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Question: Let each of the vertices of a regular $9-$gon be colored black or white. (a) Show that there are two adjacent vertices of the same color. (b) Show that there are $3$ vertices of the same color forming an isosceles triangle.

My approach: Let us assume for the sake of contradiction that $\nexists$ two adjacent vertices of the same color. Let us name the vertices of the given regular $9-$gon starting from $1$ till $9$. 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.

Now WLOG, let us assume that let $(1)$ be colored black, $(2)$ be colored white and so on. This clearly implies that $(9)$ is colored black. But since $(9)$ and $(1)$ are adjacent vertices, and $(1)$ is colored black, implies that $(9)$ must be colored white. Hence, a contradiction is obtained.

Thus there exists two adjacent vertices of the same color.

Moving on to the next part of the problem, consider any arbitrary permissible coloring. Let us assume for the sake of contradiction that $\nexists$ $3$ vertices of the same color forming an isosceles triangle. We know that there exists two adjacent vertices of the same color. Let those two vertices be $A$ and $B$. Now there exists a unique vertex $C$ such that $\Delta CAB$ is isosceles with $CA=CB$. Now WLOG let $A$ and $B$ be colored white. Then we are forced to color $C$ black. Now consider that the adjacent vertex of $A$ be $D\neq B$ and the adjacent vertex of $B$ be $E\neq A$. Observe that $\Delta CDE$ is also an isosceles triangle with $CD=CE$. Now since $C$ is colored black, at least one of $D$ and $E$ must be colored white. If $D$ is colored white, then $\Delta ADB$ is an isosceles triangle with all of its vertices colored white, which is a contradiction. Again if $E$ is colored white, then $\Delta AEB$ is an isosceles triangle with all of its vertices colored white, which is a contradiction.

Thus we can conclude that there are $3$ vertices of the same color forming an isosceles triangle.

Can someone check if my solution is correct or not? And a better solution will be appreciated.

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  • $\begingroup$ Your solution looks correct to me. $\endgroup$ – Batominovski Apr 10 at 5:11
  • $\begingroup$ I wonder what is the smallest number of mono-chromatic isosceles triangle. My guess is $3$. $\endgroup$ – Batominovski Apr 10 at 5:29
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Your proof is correct. I did a python search and found that the minimum number of monochromatic isosceles triangles is $2$. A proof of this without the use of computer search is given below. Here is one of the configurations with only two monochromatic isosceles triangles.

enter image description here


Let $ABCDEFGHI$ be the nonagon. Suppose that there exists only one monochromatic isosceles triangle, and we assume without loss of generality that this triangle is black. From the OP's attempt, this triangle must be one of these types:

  • Type I: a triangle formed by three consecutive vertices;
  • Type II: a triangle formed by two consecutive vertices and another vertex equidistant to the consecutive vertices;
  • Type III: an equilateral triangle.

Assume that we are dealing with a Type-I triangle. Assume that this triangle is $IAB$. Then, $E$ and $F$ must be white, otherwise $IAE$ or $ABF$ is a black isosceles triangle. If $D$ or $G$ is white, then $DEF$ or $EFG$ is a white isosceles triangle, which is a contradiction. Therefore, $D$ and $G$ are black. This means $ADG$ is a black equilateral triangle, leading to another contradiction.

Let now assume that we are in the Type-II situation, say, $AEF$ is our black isosceles triangle. Then, $D$ and $G$ must be white. Both $C$ and $H$ cannot be black (otherwise $ACH$ is a black isosceles triangle); they cannot both be white (otherwise $CGH$ and $CDH$ are white isosceles triangle). Thus, we may assume without loss of generality that $C$ is black and $H$ is white. Therefore, $I$ must be black (otherwise, $GHI$ is a white isosceles triangle). That is, $CFI$ is a black equilateral triangle, which is absurd.

Thus, the only situation we can end up with is the Type-III case. Let $ADG$ be our black equilateral. Note that $E$ and $F$ cannot be black simultaneously. We first assume that one of them, say, $E$ is black, whence $F$ is white. Thus, $I$ must be white (otherwise $DEI$ is a black isosceles triangle). Since $I$ and $F$ is white, $C$ must be black. Therefore, $CDE$ is a black isosceles triangle. This is a contradiction. Therefore, $E$ and $F$ must be both white.

Now, $C$ and $H$ cannot both be black (otherwise $ACH$ is a black isosceles triangle). Thus, $C$ or $H$ is white. We first assume that one of them, say, $C$, is black. Therefore, $H$ is white. Since $C$ and $D$ are black, $B$ must be white. Therefore, we obtain a white equilateral triangle $BEH$, which is absurd. Ergo, $C$ and $H$ are both white. Since $C$ and $F$ are white, $I$ must be black. Similarly, since $F$ and $H$ are white, $B$ must be black. Consequently, $IAB$ is a black isosceles triangle.

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