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Question Number 165879 by ajfour last updated on 09/Feb/22

	Answered by mr W last updated on 09/Feb/22

	Commented by ajfour last updated on 09/Feb/22
	Go ahead sir, i shall also try, n past two questions, extreme thanks to their solutions. Unbelievable presentation.I shall scrutinize them well, for sure.
	Commented by mr W last updated on 10/Feb/22

	$\tan 22.5 ° = \sqrt{2} - 1$ $\sin 22.5 ° = \frac{\sqrt{2} - 1}{\sqrt{4 - 2 \sqrt{2}}}$ $l e t t = \frac{1}{\tan 22.5 °} = \sqrt{2} + 1$ $l e t s = \frac{1}{\sin 22.5 °} = (\sqrt{2} + 1) \sqrt{2 (2 - \sqrt{2})}$ $O B = O C = O A = \frac{1}{\tan 22.5 °}$ $O K = O H = D H + O D = 1 + \frac{1}{\sin 22.5 °}$ $J B = \frac{1}{\tan 22.5 °} - R$ ${(\frac{1}{\tan 22.5 °} - R)}^{2} = {(1 + R)}^{2} - {(1 - R)}^{2}$ $R^{2} - 2 (t + 2) R + t^{2} = 0$ $\Rightarrow R = t + 2 - 2 \sqrt{t + 1}$ $C E = O K - O C - E K = 1 + \frac{1}{\sin 22.5 °} - \frac{1}{\tan 22.5 °} - r$ ${(r + 1)}^{2} = 1^{2} + {(1 + \frac{1}{\sin 22.5 °} - \frac{1}{\tan 22.5 °} - r)}^{2}$ ${(1 + s - t)}^{2} = 2 (2 + s - t) r$ $\Rightarrow r = \frac{{(1 + s - t)}^{2}}{2 (2 + s - t)}$ $\frac{R}{r} = \frac{2 (2 + s - t) (2 + t - 2 \sqrt{t + 1})}{{(1 + s - t)}^{2}} \approx 2.1989$
	Commented by mr W last updated on 09/Feb/22

	Commented by MJS_new last updated on 10/Feb/22

	$great! the solution can be simplified :$ $\frac{R}{r} = 1 - \sqrt{2} + \sqrt{2 (2 + \sqrt{2})}$
	Commented by ajfour last updated on 09/Feb/22

	$A w e s o m e s o l u t i o n s i r .$
	Commented by mr W last updated on 10/Feb/22

	$t h a n k s s i r s!$
	Commented by Tawa11 last updated on 10/Feb/22

	$Weldone sir .$
	Answered by ajfour last updated on 09/Feb/22

	Commented by ajfour last updated on 10/Feb/22
	$let quarter circle radius q. black circle radius=1 22.5°=α (q−1)sin α=1 (q−1)cos α=R+2(√R) ⇒ R+2(√R)=cot α=t (√R)=−1+(√(1+t)) R=t+2−2(√(1+t)) (q−1)cos α+r+(√((r+1)^2 −1))=q ⇒ r+1+(√((r+1)^2 −1)) =cosec α(1−cos α)+2 ⇒ (r+1)^2 −1={s−t+2−(r+1)}^2 ⇒ −1=(s−t+2)^2 −2(s−t+2)(r+1) r+1=(((s−t+2)^2 +1)/(2(s−t+2))) ⇒ r=(((s−t+1)^2 )/(2(s−t+2))) (R/r)=((2(s−t+2))/((s−t+1)^2 ))×{t+2−2(√(t+1))}$
	$l e t q u a r t e r c i r c l e r a d i u s q .$ $b l a c k c i r c l e r a d i u s = 1$ $22.5 ° = α$ $(q - 1) \sin α = 1$ $(q - 1) \cos α = R + 2 \sqrt{R}$ $\Rightarrow R + 2 \sqrt{R} = \cot α = t$ $\sqrt{R} = - 1 + \sqrt{1 + t}$ $R = t + 2 - 2 \sqrt{1 + t}$ $(q - 1) \cos α + r + \sqrt{{(r + 1)}^{2} - 1} = q$ $\Rightarrow r + 1 + \sqrt{{(r + 1)}^{2} - 1}$ $= cosec α (1 - \cos α) + 2$ $\Rightarrow {(r + 1)}^{2} - 1 = {s - t + 2 - (r + 1)}^{2}$ $\Rightarrow$ $- 1 = {(s - t + 2)}^{2} - 2 (s - t + 2) (r + 1)$ $r + 1 = \frac{{(s - t + 2)}^{2} + 1}{2 (s - t + 2)}$ $\Rightarrow r = \frac{{(s - t + 1)}^{2}}{2 (s - t + 2)}$ $\frac{R}{r} = \frac{2 (s - t + 2)}{{(s - t + 1)}^{2}} \times {t + 2 - 2 \sqrt{t + 1}}$
	Commented by Tawa11 last updated on 10/Feb/22

	$Weldone sir$
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