🚀 go-pugleaf

In article <5hs7fa$4p7@opus.vcn.bc.ca>, Scott Phung <landmark@vcn.bc.ca>
writes
>Brian McCague <brian@mccague.demon.co.uk> wrote:
>BM> In article <5hnv2g$95s@opus.vcn.bc.ca>, Scott Phung <landmark@vcn.bc.ca>
>BM> writes
>BM> >Here's a math problem that's been on my mind:
>BM> >
>BM> >  Suppose I have a circle with a radius r.  Then into it is
>BM> >inscribed a cyclic quadrilateral.  Is the area of this cyclic
>BM> >quadrilateral = 2r^2?  I know that if the cyclic quadrilateral
>BM> >was a square, then the area of the square = 2r^2.
>BM> >
>BM> >  More Generally, if a polygon is inscribed into a circle with
>BM> >a radius r, would the area of the polygon inscribed = area of
>BM> >regular polygon inscribed?
>BM> >
>BM> >  Also, does the center of the circle lie on any of the two
>BM> >diagonals of the cyclic quadrilateral?
>BM> >
>BM> >  Can anyone confirm the any of above?
>BM> >
>BM> >
>BM> Consider a quadrilateral formed by four points that are very close
>BM> together on the circle's circumference.  It follows that the answer to
>BM> the above questions is 'no'.
>BM> --
>BM> Brian McCague
>
>But does the angles diagonally opposite of each other add to 180?
>
>A ......... D
>   .     .
> B  ..... C
>
>Does <B + <D = 180
> and <A + <C = 180?
>
>Yes, it's easy to image four points close together, but do these
>four points satisfy the above?  Perhaps you can provide an
>example.
>
>If it adds to 180, my mistake and I will be beleive that
>this area is smaller than that of a inscribed square.
>
>Thanks,
>Scott Phung.

Lets deal with the area of a cyclic quadrilateral.
Let ABCD be a cyclic quadrilateral inscribed in a circle radius r.
Let WXYZ be a square inscribed in the circle such that W lies on
the arc DAB midway between D and B.
Let P be the mid point of the line segment DB.
Let K be the point on the line DB (extended if necessary) such that
AK is perpendicular to DB.  Similarly let L be the point on DB such
that CL is perpendicular to DB.
Now the area of triangle DAB = (AK x DB)/2
and the area of triangle DWB = (WP x DB)/2.
Now AK < WP unless A and W coincide.
Thus the area of triangle DAB < the area of the triangle DWB except
when A and W coincide. Similarly the area of the triangle DCB < area
of triangle DYB unless C and Y coincide.
It follows that the area of the quadrilateral ABCD < the area of the
quadrilateral WBYD unless both A and W coincide and C and Y coincide. 
Now the area of WBYD is (WP x DB)/2 + (PY x DB)/2
=(WP + PY) x DB/2 = WY x DB/2 = 2r x DB/2 = r x DB.
Now DB < ZX unless DB and ZX coincide and since ZX = 2r it follows that
the area of WBYD < r x 2r = 2r^2 unless DB and ZX coincide.
It follows that the area of ABCD < 2r^2 unless it is a square. 

Also opposite angles of a cyclic quadrilateral always add up to 180
degrees.  I can supply a proof of this if needed.
-- 
Brian McCague