Difference between revisions of "1999 AIME Problems/Problem 12"
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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. | 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. | ||
+ | __TOC__ | ||
== Solution == | == Solution == | ||
+ | === Solution 1 === | ||
[[Image:1999_AIME-12.png]] | [[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 | + | 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> | ||
== See also == | == See also == |
Revision as of 17:30, 14 October 2007
Problem
The inscribed circle of triangle is tangent to at and its radius is 21. Given that and find the perimeter of the triangle.
Contents
[hide]Solution
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
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 |