Curious triangle fact
Mike de Villiers
> At 6:34 PM 10/29/95, Peter F. Ash wrote:
> >I saw the following problem posed on the math.sci newsgroup a while back:
> >
> >Given any triangle ABC, erect isosceles triangles (ABC', BCA', CAB')
> >on each side of ABC, pointing outward and so that |A'| = |B'| = |C'|
> >= 120 degrees. Then A'B'C' is an equilateral triangle.
> >
> >I had never seen this before, and wasn't sure I believed it, but after
> >playing around with Geometer's Sketchpad for a while became less
> >skeptical.
>
> >This is a special case of a generalization of Napoleon's Theorem. In his
> >excellent book COLLEGE GEOMETRY, A DISCOVERY APPROACH, David Kay
> >states:
> >
> Mon, 30 Oct 95, Monte J. Zerger wrote
>
> >Let triangle XYZ be a given acute-angled triangle, having angles of measure
> >p,q, and r, respectibely. Given any other triangle ABC, suppose that
> >isosceles triangles PBC, QCA and RAB are constucted externally on the
> >sides of the given triangle as bases, such that the vertex angles at P, Q
> >and R have measures 2p, 2q, and 2r, respectively. Then triangle PQR has
> >angles of measure p, q and r, respectively, and triangle PQR is similar
> >to triangle XYZ, regardless of the choice of triangle ABC.
>
> Actually, there is a more general theorem about this.
>
> Take any triangle ABC. Choose any point D. Construct another two points E,
> F such that triangles ABD, CEB, and FCA are similar and these three
> triangles have same orientation for their vertices(so D can be inside or
> outside triangle ABC).
> Taking any point P, then construct two other points Q, R such that the
> relative positions of Q with respect to triangle CEB and R with respective
> to FCA are the same as the relative position of P with respect to triangle
> ABD(the special cases of corresponding P, Q, R are the corresponding
> vertices or centers of the similar triangles).
>
> Then triangle RQP will be similar to ABD, CEB, and FCA.
Johan Meyer and myself recently published an atricle entitled: "A
generalized dual of Napoleon's theorem and some further extensions", in
Int. Journ. Math. Ed. Sci. & Technol., 1995, Vol 26, No 2, 233-241 where
this result is geometrically proved using the Petersen_Schoute theorem.
Michael de Villiers
UDW
South Africa
>
>
> Now, Napolean's theorem is a special case for ABD as equilateral triangle
> and P is the center of ABD.
> The other special case(Monte J. Zergerwrote)is more general to take D
> arbitary but choose P the circumcenter of triangle ABD ( so the angle APB is
> twice the angle of ADB).
> If you use dynamic geometry software, moving constructed free points P, D
> you can see all the cases.
>
> To prove this theorm is not hard if we use geometric intepretation of
> complex number. I have not yet come up with an elementary geometric proof.
> By the way, Professor John Conway talked about the generalization of
> Napolean theorem to similar triangle several months ago in this forum. I do
> not know if this is what in his mind.
>
> Pao-Ping
>
>
> Pao-Ping Lin
> Department of Mathematics and Science Education
> Taipei Municipal Teachers College
> Taipei, Taiwan, R.O.C
> E-mail: pl...@tmtc760.tmtc.edu.tw
>
>
Mike de Villiers
published an article "A generalization of the Fermat-Torricelli point" in
The Mathematical Gazette, Vol 79 (485), 374-378. It relates to a
generalization of the point of concurrency of the line segments joining
the outer vertices of the equilateral triangles to the opposite vertex of
the base triangle in the standard version of Napoleon's theorem; Re:
Doris Schattshneider's earlier e-mail response.
Brian Hutchings
tell us what that dual of Nap's Th.is --
after we've had a few moments to think about it!
"Time is the only dimension." -RBFuller
Fermatio-Gaussian Conjecture: P^(2^N) -|- P^-(2^N) [P's golden mean;
N=0,1,2, ... ] is prime; it does work for the first five N; hmm.
Mike de Villiers
Fermat-Torricelli point and Napoleon's theorem, the following additional
information.
Regarding De Villiers, M & Meyer, J. (1995). a generalized dual of
napoleon's theorem and some further extensions. Int. Journ. Math Ed Sci.
Technol., 26(2), 233-241:
In Euclidean plane geometry, there is useful duality (although limited)
between the concepts "incentre" and "circumcentre", and between "angle
bisectors" and "perpendicular bisectors". Quite frequently, results
involving one of the above concepts have a corresponding dual. For
example,
(1) The angle bisectors of any circum polygon (a polygon circumscribed
around a circle) are concurrent at the incentre of the polygon.
(2) The perpendicular bisectors of any cyclic polygon are concurrrent at
the circumcentre of the polygon.
One generalization of Napoleon's theorem is the following (see Coxeter &
Greitzer, 1967, Geometry Revisited):
If similar triangles ADB, CBE and FAC are erected on the sides of any
triangle ABC, their circumcentres G, H and I form a triangle similar to
the three triangles.
The above duality suggests the following dual:
If similar triangles ADB, CBE and FAC are erected on the sides of any
triangle ABC, their incentres G, H and I form a triangle similar to the
three triangles.
In proving this result using the Petersen Schoute theorem, we discovered
that the result could be further generalized to (proof as a means of
discovery!):
If similar triangles ADB, CBE and FAC are erected outwardly on the sides
of any triangle ABC, and any three points P, Q and R are chosen so that
they respectively lie in the same relative positions to these triangles,
then P, Q and R form a triangle similar to the three triangles.
Another related result proved in the article is:
If similar triangles ABD, EBC and AFC are erected outwardly on the sides
of any triangle ABC, their incentres G, H and I form a triangle with
angle G = 0.5(angle DAB + angle DBA), angle H= 0.5(angle EBC + angle ECB)
and angle I = 0.5(angle FCA + angle FAC).
(Note: that for circumcentres in the above arrangement the triangle GHI
is also similar to the three triangles, is therefore an example of where
the duality breaks down).
Although the generalization of Napoleon's theorem (in respect to P, Q and
R) is not original, the proofs as far as we know are fairly distinctive.
With regard to De Villiers, M. (1995). A generalization of the
Fermat-Torricelli point. The Math Gazette, 79(485) the following result
is proved using Ceva's theorem:
If triangles DBA, ECB and FAC are constructed outwardly on the sides of
any triangle so that angle DAB = angle CAF, angle DBA = angle CBE and
angle ECB = angle FCA then DC, EA and FB are concurrent.
Although this result is also not original, the proof is distinctive as
far as I know.
Michael de Villiers
UDW, South Africa
John Conway
On Sun, 29 Oct 1995 Peter Ash wrote
> I saw the following problem posed on the math.sci newsgroup a while back:
>
> Given any triangle ABC, erect isosceles triangles (ABC', BCA', CAB')
> on each side of ABC, pointing outward and so that |A'| = |B'| = |C'|
> = 120 degrees. Then A'B'C' is an equilateral triangle.
>
>
> (2) I still have no idea WHY the proposition is true. Can anyone supply
> some geometric insight?
>
> This all leads me to believe that there is some
> larger theory lurking here that might have some bearing on question (2).
> Anyone have any ideas?
>
> --Peter
>
>
This is "Napoleon's Theorem", usually stated in the form:
the centers of equilateral triangles erected outwards on the sides
of a given triangle form an equilateral triangle.
(The attribution to Napoleon is not too likely, but it COULD
have been true - Napoleon was interested in geometry.)
As I think I've said on the net, the theorem remains true
if the "equilateral" triangles become copies of a triangle of any
fixed shape, and their "centers" are replaced by the corresponding
"copies" of any fixed point in the plane of that triangle, provided
you take the proper decision about which sides to attach them by.
There is, however, a nice proof that's valid for the equilateral
case. If your given triangle has sides a,b,c, then you can tile
the plane alternately with rotated copies of it and equilateral triangles
of sides a, b, and c, in such a way that the centers of all the
equilateral triangles are centers of rotation of the tiling. It's
then obvious that these centers form the lattice defined by an
equilateral triangle (draw the picture!).
John Conway