The largest equilateral triangle in terms of area is also the largest in edge size.
One equilateral triangle that comes to mind includes a vertex at the apex of the prolate ellipsoid, at (0,0,sqrt(3)). A second point could be on the z=0 plane as well as either the y=0 plane or the x=0 plane. Third point would also be on the z=0 plane. So, in addition to (0,0,sqrt(3)) as the first point, the second point could be (0,sqrt(3/2),0). As the length of the side is 3/sqrt(2), the third point would be this distance from (0,sqrt(3/2),0), but also on the 2x^2+2y^2 = 3 circle on the z=0 plane. Two such points exist as 3/sqrt(2) ~= 2.12 is less than the diameter of the z=0 circle, which is sqrt(6) ~= 2.45.
So that's two such triangles on the given basis. But the second point could have been (0,sqrt(3/2),0), and we'd have two more triangles, bringing the total to 4.
In addition, the second point could have been on the y=0 plane instead of the x=0 plane. That gives a total of 8 such triangles.
But we could have started with the first point's being (0,0,sqrt(3)), so double the number of triangles, making the final count 16.
So in any instance, we have two points on the z=0 plane, but one of is not on any of the other fundamental planes, and so counts as the z=0 point. One of the points is on both the z=0 plane and either the x=0 plane or the y=0 plane. That one counts as the x=0 (or the y=0) point. The point at the vertex the, although being on both the y=0 and x=0 planes, counts as the one you didn't yet have.
I don't have a proof that a larger one can't be contrived by offsetting the (0,0,sqrt(3)) somewhat, in a direction away from the second point, but I suspect that doing so would prevent the swivelling vertex of the defined equilateral triangle from meeting the z=0 plane exactly along the circle of intersection with the ellipsoid.

Posted by Charlie
on 20100818 16:55:12 