![]() From the OP's attempt, this triangle must be one of these types: Suppose that there exists only one monochromatic isosceles triangle, and we assume without loss of generality that this triangle is black. Here is one of the configurations with only two monochromatic isosceles triangles. A proof of this without the use of computer search is given below. I did a python search and found that the minimum number of monochromatic isosceles triangles is $2$. Thus we can conclude that there are $3$ vertices of the same color forming an isosceles triangle.Ĭan someone check if my solution is correct or not? And a better solution will be appreciated. Again if $E$ is colored white, then $\Delta AEB$ is an isosceles triangle with all of its vertices colored white, which is a contradiction. If $D$ is colored white, then $\Delta ADB$ is an isosceles triangle with all of its vertices colored white, which is a contradiction. ![]() Now since $C$ is colored black, at least one of $D$ and $E$ must be colored white. Observe that $\Delta CDE$ is also an isosceles triangle with $CD=CE$. Now consider that the adjacent vertex of $A$ be $D\neq B$ and the adjacent vertex of $B$ be $E\neq A$. Now WLOG let $A$ and $B$ be colored white. Now there exists a unique vertex $C$ such that $\Delta CAB$ is isosceles with $CA=CB$. We know that there exists two adjacent vertices of the same color. Let us assume for the sake of contradiction that $\nexists$ $3$ vertices of the same color forming an isosceles triangle. Moving on to the next part of the problem, consider any arbitrary permissible coloring. Thus there exists two adjacent vertices of the same color. But since $(9)$ and $(1)$ are adjacent vertices, and $(1)$ is colored black, implies that $(9)$ must be colored white. ![]() This clearly implies that $(9)$ is colored black. Now WLOG, let us assume that let $(1)$ be colored black, $(2)$ be colored white and so on. Observe that according to what we have assumed, only one type of coloring is possible, that is an alternative coloring, that is, color any pair of adjacent vertices by alternate colors. Let us name the vertices of the given regular $9-$gon starting from $1$ till $9$. ![]() My approach: Let us assume for the sake of contradiction that $\nexists$ two adjacent vertices of the same color. (b) Show that there are $3$ vertices of the same color forming an isosceles triangle. (a) Show that there are two adjacent vertices of the same color. Question: Let each of the vertices of a regular $9-$gon be colored black or white. ![]()
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