Subfield: Combinatorics


Question: There are $n$ circles drawn on a piece of paper in such a way that any two circles intersect in two points, and no three circles pass through the same point. Turbo the snail slides along the circles in the following fashion. Initially he moves on one of the circles in clockwise direction. Turbo always keeps sliding along the current circle until he reaches an intersection with another circle. Then he continues his journey on this new circle and also changes the direction of moving, i.e. from clockwise to anticlockwise or vice versa.

Suppose that Turbo's path entirely covers all circles. Prove that $n$ must be odd.

Solution:
Replace every cross (i.e. intersection of two circles) by two small circle arcs that indicate the direction in which the snail should leave the cross (see Figure 1.1). Notice that the placement of the small arcs does not depend on the direction of moving on the curves; no matter which direction the snail is moving on the circle arcs, he will follow the same curves (see Figure 1.2). In this way we have a set of curves, that are the possible paths of the snail. Call these curves snail orbits or just orbits. Every snail orbit is a simple closed curve that has no intersection with any other orbit.

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Figure 1.1

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Figure 1.2

We prove the following, more general statement.

(*) In any configuration of $n$ circles such that no two of them are tangent, the number of snail orbits has the same parity as the number $n$. (Note that it is not assumed that all circle pairs intersect.)

This immediately solves the problem.

Let us introduce the following operation that will be called flipping a cross. At a cross, remove the two small arcs of the orbits, and replace them by the other two arcs. Hence, when the snail arrives at a flipped cross, he will continue on the other circle as before, but he will preserve the orientation in which he goes along the circle arcs (see Figure 2).
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Figure 2

Consider what happens to the number of orbits when a cross is flipped. Denote by $a, b, c$, and $d$ the four arcs that meet at the cross such that $a$ and $b$ belong to the same circle. Before the flipping $a$ and $b$ were connected to $c$ and $d$, respectively, and after the flipping $a$ and $b$ are connected to $d$ and $c$, respectively.

The orbits passing through the cross are closed curves, so each of the arcs $a, b, c$, and $d$ is connected to another one by orbits outside the cross. We distinguish three cases.

Case 1: $a$ is connected to $b$ and $c$ is connected to $d$ by the orbits outside the cross (see Figure 3.1).



We show that this case is impossible. Remove the two small arcs at the cross, connect $a$ to $b$, and connect $c$ to $d$ at the cross. Let $\gamma$ be the new closed curve containing $a$ and $b$, and let $\delta$ be the new curve that connects $c$ and $d$. These two curves intersect at the cross. So one of $c$ and $d$ is inside $\gamma$ and the other one is outside $\gamma$. Then the two closed curves have to meet at least one more time, but this is a contradiction, since no orbit can intersect itself.

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Figure 3.1

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Figure 3.2
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Figure 3.3

Case 2: $a$ is connected to $c$ and $b$ is connected to $d$ (see Figure 3.2).

Before the flipping $a$ and $c$ belong to one orbit and $b$ and $d$ belong to another orbit. Flipping the cross merges the two orbits into a single orbit. Hence, the number of orbits decreases by 1 .

Case 3: $a$ is connected to $d$ and $b$ is connected to $c$ (see Figure 3.3).

Before the flipping the $\operatorname{arcs} a, b, c$, and $d$ belong to a single orbit. Flipping the cross splits that orbit in two. The number of orbits increases by 1 .

As can be seen, every flipping decreases or increases the number of orbits by one, thus changes its parity.

Now flip every cross, one by one. Since every pair of circles has 0 or 2 intersections, the number of crosses is even. Therefore, when all crosses have been flipped, the original parity of the number of orbits is restored. So it is sufficient to prove $(*)$ for the new configuration, where all crosses are flipped. Of course also in this new configuration the (modified) orbits are simple closed curves not intersecting each other.

Orient the orbits in such a way that the snail always moves anticlockwise along the circle arcs. Figure 4 shows the same circles as in Figure 1 after flipping all crosses and adding orientation. (Note that this orientation may be different from the orientation of the orbit as a planar curve; the orientation of every orbit may be negative as well as positive, like the middle orbit in Figure 4.) If the snail moves around an orbit, the total angle change in his moving direction, the total curvature, is either $+2 \pi$ or $-2 \pi$, depending on the orientation of the orbit. Let $P$ and $N$ be the number of orbits with positive and negative orientation, respectively. Then the total curvature of all orbits is $(P-N) \cdot 2 \pi$.

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Figure 4

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Figure 5

Double-count the total curvature of all orbits. Along every circle the total curvature is $2 \pi$. At every cross, the two turnings make two changes with some angles having the same absolute value but opposite signs, as depicted in Figure 5. So the changes in the direction at the crosses cancel out. Hence, the total curvature is $n \cdot 2 \pi$.

Now we have $(P-N) \cdot 2 \pi=n \cdot 2 \pi$, so $P-N=n$. The number of (modified) orbits is $P+N$, that has a same parity as $P-N=n$.
We present a different proof of $(*)$.

We perform a sequence of small modification steps on the configuration of the circles in such a way that at the end they have no intersection at all (see Figure 6.1). We use two kinds of local changes to the structure of the orbits (see Figure 6.2):

- Type-1 step: An arc of a circle is moved over an arc of another circle; such a step creates or removes two intersections.
- Type-2 step: An arc of a circle is moved through the intersection of two other circles.

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Figure 6.1

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ype-1

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Figure 6.2

We assume that in every step only one circle is moved, and that this circle is moved over at most one arc or intersection point of other circles.

We will show that the parity of the number of orbits does not change in any step. As every circle becomes a separate orbit at the end of the procedure, this fact proves (*).

Consider what happens to the number of orbits when a Type- 1 step is performed. The two intersection points are created or removed in a small neighbourhood. Denote some points of the two circles where they enter or leave this neighbourhood by $a, b, c$, and $d$ in this order around the neighbourhood; let $a$ and $b$ belong to one circle and let $c$ and $d$ belong to the other circle. The two circle arcs may have the same or opposite orientations. Moreover, the four end-points of the two arcs are connected by the other parts of the orbits. This can happen in two ways without intersection: either $a$ is connected to $d$ and $b$ is connected to $c$, or $a$ is connected to $b$ and $c$ is connected to $d$. Altogether we have four cases, as shown in Figure 7.
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Figure 7

We can see that the number of orbits is changed by -2 or +2 in the leftmost case when the arcs have the same orientation, $a$ is connected to $d$, and $b$ is connected to $c$. In the other three cases the number of orbits is not changed. Hence, Type-1 steps do not change the parity of the number of orbits.

Now consider a Type-2 step. The three circles enclose a small, triangular region; by the step, this triangle is replaced by another triangle. Again, the modification of the orbits is done in some small neighbourhood; the structure does not change outside. Each side of the triangle shaped region can be convex or concave; the number of concave sides can be $0,1,2$ or 3 , so there are 4 possible arrangements of the orbits inside the neighbourhood, as shown in Figure 8.



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Figure 8

Denote the points where the three circles enter or leave the neighbourhood by $a, b, c, d$, $e$, and $f$ in this order around the neighbourhood. As can be seen in Figure 8, there are only two essentially different cases; either $a, c, e$ are connected to $b, d, f$, respectively, or $a, c, e$ are connected to $f, b, d$, respectively. The step either preserves the set of connections or switches to the other arrangement. Obviously, in the earlier case the number of orbits is not changed; therefore we have to consider only the latter case.

The points $a, b, c, d, e$, and $f$ are connected by the orbits outside, without intersection. If $a$ was connected to $c$, say, then this orbit would isolate $b$, so this is impossible. Hence, each of $a, b, c, d, e$ and $f$ must be connected either to one of its neighbours or to the opposite point. If say $a$ is connected to $d$, then this orbit separates $b$ and $c$ from $e$ and $f$, therefore $b$ must be connected to $c$ and $e$ must be connected to $f$. Altogether there are only two cases and their reverses: either each point is connected to one of its neighbours or two opposite points are connected and the the remaining neigh boring pairs are connected to each other. See Figure 9.
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Figure 9

We can see that if only neighbouring points are connected, then the number of orbits is changed by +2 or -2 . If two opposite points are connected ( $a$ and $d$ in the figure), then the orbits are re-arranged, but their number is unchanged. Hence, Type-2 steps also preserve the parity. This completes the proof of $(*)$.
Like in the previous solutions, we do not need all circle pairs to intersect but we assume that the circles form a connected set. Denote by $\mathcal{C}$ and $\mathcal{P}$ the sets of circles and their intersection points, respectively.

The circles divide the plane into several simply connected, bounded regions and one unbounded region. Denote the set of these regions by $\mathcal{R}$. We say that an intersection point or a region is odd or even if it is contained inside an odd or even number of circles, respectively. Let $\mathcal{P}_{\text {odd }}$ and $\mathcal{R}_{\text {odd }}$ be the sets of odd intersection points and odd regions, respectively.

Claim.

$$
\left|\mathcal{R}_{\text {odd }}\right|-\left|\mathcal{P}_{\text {odd }}\right| \equiv n \quad(\bmod 2)
\tag{1}
$$

Proof. For each circle $c \in \mathcal{C}$, denote by $R_{c}, P_{c}$, and $X_{c}$ the number of regions inside $c$, the number of intersection points inside $c$, and the number of circles intersecting $c$, respectively. The circles divide each other into several arcs; denote by $A_{c}$ the number of such arcs inside $c$. By double counting the regions and intersection points inside the circles we get

$$
\left|\mathcal{R}_{\text {odd }}\right| \equiv \sum_{c \in \mathcal{C}} R_{c} \quad(\bmod 2) \quad \text { and } \quad\left|\mathcal{P}_{\text {odd }}\right| \equiv \sum_{c \in \mathcal{C}} P_{c} \quad(\bmod 2)
$$



For each circle $c$, apply EULER's polyhedron theorem to the (simply connected) regions in $c$. There are $2 X_{c}$ intersection points on $c$; they divide the circle into $2 X_{c}$ arcs. The polyhedron theorem yields $\left(R_{c}+1\right)+\left(P_{c}+2 X_{c}\right)=\left(A_{c}+2 X_{c}\right)+2$, considering the exterior of $c$ as a single region. Therefore,

$$
R_{c}+P_{c}=A_{c}+1 .
\tag{2}
$$

Moreover, we have four arcs starting from every interior points inside $c$ and a single arc starting into the interior from each intersection point on the circle. By double-counting the end-points of the interior arcs we get $2 A_{c}=4 P_{c}+2 X_{c}$, so

$$
A_{c}=2 P_{c}+X_{c} .
\tag{3}
$$

The relations (2) and (3) together yield

$$
R_{c}-P_{c}=X_{c}+1
\tag{4}
$$

By summing up (4) for all circles we obtain

$$
\sum_{c \in \mathcal{C}} R_{c}-\sum_{c \in \mathcal{C}} P_{c}=\sum_{c \in \mathcal{C}} X_{c}+|\mathcal{C}|
$$

which yields

$$
\left|\mathcal{R}_{\text {odd }}\right|-\left|\mathcal{P}_{\text {odd }}\right| \equiv \sum_{c \in \mathcal{C}} X_{c}+n \quad(\bmod 2)
\tag{5}
$$

Notice that in $\sum_{c \in \mathcal{C}} X_{c}$ each intersecting circle pair is counted twice, i.e., for both circles in the pair, so

$$
\sum_{c \in \mathcal{C}} X_{c} \equiv 0 \quad(\bmod 2)
$$

which finishes the proof of the Claim.

Now insert the same small arcs at the intersections as in the first solution, and suppose that there is a single snail orbit $b$.

First we show that the odd regions are inside the curve $b$, while the even regions are outside. Take a region $r \in \mathcal{R}$ and a point $x$ in its interior, and draw a ray $y, \operatorname{starting}$ from $x$, that does not pass through any intersection point of the circles and is neither tangent to any of the circles. As is well-known, $x$ is inside the curve $b$ if and only if $y$ intersects $b$ an odd number of times (see Figure 10). Notice that if an arbitrary circle $c$ contains $x$ in its interior, then $c$ intersects $y$ at a single point; otherwise, if $x$ is outside $c$, then $c$ has 2 or 0 intersections with $y$. Therefore, $y$ intersects $b$ an odd number of times if and only if $x$ is contained in an odd number of circles, so if and only if $r$ is odd.

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Figure 10

Now consider an intersection point $p$ of two circles $c_{1}$ and $c_{2}$ and a small neighbourhood around $p$. Suppose that $p$ is contained inside $k$ circles.



We have four regions that meet at $p$. Let $r_{1}$ be the region that lies outside both $c_{1}$ and $c_{2}$, let $r_{2}$ be the region that lies inside both $c_{1}$ and $c_{2}$, and let $r_{3}$ and $r_{4}$ be the two remaining regions, each lying inside exactly one of $c_{1}$ and $c_{2}$. The region $r_{1}$ is contained inside the same $k$ circles as $p$; the region $r_{2}$ is contained also by $c_{1}$ and $c_{2}$, so by $k+2$ circles in total; each of the regions $r_{3}$ and $r_{4}$ is contained inside $k+1$ circles. After the small arcs have been inserted at $p$, the regions $r_{1}$ and $r_{2}$ get connected, and the regions $r_{3}$ and $r_{4}$ remain separated at $p$ (see Figure 11). If $p$ is an odd point, then $r_{1}$ and $r_{2}$ are odd, so two odd regions are connected at $p$. Otherwise, if $p$ is even, then we have two even regions connected at $p$.

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Figure 11

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Figure 12

Consider the system of odd regions and their connections at the odd points as a graph. In this graph the odd regions are the vertices, and each odd point establishes an edge that connects two vertices (see Figure 12). As $b$ is a single closed curve, this graph is connected and contains no cycle, so the graph is a tree. Then the number of vertices must be by one greater than the number of edges, so

$$
\left|\mathcal{R}_{\text {odd }}\right|-\left|\mathcal{P}_{\text {odd }}\right|=1
\tag{6}
$$

The relations (1) and (6) together prove that $n$ must be odd.