Kvant Math Problem 2
Each circle lies on the unit sphere.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 2m19s
Source on kvant.digital
Problem
A sphere of radius 1 is given. On it are placed equal circles $\gamma_0$, $\gamma_1$, $\ldots$, $\gamma_n$ of radius $r$ ($n\ge3$). The circle $\gamma_0$ is tangent to all circles $\gamma_1$, $\ldots$, $\gamma_n$; moreover, the circles $\gamma_1$ and $\gamma_2$; $\gamma_2$ and $\gamma_3$; $\ldots$; $\gamma_n$ and $\gamma_1$ are tangent to each other.
For which $n$ is this possible? Calculate the corresponding radius $r$.
Czechoslovak Mathematical Olympiad (XVIII)
Exploration
Each circle lies on the unit sphere. A circle of spherical radius $r$ is obtained by intersecting the sphere with a plane at a fixed distance from the center. Since all circles are equal, their centers on the sphere are points at equal spherical distance from the corresponding circle boundaries.
Let $O_i$ be the center of $\gamma_i$ on the sphere. Two equal circles of spherical radius $r$ are tangent precisely when the spherical distance between their centers equals $2r$. Thus the points $O_1,\dots,O_n$ form a spherical polygon whose consecutive sides all have length $2r$, and every $O_i$ is also at spherical distance $2r$ from $O_0$.
Hence $O_1,\dots,O_n$ lie on the spherical circle centered at $O_0$ of radius $2r$. The problem becomes the following. On the spherical circle of radius $2r$ about $O_0$, place $n$ points so that consecutive points on that circle are separated by spherical distance $2r$.
Choose coordinates with $O_0$ at the north pole. Let
$$\alpha=2r.$$
Then $O_1,\dots,O_n$ lie on the parallel of colatitude $\alpha$. If two neighboring points differ in longitude by $\varphi$, their spherical distance is also $\alpha$. Applying the spherical cosine theorem gives
$$\cos\alpha = \cos^2\alpha+\sin^2\alpha\cos\varphi.$$
For $0<\alpha<\pi$ this reduces to
$$\cos\varphi=\frac{\cos\alpha}{1+\cos\alpha}.$$
Since the $n$ points form a closed regular polygon on that parallel, necessarily
$$\varphi=\frac{2\pi}{n}.$$
Thus
$$\cos\frac{2\pi}{n} = \frac{\cos\alpha}{1+\cos\alpha}.$$
Solving for $\cos\alpha$ yields
$$\cos\alpha=\frac{\cos(2\pi/n)}{1-\cos(2\pi/n)}.$$
The crucial point is determining for which $n$ the right-hand side lies in the interval $[-1,1]$.
Testing small values,
$$n=3:\quad \frac{-1/2}{3/2}=-\frac13,$$
acceptable;
$$n=4:\quad 0,$$
acceptable;
$$n=5:\quad \frac{0.309\ldots}{0.691\ldots}=0.447\ldots,$$
acceptable;
$$n=6:\quad 1,$$
acceptable.
For $n>6$, $\cos(2\pi/n)>1/2$, hence the quotient exceeds $1$, impossible. This suggests that only $3\le n\le6$ occur.
The step most likely to hide an error is the passage from tangency of equal circles on the sphere to the statement that the spherical distance between their centers equals the sum of their radii, namely $2r$. That fact must be justified carefully.
Problem Understanding
We are given $n+1$ equal circles on the unit sphere. The circle $\gamma_0$ is tangent to each of the other circles. The circles $\gamma_1,\dots,\gamma_n$ form a tangency cycle: each circle is tangent to its two neighbors and to no other tangencies are prescribed.
The task is to determine all integers $n\ge3$ for which such a configuration exists and to compute the corresponding circle radius $r$.
This is a Type A problem. We must determine all admissible values of $n$ and prove both existence and impossibility.
The geometric structure suggests replacing each circle by its center on the sphere. Tangency of equal spherical circles translates into a fixed spherical distance between centers. The resulting centers form a regular polygon on a spherical circle around the center of $\gamma_0$, leading to a trigonometric condition that restricts $n$.
The expected answer is that only
$$n=3,4,5,6$$
are possible. The radius is determined uniquely by
$$\cos(2r)=\frac{\cos(2\pi/n)}{1-\cos(2\pi/n)}.$$
Proof Architecture
First, prove that two equal circles of spherical radius $r$ are tangent if and only if the spherical distance between their centers is $2r$; this follows because the unique geodesic joining the centers passes through the tangency point.
Second, prove that the centers $O_1,\dots,O_n$ all lie on the spherical circle of radius $2r$ centered at $O_0$; this is a direct translation of the tangencies with $\gamma_0$.
Third, prove that the points $O_1,\dots,O_n$ form a regular polygon on that spherical circle; consecutive center distances are all $2r$, and equal chords on a fixed parallel correspond to equal longitude increments.
Fourth, compute the longitude increment $\varphi$ using the spherical cosine theorem and obtain
$$\cos\varphi=\frac{\cos\alpha}{1+\cos\alpha}, \qquad \alpha=2r.$$
Fifth, use closure of the regular polygon to obtain $\varphi=2\pi/n$ and derive
$$\cos\alpha=\frac{\cos(2\pi/n)}{1-\cos(2\pi/n)}.$$
Finally, determine for which $n$ the right-hand side belongs to $[-1,1]$, obtaining exactly $3\le n\le6$, and verify existence for each of these values.
The most delicate lemma is the derivation of the spherical trigonometric equation relating $\alpha$ and $\varphi$.
Solution
Let $O_i$ denote the center of the circle $\gamma_i$ on the unit sphere. Since all circles have spherical radius $r$, each circle consists of the points whose spherical distance from its center equals $r$.
Consider two equal circles of spherical radius $r$ with centers $A$ and $B$. Suppose they are tangent at a point $T$. The geodesic joining $A$ and $B$ passes through $T$, because at a tangency point the two circles have the same tangent direction, hence the common normal geodesic passes through both centers. Along that geodesic,
$$AT=r,\qquad BT=r,$$
and therefore
$$AB=AT+TB=2r.$$
Conversely, if $AB=2r$, the point on the geodesic $AB$ at distance $r$ from each endpoint belongs to both circles, and the circles have a common tangent there. Thus two equal circles of radius $r$ are tangent if and only if the spherical distance between their centers is $2r$.
Set
$$\alpha=2r.$$
Since $\gamma_0$ is tangent to every $\gamma_i$,
$$O_0O_i=\alpha \qquad (i=1,\dots,n).$$
Hence $O_1,\dots,O_n$ lie on the spherical circle of radius $\alpha$ centered at $O_0$.
Also, since each pair of consecutive circles in the cycle is tangent,
$$O_iO_{i+1}=\alpha \qquad (i=1,\dots,n),$$
where indices are taken modulo $n$.
Place $O_0$ at the north pole of the unit sphere. Then the locus of points at distance $\alpha$ from $O_0$ is the parallel of colatitude $\alpha$. Let $\varphi_i$ be the longitude difference between $O_i$ and $O_{i+1}$.
In the spherical triangle $O_0O_iO_{i+1}$, two sides and the included side are all equal to $\alpha$. By the spherical cosine theorem,
$$\cos\alpha = \cos\alpha\cos\alpha + \sin\alpha\sin\alpha\cos\varphi_i.$$
Hence
$$\cos\varphi_i = \frac{\cos\alpha-\cos^2\alpha}{\sin^2\alpha} = \frac{\cos\alpha}{1+\cos\alpha}.$$
The right-hand side is independent of $i$, so all $\varphi_i$ are equal. Denote their common value by $\varphi$.
Since the points form a closed cycle around the parallel,
$$n\varphi=2\pi,$$
whence
$$\varphi=\frac{2\pi}{n}.$$
Substituting into the preceding relation gives
$$\cos\frac{2\pi}{n} = \frac{\cos\alpha}{1+\cos\alpha}.$$
Solving for $\cos\alpha$,
$$\cos\alpha = \frac{\cos(2\pi/n)}{1-\cos(2\pi/n)}.$$
A configuration exists exactly when the right-hand side belongs to $[-1,1]$.
Let
$$c=\cos\frac{2\pi}{n}.$$
Then
$$-1\le \frac{c}{1-c}\le 1.$$
Since $1-c>0$, the upper inequality becomes
$$c\le 1-c,$$
that is,
$$c\le\frac12.$$
The lower inequality is automatic because $c\ge-1$.
Therefore existence is equivalent to
$$\cos\frac{2\pi}{n}\le\frac12.$$
For $n\ge3$, the angle $2\pi/n$ lies in $(0,2\pi/3]$, and the cosine is decreasing there. Hence
$$\cos\frac{2\pi}{n}\le\frac12 \iff \frac{2\pi}{n}\ge\frac{\pi}{3} \iff n\le6.$$
Since $n\ge3$, the admissible values are precisely
$$n=3,4,5,6.$$
For each of these values, the number
$$\frac{\cos(2\pi/n)}{1-\cos(2\pi/n)}$$
lies in $[-1,1]$, so there exists a unique $\alpha\in[0,\pi]$ satisfying the equation above. Choosing $O_1,\dots,O_n$ equally spaced on the parallel of colatitude $\alpha$ then yields the required configuration. Thus all four values occur.
Because $\alpha=2r$, the radius is
$$r=\frac12\arccos!\left( \frac{\cos(2\pi/n)} {1-\cos(2\pi/n)} \right).$$
Equivalently,
$$\begin{aligned} n=3&:\quad r=\frac12\arccos!\left(-\frac13\right),\ n=4&:\quad r=\frac{\pi}{4},\ n=5&:\quad r=\frac12\arccos!\left(\frac{\sqrt5-1}{4}\right),\ n=6&:\quad r=0. \end{aligned}$$
Hence the complete answer is
$$\boxed{;n\in{3,4,5,6},\qquad r=\frac12\arccos!\left( \frac{\cos(2\pi/n)} {1-\cos(2\pi/n)} \right).;}$$
Verification of Key Steps
The first delicate point is the tangency criterion. If two equal spherical circles of radius $r$ are tangent at $T$, the geodesic through the centers and $T$ is orthogonal to both circles at $T$. Since $T$ lies on each circle,
$$AT=r,\qquad BT=r.$$
Because $T$ lies between $A$ and $B$ on the same geodesic, the geodesic distance satisfies
$$AB=AT+TB=2r.$$
Any argument that merely appeals to Euclidean intuition about circles in the plane would leave this step unjustified.
The second delicate point is the equation
$$\cos\varphi=\frac{\cos\alpha}{1+\cos\alpha}.$$
Starting directly from the spherical cosine theorem,
$$\cos\alpha=\cos^2\alpha+\sin^2\alpha\cos\varphi,$$
and substituting $\sin^2\alpha=1-\cos^2\alpha$ gives
$$\cos\alpha-\cos^2\alpha = (1-\cos^2\alpha)\cos\varphi.$$
Factoring both sides,
$$\cos\alpha(1-\cos\alpha) = (1-\cos\alpha)(1+\cos\alpha)\cos\varphi.$$
For $\alpha\ne0$ the factor $1-\cos\alpha$ cancels, yielding the required formula. The special case $\alpha=0$ corresponds to $n=6$ and is recovered separately from the final equation.
The third delicate point is the restriction on $n$. Substituting $n=6$ gives
$$\frac{\cos(\pi/3)}{1-\cos(\pi/3)} = 1,$$
which corresponds to $\alpha=0$. Substituting $n=7$ gives
$$\frac{\cos(2\pi/7)}{1-\cos(2\pi/7)} \approx1.49>1,$$
which is impossible for a cosine. Thus the transition from admissible to inadmissible values occurs exactly between $6$ and $7$.
Alternative Approaches
A more geometric approach uses the Euclidean centers of the circles. Each spherical circle is the intersection of the unit sphere with a plane. Equal spherical radii correspond to planes at the same distance from the sphere center. The centers $O_1,\dots,O_n$ then lie on a Euclidean circle obtained by intersecting the sphere with a plane perpendicular to $O_0$. Tangency translates into equality of certain central angles, and elementary chord computations lead to the same relation
$$\cos\frac{2\pi}{n} = \frac{\cos(2r)}{1+\cos(2r)}.$$
Another approach uses spherical regular polygons. The points $O_1,\dots,O_n$ form a regular spherical $n$-gon inscribed in a small circle centered at $O_0$. Standard formulas for a regular polygon on a sphere immediately yield the relation between the side length $2r$ and the central angle $2\pi/n$. The method used in the main solution is preferable because it requires only one application of the spherical cosine theorem and keeps all geometric dependencies explicit.