Problem 1106

Proof: If there exists an equilateral triangle with side length 1 and vertices of the same color, the problem is solved.

If there does not exist an equilateral triangle with side length 1 and vertices of the same color, then there must exist a line segment $AB$ of length 1, with endpoints $A$ and $B$ of different colors.

Using $AB=1$ as the base, construct an isosceles triangle $\triangle ABC$ with legs of length 2. Then, point $C$ must be of a different color from either $A$ or $B$. As shown in Figure 3, without loss of generality, assume the line segment $AC$ of length 2 has endpoints of different colors.

Take the midpoint $O$ of $AC$, then $O$ must be of the same color as either $A$ or $C$ (as shown in Figure 4, without loss of generality, assume $O$ is the same color as $A$). Since there does not exist an equilateral triangle with side length 1 and vertices of the same color, the other two vertices $D$ and $E$ of the equilateral triangle with side $AO$ must be of a different color from $A$. At this point, $\triangle ECD$ is an equilateral triangle with side length $\sqrt{3}$ and vertices of the same color.

[Note] In fact, by dividing the plane into horizontal strip regions of width $\frac{\sqrt{3}}{2}$, where each region includes the lower boundary line but not the upper boundary line, and coloring adjacent strip regions with different colors, for such a two-coloring of the plane, the three vertices of any equilateral triangle with side length 1 are all of different colors, but there exists an equilateral triangle with side length $\sqrt{3}$ and vertices of the same color.

From Example 3, we can derive a more general conclusion: After two-coloring the points on the plane, either there exists an equilateral triangle with side length $a (a>0)$ and vertices of the same color, or there exists an equilateral triangle with side length $\sqrt{3}a$ and vertices of the same color.