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Circular map (Posted on 2006-11-15) Difficulty: 5 of 5
Let f be a one-to-one correspondence of the points in a plane. Prove or disprove the following statement:

"If f maps circles to circles, then it maps straight lines to straight lines."

See The Solution Submitted by JLo    
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re(6): Full Proof | Comment 25 of 29 |
(In reply to re(5): Full Proof by Bractals)

I was thinking that that would be a good idea anyway (no offense to Tristan).

Since we are fine that Line(f(A),f(B)) is a subset of f[Line(A,B)] then all we need to prove is that f[Line(A,B)] is a subset of Line(f(A),f(B)) to prove they are equal.

Suppose not.  Then there exists a C in Line(A,B) such that f(C) is not in Line(f(A),f(B)).

I stipulate that g[Circle(f(A),f(B),f(C))] is a subset of Line(A,B)

If not, there exists a D not in Line(A,B) such that f(D) is in Circle(f(A),f(B),f(C)).  But f[Circle(A,B,D)]=Circle(f(A),f(B),f(D))=Circle(f(A),f(B),f(C)) so C must not be in line(A,B) -><-

Let f(E) be a point on Line(f(A),f(B)) between f(A) and f(B), and f(F) be a point on Line(f(A),f(B)) not not between f(A) and f(B)
let G be some arbitrary point in P not in Line(E,F)=Line(A,B)
f[Circle(E,F,G)]=Circle(f(E),f(F),f(G)) must intersect Circle(f(A),f(B),f(C)) at two points f(H),f(I) (f(E) is inside the circle and f(F) is outside).  I and H must be in Line(A,B).  So, Circle(E,F,G) must intersect Line(A,B) at four points E,F,I, and H. -><-

so f[Line(A,B)] is a subset of Line(f(A),f(B))

It is still a bit complex just because of the huge number of points involved.  I suppose that H isn't really needed but I think it helps the visualization to see that it must be 4 even though 3 is enough for a contradiction.

  Posted by Joel on 2006-12-18 22:44:05

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