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 Bull's eye! (Posted on 2003-11-22)
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)

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 re(2): solution (solved numerically) -- verified | Comment 9 of 16 |
(In reply to re: solution (solved numerically) -- verified by Charlie)

Charlie, to address your query about the program I used... I used a very valuable tool in these problems.... Excel's solver :-)
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Not so easy to post an Excel spreadsheet here, but perhaps it will be useful if I explain some relationships (and each of you can reproduce this in your own Excel worksheets).

(all angular measures below are in degrees, which is kinda annoying since Excel's functions are all in radians)

Ax = .35 * cos (130) ~ -0.22497566
Ay = .35 * sin (130) ~ 0.268115566
Bx = .60 * cos (70) ~ .20521209
By = .60 * cos (70) ~ .56381557

Now, let's imagine a point P that moves around the circle. This can be described, classically, as:
Px = cos (theta)
Py = sin (theta)
where theta is the angular measure of the point where 0 degrees is at (1,0), 90 degrees at (0,1), 180 degrees at (-1,0), etc...

Then the angle, theta2, (you gotta remember your trigonometry) from point A to point P is given by:

theta2 = atan[ ( sin(theta) - .35*sin(130) ) / ( cos(theta) - .35*cos(130) ) ]

But that angle is wider than what we're after... so we need to SUBTRACT the angle of point P (from the origin's perspective (90-theta).

This will give us the 'angle of incidence' from the point A as a function of the location of point P.
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Now we do a similar thing with Point B.

the angle to point P from Point B is given by:
theta3 = atan[ ( sin(theta) - .6*sin(70) ) / (cos(theta) - .6 *cos(130) ) ]
but, again we need to subtract that from the angle of P so we have the angle of incidence (on the other side of the radial line to point P).
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Now, I 'moved point P' along the circle, until theta2 = theta 3. (I used solver to move P until the difference (theta2 - theta3) equalled zero.)

Then, I performed a similar process to identify the point P on 'the other side' of the circle.
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Whew... all that said, I'd still like to see someone's analytic solution.
 Posted by SilverKnight on 2003-11-23 21:18:05

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