Difference between revisions of "2011 AMC 10B Problems/Problem 25"
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== Problem == | == Problem == | ||
− | Let <math>T_1</math> be a triangle with side lengths <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 side lengths <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>? |
<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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<math>T_{11}</math> would have sides <math>\frac{247}{256}, \frac{503}{256}, \frac{759}{256}</math> but these lengths do not make a | <math>T_{11}</math> would have sides <math>\frac{247}{256}, \frac{503}{256}, \frac{759}{256}</math> but these lengths do not make a | ||
− | triangle as < | + | triangle as |
+ | <cmath>\frac{247}{256} + \frac{503}{256} < \frac{759}{256}</cmath> | ||
Likewise, you could create an equation instead of listing all the triangles to <math>T_{11}</math>. | Likewise, you could create an equation instead of listing all the triangles to <math>T_{11}</math>. | ||
− | The sides of a triangle <math>T_{k}</math> would be < | + | The sides of a triangle <math>T_{k}</math> would be |
− | We then have < | + | <cmath>\frac{503}{2^{k-3}} - 1, \frac{503}{2^{k-3}}, \frac{503}{2^{k-3}} + 1</cmath> |
+ | We then have | ||
+ | <cmath>503 - 2^{k-3} + 503 > 503 + 2^{k-3}</cmath> | ||
+ | <cmath>1006 - 2^{k-3} > 503 + 2^{k-3}</cmath> | ||
+ | <cmath>503 > 2^{k-2}</cmath> | ||
+ | <cmath>9 > k-2</cmath> | ||
+ | <cmath>k < 11</cmath> | ||
Hence, the first triangle which does not exist in this sequence is <math>T_{11}</math>. | Hence, the first triangle which does not exist in this sequence is <math>T_{11}</math>. | ||
− | + | Hence the perimeter is | |
− | Hence the perimeter is < | + | <cmath>\frac{375}{128} + \frac{503}{128} + \frac{631}{128} = \boxed{\textbf{(D)} \frac{1509}{128}}</cmath>. |
==Solution 2== | ==Solution 2== |
Latest revision as of 03:15, 3 November 2022
Contents
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 1
By constructing the bisectors of each angle and the perpendicular radii of the incircle the triangle consists of 3 kites.
Hence and
and
. Let
and
gives three equations:
(where for the first triangle.)
Solving gives:
Subbing in gives that has sides of
.
can easily be derived from this as the sides still differ by 1 hence the above solutions still work (now with
). All additional triangles will differ by one as the solutions above differ by one so this process can be repeated indefinitely until the side lengths no longer form a triangle.
Subbing in gives with sides
.
has sides
.
has sides
.
has sides
.
has sides
.
has sides
.
has sides
.
has sides
.
would have sides
but these lengths do not make a
triangle as
Likewise, you could create an equation instead of listing all the triangles to .
The sides of a triangle
would be
We then have
Hence, the first triangle which does not exist in this sequence is
.
Hence the perimeter is
.
Solution 2
Proceeding similarly to the first solution, we have that sides of each triangle are of the form for some number
. Also, note that the perimeter of each triangle is half of the previous one. In order for the triangle to not exist, it must not satisfy the triangle inequality, meaning that
. Then, the perimeter would be
. So, to have a proper triangle, we have
. The first triangle to not work would have perimeter
, thus the answer is
.
See Also
Identical problem to the 2011 AMC 12B Problems/Problem 22.
2011 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
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All AMC 10 Problems and Solutions |
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