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Oh, mighty LJ Brain: Geometry Quiz! - Elf M. Sternberg
elfs
elfs
Oh, mighty LJ Brain: Geometry Quiz!
Here's a very straightforward geometry problem, which I'm having trouble braining about:

I have two circles, C1, C2, both centered on the origin, but of radii r1and r2. I have a line, A that intersects both the origin and a point on on C1 defined by (r1θ). I have another point on C1 defined by (r1θi), and a line B that is parallel to A and intersects (r1θi).

Where does B intersect C2?

Current Mood: tired tired

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Comments
blaisepascal From: blaisepascal Date: August 24th, 2011 10:20 pm (UTC) (Link)
The ith point intersects C1 at (x,y)i = (r1 cos θi, r1 sin θi).

The slopes of A and B are both equal to tan θ. Therefore you can use the point-slope formula for a line y = tan θ(x-xi) + yi to get an equation for line B.

The equation defining C2 is x2+y2 = r22

You can take the first equation, substitute into the equation for C2, and get a quadratic equation in x, which can then be solved to get the 0-2 x values of the intersections.
elfs From: elfs Date: August 24th, 2011 10:22 pm (UTC) (Link)
That's what I couldn't remember, the tanθ. Thanks.
solarbird From: solarbird Date: August 24th, 2011 10:46 pm (UTC) (Link)
Dammit, I took too long and drew a diagramme. XD
elfs From: elfs Date: August 24th, 2011 10:48 pm (UTC) (Link)
I think I've got it now. Cool. I can fix a visual bug in my personal project where two thick arcs with a narrow gap between them don't seem to meet up. With this, I can create an artificial point and make the interstices perfectly rectangular.

All this geometry for a tiny visual hack.
herewiss13 From: herewiss13 Date: August 24th, 2011 10:25 pm (UTC) (Link)
At first I thought vector addition would work, but it didn't. Then, abandoning my reply and going back to my friendslist, I see you've got now a diagram, which only looks vaguely like my own mental image and scrawled diagrams. I _think_ you can use triangles. Draw the line between the origin and the point that B intersects on C1 to create a triangle and then...the further geometrical proof is too large for me to contain in this text field. :-P

...seriously, I noodled around with some scratch paper 3 with about three different theories and got nowhere. _Close_, but still nowhere. Clearly my geometrical skills have atrophied severely since high school.
elfs From: elfs Date: August 24th, 2011 10:46 pm (UTC) (Link)
Mine, too. Blaise pointed out that the slopes of the two lines are the same, that the slope can be derived from the tangent of the angle, that at least one point on the non-radial line is known, so the equation for the line can be derived. We can the solve for (x, y) by using C2's cartesian equation rather than the radial.
shockwave77598 From: shockwave77598 Date: August 25th, 2011 07:41 pm (UTC) (Link)
Drawing the diagram, line B being Parallel with line A means it crosses C2 either in two points, one point when B is r2-r1 away from A, or no points at all. Most cases it'll cross C2 at two points.

elfs From: elfs Date: August 25th, 2011 07:51 pm (UTC) (Link)
I should have stipulated that in all cases, the perpendicular distance between A and B will be less than r1. It will always cross C2 at two points.
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