Twelve identical circles are to be drawn on a spherical surface having a radius $R$ such that the circles touch one another at 30 different points i.e. each of 12 circles exactly touches other five circles thus covering up the whole sphere.
What must the radius, as a great circle arc, of each of such 12 identical circles in terms of $R$?
I have tried to calculate flat radius of circle by assuming that each of 12 identical circles is inscribed by each of 12 congruent regular pentagonal faces of a dodecahedron, but could not co-relate edge length of dodecahedron with radius of sphere $R$. Hence, unable to find out the radius as a great circle arc on the sphere.
Thanks for generous help.
Best Answer
After hard work on this problem, I could find an approach to the solution that I am posting here
In this case, let’s assume that each of 12 identical circles, with a flat radius r, is inscribed by each of 12 congruent regular pentagonal faces of $a$ regular dodecahedron with an edge length $a$ such that regular dodecahedron is concentric with the spherical surface having the center $O$ & a radius $R$.
Thus, all 30 points of tangency of the circles, lying on the spherical surface, are coincident with the mid-points of all 30 edges of a regular dodecahedron.
Now, consider one of the 12 identical circles with the center $C$ on the flat face & a flat radius r, touching five other circles at the points A, B, D, E & F (lying on the spherical surface as well as on the edges of the dodecahedron) and is inscribed by a regular pentagonal face of the dodecahedron with an edge length $a$ . (See the figure 1 below showing a regular pentagonal face of dodecahedron)
The flat radius $r$ of the circle inscribed by a regular pentagonal face with edge length $a$ is given as $$r=\frac{a}{2}\cot\frac{\pi}{5} \implies a=2r\tan\frac{\pi}{5}$$ $$\color {blue}{a=2r\sqrt{5-2\sqrt{5}}} \tag 1$$ Now, the radius $R$ of the spherical surface passing through all 12 identical vertices of a dodecahedron with edge length $a$ is given as $$\color {red}{R=\frac{\sqrt{3}(\sqrt{5}+1)a}{4}} $$ Now, the normal distance ($h=OC$) of each pentagonal face (having a circumcribed radius $\frac{a}{2}\sin\frac{\pi}{5}$) from the center O of the dodecahedron is given as $$h=\sqrt{\left(\frac{\sqrt{3}(\sqrt{5}+1)a}{4}\right)^2-\left(\frac{a}{2}\sin\frac{\pi}{5}\right)^2}$$$$=\frac{a}{2}\sqrt{\frac{25+11\sqrt{5}}{10}}$$ Now, substituting the value of $a$ in terms of $r$, we get $$h=\frac{2r\sqrt{5-2\sqrt{5}}}{2}\sqrt{\frac{25+11\sqrt{5}}{10}}$$$$\implies \color{blue} {h=OC=\frac{(1+\sqrt{5})r}{2}}$$
Draw the perpendicular OC from the centre O of the spherical surface (i.e. centre of the regular dodecahedron) to the centre C of the plane (flat) circle & join any of the points A, B, D, E & F of tangency of the plane circle say point A (i.e. mid-point of one of the edges of dodecahedron) to the centre O of the spherical surface (i.e. the centre of dodecahedron).
Thus, we obtain a right $\Delta OCA$ (as shown in the figure 2 above)
Applying Pythagoras Theorem in right $\Delta OCA$ as follows $$(OA)^2=(OC)^2+(CA)^2$$ $$\implies (R)^2=\left(\frac{(1+\sqrt{5})r}{2} \right)^2+(r)^2 $$ $$\implies \color {blue}{r=R\sqrt{\frac{5-\sqrt{5}}{10}}}$$ From the figure 2 above, we have $$\sin\theta=\frac{CA}{OA}=\frac{r}{R}=\sqrt{\frac{5-\sqrt{5}}{10}}$$ $$\implies \color{blue}{\theta=\sin^{-1}\sqrt{\frac{5-\sqrt{5}}{10}}}$$ $$\text{arc radius of each circle}=arc AC'=R\theta$$ $$\color {green}{=R\sin^{-1}\sqrt{\frac{5-\sqrt{5}}{10}}\approx 0.553574358\space R}$$