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Topic: Generalized Napolean's Theorem
Replies: 2   Last Post: Apr 15, 1996 8:53 AM

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Mike de Villiers

Posts: 38
Registered: 12/6/04
Re: Euclidean proof of a new theorem?
Posted: Apr 15, 1996 8:53 AM
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Dear John
I'm not sure whether you received the copy of my book I posted to you,
but my proof of the generalization you refer to (the concurrency of AA*,
BB* and CC* in what you call the (a,b,c)-fulcrum)) on pp. 138-140 uses
Ceva's theorem to prove it geometrically. It was also published in The
Mathematical Gazette, July 1995, 79(485), pp. 374-378, "A generalization
of the Fermat-Torricelli point" by myself. The
result that these lines are concurrent is not new as there has apparently
been a proof of this result by N. Alliston in The Mathematical Snack Bar
by W. Hoffer in 1936.
Regards
Michael D de Villiers
Univ Durban-Westville
South Africa

On Wed, 3 Apr 1996, John Conway wrote:

>
> I'm slowly organizing my thoughts on the geometry of the triangle,
> and this message concerns generalizations of Napoleon's theorem.
>
> So let me state some theorems in this area:
>
> C*------B------A* C*------B------A*
> \c b/ \b a/ \c a/ \c a/
> \ / \ / \ / \ /
> \a/ \c/ \b/ \b/
> A-------C A-------C
> \a c/ \c a/
> \ / \ /
> \b/ \b/
> B* B*
>
>
> The figures show the two nice ways of "Napoleonizing"
> a triangle with angles A,B,C by putting three mutually
> similar triangles with angles a,b,c on its sides.
> (a,b,c are supposed to be alpha, beta, gamma).
>
> The left-hand one shows "Reflecting Napoleons", and
> the right-hand one "Rotating Napoleons". There is a third
> way, "Isosceles Napoleons", in which the three outer triangles
> are copies of the same isosceles triangle, places with their
> bases on the edges of the original.
>
> Now for Rotating Napoleons, the three images of any point
> form a copy of the Napoleonizing triangle. For Reflecting
> Napoleons, this is still true if the point is the circumcenter.
> In either case, the three circumcircles concur (at the "fulcrum"
> for circumscribing an a,b,c triangle around the A,B,C one).
> For reflecting Napoleons, the three lines AA*,BB*,CC* also
> concur at this fulcrum.
>
> For isosceles Napoleons, the three lines AA*,BB*,CC* also
> concur.
>
> All those statements are old (as far as I'm concerned).
>
> A few days ago, I found a nice generalization of one of them, that
> handles both the Reflecting and Isosceles cases. Let a,b,c be any
> three angles, not necessarily adding to two right angles, and draw
> the figure
>
> C*------B------A*
> \ b/ \b /
> \ / \ /
> \a/ \c/
> A-------C
> \a c/
> \ /
> \ /
> B*
>
> in which the three "Napoleons" won't always be similar.
> Then necessarily AA*, BB*, CC* will concur, at a point it's natural to
> call the (a,b,c)-fulcrum. The "Reflecting" case is when a+b+c = 180
> degrees, and the "Isosceles" case is when a=b=c = base-angle
> of the isosceles triangles.
>
> It's a nice theorem that the (x,x,x)-fulcrums trace out a
> rectangular hyperbola as x varies, and that this passes through
> A,B,C,F,G,H ( F = Fermat point, G = barycenter, H = orthocenter ).
> The isogonal conjugate of this curve is the line OK ( O = circumcenter,
> K = Lemoine center = "symmedian point" ), while its isotomic conjugate
> is the line GK.
>
> Anyway, I could go on for a long time with more facts around here,
> but the point of this message is this : my proof that the (a,b,c)-fulcrum
> exists is an algebraic one - it actually finds that its barycentric
> coordinates are
>
> ( 1/(cot A + cot a), 1/(cot B + cot b), 1/(cot C + cot c) ).
>
> Is there a nice purely geometric proof?
>
> John Conway
>
>
>







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