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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(5): Note to Bractals | Comment 27 of 32 |
(In reply to re(4): Note to Bractals by Dej Mar)

Your principle fallacy comes here: "...the point mapped by inversion is given as the same point, then the original point must be the same as the center point of the circle of the standard clock." - no, the image of the original point must be the center of the clock's image; this is not the same.

You'll notice that I have chosen the position and size of my clock and the circle of inversion so that the image of my clock is itself (with the insides of the clock all twisted and skewed). The line does not map to this circle, but to a concentric circle centered at (13,0) with a radius of 13.

So before the mapping I have a circle of radius 5 centered at (13,0), a point at (144/13,0) and a line at x=72/13 just as the problem requires. After the mapping I have two circles of radius 5 and 13 centered at (13,0) and the point (13,0), just as the problem requires. b, in this case, is 13-(72/13) or 97/13 but the distance from my original point to my line is 72/13.

You'll see that (72/13)^2 = (97/13)^2 - (5)^2 or b^2 - a^2 though.

 


  Posted by Eric on 2006-02-25 14:13:42
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