TITLE: A question about straight lines that bisect the area of a triangle
QUESTION [5 upvotes]: Let $K$ be a convex subset of the Euclidean plane $E(2)$ whose boundary is a triangle. Is it true that there cannot exist 4 pairwise distinct concurrent straight lines in $E(2)$, each of which bisects the area of $K$? Many results in the literature strongly suggest that this is true, but I have not so far been able to find a theorem that actually states or implies that it is.

REPLY [5 votes]: As @JeanMarie has mentioned, this problem is "affine invariant"; so, if we can settle it for, say, an equilateral triangle, then we will have solved it for all triangles.
Now, given $\triangle ABC$, and a particular vertex, and a continuous parameter, we can consider the family of area-bisectors that form sub-triangles with the vertex. For instance, for vertex $A$, we get triangles $\triangle AB_t C_t$ where $$B_t = A + t\;(B-A) \quad\text{and}\quad C_t := A + \frac{1}{2t}\;(C-A) \quad\text{, for}\;\;\frac{1}{2}\leq t \leq 1 \tag{1}$$
Here, because 
$$2|\triangle AB_t C_t| = |\overline{AB_t}||\overline{AC_t}|\;\sin A = t\;|\overline{AB}|\cdot\frac{1}{2t}|\overline{AC}| \sin A = \frac{1}{2}|\overline{AB}||\overline{AC}|\sin A = |\triangle ABC|$$
we have that $\overline{B_tC_t}$ is in fact an area-bisector. (Note that the restrictions on $t$ guarantee that this bisector crosses the sides adjacent to $A$.) 
If we take our triangle to have vertices $A = (1, 0)$, $B = (\cos\frac{2\pi}{3}, \sin\frac{2\pi}{3})$, $C = (\cos\frac{4\pi}{3}, \sin\frac{4\pi}{3})$, then we compute the equation 
$$\overleftrightarrow{B_tC_t} :\qquad 
x\;(1+2 t^2) \;-\; y \;\sqrt{3}\;(1-2 t^2) \;-\; ( 1 - t )( 1 - 2t) \;=\; 0 \tag{2}$$
The "envelope" of this family of lines arises from eliminating $t$ from $(2)$ and its derivative with respect to $t$
$$4 xt\;+\;4y t\;\sqrt{3}\;+\; 3 -4 t\;=\;0 \tag{3}$$ 
to obtain this hyperbola:
$$ 8 ( x-1 )^2 - 24 y^2 = 9 \tag{4}$$
Its center is at $A$, and its asymptotes are the extended sides $\overleftrightarrow{AB}$ and $\overleftrightarrow{AC}$. The corresponding envelopes related to vertices $B$ and $C$ are appropriately-rotated copies of this hyperbola:

Observe that the three hyperbolas are pairwise tangent, and that the medians of the triangle form common tangent lines. In fact, when we restrict our attention in accordance with $(1)$, it turns out that we only care about the hyperbolic arcs determined by the points of tangency. Let's zoom in:  

This curvilinear triangle is cut into six regions by the original triangle's medians. Importantly, and quite clearly, no tangent line to any one hyperbolic arc enters either of the two regions "opposite" that arc. We see, then, that a point in the interior of any of those regions lies on three tangent lines; in this figure ...

... dots indicate how many tangents of a hyperbolic arc meet a given region. At points along the borders between regions ---that is, along the triangle's medians--- "four" tangents can meet, but two of them coincide  with the median, bringing the number of distinct tangents down to three. (The center point has ostensibly "six" concurrent tangents ---two per arc--- though obviously only three are distinct.)
Since tangents to the hyperbolic arcs are precisely the triangle's area-bisectors, we have shown that at most three distinct bisectors may concur at a point. $\square$