Difference between revisions of "2011 AMC 12B Problems/Problem 22"
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− | Now we need to find | + | Now we need to find when <math>T_n</math> fails the triangle inequality. So we need to find the last <math>n</math> such that <math>\frac{2012}{2^{n-1}} > 2</math> |
<math>\frac{2012}{2^{n-1}} > 2</math> | <math>\frac{2012}{2^{n-1}} > 2</math> |
Revision as of 00:10, 18 February 2013
Problem
Let be a triangle with sides , , and . For , if and , and are the points of tangency of the incircle of to the sides , , and , respectively, then is a triangle with side lengths , and , if it exists. What is the perimeter of the last triangle in the sequence ?
Solution
Answer: (D)
Let , , and
Then , and
Then , ,
Hence:
Note that and for , I claim that it is true for all , assume for induction that it is true for some , then
Furthermore, the average for the sides is decreased by a factor of 2 each time.
So is a triangle with side length , ,
and the perimeter of such is
Now we need to find when fails the triangle inequality. So we need to find the last such that
For , perimeter is
See also
2011 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |