The sides of a regular triangle $\triangle_1=ABC$ is bicolored(red, and blue), Do there exist three vertices on the perimeter of $\triangle_1$ three monochromatic vertices forming the corners of a rectangular triangle?
Then if no point, excluding $A$ and $B$, on the edge $AC$ and $BC$ are colored blue, then we can find three red vertices on edges $AC$ and $BC$ to form a rectangular triangle. Otherwise there exist a point $D$ on $AC$ or $BC$ colored blue. But How can we make it into a rectangular triangle?
Also, is there a good drawing software for math?
I found a solution here but I don't understand it:
Suppose there is no right triangle with vertices of the same color. Partition each side of the regular triangle by two points into three equal parts. These points are vertices of a regular hexagon. If two of its opposite vertices are of the same color, then all other vertices are of the other color, and hence there exists a right triangle with vertices of the other color. Hence opposite vertices of the hexagon are of different color. Thus there exist two neighboring vertices of different color. One pair of these bicolored vertices lies on a side of the triangle. The points of this side, differing from the vertices of the hexagon, cannot be of the first or second color. Contradiction.
My question is why the last sentence :"The points of this side, differing from the vertices of the hexagon, cannot be of the first or the second color" true?