Visualising the Minkowski sum of two triangles

Question

Consider the figure given below. I am told that the Minkowski sum of the two triangles $A,B \subset \mathbb{R}^2$ is the regular hexagon, as shown right next to them.

The Minkowski sum is defined as $$A+ B = \{a+b: a\in A, b\in B\}$$

Note that $A$ and $B$ are such that $A$ is the reflection of $B$ about the origin (and vice versa). So, knowing that the sum is a regular hexagon, I can somewhat make sense of it, but I'm not really able to see why it is a regular hexagon in the first place! Is there an intuitive way of looking at this? Finding $A+ B = \{a+b: a\in A, b\in B\}$ manually using every $a\in A, b\in B$ is obviously impossible, so there must be some special $a_1,a_2,...,a_p \in A$ and $b_1, b_2,...,b_q \in B$ we should use to construct the figure (extreme cases of some sort). Which ones are these?

As an addendum, I am wondering if there is some general intuition (at least in $\mathbb{R}$,$\mathbb{R}^2$, and $\mathbb{R}^3$) I can carry with me, when finding the Minkowski sum of two figures (sets)?

P.S.
I am not looking for algorithms to find Minkowski sums.

Answer 1

Here is a GIF I made that might help you visualize the Minkowski sum; the animation shows translates of the downward-facing triangle by points on the upward-facing triangle as they cycle around the perimeter, and how those translates fit onto the hexagon.

             

Of course, the Minkowski sum also includes translates of the downward-facing triangle by interior points of the other triangle, so this picture is not quite complete - if the second triangle were larger, the red triangles in the animation would only trace out the outer part of the resulting hexagon.

In your notation, we're looking at translates of $a$ in the boundary of $A$ by $b$ in the boundary of $B$, where each frame shows a single value of $b$. When the shapes are convex, the Minkowski sum is given by taking the convex hull of these "extreme" sums. In fact, in the case of convex polygons (or polyhedra, in the 3-dimensional case), it suffices to take just the vertices of $A$ and $B$.