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Question Number 119292 by Algoritm last updated on 23/Oct/20

	Answered by 1549442205PVT last updated on 24/Oct/20

	Commented by 1549442205PVT last updated on 25/Oct/20
	$Denote by r the radius of two small circles CQ=x,PN=OM=2n;R and R_1 −the radius of semicircle and the great circle QO=QP=m⇒R_1 =R/2 Then we have relations: { ((x(R−x)=m^2 (1)(QC.QE=QP^2 ))),((((2r)/R)=(n/m)(2)(IN//OD))),(((x−r)^2 +n^2 =r^2 (3)(ΔOSF is right at S))),(((r/R)=((x−r)/x)(4)(((MF)/(MC))=((FS)/(QC))))) :} (3)⇔x^2 −2rx+n^2 =0(5) (1)⇒x^2 =Rx−m^2 .Replace into (5)we get Rx−m^2 −2rx+n^2 =0⇒x=((m^2 −n^2 )/(R−2r))(6) (2)⇒(R/(2m))=(r/n)⇒(R^2 /(4m^2 ))=(r^2 /n^2 )=((R^2 /4−r^2 )/(m^2 −n^2 ))(7) (4)⇒rx=Rx−Rr⇒x=((Rr)/(R−r))(8) Replace into (6)we get m^2 −n^2 =(((R−2r)Rr)/(R−r)) replace into (7) we get n^2 =((4(R−2r)Rr^3 )/((R−r)(R^2 −4r^2 )))(9) (8)⇒x−r=(r^2 /(R−r))(10).Replace (9)(10) into (3)we get (r^4 /((R−r)^2 ))+((4(R−2r)Rr^3 )/((R−r)(R^2 −4r^2 )))=r^2 ⇔r^2 (R^2 −4r^2 )+4(R−2r)(R−r)Rr =(R−r)^2 (R^2 −4r^2 ) ⇔R^2 r^2 −4r^4 +4R^3 r+8Rr^3 −12R^2 r^2 =R^4 −2R^3 r+R^2 r^2 −4R^2 r^2 +8Rr^3 −4r^4 ⇔R^4 −6R^3 r+8R^2 r^2 =0 8r^2 −6Rr+R^2 =0 Δ′=9R^2 −8R^2 =R^2 ⇒r=((3R−R)/8)=(R/4) (The root r=((3R+R)/8)=(R/2) is rejected) ⇒r=(R/4),replace into (8)we get x=(R/3) replace into (9)we get n=((R(√2))/6) ⇒m=((R(√2))/3)⇒cosMCQ^(�) =(x/R)=(1/3) ⇒MCN^(�) =2cos^(−1) ((1/3))⇒S_(sectorMCN) = ((πR^2 2cos^(−1) ((1/3)))/(2π))=R^2 cos^(−1) ((1/3)) S_(ΔMCN) =(m+2n)x=((2R^2 (√2))/9) ⇒S_(segMN) =R^2 cos^(−1) ((1/3))−((2R^2 (√2))/9) Since ΔNIP⋍ΔNCM with similar ratio equal to (1/4),S_(segMO) +S_(segNP) =2×(1/(16))S_(segMN) ΔODP⋍ΔMCN with similar ratio (1/2) ⇒S_(segOP) =(1/4)S_(segMN) ⇒S_(segOEP) =S_((D)) −S_(segOP) =((πR^2 )/4)−(1/4)S_(segMN) ⇒S_(green) =S_(segMN) −(1/8)S_(segMN) −(((πR^2 )/4)−(1/4)S_(segMN) ) =(R^2 /8)(9cos^(−1) ((1/3))−2(√2)−2π) ≈0.2458R^2$
	$Denote by r the radius of two small circles$ $CQ = x, PN = OM = 2 n; R and R_{1} - the$ $radius of semicircle and the great circle$ $QO = QP = m \Rightarrow R_{1} = R / 2$ $Then we have relations :$ ${\begin{cases} x (R - x) = m^{2} (1) (QC . QE = {QP}^{2}) \\ \frac{2 r}{R} = \frac{n}{m} (2) (IN / / OD) \\ {(x - r)}^{2} + n^{2} = r^{2} (3) (Δ OSF is right at S) \\ \frac{r}{R} = \frac{x - r}{x} (4) (\frac{MF}{MC} = \frac{FS}{QC}) \end{cases}$ $(3) \Leftrightarrow x^{2} - 2 rx + n^{2} = 0 (5)$ $(1) \Rightarrow x^{2} = Rx - m^{2} . Replace into (5) we get$ $Rx - m^{2} - 2 rx + n^{2} = 0 \Rightarrow x = \frac{m^{2} - n^{2}}{R - 2 r} (6)$ $(2) \Rightarrow \frac{R}{2 m} = \frac{r}{n} \Rightarrow \frac{R^{2}}{{4 m}^{2}} = \frac{r^{2}}{n^{2}} = \frac{R^{2} / 4 - r^{2}}{m^{2} - n^{2}} (7)$ $(4) \Rightarrow rx = Rx - Rr \Rightarrow x = \frac{Rr}{R - r} (8)$ $Replace into (6) we get$ $m^{2} - n^{2} = \frac{(R - 2 r) Rr}{R - r} replace into (7) we$ $get n^{2} = \frac{4 (R - 2 r) {Rr}^{3}}{(R - r) (R^{2} - {4 r}^{2})} (9)$ $(8) \Rightarrow x - r = \frac{r^{2}}{R - r} (10) . Replace (9) (10)$ $into (3) we get$ $\frac{r^{4}}{{(R - r)}^{2}} + \frac{4 (R - 2 r) {Rr}^{3}}{(R - r) (R^{2} - {4 r}^{2})} = r^{2}$ $\Leftrightarrow r^{2} (R^{2} - {4 r}^{2}) + 4 (R - 2 r) (R - r) Rr$ $= {(R - r)}^{2} (R^{2} - {4 r}^{2})$ $\Leftrightarrow R^{2} r^{2} - {4 r}^{4} + {4 R}^{3} r + {8 Rr}^{3} - {12 R}^{2} r^{2}$ $= R^{4} - {2 R}^{3} r + R^{2} r^{2} - {4 R}^{2} r^{2} + {8 Rr}^{3} - {4 r}^{4}$ $\Leftrightarrow R^{4} - {6 R}^{3} r + {8 R}^{2} r^{2} = 0$ ${8 r}^{2} - 6 Rr + R^{2} = 0$ $Δ^{'} = {9 R}^{2} - {8 R}^{2} = R^{2} \Rightarrow r = \frac{3 R - R}{8} = \frac{R}{4}$ $(The root r = \frac{3 R + R}{8} = \frac{R}{2} is rejected)$ $\Rightarrow r = \frac{R}{4}, replace into (8) we get x = \frac{R}{3}$ $replace into (9) we get n = \frac{R \sqrt{2}}{6}$ $\Rightarrow m = \frac{R \sqrt{2}}{3} \Rightarrow \cos \hat{MCQ} = \frac{x}{R} = \frac{1}{3}$ $\Rightarrow \hat{MCN} = {2 \cos}^{- 1} (\frac{1}{3}) \Rightarrow S_{sectorMCN} =$ $\frac{π R^{2} {2 \cos}^{- 1} (\frac{1}{3})}{2 π} = R^{2} \cos^{- 1} (\frac{1}{3})$ $S_{Δ MCN} = (m + 2 n) x = \frac{{2 R}^{2} \sqrt{2}}{9}$ $\Rightarrow S_{segMN} = R^{2} \cos^{- 1} (\frac{1}{3}) - \frac{{2 R}^{2} \sqrt{2}}{9}$ $Since Δ NIP ⋍ Δ NCM with similar ratio equal$ $to \frac{1}{4}, S_{segMO} + S_{segNP} = 2 \times \frac{1}{16} S_{segMN}$ $Δ ODP ⋍ Δ MCN with similar ratio \frac{1}{2}$ $\Rightarrow S_{segOP} = \frac{1}{4} S_{segMN} \Rightarrow S_{segOEP} = S_{(D)} - S_{segOP}$ $= \frac{π R^{2}}{4} - \frac{1}{4} S_{segMN}$ $\Rightarrow S_{green} = S_{segMN} - \frac{1}{8} S_{segMN} - (\frac{π R^{2}}{4} - \frac{1}{4} S_{segMN})$ $= \frac{R^{2}}{8} (9 \cos^{- 1} (\frac{1}{3}) - 2 \sqrt{2} - 2 π) \approx 0 . {2458 R}^{2}$
	Answered by mr W last updated on 24/Oct/20

	Commented by mr W last updated on 24/Oct/20

	$O P = P C = \frac{R}{2}$ $O Q = R - r$ $P Q = \frac{R}{2} + r$ ${O E}^{2} = {O Q}^{2} - {Q E}^{2} = {(R - r)}^{2} - r^{2} = R (R - 2 r)$ $= {P Q}^{2} - {(O P - Q E)}^{2} = {(\frac{R}{2} + r)}^{2} - {(\frac{R}{2} - r)}^{2}$ $= 2 R r$ $\Rightarrow 2 R r = R (R - 2 r)$ $\Rightarrow r = \frac{R}{4}$ $\sin α = \frac{r}{R - r} = \frac{1}{3}$ $\cos γ = \frac{\frac{R}{2} - r}{\frac{R}{2} + r} = \frac{1}{3} = \cos ϕ \Rightarrow ϕ = γ = \frac{π}{2} - α$ $2 β = π - (ϕ + γ)$ $\Rightarrow β = \frac{π}{2} - ϕ = α$ $\frac{A_{g r e e n}}{2} = \frac{R^{2} ϕ}{2} - 2 \times \frac{r^{2} ϕ}{2} - \frac{{(\frac{R}{2})}^{2} (π - γ)}{2} - \frac{\frac{R}{2} \times \sqrt{R (R - 2 r)}}{2}$ $= \frac{R^{2} (9 ϕ - 2 π)}{16} - \frac{\sqrt{2} R^{2}}{8}$ $= \frac{R^{2} (9 \cos^{- 1} \frac{1}{3} - 2 π - 2 \sqrt{2})}{16}$ $A_{g r e e n} = \frac{R^{2} (9 \cos^{- 1} \frac{1}{3} - 2 π - 2 \sqrt{2})}{8} \approx 0.2459 R^{2}$
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