Math Olympiad!!!!!?

Question

"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."

What's it even asking for?

Thomasoa, I not catching on here. If regular polygons satisfy, wouldn't there be infinitely many sets?

Thanks Letao12, I finally understand.

Actually I'm confused now O.o....

I'm afraid it's not even near last week. Perhaps you mean years?

Answer 1

Take any two points in S.

Draw the perpendicular bisector of those two points.

Then the set of points one on side of the bisector are the mirror image of the set of points on the other side.

For example, the points of a regular polygon are a solution for this problem. I'm not sure if that represents all the solutions.

Part one is to realize that they all have to lie on a circle.

Give each point of S a mass 1, and let P be the center of mass for S.

Then for each A and B in S, the perpendicular bisector of AB must hit P. That's because of the symmetry around the perpendicular bisector - each point not on the perp. bisector has a corresponding point exact opposite.

So if A,B, and C are three points of S, then the perpendicular bisectors of AB, AC, and BC all hit P.

That means that P is the center of the circumscribing circle around ABC. But that means P is the same distance to A an B and C, and hence, to any point of S (since A, B, and C were chosen arbitrarily.)

So S must be on a circle.

Can you prove that S must be the points of a regular polygon? I haven't gotten that far.

ANS: Yes, there are inifinitely many sets, but the problem doesn't ask you to count them, it asks you to 'determine all' of them. That's a mathematical classification problem.

So, it is possible (I think likely) that the answer to this classification problem is: All sets S which form a regular polygon.

But there might be some odd non-regular case which still has this property.

Answer 2

It's hard to explain without drawing pictures, but I can give you an example.

Let's say S is a circle. Now choose any two points A and B on the circle (the border, not the area inside). Draw a segment between A and B. Then draw a second segment, call it P, which is perpendicular to AB and cuts directly in the middle of AB. The resulting P will always cut the circle exactly in half. In other words, the circle will be symmetric about P no matter what points A and B you chose.

The problem basically asks you to find all possible shapes (doesn't have to be continuous like a circle, just need at least 3 points) that works the same way. If you chose any two points on the shape, drew the two segments, the second segment has to be an axis of symmetry for the shape. The circle is one of those shapes, but there are others.

Answer 3

thats a real test people take. i took it last week dont be postin things like this