Perimiter and area of Minkowski sums Prove that perimiter of A+B is equal to the sum of perimiters of A and B, where + denotes Minkowski sum(or vector sum) and A, B are convex figures in a plane.
I was originally trying to prove this where one figure was a circle, but I figured that it shouldn't really make a difference. What I actually need is to find the area of A+B in terms of perimiters of A,B and their areas. I found the result, but I think i can prove the special case(where one figure is a circle) when I know the perimiter of their sum.
 A: Without loss of generality, we can assume that the shapes have smooth boundaries and no flat edges. Polygons and other convex shapes that don't match this description can be approximated arbitrarily well by ones that do.
If a convex shape has a smooth boundary and no flat edges, we can define a differentiable function $\vec v(\theta)$ that maps directions in the plane to the point on the shape furthest in that direction. Further, for any two such shapes with maps $\vec v_1(\theta), \vec v_2(\theta)$, their Minkowski sum has the map $\vec v_{1+2}(\theta) = \vec v_1(\theta) + \vec v_2(\theta)$. (Logically, to get the point farthest in direction $\theta$ in the Minkowski sum, we want to add together the points farthest in $\theta$ from shapes 1 and 2. This is true in general, we only require convexity and smoothness and no flat edges to make sure that $\vec v_1(\theta)$ and $\vec v_2(\theta)$ are differentiable.) We can compute the perimeter of a shape $s$ to be:
$$
P_s = \int_0^{2\pi}\left|\frac{d\vec v_s(\theta)}{d\theta}\right| d\theta
$$
So:
$$
P_{1+2} = \int_0^{2\pi}\left|\frac{d\vec v_1(\theta)}{d\theta}+\frac{d\vec v_2(\theta)}{d\theta}\right| d\theta
$$
Now, because $\vec v_1(\theta)$ is the point farthest in the $\theta$ direction, and the function $\vec v_1$ is differentiable, we must have that $\frac{d\vec v_1(\theta)}{d\theta}$ is perpendicular to the direction $\theta$. (If it weren't, we could walk along the edge of the shape a small distance, and get even farther in the direction $\theta$, contradicting the definition that $\vec v_1(\theta)$ is the farthest point in that direction.) A similar argument applies for $\vec v_2$.
So for a given value of $\theta$, $\frac{d\vec v_1(\theta)}{d\theta}$ and $\frac{d\vec v_2(\theta)}{d\theta}$ are both perpendicular to the direction given by $\theta$, and so must be pointing in the same direction as each other. Thus:
$$
\left|\frac{d\vec v_1(\theta)}{d\theta}+\frac{d\vec v_2(\theta)}{d\theta}\right| = \left|\frac{d\vec v_1(\theta)}{d\theta}\right|+\left|\frac{d\vec v_2(\theta)}{d\theta}\right| 
$$
So:
$$
P_{1+2}=\int_0^{2\pi}\left(\left|\frac{d\vec v_1(\theta)}{d\theta}\right|+\left|\frac{d\vec v_2(\theta)}{d\theta}\right| \right)d\theta = \int_0^{2\pi}\left|\frac{d\vec v_1(\theta)}{d\theta}\right|d\theta+\int_0^{2\pi}\left|\frac{d\vec v_2(\theta)}{d\theta}\right| d\theta = P_1+P_2
$$
EDIT: I see you're also looking for an area formula. It's not possible to compute the area of the Minkowski sum only in terms of the area and perimeters of the two shapes. Consider the shapes being identical long thin rectangles as an example. If the rectangles are pointed the same way, the sum will have a small area, while if one is rotated by a right angle relative to the other, the sum will have a much larger area. But the component shapes have the same perimeter and area in both cases.
