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

 Messages: [ Previous | Next ]
 Torsten Hennig Posts: 2,419 Registered: 12/6/04
Re: geometry puzzle
Posted: Oct 10, 2013 9:54 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(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.

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