Sauda,c~oes, O problema abaixo foi proposto numa outra lista.
[]'s Luís Friends of Hycianthos: Which is the reference and the proof of the problem: Be ABC triangle rectangle in A, are BE and CD bisecting of angles B and C. Let us considerer the segment ED and is M the point of cut of height AH1 with ED. Proof that AM measures the radius of the inscripte circle of ABC? Thanks Ricardo >From: "Nikolaos Dergiades" <[EMAIL PROTECTED]> >Reply-To: [EMAIL PROTECTED] >To: <[EMAIL PROTECTED]> >Subject: Re: [EMHL] A reference and proof >Date: Sat, 13 Jul 2002 21:52:08 +0300 > >Dear Ricardo, I don't know a reference. > >A quick proof I can think is the following: >If s is the semiperimeter, E the area of ABC >F is the point of contact of AC and the incircle >and r = inradius then >the equation of the line DE in normals is >x = y + z because D = [1,1,0], E = [1,0,1] >If x, y, z are the actual normals of the point M >(distances of M from the sides of ABC) then x = AH - AM , >y = AMcosC = AM*b/a, z = AMcosB = AM*c/a >and from x = y + z > we get a*(AH - AM) = AM*b + AM*c > >or a*AH = AM*(a+b+c) or 2*E = 2AM*s > >or 2*r*s = 2AM*s or r = AM. > >A synthetic proof must prove that the triangle AMF is >isosceles or that MF is parallel to BE > >Best regards >Nikos Dergiades ========================================================================= Instruções para entrar na lista, sair da lista e usar a lista em http://www.mat.puc-rio.br/~nicolau/olimp/obm-l.html O administrador desta lista é <[EMAIL PROTECTED]> =========================================================================