2011 AMC 10B Problems/Problem 25
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) | ||
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All AMC 10 Problems and Solutions |