Abstract. Let K 0 be a compact convex subset of the plane R 2 , and assume that K 1 ⊆ R 2 is similar to K 0 , that is, K 1 is the image of K 0 with respect to a similarity transformation R 2 → R 2 . Kira Adaricheva and Madina Bolat have recently proved that if K 0 is a disk and both K 0 and K 1 are included in a triangle with vertices A 0 , A 1 , and A 2 , then there exist a j ∈ {0, 1, 2} and a k ∈ {0, 1} such thatHere we prove that this property characterizes disks among compact convex subsets of the plane. In fact, we prove even more since we replace "similar" by "isometric" (also called "congruent"). Circles are the boundaries of disks, so our result also gives a characterization of circles.
Aim and introductionOur goal. The real plane and the usual convex hull operator on it will be denoted by R 2 and Conv, respectively. That is, for a set X ⊆ R 2 of points, Conv(X) is the smallest convex subset of R 2 that contains X. As usual, if X and Y are subsets of R 2 such that Y = ϕ(X) for a similarity transformation or an isometric transformation ϕ : R 2 → R 2 , then X and Y are similar or isometric (also called congruent), respectively. Disks are convex hulls of circles and circles are boundaries of disks. The singleton subsets of R 2 are both disks and circles. A compact subset of R 2 is a topologically closed and bounded subset. Our aim is to prove the following theorem. Theorem 1.1. If K 0 is a compact convex subset of the plane R 2 , then the following three conditions are equivalent.(i) K 0 is a disk.(ii) For every K 1 ⊆ R 2 and for arbitrary points A 0 , A 1 , A 2 ∈ R 2 , if K 1 is similar to K 0 and both K 0 and K 1 are contained in the triangle Conv({A 0 , A 1 , A 2 }), then there exist a j ∈ {0, 1, 2} and a k ∈ {0, 1} such that(iii) The same as the second condition but "similar" is replaced by "isometric".Our main achievement is that (iii) implies (i). The implication (i) ⇒ (ii) was discovered and proved by Adaricheva and Bolat [3]; see also Czédli [9] for a shorter proof. The implication (ii) ⇒ (iii) is trivial.