Difference between revisions of "2011 AMC 10B Problems/Problem 25"
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Let <math>T_1</math> be a triangle with sides <math>2011, 2012,</math> and <math>2013</math>. For <math>n \ge 1</math>, if <math>T_n = \triangle ABC</math> and <math>D, E,</math> and <math>F</math> are the points of tangency of the incircle of <math>\triangle ABC</math> to the sides <math>AB, BC</math> and <math>AC,</math> respectively, then <math>T_{n+1}</math> is a triangle with side lengths <math>AD, BE,</math> and <math>CF,</math> if it exists. What is the perimeter of the last triangle in the sequence <math>( T_n )</math>? | Let <math>T_1</math> be a triangle with sides <math>2011, 2012,</math> and <math>2013</math>. For <math>n \ge 1</math>, if <math>T_n = \triangle ABC</math> and <math>D, E,</math> and <math>F</math> are the points of tangency of the incircle of <math>\triangle ABC</math> to the sides <math>AB, BC</math> and <math>AC,</math> respectively, then <math>T_{n+1}</math> is a triangle with side lengths <math>AD, BE,</math> and <math>CF,</math> if it exists. What is the perimeter of the last triangle in the sequence <math>( T_n )</math>? | ||
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<math> \textbf{(A)}\ \frac{1509}{8} \qquad\textbf{(B)}\ \frac{1509}{32} \qquad\textbf{(C)}\ \frac{1509}{64} \qquad\textbf{(D)}\ \frac{1509}{128} \qquad\textbf{(E)}\ \frac{1509}{256}</math> | <math> \textbf{(A)}\ \frac{1509}{8} \qquad\textbf{(B)}\ \frac{1509}{32} \qquad\textbf{(C)}\ \frac{1509}{64} \qquad\textbf{(D)}\ \frac{1509}{128} \qquad\textbf{(E)}\ \frac{1509}{256}</math> | ||
− | + | ==Solution== | |
+ | {{solution}} | ||
+ | ==See Also== | ||
+ | {{AMC10 box|year=2011|ab=B|num-b=24|after=Last Problem}} |
Revision as of 15:31, 6 June 2011
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
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See Also
2011 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
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 10 Problems and Solutions |