Let us recall a well-known school task: In the (Euclidean E2) plane of a triangle ABC we draw regular triangles outward on sides of ABC, say ABC‾, BCA‾, CAB‾, respectively. Prove that the segments AA‾, BB‾, CC‾ intersect each other in a point K, that is the isogonal point of ABC and the distance sum AK + BK + CK is minimal for K among all points of the plane.Professor Kárteszi noticed that instead of regular triangles we can draw isosceles ones with all equal base angles, and the above K (called Kárteszi point) exists also in the Bolyai–Lobachevsky hyperbolic plane H2 (in the sphere S2 as well, (see also Kálmán, 1989 and Sect. 2), the orthocentre, barycentre are specific cases. There is a more general extremum problem of (Yaglom, 1968, problem 83, with modified notation):In the plane (E2) of a given triangle ABC find a point K such that the quantity αKA + βKB + γKC, where α, β, γ are given positive numbers, has the smallest possible value. This problem leads to a more general triangle configuration and to an analogous extremal point K. Moreover, as a new result of this paper, an extension onto "absolute plane" (S2, E2, H2, M2 Minkowski plane, G2 Galilei (or isotropic) plane) can be formulated and solved by three reflections theorem (see e.g. Molnár, 1978 and Sect. 4, Weiss, 2018), and geometric (Grassmann–Clifford type) algebra (Perwass et al., 2004 and Sect. 3). Open problems arise as well.