Ratio of area of triangle and hexagon

Question

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.

Answer 1

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, $n=AC/HR.$

Answer 2

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.

$\square$

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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).