I am looking for the proof of the following claim:
Claim. If the sides of the triangle are partitioned into $n$ equal segments for $n$ an even integer and each division point adjacent to the corresponding central division point is connected to the opposite vertex, a central hexagon which has an area of $\frac{32}{(3n-2)(3n+2)}$ of the area of the original triangle is still formed.
GeoGebra applet that demonstrates this claim can be found here.
I tried to adapt the proof of the Marion's theorem given on this page. The problem I encountered is the following: How to find a trilinear coordinates of the division points : $D,E,F,G,H,I$ . The only formula I am aware of is the formula for the first n-multisection.
Best Answer
We can use the Routh-Steiner theorem, which gives a method to compute the area of a sub-triangle bounded by three cevians.
The three medians of $ABC$ (meeting at the centroid $X$) divide the central hexagon into six triangles. The triangle $LMX$ is bounded by cevians defined by side ratios $x=y=1,z=(n+2)/(n-2)$. By Routh-Steiner
$$ \begin{align} \cfrac{|LMX|}{|ABC|} &= \cfrac{(xyz-1)^2}{(xy+y+1)(yz+z+1)(zx+x+1)} \\ &= \cfrac{\cfrac{16}{(n-2)^2}}{(3)\left(\cfrac{3n+2}{n-2}\right)\left(\cfrac{3n-2}{n-2}\right)} \\ &= \cfrac{16}{3(3n+2)(3n-2)} \end{align} $$ The other five triangles are built using the same ratios, so have the same area as $LMX$. Thus the area of the hexagon as a fraction of the large triangle is
$$ \cfrac{|JKLMNO|}{|ABC|}=6 \left(\cfrac{16}{3(3n+2)(3n-2)}\right)=\cfrac{32}{(3n+2)(3n-2)}. $$
Note that the construction and formula also work for non-integer values of $n.$ Here, $n=AC/HR.$