
Re: geometry puzzle
Posted:
Oct 10, 2013 1:32 PM


> > > 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 > > Maybe it's an easy argument to save your proof, but > the sine function is only strictly increasing on > [0:pi/2] ... > > Best wishes > Torsten.
Hi Torsten,
it is very easy to show that
(1) a < pi/3, using the Law of Cosines,
and
(2) (b+c)/2 < pi/2, using the relation a+b+c = pi .
Best regards, Avni

