All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Shapes
Bull's eye! (Posted on 2003-11-22) Difficulty: 3 of 5
Two points have polar coordinates as follows: θ=130,r=.35 (point A) and θ=70,r=.6 (point B). There is a surrounding circle, r=1, that acts as a mirror, and you wish to send a light ray from point A to point B by bouncing it once off the circle. What two alternative directions could you send it in (use an angular measure paralleling the θ coordinate it would have if directed from the origin)?

No Solution Yet Submitted by Antonio    
Rating: 3.6000 (5 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
first thoughts | Comment 1 of 16
My first line of thought would be to construct two quadrilaterals within our unit circle. Let us call them AOBR1 and AOBR2 where points A and B are as described above, point O is the origin of the circle and points R1 and R2 are the two points of reflection that our light rays are bouncing off with R1 at the top of the circle and R2 at the bottom. Now note that angle AR1O = OR1B and AR2O = OR2B.
Solving the top quadrilateral first, and introducing some variables(say x and y), let us define the following angles:
AOB = 60
AR1O = OR1B = x
AOR1 = y
R1OB = 60-y
R1AO = 180-x-y
R1BO = 120+y-x
By the law of sines on both triangles OAR1 and OR1B we see that sin x = .35sin(180-x-y) = .6sin(120+y-x).
Also by the law of cosines we see that the distance AB is √.2725.

I hope this helps
Edited on November 22, 2003, 12:46 pm
  Posted by Eric on 2003-11-22 12:41:16
Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (1)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information