Definitions (https://www.ias.ac.in/article/fulltext/pmsc/122/03/0459-0467):

Given a planar convex region $C$ (could be smooth or polygonal), an area bisector of $C$ is any line that partitions $C$ into 2 pieces of equal area. A 'fair bisector' is a line that partitions $C$ into 2 pieces of equal area and equal perimeter.

Thru every point on the boundary of $C$, an area bisector can be drawn (for a description of their properties, please see 'Mathematical Omnibus' by Fuchs and Tabachnikov, Lecture 11). But it can be seen that a convex planar region can have just a single fair bisector (eg. for a thin isosceles triangle, the only fair bisector is the bisector of its apex angle) or a finite number of them (in which case, their number is necessarily odd as can be seen from simple continuity arguments; see reference at the top) or infinitely many.

Observations: For regions with a center of symmetry such as a circular disk or ellipse or regular polygon with even number of sides, all fair bisectors are concurrent. But, numerically, we see that for a general convex region $C$ with finitely many fair bisectors, the fair bisectors are not necessarily concurrent but usually very close to being so. Clearly,for a general $C$ with exactly 3 fair bisectors, they determine a small triangular region deep in the interior of $C$. For $C$s with more fair bisectors, their many possible intersections will divide the interior of $C$ into many regions. Let us refer to the union of those regions which do not share the outer boundary of $C$ as the 'core' of $C$. The core must lie deep inside $C$.


  1. For which convex shape of $C$ is the area of the 'core' of $C$ the largest as a fraction of the area of $C$? Intuitively, a relatively large core is a measure of the asymmetry of $C$. Can one say (say) that such a shape is always one with exactly 3 fair bisectors?

  2. Generalizing a bit, what about lines that break off the same fraction $t$ of the area and outer boundary length of $C$? For a circular disk, it appears that only for $t=1/2$, we have such lines (any diameter). Are there $C$'s for which such lines exist for several (maybe even arbitrarily many) different values of $t$?

    Guess: All centrally symmetric convex regions (rectangles, ellipses,...) appear to give such only one single partitioning line that divides both area and outer perimeter in same ration - only for $t=1/2$. But general convex regions with no symmetry might give infinitely many such lines - one such partitioning line for each orientation - and a different value $t$ for each orientation. And the set of these lines might even have interesting envelopes.

These questions have obvious higher dimensional analogs.

  • 1
    $\begingroup$ This may be useful: Goldberg, Michael. "On area-bisectors of plane convex sets." The American Mathematical Monthly 70, no. 5 (1963): 529-531. He proves that there are convex regions with a point through which $n$ area-bisectors pass, $n \ge 4$. It is known that every convex region has a point through which at least $3$ area-bisectors pass. $\endgroup$ Apr 1, 2020 at 23:09
  • 1
    $\begingroup$ Thanks Prof. O'Rourke. The reference does give a nice intuitive property of area bisectors. The problem with 'fair' bisectors seems that they could be just a few in number so many of the structures such as envelopes which have nice properties cannot be defined in general. Still one wonders how the intersections of fair bisectors could have structures which reflect the overall nature of the convex region. $\endgroup$ Apr 2, 2020 at 19:41
  • $\begingroup$ Name of reference: Nandakumar and Ramana Rao - Fair partitions of polygons: An elementary introduction. $\endgroup$
    – LSpice
    Nov 26, 2022 at 23:03

2 Answers 2


This is not an answer, and not even that helpful, but I wanted to see the central pattern formed by the collection of perimeter bisectors.


  • $\begingroup$ Prof. O'Rourke, our experiments too indicate such patters for the envelope for perimeter bisectors - the envelope has cusps and each segment of the envelope seems to be a parabola segment (those of area bisector envelopes are hyperbola segments). For lines that cut out some other fraction t of the outer boundary length, the envelope appears to have discontinuities so on the whole they may not be as nice as envelopes that cut out a specified fraction of the area. $\endgroup$ Apr 4, 2020 at 17:10

An answer to the question 2 (written with K Sheshadri)

The guess made above that needs to be proved: For a general convex polygonal region with no symmetries, for every orientation, we have a unique line with that orientation that separates out the same fraction of both area and outer boundary length. The value of this common fraction of area and perimeter separated out will vary continuously with orientation.

Proof : Consider the convex polygonal region $C$ and a given orientation (direction). Draw both tangents to $C$ in that orientation. We assume both these tangents to touch $C$ at a single vertex (coincidences of tangents with entire edges of $C$ can be dealt with by small perturbations). Let these parallel tangents be distance $D$ apart. By sliding a line coincident with one of the tangents perpendicular to itself until it coincides with the other tangent to $C$, we get a continuous range of cutting lines. Let these cutting lines be parametrized by $d$, the perpendicular distance from the tangent from which we began sliding the cutting line.

For each value of $d$, we have a line that cuts $C$. Plot against $d$, the fraction of area (call this fraction Af) of the full $C$ that the piece separated from $C$ has and also the fraction of perimeter (call this fraction Pf) for the same piece. Obviously, as $d$ goes from 0 to $D$ both Af and Pf go from 0 to 1.

Now, we observe that the plot of Af against d has a quadratic behavior at both ends. Its plot will be continuous and made of several parabolic segments - beginning with an upward parabolic piece (where, as $d$ starts from 0, Af also starts from 0) and ending with a downward parabolic piece (when Af tends to 1 as $d$ approaches $D$). Moreover, due to convexity of $C$, the curve of Af rises monotonically.

On the other hand, Pf has a linear behaviour throughout including at ends. This graph is a continuous polyline and also rises monotonically.

From the above observations, as $d$ is increased from 0, the Af curve (quadratic) begins lower than Pf (linear) curve and as $d$ tends to $D$, Af approaches 1 from above the Pf curve. This plus the monotonically rising nature of both graphs plus their start values both being 0 and end values both being 1 guarantee that they have to necessarily intersect at some intermediate value of $d$; at these intersections, obviously, Af = Pf. It appears that convexity of $C$ also guarantees there will be only one such intersection.

Thus we have, for every orientation, a value of $d$ for which Af and Pf have same value - as claimed. If $C$ is centrally symmetric (circle, ellipse, rectangle, regular polygon with even number of sides...), the only such value of $d$ is $D$/2 and the common value of the fractions is 1/2 for all orientations. This will not be the case for asymmetric convex polygonal $C$'s - we have different common Af and Pf values for different orientations. This fraction should change continuously with orientation.

We guess that the envelope etc. of the cutting lines with common Af and Pf for each orientation might have interesting properties.


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