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Topic: geometry puzzle
Replies: 6   Last Post: Nov 12, 2013 4:55 AM

 Messages: [ Previous | Next ]
 Avni Pllana Posts: 546 Registered: 12/6/04
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(b-c)/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

Date Subject Author
10/8/13 Rich Delaney
10/8/13 Peter Scales
10/9/13 Avni Pllana
10/10/13 Torsten Hennig
10/10/13 Avni Pllana