Line diving 2 triangles in a plane in equal halves by area Can you prove or disprove that for any 2 triangles in a plane, there will always be a line passing through them both which divide each of them into 2 halves of equal area? If said line exists, how can we find it?
 A: Of course one can make appeal to the ham-sandwich-theorem. But for two triangles it should be possible to obtain, at least in principle, an explicit solution.
We first have to describe the set of lines dividing a single triangle into two parts of equal area. To this end consider the triangle $T$ with vertices $(0,0)$, $(1,0)$ and $(0,1)$. A line cutting off the origin together with half the area of this triangle has an equation of the form
$${x\over u}+{y\over v}=1\ ,$$
whereby $u,v\in\>]0,1[\>$ and $uv={1\over2}$. The family of these lines can therefore be described in the form
$$2v^2x+y-v=0,\qquad0<v<1\ .$$
The envelope of this family turns out to be an arc $\gamma$ of the hyperbola $xy={1\over8}$, whereby the endpoints of this arc are the midpoints $\bigl({1\over2},{1\over4}\bigr)$ and $\bigl({1\over4},{1\over2}\bigr)$ of the medians  through the vertices $(1,0)$ and $(0,1)$. Of course these medians themselves are tangent to the arc $\gamma$ in these points, since they are halving the area of $T$. By affinity it is not necessary to do a similar computation for cutting off another vertex of $T$: In all we obtain three arcs of hyperbolas that form a "hyperbolic triangle" with cusps in the midpoints of the medians of $T$. It is possible to rotate a line $\ell$ smoothly $180^\circ$ along these three arcs so that $\ell$ touches (at least) one of the arcs at all times.

If we now are given two arbitrary triangles $T_1$, $T_2$ then  each of them has its own "hyperbolic triangle" $H_i$. The rotating $\ell$ of $H_1$ will then at a certain moment be tangent to one of the arcs of  $H_2$ (one would have to check this). In this position the line $\ell$ halves both triangle areas simultaneously.
