2011 AMC 10B Problems/Problem 25
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
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
2011 AMC 10B (Problems • Answer Key • Resources) | ||
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