Question-198763

Question Number 198763 by mr W last updated on 24/Oct/23

Commented by mr W last updated on 24/Oct/23

$$\mathrm{8}\:{small}\:{balls}\:{and}\:{a}\:{big}\:{ball}\:{on}\:{a} \\ $$$${table}\:{touch}\:{each}\:{other}. \\ $$$${if}\:{the}\:{radius}\:{of}\:{the}\:{small}\:{balls}\:{is}\:{r}, \\ $$$${find}\:{the}\:{radius}\:{of}\:{the}\:{big}\:{ball}. \\ $$

Commented by ajfour last updated on 24/Oct/23

$${could}\:{it}\:{be}\:\:\:\frac{{R}}{{r}}=\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\:\approx\mathrm{1}.\mathrm{707} \\ $$

Commented by mr W last updated on 24/Oct/23

$${yes}.\:{i}\:{got}\:{the}\:{same}. \\ $$

Answered by MM42 last updated on 24/Oct/23

α=45^0 & (2r)^2 =(R+r)^2 +(R+r)^2 −2(R+r)^2 ×cos45 (2r)^2 =(R+r)^2 (2−(√2)) ⇒R+r=((2r)/( (√(2−(√2))))) ⇒R=((2/( (√(2−(√2)))))−1)r R=((√(4+2(√2)))−1)r ✓

$$\alpha=\mathrm{45}^{\mathrm{0}} \:\:\&\:\left(\mathrm{2}{r}\right)^{\mathrm{2}} =\left({R}+{r}\right)^{\mathrm{2}} +\left({R}+{r}\right)^{\mathrm{2}} −\mathrm{2}\left({R}+{r}\right)^{\mathrm{2}} ×{cos}\mathrm{45} \\ $$$$\left(\mathrm{2}{r}\right)^{\mathrm{2}} =\left({R}+{r}\right)^{\mathrm{2}} \left(\mathrm{2}−\sqrt{\mathrm{2}}\right) \\ $$$$\Rightarrow{R}+{r}=\frac{\mathrm{2}{r}}{\:\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}}\:\Rightarrow{R}=\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}}−\mathrm{1}\right){r} \\ $$$${R}=\left(\sqrt{\mathrm{4}+\mathrm{2}\sqrt{\mathrm{2}}}−\mathrm{1}\right){r}\:\:\checkmark \\ $$$$ \\ $$

Commented by MM42 last updated on 24/Oct/23

Commented by mr W last updated on 24/Oct/23

$${be}\:{carefull}! \\ $$$${in}\:{ground}\:{view}\:{the}\:{big}\:{circle}\:{with} \\ $$$${radius}\:{R}\:{doesnt}\:{touch}\:{the}\:{small} \\ $$$${circles}\:{with}\:{radius}\:{r}.\:{they}\:{overlap}! \\ $$

Commented by mr W last updated on 24/Oct/23

Answered by nikif99 last updated on 24/Oct/23

R+r=(a/(2 sin (((180)/n))))=((2r)/(2 sin 22.5)) ⇒R=r((1/((√(2−(√2)))/2))−1)= r((2/( (√(2−(√2)))))−1)=r(((2(√(2+(√2))))/( (√2)))−1) ⇒ R=r((√2)∙(√(2+(√2)))−1)≈1.613r

$$ \\ $$$${R}+{r}=\frac{{a}}{\mathrm{2}\:\mathrm{sin}\:\left(\frac{\mathrm{180}}{{n}}\right)}=\frac{\mathrm{2}{r}}{\mathrm{2}\:\mathrm{sin}\:\mathrm{22}.\mathrm{5}}\:\Rightarrow{R}={r}\left(\frac{\mathrm{1}}{\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}}{\mathrm{2}}}−\mathrm{1}\right)= \\ $$$${r}\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}}−\mathrm{1}\right)={r}\left(\frac{\mathrm{2}\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}}{\:\sqrt{\mathrm{2}}}−\mathrm{1}\right)\:\Rightarrow \\ $$$${R}={r}\left(\sqrt{\mathrm{2}}\centerdot\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}−\mathrm{1}\right)\approx\mathrm{1}.\mathrm{613}{r} \\ $$

Commented by nikif99 last updated on 24/Oct/23

Commented by mr W last updated on 24/Oct/23

Commented by mr W last updated on 24/Oct/23

Commented by nikif99 last updated on 24/Oct/23

$${right}.\:{A}\:{detail}\:{not}\:{noticed}. \\ $$

Answered by mr W last updated on 24/Oct/23

Commented by mr W last updated on 24/Oct/23

θ=((360°)/(8×2))=22.5° 1−2 sin^2 22.5°=cos 45°=((√2)/2) ⇒sin 22.5°=((√(2−(√2)))/2) a=(r/(sin θ))=((2r)/( (√(2−(√2)))))=(√(2(2+(√2))))r (R+r)^2 −(R−r)^2 =a^2 4Rr=2(2+(√2))r^2 ⇒(R/r)=1+((√2)/2)≈1.707 ■

$$\theta=\frac{\mathrm{360}°}{\mathrm{8}×\mathrm{2}}=\mathrm{22}.\mathrm{5}° \\ $$$$\mathrm{1}−\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\mathrm{22}.\mathrm{5}°=\mathrm{cos}\:\mathrm{45}°=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{sin}\:\mathrm{22}.\mathrm{5}°=\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}}{\mathrm{2}} \\ $$$${a}=\frac{{r}}{\mathrm{sin}\:\theta}=\frac{\mathrm{2}{r}}{\:\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}}=\sqrt{\mathrm{2}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)}{r} \\ $$$$\left({R}+{r}\right)^{\mathrm{2}} −\left({R}−{r}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$$\mathrm{4}{Rr}=\mathrm{2}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right){r}^{\mathrm{2}} \\ $$$$\Rightarrow\frac{{R}}{{r}}=\mathrm{1}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\approx\mathrm{1}.\mathrm{707} \\ $$$$\blacksquare \\ $$

Commented by MM42 last updated on 24/Oct/23

$${Sir}\:{W} \\ $$$${in}\:{the}\:{two}\:{image}\:{given}\:{by}\:{you}, \\ $$$${the}\:{value}\:{of}\:“{a}\:''\:{dose}\:{not}\:{match} \\ $$

Commented by mr W last updated on 24/Oct/23

but they do match! “a” is the projection of OA in horizontal direction. the first image is a ground view, i.e. a view from top. a≠OA=OB a=O′A′=O′B′

$${but}\:{they}\:{do}\:{match}!\:“{a}''\:{is}\:{the}\:{projection} \\ $$$${of}\:{OA}\:{in}\:\underline{{horizontal}}\:{direction}. \\ $$$${the}\:{first}\:{image}\:{is}\:{a}\:{ground}\:{view},\:{i}.{e}. \\ $$$${a}\:{view}\:{from}\:{top}. \\ $$$${a}\neq{OA}={OB} \\ $$$${a}={O}'{A}'={O}'{B}' \\ $$

Commented by mr W last updated on 24/Oct/23