
Re: Triangulating an inscribed polygon
Posted:
Mar 26, 1998 5:14 PM


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 wellknown 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

