Alhazen’s hyperbolic billiard problem

Abstract

Given two points inside a circle in the hyperbolic plane, we study the problem of finding an isosceles triangle inscribed in the circle so that the two points belong to distinct congruent sides. By means of a reduction to the corresponding result in Euclidean geometry, we prove that this problem cannot generally be solved with hyperbolic ruler and compass. In his treatise on optics, written in Arabic, the scientist and mathematician Abu Ali al-H . asan ibn al-Haytham (965‐1039) posed the problem of finding the light path between a source and an observer by way of a fixed spherical mirror, and gave a geometric solution for it. The problem may have been formulated much earlier, by the great Greek mathematicians of the Hellenistic era, but no surviving testimony confirms this. Thus it is fit that it carries al-H . asan ibn al-Haytham’s name, which was rendered as Alhazen in the Latin translation of his book — a document that played an important role in the development of modern science. Alhazen recognized that the problem is essentially two-dimensional — the path must lie in a plane determined by the center of the sphere, the source and the observer. His solution is long, in part because he is actually studying a more general problem; see [Sabra 1982] for details. It is not a ruler-and-compass construction, as it requires an auxiliary hyperbola; in fact, apart from special cases, the problem turns out not to be solvable with ruler and compass alone, though it seems this was only proved some 50 years ago ([Elkin 1965]; see also [Riede 1989; Neumann 1998]). In this paper, we study the hyperbolic version of Alhazen’s problem and relate it to its classical Euclidean counterpart. We use the following formulation of the problem: Given a circle (in the Euclidean or the hyperbolic plane) and two points A and B inside it, construct an inscribed, isosceles triangle with A on one equal leg and B on the other. The isosceles condition is equivalent to the condition that the two legs meet at equal angles the diameter of the circle that goes through their common vertex, so this is Alhazen’s problem all right. (One can also imagine a round billiard table