"The total 1-boundary is topologically a triangle, and can be flattened into the plane." Orient an ellipsoid in R^3 with symmetry planes XY, YZ, ZX, and let A, B, C be the semi-axis lengths along X,Y, and Z. Take the positive octant, and cut triangle ABC through the ellipsoid. Then define P(ABC) the curvilinear projection of ABC onto the surface (X/A)^2 + (Y/B)^2 + (Z/C)^2 = 1. Clearly the Euclidean triangle ABC satisfies the triangle inequality, and clearly the girths also satisfy the triangle inequality if P(ABC) can indeed be flattened into the plane. Now let L1, L2, L3 be the euclidean lengths along sides of ABC and L1' > L1, L2' > L2, L3' > L3 are the corresponding curvilinear lengths along P(ABC). Without loss of generality, require that L1' > L2' >= L3'. Clearly L1 < L2 + L3, but we need to prove sufficiently that L1' < L2 + L3. Now write L1' = k*L1, with at most k=(sqrt(2)/4)*pi=1.11072... Lengths L2 and L3 are minimized as the ellipsoid flattens toward a plane where L2^2 + L3^2 = L1^2 . Then write: ( k^2 - 1 )*( L2^2 + L3^2 ) < 2*L2*L3. Is this inequality always satisfied? If L2 approximately equals L3 then *obviously yes* because "k" is so close to one. Apparently in this thread, we have decided to haggle about the limit where L3 becomes small relative to L2, where also k approaches 1, but how rapidly? Can we still flatten the ellipsoidal triangles? Without loss of generality set L3<L2=1 and just graph the RHS and LHS of the inequality, ( k^2 - 1 )*(1+L3^2) < 2*L3 over 0 < L3 < 1: https://0x0.st/z2xm.png The linear red line from the RHS is always the superior, but if anyone still doubts me, they are welcome to check the series expansion of the LHS. Okay, now we have wasted a couple of hours. Why do I really think this is "obvious"? Because when you draw the triangles on the ellipse they have angles Pi/2 at every vertex. Obviously the flattening procedure does not preserve these angles, but Pi/2 is a lot of leeway, either plus or minus, and we just need to avoid going too far obtuse on the largest angel. On most days this extra 0.11072... is nothing for me to worry about. Tonight I am feeling generous. Cheers, Brad On Tue, Jun 25, 2019 at 7:59 PM Warren D Smith <warren.wds@gmail.com> wrote:

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Once I got the search working, I also found that on Tue Oct 11 19:07:21 MDT 2011 Warren Smith conjectured that the triangle inequality is obeyed for girths. Isn't this also obvious?

--obvious? Consider a triangle made of 3 quarter-girths drawn on surface of ellipsoid. Conjecture would then follow if surface metric is a metric (yes, it is) and those triangle-edge quarter-girths are geodesics and shortest paths (no, that latter aint necessarily so).

So, is it obvious to you, and why?

-- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)