Given A=(a,0), B=(0,0), and C=(0,a)
Let f(a)=the total number of unit equilateral triangles XYZ that can be formed such that the lengths AX, BY, and CZ are all 1 unit.
Give a piecewise definition by intervals for f(a)
first it should be obvious that f(a)=f(a) due to reflection about the xaxis so we can limit ourselves to considering nonnegative values of a
next if a=0 then we have A=B=C=(0,0) and and there is no such triangle thus f(0)=0
now we can place an upper bound on a for which solutions exist.
Start by drawing circle Cb around B with radius 1 and circle Cc around C with radius 1. Now we want to find a point on each of these circles such that they are a unit distance apart. The shortest possible distance between two points on these circles is a2 and this needs to be less than or equal to 1 for there to exist 2 such points. Thus we have a<=3
now a similar agrument can be used with circle Cc and Ca around A with radius 1. Here we have that the minimum distance is Sqrt((a1)^2+1)1
thus we need
Sqrt((a1)^2+1)1<=1
Sqrt((a1)^2+1)<=2
(a1)^2+1<=4
(a1)^2<=3
a1<=sqrt(3)
a<=1+sqrt(3)<3
Thus we can say that solutions can not exist for a>1+sqrt(3)
now to work on breaking down the intervals of (0,1+sqrt(3)] for the various number of solutions.

Posted by Daniel
on 20090914 18:17:15 