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.


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

$The cyclic octagon A B C D E F G H has$ $sides a, a, a, a, b, b, b, b respectively .$ $Find the radius of the circle that$ $circumscribes A B C D E F G H in terms$ $of a and b .$
	Answered by mrW1 last updated on 05/Oct/17

	$8 \times [\sin^{- 1} (\frac{a}{2 R}) + \sin^{- 1} (\frac{b}{2 R})] = 2 π$ $let α = \sin^{- 1} (\frac{a}{2 R})$ $let β = \sin^{- 1} (\frac{b}{2 R})$ $\Rightarrow 8 (α + β) = 2 π$ $\Rightarrow α + β = \frac{π}{4}$ $α = \frac{π}{4} - β$ $\sin α = \frac{\sqrt{2}}{2} (\cos β - \sin β)$ $\sqrt{2} \sin α + \sin β = \cos β$ $\sqrt{2} \times \frac{a}{2 R} + \frac{b}{2 R} = \cos β$ $\frac{1}{2 R} (\sqrt{2} a + b) = \cos β$ $\frac{{2 a}^{2} + 2 \sqrt{2} ab + b^{2}}{{4 R}^{2}} = \cos^{2} β = 1 - \sin^{2} β = 1 - \frac{b^{2}}{{4 R}^{2}}$ $\frac{{2 a}^{2} + 2 \sqrt{2} ab + {2 b}^{2}}{{4 R}^{2}} = 1$ $\frac{a^{2} + \sqrt{2} ab + b^{2}}{{2 R}^{2}} = 1$ $\Rightarrow R = \sqrt{\frac{a^{2} + b^{2} + \sqrt{2} ab}{2}}$
	Commented by mrW1 last updated on 28/Sep/17

	$it {doesn}^{'} t matter in which order the$ $sides are . a, a, a, a, b, b, b, b and a, b, a, b,$ $a, b, a, b have e . g . the same result .$ $generally {let}^{'} s say m sides with length$ $a and n sides with length b .$ $2 \times m \times \sin^{- 1} (\frac{a}{2 R}) + 2 \times n \times \sin^{- 1} (\frac{b}{2 R}) = 2 π$ $m \times \sin^{- 1} (\frac{a}{2 R}) + n \times \sin^{- 1} (\frac{b}{2 R}) = π$
	Commented by Tinkutara last updated on 06/Oct/17

	$Thank you very much Sir!$
	Commented by mrW1 last updated on 05/Oct/17

	$thank you sir!$
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