Help - Finding 3rd Point on Circle

[astro_girl_690]

If we had two points in degrees on a circle and we wanted to derive an unknown 3rd point in degrees could we do it if we knew that the third point had a known angle to the other two points?

So the three points together make a triangle, as an example ;

Point (1) is on the circle at 30 degrees

Point (2) is on the circle at 90 degrees

Point (3) exists somewhere on the circle and we know that it has a 75 degree internal angle to Point (1) and (2), how can we determine at what degree on the circle point (3) is situated at?

[jemidiah]

Nope. As it turns out, the internal angle at point 3 of the triangle in question has two possible values, +/- (a2-a1)/2 where a1 = 30 degrees in your example and a2 = 90 degrees in your example. So most of the time your third angle won't even correspond to a point on the circle, and even when it does, so do an infinite number of other points.

The proof is somewhat complex to say in text, and my pictures wouldn't be very good. You can do the following algebraically if you want: define a1, a2, and a3 to be the reference angles of the three points in question; define pi = (cos ai, sin ai) to be the points in question, taken to be on the unit circle for convenience; define v1 = p1 - p3, v2 = p2 - p3 to be vectors pointing from p3 to each of the other two points; take the identity v1 dot v2 / (|v1|*|v2|) = cos b find the internal angle, b, between the two vectors, which is the third angle you gave above; square this equation on both sides; apply a *whole bunch* of trig substitutions, notice that the a3's wonderfully cancel, and you'll end up with cos^2 ((a1-a2)/2) = cos^2 (b), which gives b = +/- (a1-a2)/2.

An alternate pseudoproof is more geometric: take the unit circle and plot the first two points wherever you want, but choose a1>=a2 by flipping the first two points if you need to. Place the third point in such a way that you make an isosceles triangle. The angle generated can be seen to be (a1-a2)/2; draw it out with a1=pi/2, a2=0 for an example. Now move the third point to be very, very close to p2 (but not touching--if it's touching, we don't have a triangle and things break). You'll see that the angle doesn't vanish at all. Draw tangent lines tangent to the circle at the first two points, draw radii from the origin to those two points, and you'll see you have a diamond. Draw a secant line between the two points to cut the diamond in half. From here, apply a bunch of triangle identities and you can find the angle you want--which will be between the tangent at p2 and the secant. This will be as expected, (a1-a2)/2. Personally I would expect any difference in the internal angle to be greatest right next to one of the original points; if that's the case, then the internal angle is probably constant across movement of p3. If we place p3 between p1 and p2, then we'll get -(a1-a2)/2, accounting for the +/- above.

I'm curious, what's this problem for?

[opus]

Maybe I'm missreading the problem, but it sounds to me like the
Thales Theorem.

You known points in degrees form a secant to the circle, and you want to contruct a triangle over or under this line having the third point on the perimeter of the circle as well. You will find that all possible triangles above that line will have the same angle in that point, the same is true for the triangles below that line, but the angle is different. If you are using the smaller part of the circle, the angle will be less then 90 degrees and more then 90 on the other side.
Sorry, but I don't know which theorem that would be.

[jemidiah]

You're definitely on the right track. It seems my result above is most of the Inscribed angle theorem. The proof Wikipedia gives seems entirely geometric, though I haven't gone through it and drawn it out myself.

[opus]

Thanks for given me the name of that Theorem.