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 Solid triangles (3) (Posted on 2010-08-18)
Three points are located on the surface of the ellipsoid: 2x2 + 2y2 + z2 = 3. One has a x coordinate of 0, another has a y coordinate of 0, and the last has a z coordinate of 0.

What is the largest possible equilateral triangle (in terms of area) that can be made using these three points as the corners? How many distinct equilateral triangles of this size are possible?

 No Solution Yet Submitted by K Sengupta No Rating

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 Parametric Thoughts | Comment 3 of 4 |
The three vertecies are located on three ellipses.  Letting t, u, and v be radian values, the vertecies can be expressed parametrically:
(sqrt(3/2)*cos t, sqrt(3/2)*sin t, 0)
(sqrt(3/2)*sin u, 0, sqrt(3)*cos u)
(0, sqrt(3/2)*cos v, sqrt(3)*sin v)

The edge length of the equilateral triangle, call it distance D, can then be expressed as:
D^2 = 3 + (3/2)*(cos u)^2 - 3*(cos t)*(sin u)
D^2 = 3 + (3/2)*(sin v)^2 - 3*(sin t)*(cos v)
D^2 = 3 + (3/2)*(cos u)^2 + (3/2)*(sin v)^2 - 6*(cos u)*(sin v)

Charlie's candidate points (0,0,sqrt(3)), (0,sqrt(3)/2, 0), (0,-sqrt(3)/2, 0) correspond to u=0, t=pi/2, v=pi.

Edited on February 20, 2016, 9:29 am
 Posted by Brian Smith on 2016-02-19 11:09:50

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