What's the connection the ratio of the arithmetic progression to the area of the triangle below?

For referce: Let us consider an equilateral triangle of $S$ area. From a point inside your triangle median, are drawn perpendicular to the sides of the equilateral triangle. these perpendicular form an arithmetic progression of ratio $r$, then:(Answer:$r < \frac{\sqrt[4]{3S^2}}{6}$)

My progress:

$S_\triangle {ABC} = \frac{l^2\sqrt3}{4}\\ S_{DEF}=\frac{l^2\sqrt3}{16}=\frac{S}{4}\\ \frac{h_1+h_3}{2} = r$

Not being able to understand the function of this median triangle...????

I find a property: $h_1+h_2+h_3 = CF$ maybe can help

Solution 1:

Suppose height of equilateral $\triangle ABC$ is $H$. It can be shown easily $H^2=\sqrt{3}S$.

The three perpendicular lengths can be taken as $h_2-r, h_2, h_2+r$. Requiring that their sum be $H$, one obtains $h_2=H/3$. One of the perpendiculars must have $1/3$rd of the total height, meaning the point $P$ is at same height as centroid $G$.

WLOG, we can take $P$ to lie on the (blue) line through $G$ parallel to $AB$. Conditioning that it is not exterior to the medial triangle $DEF$, its extremum position is (shown) on the side $DF$. Clearly for this position of $P$, $PZ/PX=PD/PF=1/2 \Rightarrow PZ=H/6$.

Thus for all allowed positions of $P$ (on the undotted portion of blue line), $$PZ=H/3-r > H/6 \Rightarrow r < H/6 $$ which using $H^2=\sqrt{3}S$ is same as $$r < \frac{(3S^2)^{1/4}}{6}$$