Let $P$ be a convex polygon (or any convex body in $\mathbb{R}^2$) with perimeter of length $1$. Call a chord $c$ of $P$ perimeter-halving if half the perimeter lies to one side of $c$ (and so half to the other side). Here are three convex polygons with many perimeter-halving chords drawn:

(Perimeter-halvings play a role in folding convex polygons to convex polyhedra.)

Define the perimeter-halving center for $P$ to be a point $x$ that minimizes the maximum distance $\delta$ of any perimeter-halving chord to $x$. So the perimeter-halving chords all nearly pass through $x$.

Q1. Does the perimeter-halving center of $P$ coincide with the centroid of $P$? Or is it located at some other natural center?

          Center of gravity (cg) marked. $\delta = 0.035$ (from the cg).

Q2. Which shapes achieve the extremes of $\delta$?

Clearly any centrally symmetric shape achieves $\delta=0$. Does any other shape realize $\delta=0$? Which shapes have the worst (largest) $\delta$?

And just out of curiosity, I would be interested to learn what are the elegant spirograph/astroid-like envelope curves visible in the figures.

Added. David Eppstein's center for an equilateral triangle, as detailed by Wolfgang.

          Perimeter-halving center (red); $\delta= \sqrt{3}/72$.

  • $\begingroup$ for the extremes of $\delta$, do you also accept the centroid inside the "criss crossed" area? If yes, maybe an isosceles triangle with the base angles close to $0$ vs one with the base angles close to $\pi/2$ would be worth investigating as extreme cases. $\endgroup$ Sep 6, 2017 at 14:45
  • $\begingroup$ @ManfredWeis: I am not at all sure the centroid is the same as the perimeter-halvings center. It would be remarkable. In the top three examples, the centroid does fall inside those dense regions. $\endgroup$ Sep 6, 2017 at 14:49
  • $\begingroup$ "Define the perimeter-halvings center for $P$ to be a point $x\dots$" The switch from "the" to "a" makes me wonder whether each $P$ has a unique perimeter-halving center. $\endgroup$ Sep 6, 2017 at 23:02
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    $\begingroup$ It is unique. It is the solution to a linear program with infinitely many constraints (find x,y,d such that x,y is within distance d of each perimeter-halving line, minimizing d) and any such LP has a basis consisting of three lines or two parallel lines from which it is equidistant. But the three-line case has a unique solution and the two-parallel-lines case can only occur in this problem when d is zero and they are the same line (because there are no two distinct parallel halving lines) in which case it's the unique point through all halving lines. $\endgroup$ Sep 7, 2017 at 20:14
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    $\begingroup$ If you can calculate its position numerically then you can look it up in the Encyclopedia of Triangle Centers and see if it matches any of them. $\endgroup$ Oct 27, 2017 at 18:49

3 Answers 3


As can be seen from the construction of the dual parabola as described here it is clear, that the astroid like curves inside convex polygons are actually composed of pieces of parabolas.

A simple proof for the claimed parabolic segments is, that the perimeter-halfings, whenever they intersect a pair of edges in inner points, resemble the construction of a quadratic Beziér curve.
To see that, simply elongate the two intersected edges to their intersection point and to a length, that equals half the perimeter of the convex polygon. The end points of the elongated edges are then the control points of a quadratic Beziér curve, along with the fact that as one endpoint of a perimeter-halfing slides towards the common intersection, the other end slides away from the intersection by the same distance, which is the same as the convex combination to obtain intermediate points on the control polygon of a Beziér curve that define the tanget to the parabolic arc. These quadratic Beziér curves also appear in String Art

In case of a regular triangle the curve collapses to a point.

  • $\begingroup$ Intriguing possibility, although I cannot say it is clear to me... $\endgroup$ Sep 6, 2017 at 15:40
  • $\begingroup$ @JosephO'Rourke I have added an explanation of why the curves are parabolic arcs; in fact they are quadratic Beziérs. $\endgroup$ Sep 6, 2017 at 18:04
  • $\begingroup$ Nice observation re quadratic Beziérs. Thanks. $\endgroup$ Sep 6, 2017 at 21:41

The answer to Q1 appears to be no.

In your figures above there is a central triangular "cusp" area (but I think in a regular pentagon the cusp would not be triangular). When this is the case, inscribe a circle within the cusp, tangent to its three sides. The perimeter-halving center must be at least as far from some halving chord as the radius of this circle, just to touch the three halving chords to which it is tangent. On the other hand, the center of this circle achieves this distance, because it is exactly this distance from the three tangent chords and closer to all other chords (as they all cross the inscribed circle. Therefore, it gives the location of the perimeter-halvings center. As can be seen in your example triangle, it can be far from the centroid.

However, from Weis' answer that the cusp boundaries are algebraic, it seems that (when there is a triangular cusp whose inscribed circle gives the location of the center) the location of the center is also algebraic.

  • $\begingroup$ "Therefore, it gives the location of the perimeter-halvings center." A convincing argument, David---Nice! $\endgroup$ Sep 6, 2017 at 22:26
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    $\begingroup$ Perhaps in light of your remarks, it would be interesting to specialize to triangles, and see if this algebraic center coincides with any of the 14,143 centers in the Encyclopedia of Triangle Centers. $\endgroup$ Sep 6, 2017 at 22:28
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    $\begingroup$ Incidentally the three tails of the cusps all point directly towards the Nagel point (en.wikipedia.org/wiki/Nagel_point), the point where the three halving chords through the triangle vertices meet. But this also appears in your example to be far from the perimeter-halvings center. $\endgroup$ Sep 7, 2017 at 4:22

I am old enough that I took a separate course in analytical geometry before taking a Calculus course. One of the lovely theorems I learned in analytical geometry was that the locus of points creating a curve where the tangent lines to the curve cut off a fixed area from the coordinate axes (which can be at right angles or a more general angle) is a hyperbola. There is a similar theorem for perimeter which gives the locus curve to be a parabola. For a triangle or more general polygon one can look for the points where several area or perimeter bisectors go through a point. This has some relation to the notion of a measure of symmetry for a convex set. See:

Branko Grünbaum. "Measures of symmetry for convex sets." Proc. Sympos. Pure Math. Vol. 7, 1963, pp. 233-270. MR 27 #6187.


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