Re: Triangles Inscribed in a Circle


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.

In news:<1156857484.474839.6880@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>
schrieb "The Qurqirish Dragon" <qurqirishd@xxxxxxx>:
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. 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. The area of
this triangle can only be increased by moving 2 points. 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.)

.

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