Re: [math-fun] conic sections
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. -----
Isn’t it just the circumcenter of the triangle formed by the three points? Steve
On Nov 8, 2018, at 9:21 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. -----
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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. -----
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Doh, my original statement about circumcenters was answering he wrong question. Here is a page showing all the curves as you vary the requested sum of distances from 3 points. http://jwilson.coe.uga.edu/emt725/Tri.Locus/Tri.Locus.html Steve On Nov 9, 2018, at 10:44 AM, Cris Moore <moore@santafe.edu<mailto:moore@santafe.edu>> wrote: 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<mailto: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<mailto: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<mailto:math-fun@mailman.xmission.com> https://urldefense.proofpoint.com/v2/url?u=https-3A__mailman.xmission.com_cg... _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com<mailto:math-fun@mailman.xmission.com> https://urldefense.proofpoint.com/v2/url?u=https-3A__mailman.xmission.com_cg...
participants (3)
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Cris Moore -
Dan Asimov -
Lucas, Stephen K - lucassk