Call a triangle with angles $30,60,90$ in some order good. We use coordinate geometry. Let $P=(0,0)$ and $Q=(1,0)$ be two red points. Then $X=(0,\frac{\sqrt{3}}{3})$ is red since $\Delta PQX$ is good, so $Y=(\frac{1}{3},0)$ is red since $\Delta PXY$ is good. Thus, for any two red points, the two points trisecting them are also red. $Z=(\frac{1}{4},\frac{\sqrt{3}}{4})$ is red since $\Delta PQZ$ is good, and so $W=(\frac{1}{4},0)$ is red since $\Delta PZW$ is good. The midpoint $M$ of $PQ$ satisfies $2WM=MQ$, so $M$ is also red. Thus, for any two red points, their midpoint is also red. For general $\Delta ABC$ with all red vertices, the midpoint $D$ of $BC$ is red, and the centroid cuts $AD$ into a $2:1$ ratio, so it is also red. $\square$

Proof without words:

5 steps is enough! This is the least steps!

I scanned my eye and bet everything on red: Statement 1: I will determine that if $X$ and $Y$ are red then their middle point, lets say $Z$, is also red . I will also state that a triangle $\triangle ABC$ where their inside angles are $90°, 60°, 30°$ in some order then $\triangle ABC$ is half an equilateral triangle. Furthermore, if $X$ and $Y$ are red then the point $L$ such that $\triangle XYL$ is half an equilateral then $L$ is red. I will set the point $S$ such that $Y\widehat{X}S=30°, X\widehat{Y}S=60° \Rightarrow X\widehat{S}Y=90°$ then $XYS$ is a half an equilateral triangle then $S$ is red. I will set the point $T$ such that $T\in XY$ and $T\widehat{Y}S=60°, Y\widehat{S}T=30° \Rightarrow S\widehat{T}Y=90°$ then $TXS$ is half an equilateral triangle then $T$ is red. I will set the point $Z$ such that $Z\in XY$ and $T\widehat{Z}S=60°, T\widehat{S}Z=30° \Rightarrow Z\widehat{T}S$ then $ZTS$ is half an equilateral triangle then $Z$ is red. Say that $TX=x$ then as a consequence of $TYS$ being an equilateral triangle I can affirm that $YS=2x$. Additionally, $Z\widehat{S}Y=30°+30°=60°, Z\widehat{S}Y=60°=T\widehat{Z}S \Rightarrow ZY=YS=2x$. Because $\triangle XYS$ is an equilateral triangle $XY=2YS=4x$ and $ZY=2x$, then $Z$ is the middle point of $XY$. Statement 2: I will demonstrate that if $X$ and $Y$ are red then the point $I$ where $XI=2XY$ is also red. I will set the point $E$ such that $Y\widehat{X}E=30°, X\widehat{Y}E=60° \Rightarrow X\widehat{E}Y=90°$ then $\triangle XYE$ is half an equilateral triangle and $E$ is red. I will set the point $F$ such that $F\in XE$ and $Y\widehat{F}E=60°, F\widehat{Y}E=30° \Rightarrow Y\widehat{E}F=90°$ the $\triangle YEF$ is half an equilateral triangle then $F$ is red. I will set the point $H$ in the same half plane that $X$ regarding to $YF$ such that $Y\widehat{F}H=30°, Y\widehat{H}F=60° \Rightarrow H\widehat{Y}F$ then $\triangle HYF$ is half an equilateral triangle then $H$ is red. I will set the point $G$ such that $G \in XY$ and $Y\widehat{H}G=30°, H\widehat{Y}G=60° \Rightarrow H\widehat{G}Y=90°$ then $\triangle HGY$ is half an equilateral triangle then $G$ is red. I will set the point $I$ in the same half plane as $X$ regarding to $HG$ such that $G\widehat{H}I=30°, H\widehat{I}G=60° \Rightarrow H\widehat{G}I=90°$ then $\triangle HGI$ is half an equilateral triangle then $I$ is red. See that $I\widehat{Y}F=90°-60°=30°=Y\widehat{I}F \Rightarrow IY=IF=x$ y $Y\widehat{I}F=180°-30°-30°=120°$. Furthermore, $X\widehat{I}F=180°-120°$ then $\triangle XIF$ is half an equilateral triangle and $XI=2IF=2x$. Using statements 1 and 2 we can set the point $Z\in BC$ such that $BZ=CZ$ and the point $I\in AZ$ such that $AI=2ZI$ or $ZI=2AI$. $I$ is the barycenter of $\triangle ABC$ and we are done.

Note that given two red points their midpoint is also red (construct a 306090 with their length as diameter). So we just need a way to construct trisections of segments; this is trivial because we can construct the center of an equilateral triangle by dividing it into six smaller (radially symmetric) 306090 triangles, and then we can drop a perpendicular.

We can construct a red midpoint by constructing a triangle with the segment as the diameter and then reflecting the 60 degree angle over the altitude from the right angle. We can also trisect stuff by using $\left( \frac{\sqrt{3}}{3} \right)^2 = \frac13$ and in particular making the segment the $\sqrt{3}$ side and then doing the same for the one opposite the 30 degree angle in this new triangle. We can trisect and bisect which makes the constrution trivial.