## Alteration to the message 16262

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• Thanks to Javier Garcia Capitán for his observations to the problems that I have raised, and also to Pavan Naidu, for his observations. Forgive the correct
Message 1 of 2 , Apr 9, 2008
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Thanks to Javier Garcia Capitán for his observations to the problems
that I have raised, and also to Pavan Naidu, for his observations.

Forgive the correct terms of reference of the problem:

in A Triángle ABC, P is on AC,C1: circle inscript in ABC y C2:
circle inscript in BPC. The straight line parallel to AC and tangent
to the circumference C2 felling AB and to BP in X and Y. C3: circle
inscript in XBY, The other one straight line parallel to AC and
tangent to the circumference C3 ¿is too tangent to the circumference
C1? Since I solve it?
• Dear Milton, It is sufficient to prove that r2 + r3 = r . . . (1) where r2, r3, r are the inradii of triangles BPC, BXY, ABC If r1 is the inradius of triangle
Message 2 of 2 , Apr 10, 2008
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Dear Milton,
It is sufficient to prove that r2 + r3 = r . . . (1)
where r2, r3, r are the inradii of triangles BPC, BXY, ABC
If r1 is the inradius of triangle BAP
BP=z, AP=x, PC=y and S is the double area of ABC then
(1) is equivalent to
Stewart's equality . . aax + ccy = zz(x+y) + xy(x+y)
since
r1 = xS/(x+y)(x+z+c)
r2 = yS/(x+y)(y+z+a)
r3 = r1(h_b-2r2)/h_b
where h_b = S/(x+y) is the B_altitude and
r = S/(a+c+x+y)
It would be nice to have a synthetic proof.
Best regards

>
> in A Triángle ABC, P is on AC,C1: circle inscript in ABC y
> C2:
> circle inscript in BPC. The straight line parallel to AC
> and tangent
> to the circumference C2 felling AB and to BP in X and Y.
> C3: circle
> inscript in XBY, The other one straight line parallel to AC
> and
> tangent to the circumference C3 ¿is too tangent to the
> circumference
> C1? Since I solve it?
>
>
>
> ------------------------------------
>