Who proposed this problem?

Borbas wrote: Who proposed this problem? Macedonia did (I don't know exactly who, but I know it was proposed by Macedonia)

This problem is too easy to be given in any mathematical Olympiads or I am doing something gravely wrong. Consider a point $K$ "extremely far away" from the regular polygon. Then it can see exactly $n$ of the edges of the polygon. Thus any $n$ consecutive sides of the polygon has to be colored in a atmost $2 $ colors. This certainly implies that two colors have been used exactly once in opposite sides of the polygon. Considering all edges to be distinct, this gives us $n\cdot 3!=6n$ distinct colorings. Here to count consider the $n$ pairs of opposite edges. Suppose the two chosen edges are $1 $ and $2$. Then there are $3 $ choices to color edge $1 $, $2 $ choices to color edge $2 $ and then the remaining choice to color all remaining edges. This gives a total of $6n$ colorings.

Solution with more details- The answear is $6n$. Note that from an external point you can see at most $n$ sides, so the problem becomes the following: In how many ways can you color $2n$ balls ($a_1, a_2, \cdots a_{2n}$) using $3$ colors - black,white,red such that any between any $n$ consecutive balls you don't have $3$ balls of different color. Now it is easy(unless i missunderstood the problem). Let $2$ balls be opposite if their indices differ by $n$. 1. Suppose there are $2$ opposite balls of different colors(let's say black and red), now it's easy to see that the rest balls are the third color, white. 2. Suppose there are $2$ opposite balls colored in same color(red), let's say $a_1$, $a_{n+1}$, now it is easy from here to achieve a contradiction: Suppose the sequence $a_1... a_n$ is white-red ( this means it's balls are colored either read, either white) and $a_{n+2}...a_{2n}, a_1$ is black-red. Now take a black ball, now we have that the only ball that can be colored in white is the ball opposite to it(because we count n balls starting from the black ball in the right and left direction, both this sequences contain a red ball so they cant contain an white ball) so by situation one we get that the rest balls are colored also in red. So we have the only possible colorings are the ones with 2 opposite balls of different colors, and the rest balls colored in third color, so the answer is straight-forward.

Yes, the answer is 6n but for n=2,3 it is not the same. We will lose 3 points if we don't check those cases.

If $n=2$ then the number of colourings is $36$ If $n=3$ then the number of colourings is $30$

There'd be lots of 37s on this JBMO then... $n=2$ is any coloring that uses all three. WLOG two red, one blue, one green (x3 later), but number of was to arrange is just 4!/2!=12, so 12x3=36. $n=3$ is the standard $6n$ ways (that applies to the higher numbers) plus the config where there are 2 of each color in succession, of which there are 12 orderings.

Borbas wrote: Who proposed this problem? The author of this problem is Viktor Simjanoski from Macedonia

Steff9 wrote: Borbas wrote: Who proposed this problem? The author of this problem is Viktor Simjanoski from Macedonia Thanks.

sqing wrote: Consider a regular 2n-gon $ P$,$A_1,A_2,\cdots ,A_{2n}$ in the plane ,where $n$ is a positive integer . We say that a point $S$ on one of the sides of $P$ can be seen from a point $E$ that is external to $P$ , if the line segment $SE$ contains no other points that lie on the sides of $P$ except $S$ .We color the sides of $P$ in 3 different colors (ignore the vertices of $P$,we consider them colorless), such that every side is colored in exactly one color, and each color is used at least once . Moreover ,from every point in the plane external to $P$ , points of most 2 different colors on $P$ can be seen .Find the number of distinct such colorings of $P$ (two colorings are considered distinct if at least one of sides is colored differently). Proposed by Viktor Simjanoski ， Macedonia JBMO 2017, Q4

Attachments: