# Cutting segments from convex objects

Let $$X$$ be some convex object in the plane. For every pair of points $$A,B$$ that are not in $$X$$, and the distance between them is at most $$1$$, we remove from $$X$$ its intersection with the segment $$AB$$. For example, in the picture below, $$X$$ is the green circle, and if the distance between $$A$$ and $$B$$ is at most $$1$$, then the entire black segment is removed from $$X$$:

Note that we first find all A,B with distance at most $$1$$ and define the to-be-removed segments, and then remove them all at once. Denote the new object (after the removals) by $$Y$$.

I am trying to understand how $$Y$$ looks like.

If $$X$$ is a disc (with diameter larger than $$1$$), then it is easy to see that $$Y$$ is a disc too: it has the same center as $$X$$ and a smaller radius.

MY QUESTION: if $$X$$ is an arbitrary convex object, is $$Y$$ convex too?

• Is this distinct from the following? "For a given convex object $C$, let $S$ denote the set of all chords of $C$ with length $1$. Find the envelope of all elements of $S$." I think this reformulation works, as long as $C$ is convex, and (for me at least) it's a bit easier to work with. Jul 6, 2023 at 11:20
• Assuming you don't cut away slice by slice (you would end up with the empty set) but first find all all such $A, B$ and define the to-be-removed segment, and then remove them all at once, the result would be convex. It's basically an intersection of convex shapes (the cutting away of a segment is just the intersection with the corresponding half-plane). Jul 6, 2023 at 11:20
• @fweth is right - intersecting convex regions means $D$ will be convex. Are you interested in the actual shape of $D$ too? Jul 6, 2023 at 11:22
• Note that it's not clear to me how you decide which side of the line you cut away, e.g. if the line cuts the shape in half. But basically you can take any set of half-planes and intersect them, whatever the process of describing the set. Jul 6, 2023 at 11:27
• Good point - my reformulation falls apart when the shape is too small. But I suppose those shapes are entirely cut away? Jul 6, 2023 at 11:48

I'll consider closed sets only, as you mentioned in your comment. Please let us denote regions with $$X,Y,Z,\dots$$ and reserve $$A,B,C,\dots$$ for points, especially let $$X$$ be the total region. If $$[A,B]$$ is a segment as described in your question, we can assume w.l.o.g. that removing it dissects $$X$$ into two parts $$Y,Z$$ (the other case is easy). My claim is that after removing all segments of your kind, either $$Y$$ or $$Z$$ vanishes. Assume not, then you have points $$y\in Y,z\in Z$$ which are not part of any such segment. In particular you can choose segments $$[C,D]\subseteq Y$$ containing $$y$$ and $$[E,F]\subseteq Z$$ containing $$z$$, both of length $$1$$ and paralell to $$[A,B]$$. Since $$X$$ is convex and $$C,D,E,F\in X$$, the segments $$[C,E]$$ and $$[D,F]$$ must be fully contained in $$X$$ as well. But also at least one of those segments must intersect with the line through $$A,B$$ outside of $$(A,B):=[A,B]\setminus\{A,B\}$$, a contradiction, as no such point is contained in $$X$$.
Hence for each line segment $$[A,B]$$ you remove, there is a set of line segments whose removal includes all points to the left or all points to the right of $$[A,B]$$. So removing all your line segments is equivalent to intersecting with a set of open half-planes, which yields a convex set. It can also be checked that the set is closed.
• I do not understand why "at least one of those segments must intersect with the line through $A,B$ outside of $(A,B)$". See this picture: i.stack.imgur.com/Rrlgk.png where both [C,E] and [D,F] intersect with $(A,B)$. Jul 7, 2023 at 10:39
• Length $1$ will do, I think. Jul 7, 2023 at 10:58
• Thanks! Based on your proof, is there a simpler way to describe the entire process of removal? I thought of "remove from $X$ the entire halfplane at one side of the line through $AB$", but it is not clear how to define that "one side". Jul 8, 2023 at 18:58