Difference between revisions of "2011 AMC 10B Problems/Problem 25"
(→Solution) |
(→Solution) |
||
Line 54: | Line 54: | ||
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 <math>\frac{503}{2^ | + | The sides of a triangle <math>T_{k}</math> would be <math>\frac{503}{2^{k-3}} - 1, \frac{503}{2^{k-3}}, \frac{503}{2^{k-3}} + 1</math> |
We then have <math>503 - 2^k / 8 + 503 > 503 + 2^k / 8</math> -> <math>1006 - 2^k / 8 > 503 + 2^k / 8</math> -> <math>503 > 2^k / 4 -> 9 > k-2</math> -> <math>k < 11</math> | We then have <math>503 - 2^k / 8 + 503 > 503 + 2^k / 8</math> -> <math>1006 - 2^k / 8 > 503 + 2^k / 8</math> -> <math>503 > 2^k / 4 -> 9 > k-2</math> -> <math>k < 11</math> | ||
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> |
Revision as of 02:26, 22 January 2017
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
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 derivied 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
(Someone please help with the LaTeX)
Hence the perimeter is
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.