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Question Number 163921 by ajfour last updated on 11/Jan/22

	Answered by mr W last updated on 12/Jan/22

	$R = r a d i u s o f b i g c i r c l e s$ $r = r a d i u s o f s m a l l c i r c l e s$ $s e m i d i a g o n a l = R + \sqrt{2} R = \sqrt{{(R + r)}^{2} - R^{2}} + \sqrt{2} r$ $(\sqrt{2} + 1) R - \sqrt{2} r = \sqrt{2 R r + r^{2}}$ $r^{2} - 2 (\sqrt{2} + 3) R r + {(\sqrt{2} + 1)}^{2} R^{2} = 0$ $\frac{r}{R} = \sqrt{2} + 3 - \sqrt{{(\sqrt{2} + 3)}^{2} - {(\sqrt{2} + 1)}^{2}}$ $\Rightarrow \frac{r}{R} = \sqrt{2} + 3 - 2 \sqrt{\sqrt{2} + 2}$ $a r e a o f p a r a l l e l o g r a m :$ $A_{p} = 2 R \sqrt{{(R + r)}^{2} - R^{2}} = 2 R \sqrt{2 R r + r^{2}}$ $a r e a o f s q u a r e :$ $A_{s} = 2 {(R + \sqrt{2} R)}^{2}$ $r a t i o$ $\frac{A_{s}}{A_{p}} = \frac{2 {(R + \sqrt{2} R)}^{2}}{2 R \sqrt{2 R r + r^{2}}} = \frac{{(1 + \sqrt{2})}^{2}}{\sqrt{(2 + \frac{r}{R}) (\frac{r}{R})}}$ $\frac{A_{s}}{A_{p}} = \frac{3 + 2 \sqrt{2}}{\sqrt{(5 + \sqrt{2} - 2 \sqrt{\sqrt{2} + 2}) (3 + \sqrt{2} - 2 \sqrt{\sqrt{2} + 2})}}$ $\frac{A_{s}}{A_{p}} \approx 4.169$
	Commented by Tawa11 last updated on 12/Jan/22

	$Great sir$
	Answered by ajfour last updated on 12/Jan/22
	$R+R(√2)=s(√2) R+r+2(√(Rr))=2s (√(r/R))+1=(√((2s)/R)) R=(2−(√2))s r=R((√(2/(2−(√2))))−1)^2 =R((√(2+(√2)))−1)^2 A_p =4R(√((r+R)^2 −R^2 )) =(4s^2 )(2−(√2))^2 {[((√(2+(√2)))−1)^2 +1]^2 −1}^(1/2) (A_p /A_s )=(2−(√2))^2 {[((√(2+(√2)))−1)^2 +1]^2 −1}^(1/2) ...$
	$R + R \sqrt{2} = s \sqrt{2}$ $R + r + 2 \sqrt{R r} = 2 s$ $\sqrt{\frac{r}{R}} + 1 = \sqrt{\frac{2 s}{R}}$ $R = (2 - \sqrt{2}) s$ $r = R {(\sqrt{\frac{2}{2 - \sqrt{2}}} - 1)}^{2}$ $= R {(\sqrt{2 + \sqrt{2}} - 1)}^{2}$ $A_{p} = 4 R \sqrt{{(r + R)}^{2} - R^{2}}$ $= (4 s^{2}) {(2 - \sqrt{2})}^{2} {{[{(\sqrt{2 + \sqrt{2}} - 1)}^{2} + 1]}^{2} - 1}^{1 / 2}$ $\frac{A_{p}}{A_{s}} = {(2 - \sqrt{2})}^{2} {{[{(\sqrt{2 + \sqrt{2}} - 1)}^{2} + 1]}^{2} - 1}^{1 / 2}$ $. . .$
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