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    Re: A triangle on the equator
    From: Antoine Couëtte
    Date: 2025 Oct 22, 16:42 -0700

    Excellent, Frank, and Thank you ! :-)

    Since I earlier submitted :

    I would then surmise that this would happen at a Latitude where the relative lengths of both 1° of Latitude and 1° of Longitude would exactly match the same ratio observed on a Sphere "

    and :

    provided that we are authorized by the Mathematician on duty to rely onto only your first published terms of such Ellipsoids infinite series, solving for:

    dLon/dLat (Sphere) = dLon/dLat (Ellipsoid)

    boils down to solving for:

    ε (1 + cos 2L ) = 0

    with immediate solutions as:

    ε = 0 , i.e. a Sphere itself (trivial case). And :

    cos 2L = -1, yielding 2L = +/- 180°, hence L = +/- 90° in the case of an Ellipsoid

    The above is a mathematical demonstration, while Page 2 only partially verifies this mathematical result albeit probably in some much more instructive manner .

    Therefore no need to publish this partial verification anywhere else then, and Thanks to u 2 David for your kind comments.

    Kermit

       
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