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Inversion Distance (Posted on 2006-02-21) Difficulty: 3 of 5
A circle (of radius a), a line, and a point are mapped by inversion into two concentric circles and the center of those concentric circles. If the distance from the given circle's center to the line is b, then what is the distance from the point to the line?

Inversion Defined:

Let O be the center of a circle of radius k. An inversion with respect to circle O is a mapping f:R2 -> R2 such that for all P in R2 (not O), P' = f(P) lies on ray OP and
|OP'||OP| = k2.

See www.geocities.com/bractals/inv.jpg

for graphical description of inversion.

See The Solution Submitted by Bractals    
Rating: 2.8000 (5 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(3): Note to Bractals | Comment 23 of 32 |
(In reply to re(2): Note to Bractals by Mindrod)

The image world is R^2, the same world we began in.

Let's say we have a standard clock (with the numbers 1-12 in their regular positions). It has a radius of say 5. Imagine we place this standard clock in the cartesian plane centered at the point (13,0). Also let us say that the circle of inversion has a radius 12 centered at the origin. So we see that the point next to the number 9 (at 8,0) maps to the point next to the number 3 (at 18,0) because 8*18=12^2. Also the points where the circle of inversion intersects the edge of the clock [(144/13,60/13) and (144/13,-60/13) i.e. points near the numbers 11 and 7 on my clock respectively] map to themselves. So if 9 and 3 map to each other and 11 and 7 map to themselves, then 12, 1,and 2 map to the arc between 9 and 11 while 4, 5, and 6 map to the arc between 7 and 9. Likewise 8 maps somewhere on the arc between 3 and 7 and 10 maps somewhere on the arc between 11 and 3. So if we say we began with some even distribution of points on the edge of the clock before the mapping, their image would be compressed on the left after the mapping and more sparse on the right.

The point at the center of my circle (at 13,0) maps to (144/13,0) and (144/13,0) maps to (13,0). You see this is directly below the 11 and above the 7.

Now if we add the line x=72/13, we see that because (144*13)/72 = 26 and lines map to circles which include the origin, it also maps to a circle centered at (13,0).

With this circle, line and point, a is 5, b is (13 - 72/13) and the distance from the point to the line is 72/13. 

b^2 - a^2 is (9409 - 4225)/169 or 5184/169 which is indeed (72/13)^2

Edited on February 25, 2006, 3:12 am
  Posted by Eric on 2006-02-25 03:08:26

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