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Let $H$ denote the union of the northern hemisphere of the unit circle $S^{1}$ with the interval $[-1,1]$ on the $x$-axis. That is, $H=\{(x,\sqrt{1-x^{2}}):-1\le x\le 1\}\cup\{(x,0):-1\le x\le 1\}$

Let us say that two copies of $H$ meet nicely if they intersect in exactly 6 points e.g. as the two pictures show below:

two half circles meeting nicely or two other circles meeting nicely

Note that the picture on the right shows two half-circles meeting nicely but whose centers do not lie inside their partner´s half disk.

Now, if we have three copies of $H$, it may not seem possible to arrange them so that they meet nicely, i.e. both of the following two conditions hold: 1. Any two meet nicely, and 2 The intersection of the three is empty.

Is this true? Or, if I am mistaken, I would appreciate that someone would show me, or describe, the desired arrangement.

EDIT: I am considering ALWAYS half-circles, i.e. copies of $H$. No need to give answers related to half-disks.

EDIT: It is possible to arrange three copies of $H$ so that both 1. and 2. hold (See the selected answer below).

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  • $\begingroup$ Where is the picture? But it is clear what you mean... $\endgroup$
    – domotorp
    Commented Sep 26, 2012 at 15:03
  • $\begingroup$ So, really, does the condition 2. mean exactly what it says? Or you look at the intersection of the half-disks instead of the copies of $H$? $\endgroup$
    – Ilya Bogdanov
    Commented Sep 26, 2012 at 17:04
  • $\begingroup$ It means exactly what it says. That is, that the intersection of the three haf-circles is empty. $\endgroup$
    – Victor
    Commented Sep 26, 2012 at 17:18
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    $\begingroup$ Could you look at areas? If when half circles meet nicely the area in common was always greater than a quarter of the area of a full circle than any three half circles that meet nicely in pairs would have a point in common. $\endgroup$
    – Kristal Cantwell
    Commented Sep 26, 2012 at 17:37

2 Answers 2

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1. This is the answer under the assumption that the condition 2. means exactly what it says.

Consider a regular triangle with side $2+\varepsilon$ and three diameters in the middles of its sides. If you construct the half-circles towards the triangle on these diameters, you obtain the desired example.

three half circles meeting nicely

Now about the four copies.

Lemma. Assume that the two copies of $H$ meet nicely. Consider their supporting half-planes determined by the diameters. Then their intersection is an acute angle, and the centers belong to its sides. (Possibly this angle is degenerate; in this case, it should be 0 but not $\pi$, which means that the intersection should be a strip but not a half-plane.)

Proof. If the two diameters do not intersect, then each of three pairs of the form (diameter, half-circle) and (half-circle,half-circle) meet at two points. Now, consider a point $A$ of intersection of the lines supporting the diameters. At least one diameter (say, $d_1$) does not contain $A$. Hence, if the angle in the Lemma statement is not acute, then the half-circle on $d_1$ cannot intersect $d_2$. Next, the center $C_1$ clearly lies on the side of this angle. Finally, the projection of $O_2$ onto $d_1$ should lie on the segment $d_1$, hence $O_2$ is also on the side of the angle (but not on its prolongation).

Finally, assume that the diameters intersect.Then each half-circle can intersect the other diameter in at most one more point, and the total number of the intersection points is less than 6. Lemma is proved.

Now we can prove that the four copies of $H$ cannot pairwise meet nicely. Let $c_{ij}$ be the angle from the lemma for $H_i$ and $H_j$. It is easy to see that $c_{12}$, $c_{13}$, $c_{23}$ should form an acute-angled triangle with the centers $C_1$, $C_2$, $C_3$ on its sides (just try to add the third diameter to $c_{12}$!). But then it is impossible to add the fourth half-plane --- these four half-planes should now form a quadrilateral with four acute angles!

2. Now let us assume that you speak on the half-disks. Then the answer is positive. From the previous paragraph, we see that the three diameters lie inside the three sides of some acute triangle $XYZ$, respectively.

Now consider the three distances between the centers. If all three are less than $\sqrt3$, then by Jung's theorem they may be covered by the unit disk, and the center of this disk belongs to all three half-disks.

Otherwise, assume that $C_1C_2\geq \sqrt3$, where $C_1$ and $C_2$ be the centers on the sides $XY$ and $XZ$, respectively. We have $d(C_1,XZ)\leq 1$, otherwise the respective half-circle and segment do not intersect. But then the projection of $C_1$ onto $XZ$ is at least $\sqrt2$ away from $C_2$, hence the first half-circle cannot intersect the second diameter twice. So in this case we also get a contradiction.

I may expand any part of the above sketch.

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  • $\begingroup$ About the paragraph right after the picture. Are you saying that it is impossible to arrange four copies of $H$ so that both 1. Any two meet nicely AND 2. Any three of more have empty intersection? $\endgroup$
    – Victor
    Commented Sep 26, 2012 at 17:49
  • $\begingroup$ Yes, I meant that the four copies are impossible. Ive tried to add some words (in fact, I've corrected some arguments...). If it is not clear (or not completely valid) --- I may expand it. $\endgroup$
    – Ilya Bogdanov
    Commented Sep 26, 2012 at 18:11
  • $\begingroup$ But, in fact, it seems that I do not use condition 2 at all... $\endgroup$
    – Ilya Bogdanov
    Commented Sep 26, 2012 at 18:13
  • $\begingroup$ What do you mean by "supporting half-planes determined by the diameters"? My understanding is that you mean planes orthogonal to the diameters passing through the centers of the half-circles. $\endgroup$
    – Victor
    Commented Sep 26, 2012 at 18:49
  • $\begingroup$ The supporting half-plane is just a half-plane bounded by the line containing the diameter and containing the copy of $H$ to which this diameter belongs. Surely, the angle may degenerate to a strip between the parallel lines, but not to the half-plane. Then the nexy arguments remain valid. Or, if you wish, you may rotate a copy a bit preserving all the properties. $\endgroup$
    – Ilya Bogdanov
    Commented Sep 26, 2012 at 19:52
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This is off the top of my head; please take it skeptically and check all assumptions, visible and hidden.

Focus on the line segment and its end points which I call p and q for a sample H. I have convinced myself that if p and q are on opposite sides of another line segment for a different half circle H', then there are at most 4 points of intersection, so H and H' do not meet nicely.

Now consider segments of two of the semicircles that meet nicely. They each lie fully in a half plane defined by the other segment. The third segment when placed lies in some intersection of two of the half planes. But in order to meet the other two semicircles nicely, you have to place the circular arc in a way to intersect both line segments. If you are in the middle, this cannot be done.

I realize this is far from rigorous, but you may be able to firm it up and use it.

Gerhard "Needs To Drink More Coffee" Paseman, 2012.09.26

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  • $\begingroup$ I just noticed Karl's suggestion, which seems cleaner. You might try both and see which can extend to semicircles of different sizes. Gerhard"Ask Me About System Design" Paseman, 2012.09.26 $\endgroup$
    – Gerhard Paseman
    Commented Sep 26, 2012 at 16:12

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