Weber problem that, at the optimum, the angles between the AD, BD, and CD straight lines must be equal to 360° / 3 = 120°. Torricelli deduced from that conclusion that: 1– if any triangle ABD, whose ∠ADB angle is equal to 120°, generates an ABDE convex quadrilateral inscribed in a circle, the ∠ABE angle of the ABE triangle must be equal to (180° − 120°)= 60°; 2– one way to determine the set of locations of D for which the ∠ADB angle is equal to 120° is to draw an equilateral ABE triangle (because each angle of an equilateral triangle is
Schwarz triangle function of the upper half plane. To see how to handle the case when an angle of Δ is a right angle, note that the inequality implies that if one of the angles is a right angle, say "a" = 2, then both "b" and "c" are greater than 2 and one of them, "b" say, must be greater than 3. In this case, reflecting the triangle across the side AB gives an isosceles hyperbolic triangle with angles /"c", /"c" and 2/"b". If 2/"b" ≤ /3, i.e. "b" is greater than 5, then all the angles of the doubled triangle are