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

Revision as of 14:57, 25 February 2022

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 , Assume three sides of triangles are , the area of the triangle is .

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

@@ Line 46: / Line 46: @@
 The perimeter is therefore <math>2\cdot\frac {245}2 + 2\cdot 23 + 2\cdot 27 = \boxed{345}.</math>
+==solution 3==
+ 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>.
+Note that <math>r=\frac{2S}{a+b+c}=21,S=1050+21x</math>
+<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>
+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>
+~bluesoul
 == See also ==

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