[ND]

> Dear Jean-Pierre

> Is it easy to explain

> how did you found this relation?

> (1-cos A)((hc-2r)R-b^2/2)+r(hc-r)=0

Dear Nikos,

There are two different questions about this

relation.

1. How JPE has constructed it ?

2. Once we have it, how to prove it?

Let's try to answer 2:

that is, to prove that:

(1-cos A)((hc-2r)R-b^2/2) = r(r-hc)

We have:

First member:

1-cosA = 2sin^2(A/2)

hc-2r = 2RsinAsinB - 8Rsin(A/2)sin(B/2)sin(C/2) =

= 8Rsin(A/2)cos(A/2)sin(B/2)cos(B/2) - 8Rsin(A/2)sin(B/2)sin(C/2)=

= 8Rsin(A/2)sin(B/2)[cos(A/2)cos(B/2) - sin(C/2)] =

= 8Rsin^2(A/2)sin^2(B/2)

(hc-2r)R - b^2/2 = 8R^2sin^2(A/2)sin^2(B/2) -

- 8R^2sin^2(B/2)cos^2(B/2)

= 8R^2sin^2(B/2)[sin^2(A/2) - cos^2(B/2)]

===>

(1-cos A)((hc-2r)R-b^2/2) = 16R^2sin^2(A/2)sin^2(B/2)*

[sin^2(A/2) - cos^2(B/2)] <1>

Second member:

r(r-hc)= 4Rsin(A/2)sin(B/2)sin(C/2)[4Rsin(A/2)sin(B/2sin(C/2) - 2RsinAsinB] =

4R^2sin(A/2)sin(B/2)sin(C/2)[4sin(A/2)sin(B/2)sin(C/2) -

- 8sin(A/2)cos(A/2)sin(B/2)cos(B/2)]=

= 16R^2sin^2(A/2)sin^2(B/2)sin(C/2)[sin(C/2) - 2cos(A/2)cos(B/2)] <2>

So we have to prove that <1> = <2>

or

sin^2(A/2) - cos^2(B/2) - sin^2(C/2)= - 2cos(A/2)cos(B/2)sin(C/2)

or

sin^2(A/2)+sin^2(B/2)-sin^2(C/2)= 1 - 2cos(A/2)cos(B/2)sin(C/2)

which is true.

APH