Difference between revisions of "2011 AMC 10B Problems/Problem 25"
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− | By constructing the bisectors of each angle and the perpendicular radii of the incircle the triangle consists of 3 kites. Hence <math>AD=AF</math> and <math>BD=BE</math> and <math>CE=CF</math>. Let <math>AD = x, BD = y</math> and <math>CE = z</math> gives three equations: | + | By constructing the bisectors of each angle and the perpendicular radii of the incircle the triangle consists of 3 kites. |
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+ | [[Image:File2011AMC10B25.png|center|260px]] | ||
+ | |||
+ | Hence <math>AD=AF</math> and <math>BD=BE</math> and <math>CE=CF</math>. Let <math>AD = x, BD = y</math> and <math>CE = z</math> gives three equations: | ||
<math>x+y = a-1</math> | <math>x+y = a-1</math> | ||
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Subbing in gives that <math>T_2</math> has sides of <math>1005, 1006, 1007</math>. | Subbing in gives that <math>T_2</math> has sides of <math>1005, 1006, 1007</math>. | ||
− | + | <math>T_3</math> can easily be derivied from this as the sides still differ by 1 hence the above solutions still work (now with <math>a=1006</math>). 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 <math>T_3</math> with sides <math>502, 503, 504</math>. | ||
<math>T_4</math> has sides <math>\frac{501}{2}, \frac{503}{2}, \frac{505}{2}</math>. | <math>T_4</math> has sides <math>\frac{501}{2}, \frac{503}{2}, \frac{505}{2}</math>. |
Revision as of 09:29, 10 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
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 |