# Ratio of area of triangle and hexagon

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.

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, $$n=AC/HR.$$

I will provide a picture and go into detail, but need my notations first - as in the following picture:

Let $$\Delta ABC$$ be the given triangle. Let $$A'$$, $$B'$$, $$C'$$ be the mid points of the sides respectively opposed to $$A$$, $$B$$, $$C$$. We denote by $$A_-$$ and $$A_+$$ the two division points ($$\ne A'$$) to the left and to the right of $$A'$$ on the side $$BC$$, then use similar notations for the other sides, so that we have \begin{aligned} A_-A'&=A'A_+=xBC=xa\ ,\\ B_-B'&=B'B_+=xCA=xb\ ,\\ C_-C'&=C'C_+=xAB=xc\ , \end{aligned} where $$x$$ is one and the same proportion. We can let it be general, in the OP it is $$\displaystyle \color{blue}{\frac1n}$$.

Let $$A^*$$ be the intersection of the cevians $$BB_+$$, $$CC_-$$, and the median $$AA'$$.

Let $$A^!$$ be the intersection of the cevians $$BB_-$$, $$CC_+$$, and the median $$AA'$$.

Similarly we construct $$B^*$$, $$B^!$$; $$C^*$$, $$C^!$$.

The OP is immediately solved if we answer the following questions:

• Which is the position of $$A^*$$ on the segment $$GA$$?
• Which is the position of $$A^!$$ on the segment $$GA'$$?

(Try to get it on your own.) Applying Menelaus in $$\Delta AGB'$$ w.r.t. the line $$BA^*B+$$, and w.r.t. the line $$BA^!B_-$$ we have: \begin{aligned} 1 &= \frac {A^*G}{A^*A} \cdot \frac {B_+A}{B_+B'} \cdot \frac {BB'}{BG} = \frac {A^*G}{A^*A} \cdot \frac {\frac 12-x}{x} \cdot \frac 32\ , &\text{ so }\qquad \frac {GA^*}{GA} &= \frac {4x}{3-2x}\ , \\ 1 &= \frac {A^!G}{A^!A} \cdot \frac {B_-A}{B_-B'} \cdot \frac {BB'}{BG} = \frac {A^!G}{A^!A} \cdot \frac {\frac 12+x}{x} \cdot \frac 32\ , &\text{ so }\qquad \frac {GA^!}{GA} &= \frac {4x}{3+2x}\ , \\ & &\text{ so }\qquad \frac {GA^!}{GA'} &= \frac {8x}{3+2x}\ . \end{aligned} Similar computations can be done for $$B^*$$, $$B^!$$, and $$C^*$$, $$C^!$$. This gives the proportions of areas: $$\frac{[\color{magenta}G\color{blue}{A^*B^!}]}{[\color{magenta}{G}AB']} = \frac {\color{magenta}{G}\color{blue}{A^*}}{\color{magenta}{G}A}\cdot \frac {\color{magenta}{G}\color{blue}{B^!}}{\color{magenta}{G}B'} = \frac {4x}{3-2x}\cdot \frac {8x}{3+2x} = \color{blue}{ \frac{32}{(3n-2)(3n+2)}} \ .$$ This is also the final answer since the medians cut the triangle $$\Delta ABC$$ in six smaller triangles of same area. This matches the claimed formula.

$$\square$$

Here is also explicitly the situation $$n=4$$:

We have placed

• on $$GA$$, $$GB$$, $$GC$$ points to illustrate $$GA^*:GA=2:5$$ (and similar proportions for the other two segments), and
• on $$GA'$$, $$GB'$$, $$GC'$$ points to illustrate $$GA^!:GA'=4:7$$ (and similar proportions for the other two segments).