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Question Number 185166 by Rupesh123 last updated on 17/Jan/23

Answered by mnjuly1970 last updated on 18/Jan/23

sin((A/2) ) = (r/(IA)) ⇒ r = IA.sin((A/2)) sin( (B/2))= (r/(IB)) ⇒ r = IB.sin((B/2)) sin((C/2))=(r/(IC)) ⇒ r= IC.sin((C/( 2)) ) −−−−−−−−− r^( 3) = (IA.IB.IC)(sin((A/2))sin((B/2))sin((C/2))) 4Rr^( 3) = IA.IB.IC .r 4Rr^( 2) = IA.IB.IC ⇒_(euler inequality) ^(r≤(R/2)) R^ ^3 ≥ IA.IB.IC

$$\:\: \\ $$$${sin}\left(\frac{{A}}{\mathrm{2}}\:\right)\:=\:\frac{{r}}{{IA}}\:\Rightarrow\:{r}\:=\:{IA}.{sin}\left(\frac{{A}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:{sin}\left(\:\frac{{B}}{\mathrm{2}}\right)=\:\frac{{r}}{{IB}}\:\Rightarrow\:{r}\:=\:{IB}.{sin}\left(\frac{{B}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:{sin}\left(\frac{{C}}{\mathrm{2}}\right)=\frac{{r}}{{IC}}\:\Rightarrow\:{r}=\:{IC}.{sin}\left(\frac{{C}}{\:\mathrm{2}}\:\right) \\ $$$$\:\:\:\:−−−−−−−−− \\ $$$$\:\:\:\:\:\:\:{r}^{\:\mathrm{3}} =\:\left({IA}.{IB}.{IC}\right)\left({sin}\left(\frac{{A}}{\mathrm{2}}\right){sin}\left(\frac{{B}}{\mathrm{2}}\right){sin}\left(\frac{{C}}{\mathrm{2}}\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{4}{Rr}^{\:\mathrm{3}} =\:{IA}.{IB}.{IC}\:.{r} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{4}{Rr}^{\:\mathrm{2}} =\:{IA}.{IB}.{IC} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\underset{{euler}\:{inequality}} {\overset{{r}\leqslant\frac{{R}}{\mathrm{2}}} {\Rightarrow}}\:\:{R}^{\:} \overset{\mathrm{3}} {\:}\geqslant\:{IA}.{IB}.{IC} \\ $$$$ \\ $$

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