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

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 Ara M Jamboulian Posts: 71 Registered: 12/6/04
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(b-c)/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(b-c)/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

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