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 Unit lengths from movable points to unit triangles. (Posted on 2009-09-14)
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)

 No Solution Yet Submitted by Jer No Rating

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 initial constraints (a start) | Comment 1 of 11

first it should be obvious that f(a)=f(-a) due to reflection about the x-axis so we can limit ourselves to considering non-negative 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 a-2 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((a-1)^2+1)-1
thus we need
Sqrt((a-1)^2+1)-1<=1
Sqrt((a-1)^2+1)<=2
(a-1)^2+1<=4
(a-1)^2<=3
a-1<=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 2009-09-14 18:17:15

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