a) Let's consider a finite number of big circles of a sphere that do not pass all from a point. Show that there exists such a point that is found only in two of the circles. (With big circle we understand the circles with radius equal to the radius of the sphere.) b) Using the result of part $a)$ show that, for a set of $n$ points in a plane, that are not all in a line, there exists a line that passes through only two points of the given set.
Problem
Source:
Tags: geometry, 3D geometry, sphere, combinatorics unsolved, combinatorics
mahanmath
23.05.2010 19:23
b) http://wapedia.mobi/en/Sylvester%E2%80%93Gallai_theorem
darij grinberg
24.05.2010 01:46
Now isn't that a retarded problem to give in a contest? Everyone with some training in olympiad geometry knows b) with a proof. And the equivalence of a) and b) is pretty much obvious (to get from a) to b), replace every big circle by the plane which contains it; these planes all pass through the center of the sphere and thus can be seen as lines on a projective plane; now, through a polar transformation you can interchange lines and points and you arrive at b), and the way back from b) to a) is similar). Let me guess: that's the way how most of the contestants have solved it?
ridgers
24.05.2010 23:16
I am a bit shamed to tell you that your guess is wrong. As far as I know no one of contestant could solve this question. I tried the a) part but my solution is quite different. I hope that the team will get better for 1 month!
oneplusone
31.05.2010 08:07
Consider an intersection $A$ whose distance to a certain circle $a$ is the smallest but more than 0. Then there must be at least 3 circles passing through it, which make 6 more intersection with $a$. Consider the 3 closest to $a$, but one of them would be closer to another circle, I hope.