Can $n$ circles be drawn such that all have a common intersection but no two intersect individually

Question

I was fiddling with plane geometry when a question came into my mind:

Can $n$ circles ($n \ge 3$, $n \in \mathbb{N}$) be drawn such that:

1. $C_1 \cap C_2 \cap C_3 \cap \ldots \cap C_n \not = 0$ but
2. $C_i \cap C_j \cap C_k \cap \ldots \not \cap C_m = 0 $
3. $C_i \cap C_j \cap C_k \not \cap \ldots \not \cap C_m = 0 $
4. $C_i \cap C_j \not \cap C_k \not \cap \ldots \not \cap C_m = 0 $

where $i,j,k \ldots m \in \mathbb{\{1,2,3,4, \ldots ,n}\}$ and $C_r$ is the area covered by the $r$th circle?

I could not find any rigorous proof for this. However, I tried to verify this for $n=3$. But I could not draw three circles such that $C_1 \cap C_2 \cap C_3 \not =0$ and $C_1 \cap C_2 \not \cap C_3=C_1 \cap C_3 \not \cap C_2=C_2 \cap C_3 \not \cap C_1=0$.

In other words, I mean in the diagram below, can we draw the circles such that the pink, yellow and sky blue (or light blue, as you call it) regions do not exist?

Can anyone prove or disprove this using rigorous mathematics and not by mere intuition? Or maybe there is a theorem or result already? Thanks.

Answer 1

Your diagram makes it evident that you are concerned about the intersections of the regions inside the circles. I'll use the term open disk to describe such a region.

Let $C_1$, $C_2$, and $C_3$ be three open disks, no two of them the same, such that $C_1 \cap C_2 \cap C_3 \neq \emptyset$.

Since $C_1 \cap C_2 \cap C_3 \neq \emptyset$, it follows that $C_1 \cap C_2 \neq \emptyset$. But since $C_1 \neq C_2$, it must be true either that $C_1 \cap C_2^\complement \neq \emptyset$ (that is, there are points in $C_1$ that are not in $C_2$) or that $C_1^\complement \cap C_2 \neq \emptyset$. Without loss of generality, assume $C_1 \cap C_2^\complement \neq \emptyset$.

The boundary between $C_1 \cap C_2$ and $C_1 \cap C_2^\complement$ is then an arc of the circle that forms the entire boundary of $C_2$. Now consider how this arc interacts with the boundary of $C_3$.

The arc cannot lie exactly on the boundary of $C_3$, because then it must have the same curvature and center as the boundary of $C_3$, which would imply that $C_2 = C_3$, which we have already assumed is false. So some part of that arc is somewhere other than exactly on the boundary of $C_3$, either inside or outside.

If any part of the arc between $C_1 \cap C_2$ and $C_1 \cap C_2^\complement$ is outside the boundary of $C_3$, then some part of $C_1 \cap C_2$ is outside $C_3$ and therefore $C_1 \cap C_2 \cap C_3^\complement \neq \emptyset$.

But if any part of that same arc is inside $C_3$, then some part of $C_1 \cap C_2^\complement$ is in $C_3$ and therefore $C_1 \cap C_2^\complement \cap C_3 \neq \emptyset$.

So if we satisfy the beginning assumptions, we will violate at least one of the conditions $C_1 \cap C_2 \cap C_3^\complement = \emptyset$ or $C_1 \cap C_2^\complement \cap C_3 = \emptyset$. So it will not be possible to achieve $C_1^\complement \cap C_2 \cap C_3 = C_1 \cap C_2^\complement \cap C_3 = C_1 \cap C_2 \cap C_3^\complement = \emptyset$.

---

Another approach, based on Jack D'Aurizio's comment (also compare the lemma in Fimpellizieri's answer):

Let $C_1$, $C_2$, and $C_3$ be open disks such that $C_1 \cap C_2 \cap C_3 \neq \emptyset$ and $C_1^\complement \cap C_2 \cap C_3 = C_1 \cap C_2^\complement \cap C_3 = C_1 \cap C_2 \cap C_3^\complement = \emptyset$. The idea is to prove that at least two of the disks are the same.

Suppose $C_1 \neq C_2$, and consider $C_1 \cap C_2$. There are two cases:

Case 1: One disk is entirely contained by the other. Without loss of generality, assume $C_1 \subsetneq C_2$. Then $C_1 \cap C_2 = C_1$, and therefore $C_1 \cap C_3^\complement = C_1 \cap C_2 \cap C_3^\complement = \emptyset$, which means $C_1 \subseteq C_3$, but $C_1^\complement \cap C_2 \cap C_3 = \emptyset$ implies that $C_3$ cannot overlap any of the parts of $C_2$ that surround $C_1$, therefore $C_3 \subseteq C_1$, and therefore $C_3 = C_1$.

Case 2: Neither disk is entirely contained in the other. Then the boundary of $C_1 \cap C_2$ consists of two circular arcs. We may consider just one of those arcs, which is the shared boundary of $C_1^\complement \cap C_2$ and $C_1 \cap C_2$, and which lies on the boundary of $C_1$. Since $C_1 \cap C_2 \cap C_3^\complement = \emptyset$, it must be that $C_1 \cap C_2 \subseteq C_3$, but $(C_1^\complement \cap C_2) \cap C_3 = \emptyset$; that is, all of $C_1 \cap C_2$ and none of $C_1^\complement \cap C_2$ lies in $C_3$, so the boundary between those two regions must coincide with part of the boundary of $C_3$. But that implies part of the boundary of $C_3$ is a circular arc with the same radius and center as $C_1$, which implies $C_3 = C_1$.

In either case $C_3 = C_1$ and we do not have three circles all distinct from each other. In order to satisfy $C_1 \cap C_2 \cap C_3 \neq \emptyset$ and $C_1^\complement \cap C_2 \cap C_3 = C_1 \cap C_2^\complement \cap C_3 = C_1 \cap C_2 \cap C_3^\complement = \emptyset$, at least two of $C_1$, $C_2$, and $C_3$ must be names of the same disk.

Answer 2

In the plane? Not if the circles are distinct.

Consider two circles $C_1$ and $C_2$, and let $A$ be their intersection. For a third cricle $C_3$ to be added to the scheme while satisfying your criteria, it must be that $A \subset C_3$ and that $C_3$ does not intersect $C_1$ or $C_2$ anywhere else. In other words, we must have that:

$$A = C_1\cap C_2=C_1\cap C_3=C_2\cap C_3=C_1\cap C_2\cap C_3$$

It's easy to see there are three possibilities for the intersection between any two circles:

• A single point, if they're tangent;
• A pointy elipse, if their boundaries intersect at two points;
• The smallest of the two circles, if the larger one contains it.

If $A$ is of the first kind, then you're looking for three circles in the plane that are tangent at a single point (remember they intersect nowhere else, so they can't be internally tangent), and this clearly cannot be.

If $A$ is of the third kind, then (without loss of generality) $C_2 \subseteq C_1$ and you 'technically' can fit your scheme by letting $C_3=C_2$ (and $C_n=C_2$ for all $n\geq 3$). Of course, it's easy to see that if you require the circles to be distinct then this fails.

Finally, if $A$ is of the second kind, then you're looking for three circles for which any two pairs intersect at the same pointy elipse.

Lemma: Given a pointy elipse in the plane, there are two unique distinct circles whose intersection is that pointy elipse.

The idea of the proof is that each side of the pointy elipse is an arc of a circle and that there is essentially one unique way of extending each such arc into a full circle.

With the lemma, this last case cannot be, for then two of $C_1,C_2$ and $C_3$ would coincide, so their intersection would not be a pointy elipse in the first place.