Equilateral Triangle: Color and distance II

[bonanova]

Suppose we add a constraint similar to the one I made in answering BMAD's

 

Four colors must be used in equal measure on each side of a unit equilateral triangle.

What is the greatest distance between two points of the same color that is unavoidable?

 

Without having given much thought, I'm wondering how this answer relates to 1/2:

with added constraint will it be greater?

[k-man]

Ignore my previous post. I got my numbers mixed up.

 

With the constraint, the largest unavoidable distance between points of the same color is 7/8.

 

First, to correct my previous post, the lower bound is sqrt(3)/2, which is the height of the equilateral triangle, not 1/sqrt(2).

 

Now, to minimize the distance to the opposite vertex, we must place all the points of the same color as close to the midpoint of the opposite side as possible. So, the 1/4 length in the middle of the opposite side is painted the same color as the opposite vertex and results in the longest distance of 7/8. Now, we need to paint the rest of the triangle without exceeding that distance. 

Paint the 1/4 length from the vertex in the same color as the vertex on both sides from the vertex. Repeat for all three vertices using 3 different colors. Now the only unpainted area is 2 segments of length 1/8 on each side. The largest distance between unpainted points is 3/4. So paint that using 4th color and we're done.

[k-man]

it must be greater than 1/2. In fact, it must be greater than 1/sqrt(2). A vertex must be of some color. With the constraint, the same color must appear somewhere on the opposite side, every point of which is at least 1/sqrt(2) away.