{2} 
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\end{eqnarray*}
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</cmath>
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We want the perimeter, which is <math>2s = 2\left(50 + \frac{245}{2}\right) = \boxed{345}</math>.
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=== 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>
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 +
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>
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==Solution 3==
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 +
Let unknown side has length as <math>x</math>, Assume three sides of triangles are <math>a,b,c</math>, the area of the triangle is <math>S</math>.
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Note that <math>r=\frac{2S}{a+b+c}=21,S=1050+21x</math>
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<math>\tan\angle{\frac{B}{2}}=\frac{7}{9}, \tan\angle{B}=\frac{63}{16}</math>. Use trig identity, knowing that <math>1+\cot^2\angle{B}=\csc^2\angle{B}</math>, getting that <math>\sin\angle{B}=\frac{63}{65}</math>
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 +
Now equation <math>(x+27)*50*\frac{63}{65}*\frac{1}{2}=1050+21x; x=\frac{245}{2}</math>, the final answer is <math>245+100=345</math>
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~bluesoul
  
 
== See also ==
 
== See also ==
−
* [[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]]
 +
{{MAA Notice}}

Latest revision as of 18:03, 21 June 2024

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 $2\cdot\frac {245}2 + 2\cdot 23 + 2\cdot 27 = \boxed{345}.$


Solution 3

Let unknown side has length as $x$, Assume three sides of triangles are $a,b,c$, the area of the triangle is $S$.

Note that $r=\frac{2S}{a+b+c}=21,S=1050+21x$

$\tan\angle{\frac{B}{2}}=\frac{7}{9}, \tan\angle{B}=\frac{63}{16}$. Use trig identity, knowing that $1+\cot^2\angle{B}=\csc^2\angle{B}$, getting that $\sin\angle{B}=\frac{63}{65}$

Now equation $(x+27)*50*\frac{63}{65}*\frac{1}{2}=1050+21x; x=\frac{245}{2}$, the final answer is $245+100=345$ ~bluesoul

See also

1999 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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All AIME Problems and Solutions

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