# The Booda hypothesis

1. Dec 17, 2006

### Loren Booda

Given a closed curve on a plane, show that there exists to any triangle a similarity whose vertices coincide with the curve.

2. Dec 17, 2006

### matt grime

That doesn't make sense. What are you asking?

3. Dec 17, 2006

### uart

Her conjecture is, that given any closed curve on a plane and given any triangle, that you can always find a triangle that is similar to the one given and whos vertices lay on the given curve.

4. Dec 17, 2006

### Loren Booda

Well said, uart! (Me's a he, he he) Can anyone prove it?

5. Dec 17, 2006

### Werg22

If it's a conjecture, I'm sure we won't prove it on this board!

6. Dec 17, 2006

### Hurkyl

Staff Emeritus
Note that Loren's conjecture is logically equivalent to:

Given a closed curve C and any three noncollinear points, we can find a curve C' similar to C that passes through those three points.

7. Dec 17, 2006

8. Dec 17, 2006

### StatusX

I don't think it's too hard to prove. Think of it like this. Label the points of the triangle A, B, and C. Pick an arbitrary point on the curve, xo, and place A at xo. Then for any other point x on the curve, we can rotate and scale the triangle ABC into AB'C' (ie, keep A fixed at xo) in such a way that B' lies on x. Then C can only lie in one of two positions, differing by a reflection through AB'.

This is a rough sketch, but it should work if the curve looks locally like a line around xo (ie, if there is a neighborhood of xo whose intersection with the curve is homeomorphic to an interval). Pick x sufficiently close to xo so that one of the choices of C' lies inside the curve. Then move x away from xo, continuously varying the choice of C', until you go all the way around and arrive back on the other side of xo. You should now have C' lying outside the curve. Thus at some point it must have crossed it, and at this point A, B', and C' all lied on the curve.

Last edited: Dec 17, 2006
9. Dec 18, 2006

### Loren Booda

Take all triangles similar to a given triangle. Label their similar vertices A, B and C.

Any closed curve can intersect similar points A and B over an infinite number of rotations.

For point C to intersect the curve containing A and B requires finding the necessary rotation and scaling of the similar triangles ABC. Thereby C and the curve will eventually intersect, since the distances between A to B, as A to C and B to C (over all rotations and scales with AB intersecting the curve), range from infinitesimal to at least equal distances (for the condition where the triangle is equilateral and the distance A to B intersecting the curve is a maximum). For a triangle of a particular rotation and scale, C can lie on the curve intersecting some A and B, with other sides equal to or less than the largest possible AB.