Take some convex polygon $P$. I'm mostly asking about the unit square, but would also appreciate thoughts on general polygons. We want to take a family of line segments inside $P$ that have the same projections as $P$ (that is, any line intersecting $P$ intersects one of the segments). Let $L$ be the minimum total length of such segments. What is $L$?

I have the following bounds:

Upper bound: $L\le\sqrt{2}+\frac{\sqrt6}{2}\approx 2.639$

Take the following construction such that the three segments meet at angles of $2\pi/3$ (the gray dahsed diagonal is not one of the segments in the construction). The sum of the lengths is $\sqrt{2}+\frac{\sqrt6}{2}$.

Best Example

Lower bound: $L\ge 2$.

The sum of the lengths of orthogonal projections onto the two diagonals is $2\sqrt{2}$. For any segment of length $a$ and angle $\theta$ to the diagonals, the sum of its projections is $a|\sin\theta|+a|\cos\theta|\le a\sqrt{2}$. Thus $L\sqrt{2}\ge 2\sqrt{2}$, hence the lower bound.

I've tried improving the lower bound by using other combinations of projections, or splitting the square into subshapes and using different projections on different subshapes, but couldn't get anything better.


2 Answers 2


This problem has been heavily studied. For example, this paper

Dumitrescu, Adrian, and Minghui Jiang. "The opaque square." In Proceedings of the 13th Annual Symposium on Computational Geometry, p. 529. ACM, 2014. ACM link. Preliminary PDF.

includes an analog of your figure:

          Dumitrescu-Jiang Fig.1 (detail).
and recites the history of the problem. One of their main results is:

Theorem 2. The length of any interior finite segment barrier for a unit square is at least $2 + 10^{−5}$.

This accords with @alesia's post.


Total length $2$ is not possible.

By integral geometry, total length $2$ is equivalent to the fact that almost every line intersecting the square intersects the union of segments $S$ exactly once. But this would mean that for almost any point $p$ in $S$, the supporting line of any segment in $S$ passes through $p$. Which is only possible if $S$ is included in a line, which contradicts the fact that $S$ intersects all lines meeting the square.


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