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

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Avni Pllana

Posts: 546
Registered: 12/6/04
Re: geometry puzzle
Posted: Oct 9, 2013 3:36 PM
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> 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,


sin(a) < (sin(b)+sin(c))/2 = sin(b+c)/2*cos(b-c)/2 < sin(b+c)/2,


a < (b+c)/2.

Best regards,

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