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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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 re(2): more precision | Comment 10 of 11 |
(In reply to re: more precision by Charlie)

BTW, for example, the likely candidate for a spurious (12th) solution for a=0.73486, would be Z near (-2.24457,0.79087), Y near (-2.31,-0.206), X near (-1.41,0.23). Just as Y approaches being 1 unit from B (as Z is moved around the circle about C), from both sides, as the X's are getting closer together also, Z gets too far from A to have both ZX and XA be equal to 1.

For 1.956, the one near X at (0.16,0.55), Y at (-0.84,0.55) and Z at (-0.34,1.42) is spurious, though there is another one very nearby which is real, and combined with two others (X near (-0.26,0.11),Y near (-0.41,0.09),Z near (-0.48,1.09), but offset in two directions slightly) make 3.

 Posted by Charlie on 2009-09-15 00:30:43

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