Choose $3n$ points on a circle, show that there are two diametrically opposite point

Question

On a circle of length $6n$, we choose $3n$ points such that they split the circle into $n$ arcs of length $1$, $n$ arcs of length $2$, $n$ arcs of length $3$. Show that there exists two chosen points which are diametrically opposite.

Source:

• Russian MO $1982$
• Swiss MO $2006$ - Final round
• IMAC $2012$
• Romania MO $2018$ - $9$. grade

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Edit: Partition of circumference into $3k$ arcs

Answer 1

Well, that's simple. If we want to avoid having opposite points, then obviously every 1-arc must be positioned diametrically opposite to the middle of some 3-arc. Consider one such pair. What is the length of the great arc connecting their edges? Apparently, $3n-2$, which means it has the same parity as $n$. But it is composed of $n-1$ odd arcs (1s or 3s) and some unknown number of 2-arcs which do not matter, so it must have the opposite parity.

That's it.

Answer 2

Assume otherwise:

On a circle of length $6n$, we choose $3n$ points such that they split the circle into $n$ arcs of length $1$, $n$ arcs of length $2$, $n$ arcs of length $3$; and no two points are diametrically opposite.

Let's remove every arc of length $1$ from the circle, and also remove its opposite segment, which is the middle of an arc of length $3$. This operation preserves pairs of opposite points, and it simplifies the setup to this:

On a circle of length $4n$, we choose $2n$ points such that they split the circle into $2n$ arcs of length $2$; and no two points are diametrically opposite.

Now the contradiction is clear: starting from any point, traverse $n$ arcs of length $2$.

Answer 3

Considering the $6n$ evenly spaces points on the circle, call a point exterior if it is one of the chosen points which splits the circle and interior if it is in the interior of one of the arcs. Let the dual splitting of the circle be the one generated by changing every interior point to exterior and vice versa. If no points are diametrically opposite, the dual overlaps the original splitting exactly if placed diametrically opposite; i.e., the splitting is self-dual. If we list the splitting by listing the arc lengths in order (e.g., $1,3,2,1,2,3$), some thought shows that $1$s and $3$s must alternate with arbitrary numbers of $2$s placed inside, and the dual is formed by changing every $1$ to a $3$ and vice versa. Since a $1$ must be diametrically opposite a $3$, $n$ must be odd; but since the same number of $2$s must appear between $1$s and $3$s as between $3$s and $1$s, $n$ must be even.