The Billiard Problem can not be solved within the confines of compass and ruler construction. However, approximations due provide some insight into fundamental principles of math and geometry.
What follows is the geometric proof supporting the Billiard Problem Approximation constructed with the GeoGebra application.
Al Hazan’s Billiard Problem Approximation Solution (Edward A. Chesky JR), Major U.S.A, (RET)
Given two points on a circle draw an isosceles triangle with the lines joining these two points as base and a point on the circle as apex.
Construction:
Given the radius of a circle C with radius r.
Points A and B are given.
With A as center draw circle C1 with radius r. This would cut the given circle at points A1 and A2.
Select the point, which makes B lie in the arc.
That is select A1 so that A1 B A is an arc of the given circle.
Draw circle C2 with A1 as center and radius r. This circle passes through the center of C.
[Proof: A1 is on C and thus the distance between the center of C and A1 is r. Hence the circle with A1 as center and radius r passes through the center of C.]
Draw another circle C3 with B as center and radius r. This circle cuts circle C at B1 and B2.
Select B1 such that B1 A B is an arc of the circle.
Draw a circle C4 with B1 as center and r as radius. This circle also passes through the center of C.
Call the center of C as point X.
Call the intersection of circles C2 and C4 other than X as Y.
The quadrilateral AXBY has sides AX, XB, BY, YA of length r.
AB is a diagonal.
XY is the other diagonal.
Let the intersection of AB and XY be Z.
Consider the triangles AXY and BXY. These are mirror image of each other along XY as AX = BX = r
and AY = BY = r. XY is common. Hence angle AXY = angle BXY.
The triangles AXZ and BXZ are congruent. AX = BX = r; Angle AXZ = Angle BXZ; side XZ is common.
Therefore AZ = BZ. Z is a mid point of AB. Angle AZX = Angle BZX = 90 degrees.
Thus the line XZ is a perpendicular bisector of the line AB.
Extend the line XZ to intersect C at T1 and T2.
The triangle A T1 B is an isosceles triangle. So also the triangle A T2 B.
QED.
