Question-225703

Question Number 225703 by ajfour last updated on 07/Nov/25

Answered by mahdipoor last updated on 07/Nov/25

2×(R−r)^2 =(R+r)^2 ⇒r=R(3−(√8)) center of balls (r) in R_c =R−r θ_(each ball) =2tan^(−1) (r/R_c )=2tan^(−1) (((3−(√8))/( (√8)−2))) ((2π)/θ)=15.38 ⇒ 15

$$\mathrm{2}×\left(\mathrm{R}−\mathrm{r}\right)^{\mathrm{2}} =\left(\mathrm{R}+\mathrm{r}\right)^{\mathrm{2}} \Rightarrow\mathrm{r}=\mathrm{R}\left(\mathrm{3}−\sqrt{\mathrm{8}}\right) \\ $$$$\mathrm{center}\:\mathrm{of}\:\mathrm{balls}\:\left(\mathrm{r}\right)\:\mathrm{in}\:\mathrm{R}_{\mathrm{c}} =\mathrm{R}−\mathrm{r}\:\: \\ $$$$\theta_{\mathrm{each}\:\mathrm{ball}} =\mathrm{2tan}^{−\mathrm{1}} \left(\mathrm{r}/\mathrm{R}_{\mathrm{c}} \right)=\mathrm{2tan}^{−\mathrm{1}} \left(\frac{\mathrm{3}−\sqrt{\mathrm{8}}}{\:\sqrt{\mathrm{8}}−\mathrm{2}}\right) \\ $$$$\frac{\mathrm{2}\pi}{\theta}=\mathrm{15}.\mathrm{38}\:\Rightarrow\:\mathrm{15}\: \\ $$

Commented by ajfour last updated on 07/Nov/25

$${yeah}!\:{Indeed}\:{v}\:{nice}\:{solution}. \\ $$

Answered by fantastic last updated on 08/Nov/25

from Sir mr W′s first picture we get 2(R−r)^2 =(R+r)^2 ⇒r=R(((√2)−1)/( (√2)+1))=R(3−2(√2)) α_(taken by each ball) =2tan^(−1) ((r/(R−r))) The R−r came because if you take the ball and move it vertically till r_(center) at z=R[x,y are not changed] and cut the whole thing at z=R you will notice the angle subtended by the r_(circle) is=2tan^(−1) ((r/(R−r)))[2nd picture of mrW sir] maximum you can put is= 2{((2π)/(2tan^(−1) ((r/(R−r)))))}=30[you can put balls on top side too]

$${from}\:{Sir}\:{mr}\:{W}'{s}\:{first} \\ $$$${picture}\:{we}\:{get} \\ $$$$\mathrm{2}\left({R}−{r}\right)^{\mathrm{2}} =\left({R}+{r}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{r}={R}\frac{\sqrt{\mathrm{2}}−\mathrm{1}}{\:\sqrt{\mathrm{2}}+\mathrm{1}}={R}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right) \\ $$$$\alpha_{{taken}\:{by}\:{each}\:{ball}} =\mathrm{2tan}^{−\mathrm{1}} \left(\frac{{r}}{{R}−{r}}\right) \\ $$$${The}\:{R}−{r}\:{came}\:{because} \\ $$$${if}\:{you}\:{take}\:{the}\:{ball}\:{and} \\ $$$${move}\:{it}\:{vertically}\:{till}\:{r}_{{center}} \:{at}\:{z}={R}\left[{x},{y}\:{are}\:{not}\:{changed}\right] \\ $$$${and}\:{cut}\:{the}\:{whole}\:{thing}\:\:{at}\:{z}={R} \\ $$$${you}\:{will}\:{notice}\:{the}\:{angle}\:{subtended} \\ $$$${by}\:{the}\:{r}_{{circle}} \:{is}=\mathrm{2tan}^{−\mathrm{1}} \left(\frac{{r}}{{R}−{r}}\right)\left[\mathrm{2}{nd}\:{picture}\:{of}\:{mrW}\:{sir}\right] \\ $$$${maximum}\:{you}\:{can}\:{put} \\ $$$${is}=\:\mathrm{2}\left\{\frac{\mathrm{2}\pi}{\mathrm{2tan}^{−\mathrm{1}} \left(\frac{{r}}{{R}−{r}}\right)}\right\}=\mathrm{30}\left[{you}\:{can}\:{put}\:{balls}\:{on}\:{top}\:{side}\:{too}\right] \\ $$$$ \\ $$

Answered by mr W last updated on 07/Nov/25

Commented by mr W last updated on 09/Nov/25

(√2)R−R−r=(√2)r ⇒(R/r)=(((√2)+1)/( (√2)−1))=3+2(√2) sin φ=(r/(R−r))=(1/((R/r)−1)) =(1/(3+2(√2)−1))=(((√2)−1)/2) φ=sin^(−1) (((√2)−1)/2) 2nφ≤2π ⇒n≤(π/(sin^(−1) (((√2)−1)/2)))≈15.06 that means 15 marble balls can be placed inside.

$$\sqrt{\mathrm{2}}{R}−{R}−{r}=\sqrt{\mathrm{2}}{r} \\ $$$$\Rightarrow\frac{{R}}{{r}}=\frac{\sqrt{\mathrm{2}}+\mathrm{1}}{\:\sqrt{\mathrm{2}}−\mathrm{1}}=\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\mathrm{sin}\:\phi=\frac{{r}}{{R}−{r}}=\frac{\mathrm{1}}{\frac{{R}}{{r}}−\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{1}}=\frac{\sqrt{\mathrm{2}}−\mathrm{1}}{\mathrm{2}} \\ $$$$\phi=\mathrm{sin}^{−\mathrm{1}} \frac{\sqrt{\mathrm{2}}−\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{2}{n}\phi\leqslant\mathrm{2}\pi \\ $$$$\Rightarrow{n}\leqslant\frac{\pi}{\mathrm{sin}^{−\mathrm{1}} \frac{\sqrt{\mathrm{2}}−\mathrm{1}}{\mathrm{2}}}\approx\mathrm{15}.\mathrm{06} \\ $$$${that}\:{means}\:\mathrm{15}\:{marble}\:{balls}\:{can}\:{be}\: \\ $$$${placed}\:{inside}. \\ $$

Commented by ajfour last updated on 07/Nov/25

$${Thank}\:{you}\:{sir}.\:{The}\:{question} \\ $$$${occured}\:{to}\:{me}\:{on}\:{itself}. \\ $$

Commented by mr W last updated on 08/Nov/25

Commented by mr W last updated on 09/Nov/25

sin φ=(r/(R−r)) ⇒φ=sin^(−1) (r/(R−r)) angle taken by each ball is α=2φ=2× sin^(−1) (r/(R−r))=2 sin^(−1) (r/(R−r)) so n=((2π)/α)=(π/(sin^(−1) (r/(R−r)))) =(π/(sin^(−1) (((√2)−1)/2)))≈15.06 since sin^(−1) (r/(R−r))≈(r/(R−r)), n≈(π/(r/(R−r)))=((R/r)−1)π =2((√2)+1)π=15.17

$$\mathrm{sin}\:\phi=\frac{{r}}{{R}−{r}}\:\Rightarrow\phi=\mathrm{sin}^{−\mathrm{1}} \frac{{r}}{{R}−{r}} \\ $$$${angle}\:{taken}\:{by}\:{each}\:{ball}\:{is} \\ $$$$\alpha=\mathrm{2}\phi=\mathrm{2}×\:\mathrm{sin}^{−\mathrm{1}} \frac{{r}}{{R}−{r}}=\mathrm{2}\:\mathrm{sin}^{−\mathrm{1}} \frac{{r}}{{R}−{r}} \\ $$$${so} \\ $$$${n}=\frac{\mathrm{2}\pi}{\alpha}=\frac{\pi}{\mathrm{sin}^{−\mathrm{1}} \frac{{r}}{{R}−{r}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{\pi}{\mathrm{sin}^{−\mathrm{1}} \frac{\sqrt{\mathrm{2}}−\mathrm{1}}{\mathrm{2}}}\approx\mathrm{15}.\mathrm{06} \\ $$$${since}\:\mathrm{sin}^{−\mathrm{1}} \frac{{r}}{{R}−{r}}\approx\frac{{r}}{{R}−{r}},\: \\ $$$${n}\approx\frac{\pi}{\frac{{r}}{{R}−{r}}}=\left(\frac{{R}}{{r}}−\mathrm{1}\right)\pi \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)\pi=\mathrm{15}.\mathrm{17} \\ $$

Commented by mr W last updated on 09/Nov/25