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)?

(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.