I don't know how to go about such questions. It's not exactly my area, so maybe it is stupid, but curiosity is winning.

So I have a piece of bread $P$ of a really non-regular shape (let's make it convex though), and I want to make a sandwich from it, i.e. two pieces of bread with some stuff between them. Obviously I have to cut my piece into two pieces. My knife is straight and I can only make one cut, otherwise the cooking will be too messy. In other words the two pieces that I obtain are the intersections $P^l_1$ and $P^l_2$ of $P$ with half-planes with respect to a certain line $l$.

Then, in order to make a sandwich I have to place $P_1$ above $P_2$ and put stuff in between. The most effective use of bread happens when these pieces are approximately aligned, i.e. only most $P_2$ is covered with $P_1$ and another way around. How to make it that way?

Let's state the questions in the mathematical language.

Let $P$ be a convex body in $\mathbb{R}^{n}$, let $l$ be a hyperplane and let $P^l_1$ and $P^l_2$ be the pieces in which $l$ cuts $P$. Let $\lambda_n$ be the usual volume in $\mathbb{R}^{n}$.

The quantity we are interested in is $$\sup\{ \frac{\lambda_n(P^l_1\bigcap f(P^l_2))}{\lambda_n(P)},f\mbox{ - isometry of }\mathbb{R}^{n},~l\mbox{ - hyperplane}\}.$$

Q1: Is it bounded from below by some absolute constant? EDIT: As suggested by UriBader John's inscribed ellipsoid provides with an estimate from below with an absolute constant that depends on $n$. Thus, I am leaving the stronger version of the question, i.e. if there is an absolute constant that does not depend on the dimension.

Q2: Is there an algorithm to construct an optimal, or almost optimal $l$ and $f$?

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    $\begingroup$ John Elipsoid theorem states that $P$ contains a large ellipsoid. This will give you a lower bound by an absolute constant (depending on the dimension). Bon appetit. $\endgroup$
    – Uri Bader
    Feb 23, 2017 at 10:05
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    $\begingroup$ Many people eat open sandwiches. $\endgroup$
    – user50229
    Feb 23, 2017 at 10:06
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    $\begingroup$ @UriBader: Another bound, better than that produced by the John ellipsoid, is obtained by intersecting $P$ with its reflection in a suitable point in $P$ yielding the maximum-volume centrally symmetric convex subset of $P$. This (common) bound still depends on the dimension. $\endgroup$ Feb 23, 2017 at 17:47
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    $\begingroup$ I applaud this question with one hand. $\endgroup$
    – Marty
    Feb 23, 2017 at 22:06
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    $\begingroup$ Why don't you cut it into two thinner slices of bread? (I know, I know). $\endgroup$ Jun 21, 2018 at 9:55

2 Answers 2


This won't be a complete answer to Q2, but something of a starting point, at least for all two-dimensional convex $P$. Assuming for simplicity that the area of $P$ is 1, $P$ contains a parallelogram of area at least $1/2$ (triangular $P$-s are extreme in this respect). This is quite easy to prove, and for polygonal $P$, an algorithm can be produced to find such a parallelogram. Cut $P$ along one of the parallelogram's diagonals, and put the two pieces together so that the two (congruent) halves of the parallelogram fit together. This way the common area between the pieces is at least $1/4$. I am sure this is not the best constant for the planar $P$, and by some elaboration on this technique a larger constant can be obtained. In fact, an old theorem of Kovner and Besicovitch states that every planar convex $P$ of area $1$ contains a centrally symmetric convex set of area at least $2/3$, which implies that the constant in question, for planar $P$, is at least $1/3$.

To this answer I would add narrower question, restricted to the special case when $P$ is triangular:

Is there a constant $c>1/3$ such that each triangle of area $1$ can be cut into two pieces, producing a sandwich measuring at least $c$ in the common area?

This already seems to be non-trivial.

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    $\begingroup$ Estimating maximum-area of axially symmetric (along with centrally symmetric) subsets of $P$ may lead to some better bounds, at least for the plane. It would be already interesting to find the best constant just for triangular $P$. $\endgroup$ Mar 7, 2017 at 17:14

Restricting to the case where $P$ is triangular, if the angles of the triangle are $A$, $B$, $C$ and you fold the triangle along the bisector to the angle $C$, then assuming wlog that $A<B$, you obtain an overlapping area whose ratio to that of the full triangle is:

$$r(A,B)=\frac{\sin A}{\sin A + \sin B}$$

Choosing to fold along the bisector of the angle represents a local maximum for the overlapping area, among all choices of folding lines that pass through the same vertex.

You can then choose which vertex to bisect, of the three, as the one that yields the largest value of $r(A,B)$:


A lower bound for $r_{max}$ occurs when two of the vertex choices give the same result, and this lower bound gets smaller in the limit that the two angles $A, B$ are both small. Set $B = f A$ and express the equality of two solutions with the function:

$$P(f,A) = r(A, f A)-r(f A, \pi-A-f A)\\ = \frac{\sin A}{\sin A + \sin (f A)} - \frac{\sin (f A)}{\sin (f A) + \sin (\pi-A-f A)}$$

To first order in $A$, this becomes:

$$P(f,A) \approx \frac{1}{1+f} - \frac{f}{1+2f}$$

This becomes zero (i.e. the two ratios for the different choice of vertex match) when:

$$f = \Phi= \frac{1+\sqrt{5}}{2}$$

So $B/A$ is equal to the golden ratio. Inserting this value for $f$ back into either area ratio gives a lower bound of:

$$r_{min}=\frac{1}{1+\Phi}\approx 0.381966$$

Edited to add:

A slight improvement can be achieved by allowing the additional possibility of folding along the perpendicular bisector to any of the sides. If the two angles at the endpoints of the side whose bisector is used are $A$ and $B$ and $A \lt B$, then the area ratio is:

$$r_{p}(A,B)=\frac{1}{4} (1 + \tan A \cot B)$$

An approach that included only these perpendicular bisectors would do terribly, but adding three more possibilities due to $r_p$ to the original three used to calculate $r_{max}$ yields a slight improvement.

In this case, the lower bound is found by setting:

$$P(f,A) = r_p(A, f A)-r(f A, \pi-A-f A)\\ = \frac{1}{4} (1 + \tan A \cot f A) - \frac{\sin (f A)}{\sin (f A) + \sin (\pi-A-f A)}$$

To first order in $A$, this becomes:

$$P(f,A) \approx \frac{1+f}{4 f} - \frac{f}{1+2f}$$

This is zero when:

$$f = \frac{3+\sqrt{17}}{4} \approx 1.78078$$

and the ratios here are:

$$r_{min} = \frac{\sqrt{17}-1}{8} \approx 0.390388$$

Edited to add:

There is a further improvement possible. While using the bisector of an angle as a folding line yields a local maximum for the overlapping area (among all lines passing through a given vertex), using the perpendicular bisector of a side is not a local maximum (among all lines perpendicular to that side).

When the folding line is displaced from the midpoint of the side to lie further away from the smaller of the angles incident on that side, the portion of the original triangle on the small-angle side of the folding line will, after folding, have a non-overlapping piece that needs to be excised, leading to a quadrilateral overlapping region. But the increasing base and altitude of the triangle more than compensate for the excision, up until a scaled position for the folding line of:

$$\alpha(A,B) = \frac{2}{3+\tan A \cot B}$$

where as usual we assume $A\lt B$, and $\alpha$ ranges from $0$ to $1$ as the point where the folding line crosses the sides moves from angle $A$ to angle $B$. At $\alpha(A,B)$, the ratio for the overlapping area has its peak of:

$$r_{dp}(A,B) = 1 - \frac{2}{3+\tan A \cot B}$$

In principle, we might worry that this optimal folding line lies past the vertex of the original triangle, complicating the geometry, but that turns out to be impossible. So one of these displaced-perpendicular folding lines can be used whenever the $r_{dp}$ for a side gives a better result than all the angle-bisectors.

The new lower bound can be found by the same kind of methods used previously, in the limit where angles $A$ and $B$ both get smaller but maintain a fixed ratio, $B = f A$. Using $r_{dp}$ rather than $r_p$ in the previous calculations, we get:

$$f = 1 + \sqrt{2}$$

and a new lower bound of:

$$r_{min} = \sqrt{2} - 1 \approx 0.414214$$

The figure below shows how the three approaches described (folding along angle bisectors, along perpendicular bisectors to sides, and along the optimal displaced perpendicular) can allow successively greater overlapping areas, for one particular triangle. But for other triangles, folding along the angle bisectors will give the best result.

Different ways of folding the same triangle

  • $\begingroup$ Super! Is this bound the best possible for triangles? I expect you could obtain the same bound for quadrilaterals, perhaps even for all convex $P$. $\endgroup$ Jun 21, 2018 at 15:20
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    $\begingroup$ It turns out this can be improved upon (very slightly) if you also allow folding along perpendicular bisectors to the sides of the triangle. I’ve edited the answer to include this. $\endgroup$
    – Greg Egan
    Jun 21, 2018 at 23:10
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    $\begingroup$ There’s another improvement if you allow the folding lines perpendicular to the triangle sides to be displaced from perp. bisectors. When these folding lines are used, the overlapping area is a quadrilateral. This raises the lower bound to $\sqrt{2}-1 \approx 0.414214$. $\endgroup$
    – Greg Egan
    Jun 22, 2018 at 5:16

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