Let (c) be a compact smooth convex curve in E2 and A,B are interior points. The point M is on (c) and we denote e(M) the support line of the point M and e0 the unit tangent vector at the point M. We suppose that: angle (AM,e0)=angle(BM,-e0).
Let (k) be a rotund closed convex curve and A,B interior points. We consider, from an external point M the two support lines MM1,MM2 and we assume that for every point M, the angles AMM1 and BMM2 are equal. In this note we answer to the question whether (k) is an ellipse.
Theorem: If a closed smooth convex curve (c) in E2 and an interior point O have the property that (c) possesses parallel supporting lines at the endpoints of every chord through O, then the (c) is centrosymmetric and O is the center.
Let L2 be an orthogonal lattice in a plane q and F a figure in q. We define that F has the l.p.c.p.(lattice point covering property) if for every position of F in the plane, the figure F includes at least one lattice point.
It is very simple to see that the circle c0 of diameter 2R=21/2 has the l.p.c.p. So every figure including c0 has the l.p.c.p. For example the equilateral triangle including c0 has side equal to 2.449 and obviously has the l.p.c.p. Nevertheless we can prove that an equilateral triangle with side 2.154 has l.p.c.p. That is, the problem is to find the minimum conditions between the elements of the figure F, so that the figure F to be able for the l.p.c.p.
Another example for L3. A cube of side 31/2 has obviously the l.p.c.p. because includes a sphere of diameter 31/2. But it has been proved by M. Henk that a cube of side 21/2 has the l.p.c.p. Remains open the problem for n-cube. M. Henk and the author solved the problem for the ellipsoid in Ln and here is the solution for l.p.c.p. of the ellipse in the plane.
Every trangle including a unit square in an orthogonal lattice has the lattice point covering property.