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Suppose you are given a convex unit area region of the plane, and you wish to minimize the length of a line segment that cuts the shape into two regions of equal area. Which shape results maximizes this length, i.e., is the hardest to slice into two equal regions. Is it the circle?

For generalizations, does anything change if we allow the slice to be any curve? What if we want to chop it into $n$ equal regions for $n>2$? What if we are minimizing the area of planes that slice higher dimensional regions in half? I'm curious in any generalizations along these lines.

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  • $\begingroup$ If you made a rectangle of length $x$ by $\frac{1}{x}$, the cut length could be equal to $x$, which is arbitrarily large (approaching $\infty$). $\endgroup$ Sep 28, 2020 at 21:10
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    $\begingroup$ But the minimum length cut in your example would be length $1/x$ which is arbitrarily small. $\endgroup$
    – Alex
    Sep 30, 2020 at 14:05

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Edit: This question is discussed at length in Chords halving the area of a planar convex set (Grüne et al., 2007), along with the question of the longest possible splitting chord. However, the paper (while establishing many other bounds) leaves the relation between area and minimal halving distance open, except to note that the $\sqrt[4]{3}$ upper bound is not tight.

Notably, the circle is not optimal, and any (non-circular) convex Zindler curve serves as a counterexample. The aforementioned paper cites a 2005 preprint On geometric dilation of closed curves, graphs and point sets (Dumitrescu et al., 2005) for this fact, where perusing Section 7 we can see that the "rounded triangle" $C_\triangle$ is conjectured to have maximal cutting length $h$ for a fixed area, at $$A(C_\triangle)=\sqrt{3}\left(\log(3) - \frac{\log(3)^2}{8} - \frac12\right)h^2$$ so $h\approx 1.135547\sqrt{A}$.

The curve $C_\triangle$ is pictured below, from Dumitrescu et al.'s paper. (The central region $M$ is not a hole, but the locus of the midpoints of area-splitting chords; we have $A(Z)=\frac\pi4h^2-2A(M)$ for any Zindler curve $Z$.)

                                    enter image description here

My original answer, written before locating the above papers, is given below.


Let our convex set be $C$. Note that such a cut is at most the width $w$ of the shape, because if we place $C$ between two parallel lines, then some cut orthogonal to those lines will cut $C$ in half, and its length must be at most the distance between the lines. (This argument extends to the same question for dividing $C$ into regions of equal perimeter, or for that matter most natural measures of "size" of a convex set.)

We can then use the fact that $w^2\le \sqrt{3}A$ among bounded convex planar sets; see for instance Inequalities For Convex Sets (Scott and Awyong, 2000), which cites Yaglom and Boltyanski's book Convex Figures. This gives us a minimal cut width of at most $\sqrt[4]{3}\approx 1.316$.

Note that since the extremal convex set for this inequality is the equilateral triangle, which has a cut width of $\sqrt{2}/3^{1/4}\approx 1.075$, it does not in fact surpass the circle's $\frac{2}{\sqrt{\pi}}\approx 1.128$, but at least provides a reasonable upper bound.

As an unrelated piece of evidence, I believe I can get a cut of length around $1.1038$ on the Releaux triangle of unit area, so that particular candidate (worth checking for any question about convex sets) seems to not quite beat the circle.

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    $\begingroup$ I think the shortest divider for an equilateral triangle is parallel to a side. Still shorter than for the circle. $\endgroup$
    – orangeskid
    Nov 14, 2020 at 8:10
  • $\begingroup$ Yes, that yields the $3^{-1/4}$ mentioned. For a square it is 1, and in general for a centrally symmetric body it will be at most that of the circle (with equality only when equal to the circle). $\endgroup$ Nov 14, 2020 at 8:16
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    $\begingroup$ Funny, I got $\frac{\sqrt{2}}{3^{\frac{1}{4}}}= 1.07...$ $\endgroup$
    – orangeskid
    Nov 14, 2020 at 9:33
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    $\begingroup$ Oops, thanks for the correction! Fixed now, apologies for the miscalculation. $\endgroup$ Nov 14, 2020 at 9:54
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    $\begingroup$ A very informative answer! Great! $\endgroup$
    – orangeskid
    Nov 15, 2020 at 15:49
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I'll only consider the particular case of an acute angled triangle with area $1$ and with angles $\alpha \le \beta \le \gamma$.

Let's first find the arc of circle inside an angle of angle $\phi$ and of area $\frac{1}{2}$.

If the radius of the arc is $r$ then we have $$\frac{1}{2} r^2 \phi = \frac{1}{2}$$ so $r= \frac{1}{\sqrt{\phi}}$, and therefore the length of the arc is $$r \phi = \sqrt{\phi}$$

Let us see when the arc corresponding to the angle $\alpha$ lies inside the triangle. This is true if and only if the distance from the vertex to the opposite side $h$ is $\ge \frac{1}{\sqrt{\alpha}}$.

Now, we calculate the area of the triangle $$1 = \frac{1}{2} h^2 ( \cot \beta + \cot \gamma)$$ and therefore $$h = \sqrt{\frac{2}{\cot \beta + \cot \gamma}}$$ and so the condition is $$\frac{\cot \beta + \cot \gamma}{2}\le \alpha$$

Now, since the function $\cot $ is decreasing, we get $$\frac{\cot \beta + \cot \gamma}{2}\le \cot \alpha$$ Therefore, if $$\cot \alpha \le \alpha$$ (equivalently, $\alpha \ge \approx 49^{\circ}.\ldots$) then we have the inequality.

Therefore, if the angle $\alpha$ is not too small, the shortest dividing line is of length $\sqrt{\alpha}$.

In the particular case of an equilateral triangle, the shortest line is $\sqrt{\frac{\pi}{3}} = 1.0233\ldots$

For the equilateral triangle, the radius halving the area equals $$\frac{\sqrt{\pi}}{2 \sqrt[4]{3}}\cdot l= 0.6733\ldots \cdot l$$ where $l$ is the side of the triangle. division into equal area parts

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    $\begingroup$ I think the OP is focused on line segment cuts? Though this seems like a natural variant to ask - I suppose if the circle is optimal for the linear case, it must be for general cuts as well. $\endgroup$ Nov 14, 2020 at 20:33
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    $\begingroup$ @RavenclawPrefect: Yes he does starts with straight cut. It seems that the optimal cuts are circular or straight (infinite radius). An interesting question... $\endgroup$
    – orangeskid
    Nov 15, 2020 at 5:35

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