# Shape whose translated and scaled copies are closed under intersection

The translated and scaled copies of an equilateral triangle with fixed orientation are closed under intersection - the intersection is again an equilateral triangle with the same orientation.

What other convex shapes in 2D have this property?

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Don't know if these count as a shapes but this also holds for single points, lines and line segments. – Dirk Mar 4 at 19:07
it holds for any triangle not just equilateral ones. – user35593 Mar 4 at 19:19
@Dirk I would prefer shapes with non-zero area. – Mangara Mar 4 at 19:47
@user35593 You're right! I don't know how I overlooked that... – Mangara Mar 4 at 19:48
It seems to be (quite vaguely) related to the concept of generating sets in differential games --- namely, such sets that any (nonempty) their translates' intersection is a Minkowski summand of the set itself; see, e.e., this paper: mathnet.ru/php/… – Ilya Bogdanov Mar 4 at 23:04

## 3 Answers

I think triangles (degenerated ones included) are the only such convex shapes. The idea is: If $C$ is such a convex shape, let $p_i\in\partial C$ for $i=1,2,3$ be three smooth points. One can obtain an approximate triangle as an intersection of three large copies $C$, say $m(C-p_i)+p_i$ for large $m$. Since these intersections are similar to $C$, and converge to a triangle as $m\to\infty$, $C$ itself is a triangle.

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(Note that since ∂C is convex, all points but at most countably many are smooth. Also note, the analogous statement and proof hold true in any dimension ) – Pietro Majer Mar 5 at 6:41

Since the shape $A$ is convex its boundary is differentiable almost everywhere. Take a point $p$ where it is differentiable. A line $l$ is tangent to A in $p$. Take two more points $q_1$ and $q_2$ with tangent lines $l_1$ and $l_2$, translate the shape by vectors $q-q_1$ and $q-q_2$ and intersect results. Suppose we get a shape $B$. The only point which can possibly have a tangent line parallel to $l$ will be $p$, but $\partial B$ is not differentiable there (unless all lines are parallel). Therefore, $B$ is not a homothetic image of $A$ and that means $B$ is at most one-dimensional.

$B$ being one-dimensional means $\angle (l_2, l) + \angle (l, l_1) \le \pi$ (we take angles which "look" at $A$). If there is at least 5 different directions for tangent lines we get a contradiction since in a convex pentagon the sum of angles is $3\pi$ but from our inequality it must be no more than $5\pi/2$. Therefore our shape is either a triangle (for which it is true) or parallelogram (false).

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Because you do not provide a motivation for your question I do not know if the following reference is appropriate; however, you might want to look at George Stiny's book Shape: Talking about Seeing and Doing (Cambridge, Massachusetts: The MIT Press, 2006), in particular his discussion of closure on pp. 285-87 and p. 306.

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