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

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John Conway

Posts: 2,238
Registered: 12/3/04
Generalized Napolean's Theorem
Posted: Apr 3, 1996 10:00 AM
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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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