Question-226246

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Question Number 226246 by ajfour last updated on 23/Nov/25

Commented by fantastic2 last updated on 23/Nov/25

$$\alpha\:{is}\:\mathrm{30}^{\mathrm{0}} \\ $$

Commented by fantastic2 last updated on 23/Nov/25

Commented by fantastic2 last updated on 23/Nov/25

Commented by ajfour last updated on 23/Nov/25

If sector angle is 2α. Find r/R. Smaller ones are equal in size. Lets say tan α=m

$${If}\:{sector}\:{angle}\:{is}\:\mathrm{2}\alpha. \\ $$$${Find}\:{r}/{R}.\:{Smaller}\:{ones}\:{are}\:{equal} \\ $$$${in}\:{size}.\:\:{Lets}\:{say}\:\mathrm{tan}\:\alpha={m} \\ $$

Commented by mr W last updated on 23/Nov/25

$$\alpha\:{must}\:{be}\:{equal}\:{to}\:\mathrm{30}°? \\ $$

Answered by fantastic2 last updated on 23/Nov/25

Commented by fantastic2 last updated on 24/Nov/25

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Commented by ajfour last updated on 24/Nov/25

Commented by ajfour last updated on 24/Nov/25

$${yes}\:{all}\:{correct}.\:{Great}. \\ $$

Commented by fantastic2 last updated on 23/Nov/25

(R−r)^2 =(2r)^2 +(2r)^2 −2(2r)(2r)cos 150^0 (R−r)^2 =8r^2 +4(√3)r^2 (R−r)^2 =4r^2 (2+(√3)) R=r((√(8+4(√3)))+1) (R/r)=((√(8+4(√3)))+1)

$$\left({R}−{r}\right)^{\mathrm{2}} =\left(\mathrm{2}{r}\right)^{\mathrm{2}} +\left(\mathrm{2}{r}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{2}{r}\right)\left(\mathrm{2}{r}\right)\mathrm{cos}\:\mathrm{150}^{\mathrm{0}} \\ $$$$\left({R}−{r}\right)^{\mathrm{2}} =\mathrm{8}{r}^{\mathrm{2}} +\mathrm{4}\sqrt{\mathrm{3}}{r}^{\mathrm{2}} \\ $$$$\left({R}−{r}\right)^{\mathrm{2}} =\mathrm{4}{r}^{\mathrm{2}} \left(\mathrm{2}+\sqrt{\mathrm{3}}\right) \\ $$$${R}={r}\left(\sqrt{\mathrm{8}+\mathrm{4}\sqrt{\mathrm{3}}}+\mathrm{1}\right) \\ $$$$\frac{{R}}{{r}}=\left(\sqrt{\mathrm{8}+\mathrm{4}\sqrt{\mathrm{3}}}+\mathrm{1}\right) \\ $$

Commented by fantastic2 last updated on 23/Nov/25

or (r/(R−r))=sin 15^0 =(((√6)−(√2))/4) 4r=((√6)−(√2))(R−r)=R((√6)−(√2))−r((√6)−(√2)) r(4+(√6)−(√2))=R((√6)−(√2)) (R/r)=((4+(√6)−(√2))/( (√6)−(√2))) example if R=1 r≈0.205604

$${or} \\ $$$$\frac{{r}}{{R}−{r}}=\mathrm{sin}\:\mathrm{15}^{\mathrm{0}} =\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$$$\mathrm{4}{r}=\left(\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}\right)\left({R}−{r}\right)={R}\left(\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}\right)−{r}\left(\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}\right) \\ $$$${r}\left(\mathrm{4}+\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}\right)={R}\left(\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}\right) \\ $$$$\frac{{R}}{{r}}=\frac{\mathrm{4}+\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}} \\ $$$${example} \\ $$$${if}\:{R}=\mathrm{1}\: \\ $$$${r}\approx\mathrm{0}.\mathrm{205604} \\ $$

Answered by Ghisom_ last updated on 24/Nov/25

let the radius of the sector R=1 the angle = 2α I rotate the image by −90° and use the upper half of the sector for calculations 1. circle touching both y=xtan α and x^2 +y^2 =1 C_1 : (x−p)^2 +(y−r_1 )^2 =r_1 ^2 ⇒ p=((1−sin (α/2))/(cos (α/2)))∧r_1 =ptan (α/2) 2. circle touching both y=xtan α and C_1 C_2 : (x−q)^2 +y^2 =r_2 ^2 ⇒ q=((1−sin (α/2))/(cos (α/2) (1+(√2)sin (α/2))^2 ))∧r_2 =2qcos (α/2) sin (α/2) α=30° ⇒ r_1 =r_2 =3(√6)+4(√3)−5(√2)−7 (r_(1, 2) /R)=(√6)+(√2)+1

$$\mathrm{let}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sector}\:{R}=\mathrm{1} \\ $$$$\mathrm{the}\:\mathrm{angle}\:=\:\mathrm{2}\alpha \\ $$$$\mathrm{I}\:\mathrm{rotate}\:\mathrm{the}\:\mathrm{image}\:\mathrm{by}\:−\mathrm{90}°\:\mathrm{and}\:\mathrm{use}\:\mathrm{the} \\ $$$$\mathrm{upper}\:\mathrm{half}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sector}\:\mathrm{for}\:\mathrm{calculations} \\ $$$$ \\ $$$$\mathrm{1}. \\ $$$$\mathrm{circle}\:\mathrm{touching}\:\mathrm{both}\:{y}={x}\mathrm{tan}\:\alpha\:\mathrm{and}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{1} \\ $$$${C}_{\mathrm{1}} :\:\left({x}−{p}\right)^{\mathrm{2}} +\left({y}−{r}_{\mathrm{1}} \right)^{\mathrm{2}} ={r}_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$${p}=\frac{\mathrm{1}−\mathrm{sin}\:\frac{\alpha}{\mathrm{2}}}{\mathrm{cos}\:\frac{\alpha}{\mathrm{2}}}\wedge{r}_{\mathrm{1}} ={p}\mathrm{tan}\:\frac{\alpha}{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{2}. \\ $$$$\mathrm{circle}\:\mathrm{touching}\:\mathrm{both}\:{y}={x}\mathrm{tan}\:\alpha\:\mathrm{and}\:{C}_{\mathrm{1}} \\ $$$${C}_{\mathrm{2}} :\:\left({x}−{q}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} ={r}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$${q}=\frac{\mathrm{1}−\mathrm{sin}\:\frac{\alpha}{\mathrm{2}}}{\mathrm{cos}\:\frac{\alpha}{\mathrm{2}}\:\left(\mathrm{1}+\sqrt{\mathrm{2}}\mathrm{sin}\:\frac{\alpha}{\mathrm{2}}\right)^{\mathrm{2}} }\wedge{r}_{\mathrm{2}} =\mathrm{2}{q}\mathrm{cos}\:\frac{\alpha}{\mathrm{2}}\:\mathrm{sin}\:\frac{\alpha}{\mathrm{2}} \\ $$$$ \\ $$$$\alpha=\mathrm{30}°\:\Rightarrow\:{r}_{\mathrm{1}} ={r}_{\mathrm{2}} =\mathrm{3}\sqrt{\mathrm{6}}+\mathrm{4}\sqrt{\mathrm{3}}−\mathrm{5}\sqrt{\mathrm{2}}−\mathrm{7} \\ $$$$\frac{{r}_{\mathrm{1},\:\mathrm{2}} }{{R}}=\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}+\mathrm{1} \\ $$

Commented by fantastic2 last updated on 25/Nov/25

$${good}\:{method} \\ $$

Commented by ajfour last updated on 24/Nov/25

$${yeah},\:{i}\:{appreciare}\:{this}\:{more}. \\ $$$${Thanks}. \\ $$

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