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Question Number 169655 by Shrinava last updated on 05/May/22

Answered by mr W last updated on 07/May/22

Commented by Shrinava last updated on 07/May/22

$$\mathrm{Professor},\mathrm{the}\mathrm{picture}\mathrm{is}\mathrm{perfect}$$

Commented by Shrinava last updated on 07/May/22

$$\mathrm{Professof},\mathrm{I}\mathrm{look}\mathrm{forward}\mathrm{to}\mathrm{your}\mathrm{perfect}$$$$\mathrm{solution},\mathrm{thank}\mathrm{you}$$

Commented by mr W last updated on 08/May/22

$$i{didn}^{\prime }tgetthesameresultasinthe$$$$question.$$

Commented by Shrinava last updated on 08/May/22

$$\mathrm{Is}\mathrm{interesting},\mathrm{thank}\mathrm{you}\mathrm{so}\mathrm{much}\mathrm{Professor}$$

Commented by mr W last updated on 09/May/22

$$\left[I_a{I}_b{I}_c\right]=8R^2\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}$$$$$$$$\left[ABC\right]=2R^2\sin A\sin B\sin C$$$$\left[ABC\right]=16R^2\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}$$$$$$$$\left[I_a{I}_b{I}_c\right]-\left[ABC\right]=8R^2\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}(1-2\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2})$$$$$$$$AE=AF=s-a=\frac{-a+b+c}{2}$$$$r_1=\frac{(s-a)\cos \frac{A}{2}}{1+\frac{1}{\sin \frac{A}{2}}}=\frac{(s-a)\sin \frac{A}{2}\cos \frac{A}{2}}{1+\sin \frac{A}{2}}$$$$r_1=\frac{R(-\sin A+\sin B+\sin C)\sin \frac{A}{2}\cos \frac{A}{2}}{1+\sin \frac{A}{2}}$$$$r_1=\frac{4R\cos \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\sin \frac{A}{2}\cos \frac{A}{2}}{1+\sin \frac{A}{2}}$$$$r_1=\frac{4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}{\cos }^2\frac{A}{2}}{1+\sin \frac{A}{2}}$$$$r_1=4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}(1-\sin \frac{A}{2})$$$$r_1+r_2+r_3=4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}(3-\sin \frac{A}{2}-\sin \frac{B}{2}-\sin \frac{C}{2})$$$$this{doesn}^{\prime }tprove$$$$r_1+r_2+r_3=\frac{\left[I_a{I}_b{I}_c\right]-\left[ABC\right]}{2R(\cos \frac{A}{2}+\cos \frac{B}{2}+\cos \frac{C}{2})}$$

Commented by mr W last updated on 09/May/22

$$canyoucheckwhereisthemistake?$$

Commented by Shrinava last updated on 09/May/22

$$\mathrm{Dear}\mathrm{Professor},\mathrm{l}\mathrm{could}\mathrm{not}\mathrm{find}\mathrm{exactly},$$$$\mathrm{but}\mathrm{the}\mathrm{solution}\mathrm{is}\mathrm{perfect}$$$$$$

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