Solution to monochromatic equilateral triangle puzzle

This is a solution to the math puzzle posted earlier.

We will post a solution to the variant (which also solves the main puzzle).

The problem:

Each point of 2D plane is coloured either black or white. Given three positive angles $\alpha + \beta + \gamma = 180^{\circ}$, show that there are there points $A,B,C$ of the same colour such that the triangle formed by those three has the angles $\alpha, \beta, \gamma$.

Solution

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The solution is quite similar to the one posted in the comments by 226.

First we show that there are three collinear points $X,Y,Z$ which are coloured the same, and $Y$ is the midpoint of $X$ and $Z$.

Assume that is not true.

Consider two points which are coloured the same (say black, wlog). Say they are $(0,0)$ and $(2a, 0)$.

Now $(a,0)$, $(-2a, 0)$ and $(4a, 0)$ must be coloured white. These three points satisfy the requirement.

Say $X,Y,Z$ are coloured black.

Let $\gamma = \max \{\alpha, \beta, \gamma\}$

Let $P,Q$ be such that $\angle{YXP} = \angle{ZYQ} = \alpha$ and $\angle{XYP} = \angle{YZQ} = \beta$. Both $\triangle{XYP}$ and $\triangle{YZQ}$ have the angles $\alpha, \beta, \gamma$.

Ths $P$ and $Q$ must both be coloured white.

Now consider point $R$ such that $\angle{QPR} = \angle{ZXR} = \alpha$ and $\angle {PQR} = \angle{XZR} = \beta$

Triangle $PQR$ also has the given angles, and so $R$ must be black. But $\triangle{XZR}$ also has the given angles, and all vertices are black.

Basically, $XZR$ is an $\alpha, \beta, \gamma$ triangle, and $P,Q,Y$ are the midpoints of the sides of $\triangle{XZR}$.