Experiments with the
Wallace-Simson lines of a triangle. 
A reminder: The Wallace-Simson theorem.

Let ABC be an arbitrary triangle and K its circumcircle. For an arbitrary point P of K draw its orthogonal projections over the three sides, U on a, V on b, W on c. Then U,V,W are collinear and the straight line UVW is called the Wallace-Simson line of P with respect to ABC.
 
 
 
 


Motivated by a certain extension I had obtained of the Wallace-Simson theorem (American Mathematical Monthly, June-July 1999) I had assembled a few DERIVE functions to draw the Wallace-Simson line corresponding to a point P.

For any arbitrary triangle inscribed in the unit circle and for any arbitrary point on the circle DERIVE was giving me the corresponding W-S line.


Having this tool at hand, it was tempting to think about the envelope of all Wallace-Simson lines of the triangle.

I made some experiments with very different and very irregular triangles, by drawing for each one 30 W-S lines.

Here are the corresponding figures (the numbers in parentheses correspond to the measure, in radians, of the angles determining the vertices) to two of these experiments.


The envelope for a rather "normal" triangle


The envelope for a rather "abnormal" triangle


From these experiments it seemed "quite clear" that:

(1) The envelope was always the same curve in shape, independently of the triangle. This shape ressembled very much the tricuspidal hypocycloid which was familiar to me.

(The tricuspidal hypocycloid is the curve generated by a point of a circle of radius r when it rotates inside another circle of radius 3r).
 

(2) The size of the curve seemed to be always the same.

(According to measurements over the figure the distance from one cuspid to the opposite "side" was equal to the diameter of the circumscribed circle).

(3) The position of the curve depended of the triangle in a strange way. But even for very irregular triangles it was never far from the circle.

All this seemed to be a very nice theorem... of whose proof I had no idea.

If it was true it should have been previously discovered...

And so it was. 


In 1856 Jakob Steiner showed that the envelope of the Wallace-Simson lines of the points of K with respect to ABC is a curve of third degree which has three cuspidal points and many interesting properties.


Many more have been later on discovered:

The curve is in fact a tricuspidal hypocycloid.

It has the same center as the Feuerbach circle (the nine-point circle) of the triangle ABC and this circle is tangent to the deltoid.

Its orientation is the same as the orientation of the Morley triangle.
.....



A CHALLENGE

There seemed to be no easy sinthetic (nor analytic) proof of all these facts. (One can look at the Notes 1 and 2 at the end of of H.F. Baker, An introduction to plane geometry, Cambridge University Press, 1943)

I tried my hand since I had the opportunity of experimenting with the tools I had programmed with DERIVE around the theorems of Morley, Wallace-Simson,...



Was there any way to make clear why all the triangles inscribed in the same circle should have the same envelope for their W-L lines with just a change in position?

I started by thinking of possible transformations of a triangle keeping the same circumcircle...

Several reasonable possibilities...

Changing one vertex and keeping the opposite side,..
but there seemed to be no easy relation between the corresponding W-L lines, the Feuerbach circle...


Until I used an easy transformation ABC->A'B'C' which seemed to shed a strong light on the problem. I thought of keeping a vertex A=A' of the triangle and of shifting the side BC to B'C' parallel to BC, B' and C' being on the same circumcircle. This turned out to be the key of an easy proof.
 



The W-S line of P with respect to A'B'C' seemed to be parallel to the W-S line of P with respect to ABC!
It is rather easy to prove that it is so and that the W-S line for A'B'C' is obtained from the other by the same shift which takes BC to B'C'!
Since the radius of the Feuerbach circle (nine-point circle) is half the radius of the circumcircle it is also easy to see that the Feuerbach circle for A'B'C' is obtained from the one for ABC by the same shift.
A set of relationships which confirmed that this transformation was the right one for the study of the envelope.

ALL THIS MADE CLEAR THAT THE ENVELOPE OF ALL W-S LINES FOR A'B'C' IS OBTAINED BY THIS PARALLEL SHIFT FROM THE ENVELOPE OF ALL W-S LINES FOR ABC.



Now there remained the question:

Could we use another transformation of this type that takes A'B'C' into a triangle A''B''C'' for which the question of the envelope might be really easy to handle?

Would it be possible to obtain A''B''C'' equilateral and inscribed in the same circle?



YES!! And one can determine exactly what are the right shifts to do it!
 

It is easy to see that we can choose m and n such that  A''=B''=C''=60º (for that it is enough to take m = (180-B-2C)/3, n = (B - C)/3). Then the value of the angle between  B'' C'' and BC is exactly (C-B)/3.
 
 
 
 
 


Now it was evident why for all  triangles inscribed in the same circle the envelope of their W-S lines is the same curve in shape and size and it only changes in position.

The proof of the fact that it is a tricuspidal hypocycloid was reduced to do it for the equilateral triangle and so was the fact that it is concentric and tangent to the Feuerbach circle. This is not a difficult task.


Several questions were still pending:

(1) The center of the envelope is the center of the Feuerbach circle, but how is its position determined?

(2) Could one say something in order to explain the strange relations of Steiner's deltoid with the Morley triangle?
(One can consult the Notes 1 and 2 at the end of H.F. Baker, An introduction to plane geometry, Cambridge University Press, 1943)



An experiment on how the transformation we have introduced (parallel shift of one side) affects the Morley triangle makes one conjecture the answer:

This experiment makes one conjecture that by our transformation the Morley triangles preserve the same orientation. Can one prove this fact?

YES!
One can show (see the classical proofs of Morley theorem)  that the small angle that the side NP of the Morley triangle of ABC  forms with BC is (C-B)/3 and this is the same value for all the triangles in the figure.



This observations answers our both last questions:

The angle we need for converting our original triangle ABC into an equilateral triangle was also (C-B)/3 and it is clear by symmetry what is the position of the envelope of the W-S lines and the Morley triangle for an equilateral triangle .


All this gives us a rather simple synthetic proof of the different properties of the envelope of the W-S lines that can be seen here.


++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

Many other interesting properties of the hypocycloid can be obtained by appropriate experiments and calculations with DERIVE. For example:

(1) The midpoint of the segment of tangent to the deltoid (Wallace-Simson line) determined by the two other intersections with the deltoid is in the Feuerbach circle.

(2) A point moves on the axis Ox and has coordinates (cos t,0). A straight line is attached to this point and rotates so that forms an angle t/2 with the axis Oy. Then the envelope of all these lines when t varies is a tricuspidal hypocycloid whose main elements can be easily determined.

(3) A point has coordinates (cost, sint). A straight line is attached to it and rotates so that its angle wit the axis Ox is t/2. Then the envelope of all these lines when t varies is also a tricuspidal  hypocycloid.

(4) A point (cost, sint) moves on a circle. A straight line is attached to it and rotates so that its angle with axis Ox is t/4. Then the envelope of this line is also a hypocycloid.

A figure corresponding to this experiment