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$$\frac{R(\sin 2A+\sin 2B+\sin 2C)}{a+b+c}$$ $$=\frac{R(4\sin A \sin B\sin C)}{2R(\sin A+\sin B+\sin C}$$ $$=\frac{8R\sin \frac A2 \sin \frac B2 \sin \frac C2 \cos \frac A2 \cos \frac B2 \cos \frac C2}{8R\cos \frac A2 \cos \frac B2 \cos \frac C2}$$ $$=\frac{2r}{8R}$$ $$=\frac {r}{4R}$$ But the answer given is $\frac{r}{R}$

R means the circumradius of the triangle

r is the inradius

Aditya
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  • @TheDemonix_Hermit my bad, fixed it. – Aditya Dec 09 '19 at 11:51
  • Can you test the right factor for the expression in the title (that should also be in the text of the question, with a mention that we want to compute it in terms of $r,R$, if we do indeed have to do this) in the special case $a=b=c$? (An edit to make things clear is welcome.) – dan_fulea Dec 09 '19 at 11:51

1 Answers1

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In the second line, you have a factor of $4$ in the numerator, so the numerator in the third line should be $4 \times (2 \times 2 \times 2) = 32$. The rest is fine.

Toby Mak
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