All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Shapes > Geometry
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: Note to Bractals | Comment 19 of 32 |
(In reply to Note to Bractals by Mindrod)

You must be reading it wrong; Bractals' images are right.

When we say the circle maps to itself, we mean the set of points which make up the edge of the circle all map to other places on the surface of that circle. It says nothing of the points on the interior of the circle. Those points inside the magenta circle which are outside the black circle will map to points inside both circles and vice versa.

The easiest way to check that Bractals' image is right is to note that the intercections of any circle which maps to itself with the circle of inversion must be perpendicular to the circle of inversion. The only way to do this and have the center on the circle of inversion is to have a circle of radius 0. Obviously, if these circles intersect in non-perpendiculars, there would be a point on one side of the inversion circle which wouldn't have a corresponding image across the circle of inversion from it remember all points and their images are collinear with the origin.

To put it another way, if I were a point wandering around in R^2 when I approach the circle of inversion I should see my image just on the other side. No different from a curved mirror.


  Posted by Eric on 2006-02-23 23:34:25
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (14)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information