Difference between revisions of "1999 AIME Problems/Problem 12"
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== Problem == | == Problem == | ||
− | The inscribed circle of triangle <math> | + | 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. |
== Solution == | == Solution == | ||
+ | [[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, | ||
+ | |||
+ | <math></math>\begin{eqnarray*} | ||
+ | [21(50+x)]^2 &=& (50+x)(x)(621)\ | ||
+ | 441(50+x) &=& 621x\ | ||
+ | 180x = 441 \cdot 50 \Longrightarrow x <math>=</math> \frac{245}{2} | ||
+ | <math></math> | ||
+ | |||
+ | We want the perimeter, which is <math>2s = 2\left(50 + \frac{245}{2}\right) = \boxed{345}</math>. | ||
== See also == | == See also == | ||
− | + | {{AIME box|year=1999|num-b=11|num-a=13}} | |
− | + | ||
− | + | [[Category:Intermediate Geometry Problems]] |
Revision as of 17:22, 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.
Solution
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,
$$ (Error compiling LaTeX. Unknown error_msg)\begin{eqnarray*} [21(50+x)]^2 &=& (50+x)(x)(621)\ 441(50+x) &=& 621x\ 180x = 441 \cdot 50 \Longrightarrow x \frac{245}{2} $$ (Error compiling LaTeX. Unknown error_msg)
We want the perimeter, which is .
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 |