3 ms·

You should take a look at the IMO problems - they are generally very different from any sort of math problem you’ve ever seen before. For instance: Determine

by elil17 7y ago

You should take a look at the IMO problems - they are generally very different from any sort of math problem you’ve ever seen before. For instance: Determine all finite sets S of at least three points in the plane which satisfy the following condition: for any two distinct points A and B in S, the perpendicular bisector of the line segment AB is an axis of symmetry for S. This is a deeply abstract and complex problem.

MauranKilom 7y ago
That was an IMO problem? How long ago was this? I'm just surprised because I found the solution set + proof in like 5 minutes of writing elementary observations on a sheet of paper... (Very much unlike the IMO qualifying test I took a few years ago. Couldn't even produce a meaningful sketch of the geometry in question there...) --- quick solution sketch --- - every regular polygon is an obvious solution - no three points can be collinear (can't have distinct parallel symmetry axes) - for a given solution, look at all (3+) symmetry axes - they have to meet at the same point (easy to prove by contradiction) - thus the angles between "neighboring" axes must be constant, namely PI / n for a solution with n distinct axes - since any solution has to fulfill the above, for a given n > 2 we can draw all the axes (intersection point and "phase" are degrees of freedom). Add one point and mirror it on all axes - if no new axes are added (and you now have > 2 points, i.e. didn't put the initial point at the center) it's a valid solution. (You can't add another point afterwards - it would form a new axis with the first one). - some footwork left to conclude that only "starting points" that produce regular polygons fulfill that. A proof by contradiction is pretty obvious in a sketch, but takes some effort to get even/odd n-gons covered correctly - there's probably something more elegant, but at this point you already know you've got the proof down. While writing this out I realized that all the points must lie on a circle (apply circle chord perpendicular bisector theorem to "successive" point pairs) which should help simplify the proof. (Please point out any mistake/misunderstanding - the solution seems way too trivial for an IMO problem. Well, a recent one at least.)
- saagarjha 7y ago
  It’s from 1999. I’d agree that this is a relatively easy problem; I think I’ve seen harder questions on the AMC…

by elil17 7y ago

MauranKilom 7y ago

That was an IMO problem? How long ago was this? I'm just surprised because I found the solution set + proof in like 5 minutes of writing elementary observations on a sheet of paper... (Very much unlike the IMO qualifying test I took a few years ago. Couldn't even produce a meaningful sketch of the geometry in question there...) --- quick solution sketch --- - every regular polygon is an obvious solution - no three points can be collinear (can't have distinct parallel symmetry axes) - for a given solution, look at all (3+) symmetry axes - they have to meet at the same point (easy to prove by contradiction) - thus the angles between "neighboring" axes must be constant, namely PI / n for a solution with n distinct axes - since any solution has to fulfill the above, for a given n > 2 we can draw all the axes (intersection point and "phase" are degrees of freedom). Add one point and mirror it on all axes - if no new axes are added (and you now have > 2 points, i.e. didn't put the initial point at the center) it's a valid solution. (You can't add another point afterwards - it would form a new axis with the first one). - some footwork left to conclude that only "starting points" that produce regular polygons fulfill that. A proof by contradiction is pretty obvious in a sketch, but takes some effort to get even/odd n-gons covered correctly - there's probably something more elegant, but at this point you already know you've got the proof down. While writing this out I realized that all the points must lie on a circle (apply circle chord perpendicular bisector theorem to "successive" point pairs) which should help simplify the proof. (Please point out any mistake/misunderstanding - the solution seems way too trivial for an IMO problem. Well, a recent one at least.)

saagarjha 7y ago
It’s from 1999. I’d agree that this is a relatively easy problem; I think I’ve seen harder questions on the AMC…