Let $P_1$ be a regular $n$-gon, where $n\in\mathbb{N}$. We construct $P_2$ as the regular $n$-gon whose vertices are the midpoints of the edges of $P_1$. Continuing analogously, we obtain regular $n$-gons $P_3,P_4,\ldots ,P_m$. For $m\ge n^2-n+1$, find the maximum number $k$ such that for any colouring of vertices of $P_1,\ldots ,P_m$ in $k$ colours there exists an isosceles trapezium $ABCD$ whose vertices $A,B,C,D$ have the same colour. Radu Ignat
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Tags: geometry, trapezoid, ceiling function, combinatorics proposed, combinatorics
09.10.2012 11:01
WakeUp wrote: Let $P_1$ be a regular $n$-gon, where $n\in\mathbb{N}$. We construct $P_2$ as the regular $n$-gon whose vertices are the midpoints of the edges of $P_1$. Continuing analogously, we obtain regular $n$-gons $P_3,P_4,\ldots ,P_m$. For $m\ge n^2-n+1$, find the maximum number $k$ such that for any colouring of vertices of $P_1,\ldots ,P_m$ in $k$ colours there exists an isosceles trapezium $ABCD$ whose vertices $A,B,C,D$ have the same colour. Radu Ignat The required number is $n - 1$ If we use at most $n - 1$ colors then we can find for each polygon two vertices colored in the same way, so we have $n^2- n + 1$ monochromatic segments with endpoints the vertices of the polygons. We observe that the sides and the diagonals of each of the polygons are parallel to $n$ different straight lines and these lines are the same for all $m$ of them. From the $PHP$ it follows that at least two of the monochromatic segments, say $(AB)$ and $(CD)$ must have the same color and must be parallel. Since the $m$ polygons are inscribed in concentric circles and $(AB), (CD)$ cannot be diameters So $ABCD$ is an isosceles trapezoid. We will show now that if we use $n$ colors $C_1,C_2,......C_n$ then it is possible to color the vertices so that no two segments which are colored in the same way form an isosceles trapezoid. This can be easily done by coloring all vertices which are placed on the straight line $OA_k$ with the color $C_k$ where $O$ is the common centre of the polygons and $P_1 = A_1A_2......A_n$. The only possibility to find four points of the same color is when they are collinear, so they cannot be the vertices of a trapezoid.
04.01.2013 23:35
More succintly, $\left\lceil\frac{m}{2}\right\rceil>\binom{n}{2}$ implies that if we have fewer than $n$ colors, then two monochromatic diagonals (including sides) are parallel. The endpoints of the diagonals form an isosceles trapezium. However, the problem should include the possibility of having degenerate trapezium. If $n$ is even, it is possible to assign the colors so that no nondegenerate monichromatic trapeziums exist.