Steinhaus’ Problem on Partition of a Triangle

H. Steinhaus has asked whether inside each acute triangle there is a point from which perpendiculars to the sides divide the triangle into three parts of equal areas. We present two solutions of Steinhaus’ problem. The n-dimensional case of Theorem 1 below was proved in [6], see also [2] and [4, Theorem 2.1, p. 152]. For an earlier mass-partition version of Theorem 1, for bounded convex masses in Rn and r1 = r2 = ... = rn+1, see [7]. Theorem 1 (Kuratowski-Steinhaus). Let T ⊆ R2 be a bounded measurable set, and let |T| be the measure of T. Let α1, α2, α3 be the angles determined by three rays emanating from a point, and let α1 < π, α2 < π, α3 < π. Let r1, r2, r3 be nonnegative numbers such that r1 + r2 + r3 = |T|. Then there exists a translation λ : R2 → R2 such that |λ(T) ∩ α1| = r1, |λ(T) ∩ α2| = r2, |λ(T) ∩ α3| = r3. H. Steinhaus asked ([10], [11]) whether inside each acute triangle there is a point from which perpendiculars to the sides divide the triangle into three parts with equal areas. Long and elementary solutions of Steinhaus’ problem appeared in [8, pp. 101–104], [9, pp. 103–105], [12, pp. 133–138] and [13]. For some acute triangles with rational coordinates of vertices, the point solving Steinhaus’ problem is not constructible with ruler and compass alone, see [15]. Following article [14], we will present two solutions of Steinhaus’ problem.