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Question Number 21574 by Tinkutara last updated on 27/Sep/17

The cyclic octagon ABCDEFGH has sides a, a, a, a, b, b, b, b respectively. Find the radius of the circle that circumscribes ABCDEFGH in terms of a and b.

$$\mathrm{The}\:\mathrm{cyclic}\:\mathrm{octagon}\:{ABCDEFGH}\:\mathrm{has} \\ $$$$\mathrm{sides}\:{a},\:{a},\:{a},\:{a},\:{b},\:{b},\:{b},\:{b}\:\mathrm{respectively}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{that} \\ $$$$\mathrm{circumscribes}\:{ABCDEFGH}\:\mathrm{in}\:\mathrm{terms} \\ $$$$\mathrm{of}\:{a}\:\mathrm{and}\:{b}. \\ $$

Answered by mrW1 last updated on 05/Oct/17

8×[sin^(−1) ((a/(2R)))+sin^(−1) ((b/(2R)))]=2π let α=sin^(−1) ((a/(2R))) let β=sin^(−1) ((b/(2R))) ⇒8(α+β)=2π ⇒α+β=(π/4) α=(π/4)−β sin α=((√2)/2)(cos β−sin β) (√2)sin α+sin β=cos β (√2)×(a/(2R))+(b/(2R))=cos β (1/(2R))((√2)a+b)=cos β ((2a^2 +2(√2)ab+b^2 )/(4R^2 ))=cos^2 β=1−sin^2 β=1−(b^2 /(4R^2 )) ((2a^2 +2(√2)ab+2b^2 )/(4R^2 ))=1 ((a^2 +(√2)ab+b^2 )/(2R^2 ))=1 ⇒R=(√((a^2 +b^2 +(√2)ab)/2))

$$\mathrm{8}×\left[\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{a}}{\mathrm{2R}}\right)+\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{b}}{\mathrm{2R}}\right)\right]=\mathrm{2}\pi \\ $$$$\mathrm{let}\:\alpha=\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{a}}{\mathrm{2R}}\right) \\ $$$$\mathrm{let}\:\beta=\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{b}}{\mathrm{2R}}\right) \\ $$$$\Rightarrow\mathrm{8}\left(\alpha+\beta\right)=\mathrm{2}\pi \\ $$$$\Rightarrow\alpha+\beta=\frac{\pi}{\mathrm{4}} \\ $$$$\alpha=\frac{\pi}{\mathrm{4}}−\beta \\ $$$$\mathrm{sin}\:\alpha=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{cos}\:\beta−\mathrm{sin}\:\beta\right) \\ $$$$\sqrt{\mathrm{2}}\mathrm{sin}\:\alpha+\mathrm{sin}\:\beta=\mathrm{cos}\:\beta \\ $$$$\sqrt{\mathrm{2}}×\frac{\mathrm{a}}{\mathrm{2R}}+\frac{\mathrm{b}}{\mathrm{2R}}=\mathrm{cos}\:\beta \\ $$$$\frac{\mathrm{1}}{\mathrm{2R}}\left(\sqrt{\mathrm{2}}\mathrm{a}+\mathrm{b}\right)=\mathrm{cos}\:\beta \\ $$$$\frac{\mathrm{2a}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{2}}\mathrm{ab}+\mathrm{b}^{\mathrm{2}} }{\mathrm{4R}^{\mathrm{2}} }=\mathrm{cos}^{\mathrm{2}} \:\beta=\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:\beta=\mathrm{1}−\frac{\mathrm{b}^{\mathrm{2}} }{\mathrm{4R}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{2a}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{2}}\mathrm{ab}+\mathrm{2b}^{\mathrm{2}} }{\mathrm{4R}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{\mathrm{a}^{\mathrm{2}} +\sqrt{\mathrm{2}}\mathrm{ab}+\mathrm{b}^{\mathrm{2}} }{\mathrm{2R}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\Rightarrow\mathrm{R}=\sqrt{\frac{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\sqrt{\mathrm{2}}\mathrm{ab}}{\mathrm{2}}} \\ $$

Commented by mrW1 last updated on 28/Sep/17

it doesn′t matter in which order the sides are. a,a,a,a,b,b,b,b and a,b,a,b, a,b,a,b have e.g. the same result. generally let′s say m sides with length a and n sides with length b. 2×m×sin^(−1) ((a/(2R)))+2×n×sin^(−1) ((b/(2R)))=2π m×sin^(−1) ((a/(2R)))+n×sin^(−1) ((b/(2R)))=π

$$\mathrm{it}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{matter}\:\mathrm{in}\:\mathrm{which}\:\mathrm{order}\:\mathrm{the} \\ $$$$\mathrm{sides}\:\mathrm{are}.\:\:\mathrm{a},\mathrm{a},\mathrm{a},\mathrm{a},\mathrm{b},\mathrm{b},\mathrm{b},\mathrm{b}\:\mathrm{and}\:\mathrm{a},\mathrm{b},\mathrm{a},\mathrm{b}, \\ $$$$\mathrm{a},\mathrm{b},\mathrm{a},\mathrm{b}\:\mathrm{have}\:\mathrm{e}.\mathrm{g}.\:\mathrm{the}\:\mathrm{same}\:\mathrm{result}. \\ $$$$ \\ $$$$\mathrm{generally}\:\mathrm{let}'\mathrm{s}\:\mathrm{say}\:\mathrm{m}\:\mathrm{sides}\:\mathrm{with}\:\mathrm{length} \\ $$$$\mathrm{a}\:\mathrm{and}\:\mathrm{n}\:\mathrm{sides}\:\mathrm{with}\:\mathrm{length}\:\mathrm{b}. \\ $$$$\mathrm{2}×\mathrm{m}×\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{a}}{\mathrm{2R}}\right)+\mathrm{2}×\mathrm{n}×\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{b}}{\mathrm{2R}}\right)=\mathrm{2}\pi \\ $$$$\mathrm{m}×\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{a}}{\mathrm{2R}}\right)+\mathrm{n}×\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{b}}{\mathrm{2R}}\right)=\pi \\ $$

Commented by Tinkutara last updated on 06/Oct/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

Commented by mrW1 last updated on 05/Oct/17

thank you sir!

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}! \\ $$

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