Difference between revisions of "2011 AMC 12B Problems/Problem 22"
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==Problem== | ==Problem== | ||
− | Let <math>T_1</math> be a triangle with | + | Let <math>T_1</math> be a triangle with side lengths <math>2011</math>, <math>2012</math>, and <math>2013</math>. For <math>n \geq 1</math>, if <math>T_n = \Delta ABC</math> and <math>D, E</math>, and <math>F</math> are the points of tangency of the incircle of <math>\Delta ABC</math> to the sides <math>AB</math>, <math>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>\left(T_n\right)</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> | <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> | ||
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== See also == | == See also == | ||
+ | Identical problem to the [[2011 AMC 10B Problems/Problem 25]]. | ||
+ | |||
{{AMC12 box|year=2011|num-b=21|num-a=23|ab=B}} | {{AMC12 box|year=2011|num-b=21|num-a=23|ab=B}} | ||
+ | {{MAA Notice}} |
Latest revision as of 17:24, 29 July 2021
Problem
Let be a triangle with side lengths
,
, 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
Identical problem to the 2011 AMC 10B Problems/Problem 25.
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 |
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