Difference between revisions of "2011 AMC 10B Problems/Problem 25"
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Hence the perimeter is <math>\frac{375}{128} + \frac{503}{128} + \frac{631}{128} = \boxed{\textbf{(D)} \frac{1509}{128}}</math> | Hence the perimeter is <math>\frac{375}{128} + \frac{503}{128} + \frac{631}{128} = \boxed{\textbf{(D)} \frac{1509}{128}}</math> | ||
| + | <math>\blacksquare</math> | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2011|ab=B|num-b=24|after=Last Problem}} | {{AMC10 box|year=2011|ab=B|num-b=24|after=Last Problem}} | ||
Revision as of 18:58, 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
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 indefinately 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 length do not make a
triangle as 
.
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 | ||
