Very nice. (I cheated and graphed it.) At first I thought it was the curvy triangle with constant width. But the pieces are not quite circular arcs, right? - Cris
On Nov 8, 2018, at 7:21 PM, Dan Asimov <dasimov@earthlink.net> wrote:
So I was idly wondering what the locus is for the sum of distances to *three* points to be constant looks like.
For convenience I took (-1,0), (1,0) and (0, sqrt(3)) as the vertices of an equilateral triangle, and the sum-of-distances equal to 4.
Anyone care to guess what this looks like before graphing it?
—Dan
----- On Nov 8, 2018, at 10:06 AM, Mike Speciner <ms@alum.mit.edu> wrote:
And the focus to focus reflection of ellipses is easy to see when using the constant sum of distances to foci property.
I guess I need to sketch a proof that at each point x on the ellipse, the curve is perpendicular to the bisector of the lines connecting x to the two foci. -----
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun