 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Light Beam Reflection (Posted on 2016-03-01) PQRS and QRTU are two faces of a cube with PQ = 12.

A beam of light emanates from vertex P and reflects off face QRTU at point X, which is 7 units from QU and 5 units from QR. The beam continues to be reflected off the faces of the cube.

The length of the light path from the time it leaves point P until it next reaches a vertex of the cube is given by A√B, where A and B are integers and B is not divisible by the square of any prime.

Find A and B.

 No Solution Yet Submitted by K Sengupta No Rating Comments: ( Back to comment list | You must be logged in to post comments.) solution Comment 1 of 1
The components of the beam's travel when it reaches X are 12, 5 and 7, making that first leg, before reflection have length sqrt(218).

As in a hall of mirrors, we can consider the problem as being one straight line in an infinite array of tessellated cubes.

Each passage through a plane parallel to QRTU has a motion with a 12 component, a 5 component and a 7 component. But also the multiple of 5 and the multiple of 7 must also be a multiple of 12 so that all axes have experienced a multiple of 12 units since leaving P. The numbers 5, 7 and 12 are relatively prime, so the number of segments of motion will be 5*7*12 = 420, each having length sqrt(218).

So the answer is 420 * sqrt(218) ~= 6201.22568529803.

The only fly in the ointment is that the beam will pass through edges (in the actual world, hit a 2-plane corner several times). Ideally it would bounce straight back vis-a-vis the two planes without affecting the results.  Real world mirrors might just absorb light where two mirrors join.

 Posted by Charlie on 2016-03-01 15:22:28 Please log in:

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