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Notice middle-points of all diagonals and sides of a $2010$-gon. What's the maximum number of those points which can lie on a single circle?

The solution goes like this:

Note point $O$ which is the center of circumcircle. It's obvious that middle-points of all diagonals (or sides) with same length lie on a circle with center $O$. There are 1005 of these circles, and it's obvious that a circle with maximum number of points will be one of these or will contain at most two points from each of them, that is, $2\cdot1005=2010$ in total. Since every circle with center $O$ contains at most 2010 points, and the circle which contains middle-points contains exactly 2010 points, the answer is 2010.

Can someone explain me the part in bold and after that? Also, what would the answer be for $(2k+1)$-gon?

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I take it that one is considering a regular $n$-gon (for $n=2010$). It would be good to say so. –  Marc van Leeuwen Feb 11 '13 at 15:03
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2 Answers 2

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You've got a bunch of $1005$ concentric circles with center $O$, all of them containing $2010$ marked points (except the circle with diameter $0$ at $O$, which contains a single point). The bold part serves to exclude the possibility of any other circle $C$ containing (more than) $2010$ marked points. This is because $C$ intersects each of those $1005$ circles at most twice, which gives at most $2010$ points of intersection, and any marked point on $C$ must be one of those points of intersection. In fact the intersection with the trivial circle at $O$ can have at most $1$ point, so circles not centered at $O$ cannot contain more than $2009$ points.

Note that if you count $O$ as a point marked with multiplicity $1005$ (for that number of maximal diagonals that each have their center at $O$), then this argument no longer works. Indeed all the centers of all diagonals (or sides) involving a single chosen vertex $V$ of the $2010$-gon lie on a circle $C$ passing through $V$ and through $O$; it contains $2009$ such centers (marked points), but if you count $O$ with multiplicity $1005$, that would give $3013$ marked points on $C$. By the way this shows that the bound $2009$ above is sharp.

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A circle containing at least three points of one type (i.e. the centers of three diagonals of one fixed length) must be the circumscribed circle of the triangle formed by these three points, hence must be the common circumscribed circle of all these points. Note that one has to be a bit careful with $O$ itself, where 1005 middle-points of diametric diagonals coincide.

For a $(2k+1)$-gon, there are $k$ different lengths of (sides or) diagonals, with each length occuring exactly $2k+1$ times. Thus we find $k$ circles with center $O$, each having $2k+1$ points on it. Once again, any other circle has at most two points per length, hence in total at most $2k$ points. So here the answer is also $n$.

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I'm still not sure I understand the "will contain at most two points from each of them". Can you show me some example? –  Lazar Ljubenović Feb 11 '13 at 14:20
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