Twelve identical circles touching one another on the surface of a sphere

Question

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.

Answer 1

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 $$\sin\theta=\frac{CA}{OA}=\frac{r}{R}=\sqrt{\frac{5-\sqrt{5}}{10}}$$ $$\implies \color{blue}{\theta=\sin^{-1}\sqrt{\frac{5-\sqrt{5}}{10}}}$$ $$\text{arc radius of each circle}=arc AC'=R\theta$$ $$\color {green}{=R\sin^{-1}\sqrt{\frac{5-\sqrt{5}}{10}}\approx 0.553574358\space R}$$

Answer 2

I computed numerically the radius of these circles, for a sphere of radius $1$. I obtained $R = 0.5257$. I'm not sure how you could approach the problem analytically.

The numerical procedure is quite straightforward. Choose the vertices of a dodecahedron, for every face, compute the midpoints, project them on the sphere, and compute the radius of the circle going through them.

Here's a picture of the circle configuration:

Answer 3

The dihedral angle between two faces of the dodecahedron is $\pi-\arctan(2)$; since this angle forms a kite-shaped quadrilateral with the interior angle from the center of the sphere to the two faces, with both other angles being right angles, then that interior angle is $\arctan(2)$, and therefore the great-circle radius is the radius of the sphere times one-half this angle; $\frac12R\arctan(2)$.

Answer 4

The twelve circles are inscribed in the faces of a dodecahedron. The midpoint of each edge of that dodecahedron is tangent to the sphere.

Let the center of the sphere (which is also the centroid of the dodecahedron) be the origin of a Cartesian coordinate system. The following set of $20$ coordinates, $$(\pm a, \pm a, \pm a),$$ $$(\pm a\phi, \pm a\phi^{-1}, 0),$$ $$(0, \pm a\phi, \pm a\phi^{-1}),$$ $$(\pm a\phi^{-1}, 0, \pm a\phi),$$ including every possible choice of sign for each $\pm$ sign in each triple, where $\phi = \frac12(1+\sqrt5)$ is the golden ratio, are the vertices of a regular dodecahedron. By orienting the axes appropriately, we can make these be the coordinates of the vertices of the dodecahedron on which the circles are inscribed. (Usually these coordinates are given without the factor $a$, but we may need to scale the dodecahedron larger or smaller in order to fit the circles on the sphere of radius $R$.)

We could now use these coordinates to find the length of an edge of the dodecahedron in terms of $R$, then use the ratio of the inradius of a regular pentagon and its edge to find the radius of one of the circles in terms of $R$, and take an arc sine to convert this to an angle of arc on the surface of the sphere. But instead, let's consider the face of the dodecahedron with vertices at $(a\phi, a\phi^{-1}, 0)$, $(a\phi, -a\phi^{-1}, 0)$, $(a,-a,a)$, $(a\phi^{-1}, 0, a\phi)$, and $(a, a, a)$. By symmetry, the centroid of that pentagon, which is also the center of one of the circles, is at the aritmetic mean of those five points, that is,

$$\begin{multline} \frac{(a\phi, a\phi^{-1}, 0) + (a\phi, -a\phi^{-1}, 0) + (a,-a, a) + (a\phi^{-1}, 0, a\phi) + (a, a, a)}{5} \\ = \left(\frac15\left(2a\phi + 2a + a\phi^{-1}\right), 0, \frac15\left(2a + a\phi\right) \right) = \left( \frac{1}{10}(5+3\sqrt5)a, 0, \frac{1}{10}(5+\sqrt5)a \right). \end{multline}$$

But the point $(a\phi,0,0)$ is the midpoint of one of the edges of that same pentagon, which is also a point on the circle whose center is at $\left( \frac{1}{10}(5+3\sqrt5)a, 0, \frac{1}{10}(5+\sqrt5)a \right)$, and the angle subtended by those two points from the center of the sphere is just the angle adjacent to the leg of length $\frac{1}{10}(5+3\sqrt5)a$ of a right triangle with legs $\frac{1}{10}(5+3\sqrt5)a$ and $\frac{1}{10}(5+\sqrt5)a$. You can find that angle by taking the appropriate arc tangent.

Answer 5

Here is another take on this probelm.

You are given twelve circles of equal radii on the surface of sphere $S_R$ of radius $R$, so that every two circles touch externally at exactly one point. Two such touching circles, say $c_1$ and $c_2$ touching at point $T_{12}$, lie in pair of planes, so that each circle is the intersection of the corresponding plane with the sphere $S_R$. The intersection of the planes of circles $c_1$ and $c_2$ is a line $l_{12}$ tangent to the sphere at point $T_{12}$. If you take all such lines tangent to the sphere, they form the edges of a regular dodecahedron, such that the sphere $S_R$ is tangent to the edges of the dodecahedron at their midpoints (e.g. $T_{12}$) and each circle is the incircle of its corresponding regular pentagonal face.

Now, for each tangent point $T_{12}$ between a pair of touching circles $c_1$ and $c_2$, there exists a unique plane tangent to the sphere $S_{R}$ at the point $T_{12}$ and the line $l_{12}$ lies in ti. In this tangent plane, take the unique line $t_{12}$ that passes through point $T_{12}$ and is perpendicular to the tangent line $l_{12}$. By construction, since the line $t_{12}$ lies in the tangent plane to the sphere at the point $T_{12}$, it is tangent to the sphere $S_{R}$. The set of all these tangent lines $t_{12}$ form the edges of a regular icosahedron such that the sphere $S_R$ is tangent to the edges of the icosahedron at their midpoints (i.e. $T_{12}$). In this construction, the two polyhedra, the dodecahedron and the icosahedron are dual, with pairwise perpendicular edges that are also tangent to the surface of the sphere. Observe that the faces of the icosahderon are equilateral triangles, tangent to the sphere.

Let's focus on one regular pentagonal face of the dodecahedron. Then the incircle of the face, which is one of the twelve circle on the sphere, is tangent to the midpoints of the edges of the pentagon. At the same time, there are five half-edges of the icosahedron, that pass through these five midpoints, are perpendicular to the edges of the pentagon and meet at a common vertex (of the icosahedron) right above the center of the circle (i.e. the vertex projects orthogonally onto the center of the circle).

Thus, the five midpoints of the edges of the regular pentagonal faceof the dodecahedron, together with the vertex of intersection of the five half-edges of the icosahedron passing through these five midpoints, form a regular pentagonal pyramid, whose base is a regular pentagon inscribed in the circle. Each triangular face of this pyramid is isosceles triangle. However, the angle of each triangular face of this pyramid at the common vertex of the icosahedron, which we will call the apex of the pyramid, is $60^{\circ}$. Hence all of the triangular faces of the pyramid are equilateral triangles.

Now, let us focus on this regular pentagonal pyramid. Let the circle on the sphere $S_R$, which is the circumcircle of the pyramid's pentagonal base, have euclidean radius $r$ (radius of the circle in the plane that contains it). If you draw the triangle formed by one edge of the pentagonal face of the pyramid and the center of the base's circumcircle, the said triangle is isosceles with two equal edges of length $r$ and angle between them $\frac{2\pi}{5}$. In this triangle, the perpendicular from the center of the circle to the edge of the regular pentagonal base is the orthogonal bisector of that edge, so this perpendicular splits the isosceles triangle into two mirror-symmetric right-angle triangles with hypotenuse $r$ and angle at the center of the circle $\frac{\pi}{5}$. Thus, the half-edge of the pentagonal base of the regular pyramid is $r\sin\left(\frac{2\pi}{5}\right)$ and consequently, the edges of the pyramid's regular pentagonal base are all equal to $2\,r\sin\left(\frac{2\pi}{5}\right)$.

However, the rest of the faces of the pyramid are equilateral triangular faces, so the non-base edges of the pyramid have length $2\,r\sin\left(\frac{2\pi}{5}\right)$.

Now, let us take the triangle formed by the apex, call it $A$, of the regular pentagonal pyramid, the center, call it $O$, of the sphere $S_R$, and one fixed base vertices, call it $B$, of the regular pentagonal pyramid. Since $AB$ is in fact one of the half-edges of the icosahedron, by construction it is tangent to the sphere $S_R$ at the point $B$ (in fact $B$ is one of the points of tangency $T_{12}$ between the pairs of equal circle on the sphere, which were introduced in the beginning), which means that the radius $OB$ of the sphere $S_R$ is perpendicular to $AB$. Therefore triangle $\Delta\, ABO$ is right-angled at vertex $B$. Also, its hypotenuse $AO$ is the unique perpendicular to the pentagonal base of the regular pentagonal pyramid through the apex $A$, so $AO$ passes through the center (call it $Q$) of the circle circumscribed around the pyramid's base. Hence, $BQ$ is a radius of the circle and at the same time it is the height of the triangle $\Delta\, ABO$ through vertex $B$. Consequently, we have that $$AB = 2\,r\sin\left(\frac{2\pi}{5}\right), \,\,\, BO = R, \,\,\, BQ = r$$ The two right-angled triangles $\Delta \, ABO$ and $\Delta\, AQB$ share a common angle at vertex $A$, so they are similar and therefore $$\theta = \angle \, AOB = \angle \, ABQ$$ and $$\cos(\theta) = \frac{BO}{AO} = \frac{BQ}{AB}$$ so after substituting in the first equality from the above formula, we find that $$\cos(\theta) = \frac{BQ}{AB} = \frac{r}{2\,r\sin\left(\frac{2\pi}{5}\right)} = \frac{1}{2\,\sin\left(\frac{2\pi}{5}\right)}$$ Thus, $$\theta = \arccos\left(\frac{1}{2\,\sin\left(\frac{2\pi}{5}\right)}\right)$$ and therefore, the spherical radius of any of the twelve congruent circles on the sphere $S_R$ of radius $R$ is $$\rho = R\,\theta = R\,\arccos\left(\frac{1}{2\,\sin\left(\frac{2\pi}{5}\right)}\right) = 0.5535744\, R$$