
Generalized Napolean's Theorem
Posted:
Apr 3, 1996 10:00 AM


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

