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.
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Posted by Charlie
on 2010-08-18 16:55:12 |
a candidate
