sborník

Světlana Tomiczková
follows:
1. Both edges c i k
and ci+1
k+1
are parallel to the edges of the polygon
A and the parallelograms R i and R i+1 have a common edge.
2. The edge c i k
is parallel to the edge of edge of the polygon A and
c j k+1 is parallel to the edge of edge of the polygon B (or cj k ‖ b j
and c i k+1 ‖ a i). Then the parallelogram and the triangle have
a common edge.
3. Both edges c j k
and cj+1
k+1
are parallel to the edges of the polygon
B and then the triangles T j and T j+1 have a common edge.
As the polygon C is convex, the triangles and parallelograms do
not overlap.
We can express the area of the polygon C as the sum of the areas
of the polygons A and B and the areas of the parallelograms R i .
Thus S(C) = S(A) + S(B) + ∑ n
i=1 |a i||v i |.
Remark: Since A s ⊕ B t = (A ⊕ B) s+t , the relation for the computation
of the area of Minkowski sum holds for arbitrarily placed
polygons.
2.1 Area of the Minkovski sum of two convex sets
bounded by closed curves
Theorem 2: (see fig. 3) Let A, B be the convex, bounded and closed
sets in E 2 and C = A ⊕ B.
Let the curve C 1 (t) = (x 1 (t), y 1 (t)), t ∈ I be the boundary of the
set A and the curve C 2 (s) = (x 2 (s), y 2 (s)), s ∈ J be the boundary of
the set B and let the transformation s(t) : I → J of the parametr s be
such that (dx 1 (t), dy 1 (t)) = k(dx 2 (s(t)), dy 2 (s(t))), t ∈ I and k > 0.
Then the area S(C) of the set C is
∫
S(C) = S(A)+S(B)+ ‖(x 2 (s(t)), y 2 (s(t)), 0)×(dx 1 (t), dy 1 (t), 0)‖ dt .
I
(1)
Proof: We know that an extreme point in direction d on the set
C = A ⊕ B is the sum of the extreme points in direction d on the sets
A and B (see [1]).
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