Difference between revisions of "1999 AIME Problems/Problem 12"
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The perimeter is therefore <math>2\cdot\frac {245}2 + 2\cdot 23 + 2\cdot 27 = \boxed{345}.</math> | The perimeter is therefore <math>2\cdot\frac {245}2 + 2\cdot 23 + 2\cdot 27 = \boxed{345}.</math> | ||
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== See also == | == See also == |
Revision as of 14:58, 25 February 2022
Problem
The inscribed circle of triangle is tangent to at and its radius is . 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 |
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