Re: Triangles Inscribed in a Circle

In news:<1157468330.323120.11960@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>
schrieb "The Qurqirish Dragon" <qurqirishd@xxxxxxx>:
Thomas Mautsch wrote:
Maury Barbato wrote:
consider an isosceles triangle ABC, with AC=BC, inscribed
in a circle C. Then move the vertex A (or B) on the
circle to obtain A'C=A'B (respectively B'A= B'C). The new
triangle A'BC (or AB'C, but these two triangles are
equal) has a greater area and and a longer perimeter than
ABC. Now you can repeat the procedure, moving B or C
(respectively A or C), to obtain a new isosceles
triangle, and so on. The sequence of inscribed triangles
has increasing area and perimeter, and I think it
converges to an equilateral triangle inscribed in the
circle. What do you think?

This result would allow to give a pure geometric proof
of the well-known fact that the equilateral triangle
is the triangle of maximum area and perimeter among all
the triangle inscribed in a given circle.

"The Qurqirish Dragon" <qurqirishd@xxxxxxx> schrieb:
My previous post, which asserted the conclusion actually
proved that the equilateral triangle has the maximum area.

No. You only showed
that to every inscribed triangle that is not equilateral,
there exists an inscribed triangle with larger area.
But you did not mention that one also has to assert
that there *is* an inscribed triangle
for which the area takes on a maximum value.

Otherwise, you could apply your style of "proof"
to show that 0 is the maximum number in the half-open interval [0,1):

For every value x in (0,1), the number x*(2-x) also lies in (0,1),
and is larger than x itself - so it can't be the maximal value;
on the other hand for the value x = 0, x*(2-x) stays at 0.
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Would you conclude that 0 will be the maximum in the interval [0,1)?

No, since your proof only considers the open interval. Nowhere did you
state anything about 0 itself.

You'd better learn how to read!
I *did* write:

"for the value x = 0, x*(2-x) stays at 0."


So, I did *exactly* as you did in your proof:

I showed that my procedure - replacing x by x*(2-x) -
leaves the "maximum" 0 fixed,
while it increases every other number x in the interval [0,1).

You showed that your procedure - replacing the triangle T
by the largest amongst those which have one chord in common with T -
keeps the equilateral triangle fixed, while it enlarges every other
inscribed triangle.

Correct or not?!

My triangle statement, now that you use
this example so I can see the flaw, does miss the degenerate case where
all 3 vertices of the original triangle are the same point.

You seem to have some compactification of the space
of real inscribed triangles in mind here. -
This will be necessary to turn your pseudo-proof into a real proof,
but I don't see where it is important in your current "proof".

The area of
this triangle can only be increased by moving 2 points.

If you don't consider degenerate triangles,
you don't have worry about this case.

For any other
case, my proof works. (any non-equilateral triangle can have its area
increased by moving a vertex not on the perpendicular bisector of the
opposite side to such a point. There are two such points, but my
algorithm chooses the one such that the center of the circle lies
inside the triangle.)

This proof goes *exactly* along the same lines as my pseudo-proof
of max [0,1) = 0 above. - It is simply wrong (i.e., incomplete).
.

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