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The problem to solve is this. Imagine a circle. We know two points on the circumference, anchor A and anchor B, they could be anywhere on the circumference of the circle. Draw a line between these two. Pretend this line is a mirror. Obviously it can be different lengths depending on the anchor points. Then a third point C is added to the circle.
Pretend point C emits a beam of light which always starts its journey travelling horizontally across the circle, so if C is 100 degrees, it takes aim at 260 degrees. The beam encounters the mirror between points A & B. It bounces back at an equal angle from the mirror, what is the angle of the point D where it bounces back and hits the circumference. The size of the circle is immaterial. All we know is the exact angles of A B C around the circle, and that C always travels horizontally; we need a formula to calculate the angle of D on the circumference. Thanks!!

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I think I puzzled out what you meant by 100 degrees and 260, but obviously you don't mean "horizontally". It looks like you mean "diametrically". "From C towards the center" would be a much less convoluted way to say it, too. –  rschwieb Sep 12 '12 at 16:46
    
I think he means that $C = (R \sin 100^o, R \cos 100^o)$ and the ray aims at $(R \sin 260^o, R \cos 260^o)$. In ither words, the ray aims from $C=(x,y)$ at $(-x,y)$, aka. horizontally. –  Hagen von Eitzen Sep 12 '12 at 16:47
    
@hagen Yeah I think you're right. In English, anyhow, "horizontally" cannot be used this way. –  rschwieb Sep 12 '12 at 16:53

1 Answer 1

Rotate your center so that the chord formed by the mirror is horizontal.

Compute the slope of the line through the point C on the circumference and the center O of the circle.

Multiplying this slope by -1 yields the reflected slope, if the line is in fact pointed at the mirror.

You will have to discern carefully on your picture whether or not the beam of light is traveling toward your mirror or not. There are configurations where the beam is "going away" from the mirror, or parallel to it. (There are also configurations where the beam goes toward the mirror but meets the circumference first, but I don't know what you intended to happen in that case.)

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