D is circle ,with radius 1/00001 ?

I mean 1.0001 excuse me

I want to give a hint from the proposer of problem u can use {na} is dense in [0,1]

any sol?noone have no idea?

I give up Anyone else ? I wud like 2 see the solution. I thought a little. btw, {na} is dense only for a - irrational

I would like to find out the solution. I thinks {na} is dense only for a - irrational

I think I have a solution for a,b. I didn't understand c. But I see not many people visit this forum last days. If anyone except DusT wants to see my solution, please send me a message or reply. DusT I will show you at school

iura you may rest asure that people will see your post. there are aproximatively 150 registred users checking in each day, 100 checking in once a week, plus many other guests that do not log on

Well, I think the idea for (c) is this one (I'm just explaining it; I don't have a soln or anything like that ): You are given P=a regular pentagon somewhere in the plane, s.t. its circumradius is 1. You are also given a circle of radius r>1 (I don't think it's important that it's 1.0001 or 1.0000000001 or anything else, as long as it's >1). Is there a sequence of moves S (defined at the beginning) s.t. the image of the regular pentagon, when it's moved in the way described in the text, is completely inside the given circle?

I will think about c). The proof of a) is long and tricky, some things can't be explained without a good picture. So I present it briefly. Lemma 1) Let ABC.. be a position of the polygon, P a point in the plane and K on [AB], PK perp AB .Let PK meet the polygon at L too. Then we can find c such that KL>=c any P,AB.. Proof. Just suppose l is the shortest side u is the closest to 90 angle of the polygon( but not equal to 90). Then we can take c=l cos u. If PK contans a side of the polygon, we put L the other extremity. Lemma 2) Let ABC.. be a position of the polygon, d(P) the smallest distance from P to a point on the perimeter. Then for a fixed P, we can perform a move such that the polygon either contains P, or d(P) decreases with at least C, for some C>0. Proof. Let PK =d(P). if K is not a vertex, then it satisfies conditions in lemma 1 and we just reflect by syde containing K. Now suppose K=A is a vertex. (this is the part where you need a picture). If the angle at A is 120 deg, we can reflect by one of sides containing A. We see that there is a constant c in this case. If A is not 120 deg, then we reflect by a fixed side containing A a lot of times. If B is the other extremity of the side, B will describe either a regular polygon with at least 4 sides, or will rotate densely on a circle. In each case, polygon will either contain P or the distance from P to B will decrease by at least a constant c.( Sorry I can't give a good picture to explain this) Lemma is proved. Lemma 3) For any P, we can make some moves the polygon to contain P in its interior. Proof. By Lemma 2, we can perform moves such that either d(P) decreases by at least C, or P gets into the polygon. Since d(P) is finite, sometime we will reach the second case. Now we can choose u very small, divide the plane into u-sided squares. We can move the polygon to contain the center of any square, the if u is sufficiently small, we can reflect in such a way the polygon to pass thrpugh all points of square. Now as N*N is countabel(in bijection with N), we can label the squares u1,u2..un. Now we visit u1,u2 etc. and here is our desired sequence of moves. b)Suppose Sn is periodic. Suppose the period is even. By any even sequence of moves, the plygon will change by a translation and a rotation. If we fix K a vertex of polygon and pass to complex numbers, K to have afix z, then z will change to z+d, then z+d+dw, the z+d+dw+dw^2 etc., where d is the length of translation, w is a number of modulus 1 corresponding to the angle of rotation. Obviously w=1 is not the case So if w<>1 then after kT moves (T-period) K will have afix z[k]=z+d(w^k-1)/(w-1). So |z[k]-z|<d*2/(w-1) so k will dever exit some disc and the polygon can not visit all the plane

As I read your solution for b, you say that if there is a cycle, then the polygon doesn't exit a disk, which is wrong. Take a square, and reflect it by 2 opposite sides.

Dust I said w=1 is not the case and your example matches w=1. I also found a sol for c). Not that all even number of mover are just translations of polygon. Lemma: Suppose we can translate a point with vectors (+-a,+-b),(+-c,+-d), where a,b,c,d are non-zero real numbers with a/c,b/d irrational. Then we can move the point as close to a fixed point as we wish. Proof of lemma. By Kronecker's Theorem, if the fixed point has x,y coordinates, we can find ma+nc close to x/2, pb+qd close to y/2, where m,n,p,q are integers. As + and - cancel each other, we can easily construct some moves to reach (2(ma+nc),2(pb+qd)). The lemma is proven. To prove the problem, we can move the polygon by vector u, taking sides 1,3. By radial symmetry, we can also move by 72,144,216,288 rotations of u. Moveing backwards( i.e. 3 1) we can move to 180, 36,108 etc. rotations of u(backwards means rotation by 180. please draw a picture with trigonometric circle to understand better).The taking OX||u, we can easily find two quadruples of vectors satisfying the lemma. So we can move the center of polygon as close to the center of given disk as we wish, and so we are done

Definitely, this problem is far from beautiful for me. It is just opinion.