Problem:
Let T1 be a triangle with sides 2011,2012 , and 2013. For n≥1, if Tn=△ABC and D,E, and F are the points of tangency of the incircle of △ABC to the sides AB,BC, and AC, respectively, then Tn+1 is a triangle with side lengths AD,BE, and CF, if it exists. What is the perimeter of the last triangle in the sequence (Tn) ?
Answer Choices:
A. 81509
B. 321509
C. 641509
D. 1281509
E. 2561509
Solution:
Let Tn=△ABC. Suppose a=BC,b=AC, and c=AB. Because BD and BE are both tangent to the incircle of △ABC, it follows that BD=BE. Similarly, AD=AF and CE=CF. Then
2BE=BE+BD=BE+CE+BD+AD−(AF+CF)=a+c−b
that is, BE=21(a+c−b). Similarly AD=21(b+c−a) and CF=21(a+b−c). In the given △ABC, suppose that AB=x+1,BC=x−1, and AC=x. Using the formulas for BE,AD, and CF derived before, it must be true that
BE=21((x−1)+(x+1)−x)=21xAD=21(x+(x+1)−(x−1))=21x+1, and CF=21((x−1)+x−(x+1))=21x−1
Hence both (BC,CA,AB) and (CF,BE,AD) are of the form (y−1,y,y+1). This is independent of the values of a,b, and c, so it holds for all Tn. Furthermore, adding the formulas for BE,AD, and CF shows that the perimeter of\
Tn+1 equals 21(a+b+c), and consequently the perimeter of the last triangle TN in the sequence is
2N−11(2011+2012+2013)=2N−31509
The last member TN of the sequence will fail to define a successor if for the first time the new lengths fail the Triangle Inequality, that is, if
−1+2N2012+2N2012≤1+2N2012
Equivalently, 2012≤2N+1 which happens for the first time when N=10. Thus the required perimeter of TN is 271509=1281509.
The problems on this page are the property of the MAA's American Mathematics Competitions