We have a string of integers with $n$ each of 1s, 2s, and 3s. We need to prove that a substring has a sum of $3n$. Ideas?

Note, that for every arc with length $1$ should be arc with length $3$ such that their midpoints lies on the diameter of circle.

I think php will help

Consider all possible diameters of the circle divided by arcs of length 1, so that of the $3n$ points on the circle lie on at least one of these diameters. There are $3n$ diameters because the total circle has length $6n$. Since there are $3n$ diameters and $3n$ points, each diameter has an average of one point. We want to prove that there is at least one of these diameters that contain two points. Assume the contrary. This means that every single diameter contains exactly one point. Now, consider one arc of length three. There are two diameters that pass through that arc, and both must have exactly one point, which means that every arc of length three must be opposed by an arc of length 1. Since the number of arcs of length 3 and 1 are equal, they must all lie on the circle in opposite pairs. Now, consider one pair of two opposite arcs which divide the circle into two equal arcs of length $\frac{6n-4}{2}=3n-2$ Call these arcs of length $3n-2$ $A$ and $B$. Let's say $A$ contains $x$ arcs of length $3$. This means that $A$ also contains $n-1-x$ arcs of length $1$. Note that all arcs not yet accounted for must be arcs of length $2$. Let's say $A$ has $y$ arcs of length $2$. Then, the total length of $A$ is $$1\cdot (n-1-x)+2\cdot y+3\cdot x=2x+2y+n-1\equiv n-1 \mod 2$$ But we already know the length of $A$ is $3n-2\equiv n\mod 2$ Since it's impossible for $n$ to equal $n-1$ modulo 2, we have a contradiction, so there must exist a diameter containing two points.

Assume contradiction Let $6n$ pieces on circle Colour all arc with red,reflection them through center, and colour the image with blue So there are no 2 arc red and blue have common rest point Then blue arc length 3 must have red arc lenght 1 in midpoint,there are $n$ arc length 3,$n$ arc length 1 so there is $2n$ blue pieces length 1 need to be covered But note that red arc length 3 and blue arc length 3 can't have common piece, so those blue pieces must be covered by red arc length 2,there are $n$ arc length 2 so they are perfectly covered, but this can happen only when all blue arc length 3 must cover all the circle (contradiction). Done

Let the circle be divided into $6n$ arcs of lenght 1(the arcs are delimited by $6n$ points which we will color black) . The $3n$ points placed on the circle , we will color green . We will call the arc determined by 2 consecutive green points $full$ $green$ . Suppose that it is possible to have a configuration such that there aren't any 2 green points that determine a diameter $\implies$ from the $3n$ diameters with black end-points , each one will have EXACTLY 1 green point at one of its ends. This implies that for every $full$ $green$ arc of lenght 3 , there will be a $full$ $green$ arc of lenght 1. (which is made up from 2 green points that cannot be part of some other $full$ $green$ arc) For the construction of the $n$ $full$ $green$ arcs of lenght 3 we will need atleast $n+1$ green points . For the construction of the $n$ $full$ $green$ arcs of lenght 1 , we will need exactly $2n$ green points . Summing up the points needed , we get $3n+1$ > $3n$ - contradiction