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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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 more precision | Comment 6 of 11 |

6 digits ran much faster then I thought it would.
Here are the more precise intervals
on [0,0.491092) F(a)=0
on [0.491092,0.517638) F(a)=4
on [0.517638,0.63507) F(a)=6
on [0.63507,0.703821) F(a)=8
on [0.703821,1.06547) F(a)=12
on [1.06547,1.15271) F(a)=8
on [1.15271,1.41421) F(a)=6
on [1.41421,1.93185) F(a)=2
on [1.93185,2.) F(a)=4
on [2., Infinity) F(a)=0

So it would appear that the upper bound for which solutions exist is 2.  Now to attempt to determine exact values for each of these.

 Posted by Daniel on 2009-09-14 23:10:21

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