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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

Professor, the picture is perfect

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

Commented by Shrinava last updated on 07/May/22

Professof, I look forward to your perfect solution, thank you

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

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

i didn′t get the same result as in the question.

$${i}\:{didn}'{t}\:{get}\:{the}\:{same}\:{result}\:{as}\:{in}\:{the} \\ $$$${question}. \\ $$

Commented by Shrinava last updated on 08/May/22

Is interesting, thank you so much Professor

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

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

[I_a I_b I_c ]=8R^2 cos (A/2) cos (B/2) cos (C/2) [ABC]=2R^2 sin A sin B sin C [ABC]=16R^2 sin (A/2) sin (B/2) sin (C/2) cos (A/2) cos (B/2) cos (C/2) [I_a I_b I_c ]−[ABC]=8R^2 cos (A/2) cos (B/2) cos (C/2)(1−2 sin (A/2) sin (B/2) sin (C/2)) AE=AF=s−a=((−a+b+c)/2) r_1 =(((s−a)cos (A/2))/(1+(1/(sin (A/2)))))=(((s−a)sin (A/2) cos (A/2))/(1+sin (A/2))) r_1 =((R(−sin A+sin B+sin C) sin (A/2) cos (A/2))/(1+sin (A/2))) r_1 =((4R cos (A/2) sin (B/2) sin (C/2) sin (A/2) cos (A/2))/(1+sin (A/2))) r_1 =((4R sin (A/2) sin (B/2) sin (C/2) cos^2 (A/2))/(1+sin (A/2))) r_1 =4R sin (A/2) sin (B/2) sin (C/2) (1−sin (A/2)) r_1 +r_2 +r_3 =4R sin (A/2) sin (B/2) sin (C/2) (3−sin (A/2)−sin (B/2)−sin (C/2)) this doesn′t prove r_1 +r_2 +r_3 =(([I_a I_b I_c ]−[ABC])/(2R(cos (A/2)+cos (B/2)+cos (C/2))))

$$\left[{I}_{{a}} {I}_{{b}} {I}_{{c}} \right]=\mathrm{8}{R}^{\mathrm{2}} \mathrm{cos}\:\frac{{A}}{\mathrm{2}}\:\mathrm{cos}\:\frac{{B}}{\mathrm{2}}\:\mathrm{cos}\:\frac{{C}}{\mathrm{2}} \\ $$$$ \\ $$$$\left[{ABC}\right]=\mathrm{2}{R}^{\mathrm{2}} \mathrm{sin}\:{A}\:\mathrm{sin}\:{B}\:\mathrm{sin}\:{C} \\ $$$$\left[{ABC}\right]=\mathrm{16}{R}^{\mathrm{2}} \:\mathrm{sin}\:\frac{{A}}{\mathrm{2}}\:\mathrm{sin}\:\frac{{B}}{\mathrm{2}}\:\mathrm{sin}\:\frac{{C}}{\mathrm{2}}\:\mathrm{cos}\:\frac{{A}}{\mathrm{2}}\:\mathrm{cos}\:\frac{{B}}{\mathrm{2}}\:\mathrm{cos}\:\frac{{C}}{\mathrm{2}} \\ $$$$ \\ $$$$\left[{I}_{{a}} {I}_{{b}} {I}_{{c}} \right]−\left[{ABC}\right]=\mathrm{8}{R}^{\mathrm{2}} \mathrm{cos}\:\frac{{A}}{\mathrm{2}}\:\mathrm{cos}\:\frac{{B}}{\mathrm{2}}\:\mathrm{cos}\:\frac{{C}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{2}\:\mathrm{sin}\:\frac{{A}}{\mathrm{2}}\:\mathrm{sin}\:\frac{{B}}{\mathrm{2}}\:\mathrm{sin}\:\frac{{C}}{\mathrm{2}}\right) \\ $$$$ \\ $$$${AE}={AF}={s}−{a}=\frac{−{a}+{b}+{c}}{\mathrm{2}} \\ $$$${r}_{\mathrm{1}} =\frac{\left({s}−{a}\right)\mathrm{cos}\:\frac{{A}}{\mathrm{2}}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{sin}\:\frac{{A}}{\mathrm{2}}}}=\frac{\left({s}−{a}\right)\mathrm{sin}\:\frac{{A}}{\mathrm{2}}\:\mathrm{cos}\:\frac{{A}}{\mathrm{2}}}{\mathrm{1}+\mathrm{sin}\:\frac{{A}}{\mathrm{2}}} \\ $$$${r}_{\mathrm{1}} =\frac{{R}\left(−\mathrm{sin}\:{A}+\mathrm{sin}\:{B}+\mathrm{sin}\:{C}\right)\:\mathrm{sin}\:\frac{{A}}{\mathrm{2}}\:\mathrm{cos}\:\frac{{A}}{\mathrm{2}}}{\mathrm{1}+\mathrm{sin}\:\frac{{A}}{\mathrm{2}}} \\ $$$${r}_{\mathrm{1}} =\frac{\mathrm{4}{R}\:\mathrm{cos}\:\frac{{A}}{\mathrm{2}}\:\mathrm{sin}\:\frac{{B}}{\mathrm{2}}\:\mathrm{sin}\:\frac{{C}}{\mathrm{2}}\:\mathrm{sin}\:\frac{{A}}{\mathrm{2}}\:\mathrm{cos}\:\frac{{A}}{\mathrm{2}}}{\mathrm{1}+\mathrm{sin}\:\frac{{A}}{\mathrm{2}}} \\ $$$${r}_{\mathrm{1}} =\frac{\mathrm{4}{R}\:\mathrm{sin}\:\frac{{A}}{\mathrm{2}}\:\mathrm{sin}\:\frac{{B}}{\mathrm{2}}\:\mathrm{sin}\:\frac{{C}}{\mathrm{2}}\:\mathrm{cos}^{\mathrm{2}} \:\frac{{A}}{\mathrm{2}}}{\mathrm{1}+\mathrm{sin}\:\frac{{A}}{\mathrm{2}}} \\ $$$${r}_{\mathrm{1}} =\mathrm{4}{R}\:\mathrm{sin}\:\frac{{A}}{\mathrm{2}}\:\mathrm{sin}\:\frac{{B}}{\mathrm{2}}\:\mathrm{sin}\:\frac{{C}}{\mathrm{2}}\:\left(\mathrm{1}−\mathrm{sin}\:\frac{{A}}{\mathrm{2}}\right) \\ $$$${r}_{\mathrm{1}} +{r}_{\mathrm{2}} +{r}_{\mathrm{3}} =\mathrm{4}{R}\:\mathrm{sin}\:\frac{{A}}{\mathrm{2}}\:\mathrm{sin}\:\frac{{B}}{\mathrm{2}}\:\mathrm{sin}\:\frac{{C}}{\mathrm{2}}\:\left(\mathrm{3}−\mathrm{sin}\:\frac{{A}}{\mathrm{2}}−\mathrm{sin}\:\frac{{B}}{\mathrm{2}}−\mathrm{sin}\:\frac{{C}}{\mathrm{2}}\right) \\ $$$${this}\:{doesn}'{t}\:{prove} \\ $$$${r}_{\mathrm{1}} +{r}_{\mathrm{2}} +{r}_{\mathrm{3}} =\frac{\left[{I}_{{a}} {I}_{{b}} {I}_{{c}} \right]−\left[{ABC}\right]}{\mathrm{2}{R}\left(\mathrm{cos}\:\frac{{A}}{\mathrm{2}}+\mathrm{cos}\:\frac{{B}}{\mathrm{2}}+\mathrm{cos}\:\frac{{C}}{\mathrm{2}}\right)} \\ $$

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

can you check where is the mistake?

$${can}\:{you}\:{check}\:{where}\:{is}\:{the}\:{mistake}? \\ $$

Commented by Shrinava last updated on 09/May/22

Dear Professor, l could not find exactly, but the solution is perfect

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

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