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Question Number 181852 by mnjuly1970 last updated on 01/Dec/22

Answered by mr W last updated on 01/Dec/22

(a/(sin α))=(b/(sin β))=(c/(sin γ))=2R S_(ABC) =((bc sin α)/2)=2R^2 sin α sin β sin γ S_(ABC) =R(sin α+sin β+sin γ)r S_(SPQ) =2r^2 sin ∠Q sin ∠S sin ∠P =2r^2 sin ((π−α)/2) sin ((π−β)/2) sin ((π−γ)/2) =2r^2 cos (α/2) cos (β/2) cos (γ/2) (S_(SPQ) /S_(ABC) )=((2r^2 cos (α/2) cos (β/2) cos (γ/2))/(R(sin α+sin β+sin γ)r)) =((2r cos (α/2) cos (β/2) cos (γ/2))/(R×4 cos (α/2) cos (β/2) cos (γ/2))) =(r/(2R)) ✓

$$\frac{{a}}{\mathrm{sin}\:\alpha}=\frac{{b}}{\mathrm{sin}\:\beta}=\frac{{c}}{\mathrm{sin}\:\gamma}=\mathrm{2}{R} \\ $$$${S}_{{ABC}} =\frac{{bc}\:\mathrm{sin}\:\alpha}{\mathrm{2}}=\mathrm{2}{R}^{\mathrm{2}} \mathrm{sin}\:\alpha\:\mathrm{sin}\:\beta\:\mathrm{sin}\:\gamma \\ $$$${S}_{{ABC}} ={R}\left(\mathrm{sin}\:\alpha+\mathrm{sin}\:\beta+\mathrm{sin}\:\gamma\right){r} \\ $$$${S}_{{SPQ}} =\mathrm{2}{r}^{\mathrm{2}} \mathrm{sin}\:\angle{Q}\:\mathrm{sin}\:\angle{S}\:\mathrm{sin}\:\angle{P} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{2}{r}^{\mathrm{2}} \mathrm{sin}\:\frac{\pi−\alpha}{\mathrm{2}}\:\mathrm{sin}\:\frac{\pi−\beta}{\mathrm{2}}\:\mathrm{sin}\:\frac{\pi−\gamma}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{2}{r}^{\mathrm{2}} \mathrm{cos}\:\:\frac{\alpha}{\mathrm{2}}\:\mathrm{cos}\:\frac{\beta}{\mathrm{2}}\:\mathrm{cos}\:\frac{\gamma}{\mathrm{2}} \\ $$$$\frac{{S}_{{SPQ}} }{{S}_{{ABC}} }=\frac{\mathrm{2}{r}^{\mathrm{2}} \mathrm{cos}\:\:\frac{\alpha}{\mathrm{2}}\:\mathrm{cos}\:\frac{\beta}{\mathrm{2}}\:\mathrm{cos}\:\frac{\gamma}{\mathrm{2}}}{{R}\left(\mathrm{sin}\:\alpha+\mathrm{sin}\:\beta+\mathrm{sin}\:\gamma\right){r}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2}{r}\:\mathrm{cos}\:\:\frac{\alpha}{\mathrm{2}}\:\mathrm{cos}\:\frac{\beta}{\mathrm{2}}\:\mathrm{cos}\:\frac{\gamma}{\mathrm{2}}}{{R}×\mathrm{4}\:\mathrm{cos}\:\:\frac{\alpha}{\mathrm{2}}\:\mathrm{cos}\:\frac{\beta}{\mathrm{2}}\:\mathrm{cos}\:\frac{\gamma}{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{{r}}{\mathrm{2}{R}}\:\checkmark \\ $$

Commented by mnjuly1970 last updated on 01/Dec/22

very nice proof ...thanks alot

$$\:\:{very}\:{nice}\:{proof}\:...{thanks}\:{alot} \\ $$

Answered by mnjuly1970 last updated on 01/Dec/22

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