Difference between revisions of "1999 AIME Problems/Problem 12"
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== Problem == | == Problem == | ||
− | The | + | 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__ | ||
== 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))); | ||
+ | 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></center> | ||
− | 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 | + | === Solution 1 === |
+ | 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, | ||
− | < | + | <cmath> |
+ | \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\\ | ||
− | 180x = 441 \cdot 50 \Longrightarrow x | + | 180x = 441 \cdot 50 &\Longrightarrow & x = \frac{245}{2} |
− | + | \end{eqnarray*} | |
+ | </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 === | ||
+ | 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 | ||
+ | <cmath> | ||
+ | \begin{eqnarray*} \tan \alpha & = & \frac {21}{27} \\ | ||
+ | \tan \beta & = & \frac {21}{23} \\ | ||
+ | \tan \gamma & = & \frac {21}x. \end{eqnarray*} | ||
+ | </cmath> | ||
+ | So naturally we look at <math>\tan \gamma.</math> But since <math>\gamma = \frac \pi2 - (\beta + \alpha)</math> we have | ||
+ | <cmath> | ||
+ | \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*} | ||
+ | </cmath> | ||
+ | Doing the algebra, we get <math>x = \frac {245}2.</math> | ||
+ | |||
+ | 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 == | == See also == | ||
Line 19: | Line 63: | ||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 19:03, 21 June 2024
Problem
The inscribed circle of triangle is tangent to
at
and its radius is
. Given that
and
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]](http://latex.artofproblemsolving.com/c/d/1/cd1d3a517ae3f4ddc9470481bb6e69b4808fbfaf.png)
Solution 1
Let be the tangency point on
, and
on
. By the Two Tangent Theorem,
,
, and
. Using
, where
, we get
. By Heron's formula,
. Equating and squaring both sides,
We want the perimeter, which is .
Solution 2
Let the incenter be denoted . It is commonly known that the incenter is the intersection of the angle bisectors of a triangle. So let
and
We have that
So naturally we look at
But since
we have
Doing the algebra, we get
The perimeter is therefore
Solution 3
Let unknown side has length as , Assume three sides of triangles are
, the area of the triangle is
.
Note that
. Use trig identity, knowing that
, getting that
Now equation , the final answer is
~bluesoul
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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