We have a billiard table in the shape of a right triangle ABC with B the right angle. A cue ball is struck at vertex A and bounces off sides BC, AC, AB, and AC at points D, E, F, and G respectively ending up at vertex B. Assume the angle of incidence equals the angle of reflection at each bounce. If the path segments AD, EF, and GB are concurrent, then what is the tangent of angle BAD?
The best way to solve reflection problems is usually to set up mirrorimage triangles (or other shapes for other problems) with the reflecting line as the axis of reflection.
For clarity of exposition, consider the base as AB with BC vertical. The first reflection is on BC and so a reflection of the triangle is made by placing A' an equal distance on the other side of B from A, along the extension of AB. The next reflection is off AC, which now is represented by A'C, so reflect point B into B' about AC, remembering that angle CA'B' is to be equal to CAB, and not its complement (the mistake I first made).
Then C' is the reflection of C about A'B'. Finally B'' is the reflection of B' about A'C'. Draw line AB'' and label the intersections in turn D, E, F, G.
Now the mathematical problem is to arrange this so that when ultimately reflected back to the original triangle ABC, segments EF and GB'' are concurrent with AD.
Edited on May 12, 2005, 7:13 pm

Posted by Charlie
on 20050512 19:12:51 