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Question-136597

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Question Number 136597 by Dwaipayan Shikari last updated on 23/Mar/21
Answered by mr W last updated on 23/Mar/21
Commented by otchereabdullai@gmail.com last updated on 25/Mar/21
The Gifted prof W!
$$\mathrm{The}\:\mathrm{Gifted}\:\mathrm{prof}\:\mathrm{W}! \\ $$
Commented by mr W last updated on 23/Mar/21
A_0 A_1 =A_1 A_2 =...=A_6 A_7 =a=2×sin (π/(14)) A_0 A_7 =2 A_0 A_6 =2×cos (π/(14)) A_0 A_5 =2×cos ((2π)/(14)) A_0 A_4 =2×cos ((3π)/(14)) A_0 A_3 =2×cos ((4π)/(14)) A_0 A_2 =2×cos ((5π)/(14)) (1/2)r_6 (A_0 A_7 +A_0 A_6 +A_6 A_7 )=(1/2)A_0 A_7 ×A_0 A_6 ×sin (π/(14)) r_6 (1+cos (π/(14))+sin (π/(14)))=2 cos (π/(14)) sin (π/(14)) r_5 (cos (π/(14))+cos ((2π)/(14))+sin (π/(14)))=2 cos ((2π)/(14))cos (π/(14)) sin (π/(14)) r_4 (cos ((2π)/(14))+cos ((3π)/(14))+sin (π/(14)))=2 cos ((3π)/(14))cos ((2π)/(14)) sin (π/(14)) r_3 (cos ((3π)/(14))+cos ((4π)/(14))+sin (π/(14)))=2 cos ((4π)/(14))cos ((3π)/(14)) sin (π/(14)) r_2 (cos ((4π)/(14))+cos ((5π)/(14))+sin (π/(14)))=2 cos ((5π)/(14))cos ((4π)/(14)) sin (π/(14)) r_1 (cos ((5π)/(14))+cos ((6π)/(14))+sin (π/(14)))=2 cos ((6π)/(14))cos ((5π)/(14)) sin (π/(14)) sum area of small circles: S=πΣ_(n=1) ^6 r_n ^2 =0.4076 5528 1334 5393 ⌊10^4 S⌋=4076 5528 1334 53 sum of digits=57
$${A}_{\mathrm{0}} {A}_{\mathrm{1}} ={A}_{\mathrm{1}} {A}_{\mathrm{2}} =…={A}_{\mathrm{6}} {A}_{\mathrm{7}} ={a}=\mathrm{2}×\mathrm{sin}\:\frac{\pi}{\mathrm{14}} \\ $$$${A}_{\mathrm{0}} {A}_{\mathrm{7}} =\mathrm{2} \\ $$$${A}_{\mathrm{0}} {A}_{\mathrm{6}} =\mathrm{2}×\mathrm{cos}\:\frac{\pi}{\mathrm{14}} \\ $$$${A}_{\mathrm{0}} {A}_{\mathrm{5}} =\mathrm{2}×\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{14}} \\ $$$${A}_{\mathrm{0}} {A}_{\mathrm{4}} =\mathrm{2}×\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{14}} \\ $$$${A}_{\mathrm{0}} {A}_{\mathrm{3}} =\mathrm{2}×\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{14}} \\ $$$${A}_{\mathrm{0}} {A}_{\mathrm{2}} =\mathrm{2}×\mathrm{cos}\:\frac{\mathrm{5}\pi}{\mathrm{14}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{r}_{\mathrm{6}} \left({A}_{\mathrm{0}} {A}_{\mathrm{7}} +{A}_{\mathrm{0}} {A}_{\mathrm{6}} +{A}_{\mathrm{6}} {A}_{\mathrm{7}} \right)=\frac{\mathrm{1}}{\mathrm{2}}{A}_{\mathrm{0}} {A}_{\mathrm{7}} ×{A}_{\mathrm{0}} {A}_{\mathrm{6}} ×\mathrm{sin}\:\frac{\pi}{\mathrm{14}} \\ $$$${r}_{\mathrm{6}} \left(\mathrm{1}+\mathrm{cos}\:\frac{\pi}{\mathrm{14}}+\mathrm{sin}\:\frac{\pi}{\mathrm{14}}\right)=\mathrm{2}\:\mathrm{cos}\:\frac{\pi}{\mathrm{14}}\:\mathrm{sin}\:\frac{\pi}{\mathrm{14}} \\ $$$${r}_{\mathrm{5}} \left(\mathrm{cos}\:\frac{\pi}{\mathrm{14}}+\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{14}}+\mathrm{sin}\:\frac{\pi}{\mathrm{14}}\right)=\mathrm{2}\:\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{14}}\mathrm{cos}\:\frac{\pi}{\mathrm{14}}\:\mathrm{sin}\:\frac{\pi}{\mathrm{14}} \\ $$$${r}_{\mathrm{4}} \left(\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{14}}+\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{14}}+\mathrm{sin}\:\frac{\pi}{\mathrm{14}}\right)=\mathrm{2}\:\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{14}}\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{14}}\:\mathrm{sin}\:\frac{\pi}{\mathrm{14}} \\ $$$${r}_{\mathrm{3}} \left(\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{14}}+\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{14}}+\mathrm{sin}\:\frac{\pi}{\mathrm{14}}\right)=\mathrm{2}\:\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{14}}\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{14}}\:\mathrm{sin}\:\frac{\pi}{\mathrm{14}} \\ $$$${r}_{\mathrm{2}} \left(\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{14}}+\mathrm{cos}\:\frac{\mathrm{5}\pi}{\mathrm{14}}+\mathrm{sin}\:\frac{\pi}{\mathrm{14}}\right)=\mathrm{2}\:\mathrm{cos}\:\frac{\mathrm{5}\pi}{\mathrm{14}}\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{14}}\:\mathrm{sin}\:\frac{\pi}{\mathrm{14}} \\ $$$${r}_{\mathrm{1}} \left(\mathrm{cos}\:\frac{\mathrm{5}\pi}{\mathrm{14}}+\mathrm{cos}\:\frac{\mathrm{6}\pi}{\mathrm{14}}+\mathrm{sin}\:\frac{\pi}{\mathrm{14}}\right)=\mathrm{2}\:\mathrm{cos}\:\frac{\mathrm{6}\pi}{\mathrm{14}}\mathrm{cos}\:\frac{\mathrm{5}\pi}{\mathrm{14}}\:\mathrm{sin}\:\frac{\pi}{\mathrm{14}} \\ $$$${sum}\:{area}\:{of}\:{small}\:{circles}: \\ $$$${S}=\pi\underset{{n}=\mathrm{1}} {\overset{\mathrm{6}} {\sum}}{r}_{{n}} ^{\mathrm{2}} =\mathrm{0}.\mathrm{4076}\:\mathrm{5528}\:\mathrm{1334}\:\mathrm{5393} \\ $$$$\lfloor\mathrm{10}^{\mathrm{4}} {S}\rfloor=\mathrm{4076}\:\mathrm{5528}\:\mathrm{1334}\:\mathrm{53} \\ $$$${sum}\:{of}\:{digits}=\mathrm{57} \\ $$
Commented by Dwaipayan Shikari last updated on 24/Mar/21
Great sir! thanks!
$${Great}\:{sir}!\:{thanks}!\: \\ $$

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