Three sides of a regular triangle is bicolored, are there three points with the same color forming a rectangular triangle

contest-mathgeometrypigeonhole-principle

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?

So here is my attempt: Two of the tree vertices, say $A,B$ of $\triangle_1$ must be colored the same, say blue, by pigeonhole principle.
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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?

Best Answer

(Note: A first version of this answer was given before your edit.) I claim that a monochromatic right triangle cannot be avoided.

Proof. The smaller triangle $\triangle(DEF)$ in the following figure has at least two vertices of equal color. We may assume that $D$ and $E$ are red. This enforces $P$ and $Q$ both to be blue, or there would be a monochromatic red triangle. Now the point $C$ can be colored neither red nor blue without creating a monochromatic right triangle.

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