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Re: Euclidean proof of a new theorem?
Posted:
Apr 15, 1996 8:53 AM


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. 138140 uses Ceva's theorem to prove it geometrically. It was also published in The Mathematical Gazette, July 1995, 79(485), pp. 374378, "A generalization of the FermatTorricelli 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 DurbanWestville 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*BA* C*BA* > \c b/ \b a/ \c a/ \c a/ > \ / \ / \ / \ / > \a/ \c/ \b/ \b/ > AC AC > \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 lefthand one shows "Reflecting Napoleons", and > the righthand 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*BA* > \ b/ \b / > \ / \ / > \a/ \c/ > AC > \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 = baseangle > 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 > > >



