# A question about straight lines that bisect the area of a triangle

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.

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. $\square$

• What a wonderful answer ! May I ask you which software you have used for your animations ? Geogebra ? Have you seen the animations (demonstrations.wolfram.com/TriangleAreaBisectors) ? Commented Oct 1, 2016 at 10:48
• @JeanMarie: Thanks. :) I usually use GeoGebra, but this time I used Mathematica. I had not seen the Demonstration; it's interesting.
– Blue
Commented Oct 1, 2016 at 17:45

Suppose you have an angle of vertex $V$ and a point $P$ inside it. If $r$ is a line through $P$ forming a chord $QR$, then the area $s$ of triangle $VQR$ is a function of the angle $\theta=\angle VPR$. Let $\alpha=\angle QVP$ and $\beta=\angle RVP$: $\theta$ can take any value between $\beta$ and $\pi-\alpha$, and the area of triangle $VQR$ can be written as $s(\theta)=(\overline{VP}{}^2/2)\,f(\theta)$, where: $$f(\theta)= {\sin\alpha\sin\theta\over \cos\alpha\sin\theta+\sin\alpha\cos\theta}+ {\sin\beta\sin\theta\over \cos\beta\sin\theta-\sin\beta\cos\theta}.$$ Substituting here $t=\tan\theta$, $u=\tan\alpha$ and $v=\tan\beta$ this can be rewritten in the simpler form $$f(t)={tu\over t+u}+{tv\over t-v},$$ with $t<-u$ or $t>v$. Notice that $\lim\limits_{t\to-u}f(t)=\lim\limits_{t\to v}s=+\infty$, while $\lim\limits_{t\to\pm\infty}f(t)=u+v$.

$f'(t)$ vanishes, in the allowed range, for $t=2uv/(u-v)$, which then corresponds to the only minimum $f_\min$ of $f(t)$ and of $f(\theta)$. For every value $y>f_\min$ the equation $f(\theta)=y$ has then exactly two solutions. See below for a typical plot of $f(\theta)$, for $\alpha=\pi/6$ and $\beta=\pi/4$.

Consider now a triangle $ABC$ of area $S$ and a point $P$ inside it. Let $A'$, $B'$ and $C'$ be the points where lines $PA$, $PB$ and $PC$ intersect $BC$, $AC$ and $AB$ respectively. A chord $QR$ of $\angle BAC$ passing through $P$ (with $Q\in AB$ and $R\in AC$) is inside the triangle only if $\angle APB'<\angle APR<\angle APC$. From the above discussion it follows that the area of $AQR$ can be $S/2$ only if $Area_{ABB'}>S/2$ or $Area_{ACC'}>S/2$ (I'm not considering here the special case when both areas are $S/2$, that is when $P$ is the centroid of $ABC$). In addition, there can be two different chords through $P$ bisecting the area of triangle $ABC$ only if both inequalities hold.

The same reasoning can be repeated for angles $\angle ACB$ and $\angle ABC$. In the end we find that the number of chords bisecting the area of $ABC$ is bounded by the number of triangles among $ABB'$, $ACC'$, $BAA'$, $BCC'$, $CAA'$, $CBB'$ whose area is larger than $S/2$. But those six triangles can be paired so that the sum of the triangles in each couple is $ABC$: it follows that only three among them can have an area greater than $S/2$. Hence no more than three lines passing through $P$ and bisecting the area of $ABC$ can exist.

The "ratio of areas" being an affine concept (i.e., invariant under affine transforms), we can assume that the triangle is equilateral.

Edit (almost 5 years later) in connection with the answer of @Blue:

If we are looking for an area ratio which is not exactly $$1:1$$ but slightly different, we have a kind of "defocalization" of the triangle, still with hyperbolic arcs, as illustrated in the following figure:

extracted from the article by D. Ball, “Halving Envelopes.” The Mathematical Gazette, vol. 64, no. 429, 1980, pp. 166–173, on JSTOR,www.jstor.org/stable/3615118 which gives further details.

• But are there four concurrent area-bisectors? (There are certainly three: the medians.)
– Blue
Commented Sep 28, 2016 at 22:05
• It's not my question. ;)
– Blue
Commented Sep 29, 2016 at 6:45
• These proofs are very nice, and much more than I expected to get in a response to my question. Jean Marie is right. My question was badly phrased. I was trying to express it all in one sentence. Luckily for me, the responders who provided these proofs understood what I meant. Commented Sep 30, 2016 at 15:00