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?

Just playing around with Geometer's sketchpad.  The value of the tan(BAD) seems to be able to take on any value from 0 to about .42

The former when tan(BAC) is 0 the latter when tan(BAC) is around .61

I don't see any connection with the way tan(BAD) changes and any other quantity.  Am I missing something?

Oops!  I am.  I wasn't forcing the ball to end up at B.

Playing some more.   tan(BAD) is about .58 and tan(BAC) is around 1/3

I'll try for a proof, but I don't know if I can.

Posted by Jer on 2005-05-12 18:44:13