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Re: Triangulating an inscribed polygon
Posted:
Mar 26, 1998 5:14 PM
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On 26 Mar 1998 wolk@ccm.umanitoba.c quoted:
> famedini@tin.it wrote:
>
> >If you triangulate a polygon inscribed in a circle, it doesn't matter
>
> >the way you do it, the sum of the radii of the circles inscribed in
> >the triangles will be constant, not depending on the way you
> >triangulated the polygon.
> >That seems to be true, but I don't know how to proove it.
> >Help me!
>
> I first saw that problem about five years ago, in a book review
> published in the American Mathematical Monthly. The review quoted
> a few problems from the book, including this problem. The book
> title was something like "The Chinese Garden Geometry Problems"
> -- I hope someone in this group can provide the correct title,
> because that book is probably where you want to look this up
> to find a geometric proof.
It seems to me that one should be able to get a proof using
the well-known formulae
r = DELTA/s, R = abc/4DELTA
for the inradius r and circumradius R of a triangle with sides a,b,c,
area DELTA, and semiperimeter s = (a+b+c)/2.
Combining them, we have r = abc/2R(a+b+c), so that 2R times the
desired sums (for a quadrilateral, which suffices for the general
theorem) are, for the quadrilateral of the figure below:
abe/(a+b+e) + cde/(c+d+e) =?= bcf/(b+c+f) + daf/(d+a+f).
But I'm having trouble showing that this is true. I thought it
had to follow just from Ptolemy's theorem ac + bd = ef, but now
see that it needn't.
___a__
|\ /|
|e\ /f|
d| \/ |b
| /\ |
| / \ |
|/____\|
c
John Conway
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