Difference between revisions of "1999 AIME Problems/Problem 12"
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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, | + | 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> | <cmath> |
Revision as of 15:39, 19 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 |