
Re: geometry puzzle
Posted:
Nov 12, 2013 4:55 AM


> > Given a triangle, with sides A, B, C, and > > opposite angles a, b, c. > > > > Prove: if A < (B + C)/2, then a < (b + c)/2 > > > >  > > Rich > > > > Hi Rich, > > using the Law of Sines > > A/sin(a) = B/sin(b) = C/sin(c) = 2*R, > > we have > > A < (B + C)/2, > > or > > sin(a) < (sin(b)+sin(c))/2 = sin(b+c)/2*cos(bc)/2 < > sin(b+c)/2, > > or > > a < (b+c)/2. > > Best regards, > Avni
1. Could you please explain how you got the formula (sin(b)+sin(c))/2 = sin(b+c)/2*cos(bc)/2
2. Also, for your argument to work you need sin(a) < sin((b+c)/2)
3. sin(a) < sin(b+c)/2 is wrong since sin(b+c)=sin(a)
Thanks

