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.

#Combinatorics #Coloring #CosinesGroup #Goldbach'sConjurersGroup #TorqueGroup

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Comments

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 - 7 years, 1 month ago

Nice job!!

Eddie The Head - 7 years, 1 month ago

Thank you!

Nanayaranaraknas Vahdam - 7 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$