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Colored nonagon problem

Let the vertices of a regular 9-gon be colored black or white.

(A)Show that there are two adjacent vertices of same color

(B)Show that there are three vertices of same color forming an isoceles triangle.

Note by Eddie The Head
3 years ago

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For (A), if two consecutive vertices are not found having the same colour, the worst case scenario is when \(8\) vertices are alternating between black and white. In this case, the \(9^{th}\) vertex will share a colour with one of its two neighboring vertices.

For (B), let a point \(A\) be the reference vertex. From \(A\), no point on its left must be equidistant to it as a point from its right, otherwise the two points and \(A\) will form an isosceles triangle.Thus, let us assume that the vertex adjacent to \(A\), to the left, is the same colour as \(A\). Thus, a point to the right of \(A\) can be at least \(2\) vertices away. This pattern continues until the \(4^{th}\) and \(5^{th}\) vertices from \(A\). By the pattern, they share the same colour as \(A\), and thus, form an isosceles triangle. Also, in the case of \(3\) or more consecutively coloured points, there is always an isosceles triangle of the same coloured vertices. Nanayaranaraknas Vahdam · 3 years ago

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@Nanayaranaraknas Vahdam Nice job!! Eddie The Head · 3 years ago

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@Eddie The Head Thank you! Nanayaranaraknas Vahdam · 3 years ago

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