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POLYA019: Plane set containing reflections of points in lines (proposed by Jean-Pierre Ehrmann, 12/4/01).

The following exercise comes from a French examination (ENS 1988) : S is a nonempty part of the euclidean plane; suppose that S is not included in a line and that for every triple A, B, C of points of S, the reflection of A wrt the line BC lies in S.Prove that S is dense in the plane.

Discussion. [SCL,12/4/01]: I would like some clarification on this problem. If S is a set of points with no other conditions other than the stated one ones, I think I will have trouble proving the result. For example, Suppose I take S to be the set of vertices of the standard infinite triangular grid. It certainly looks to me like this set S satisfies the hypotheses, but not the conclusion.
I must be misinterpreting the intent of the question.

[JPE, 12/4/01]:  I don't think that such a grid verifies the hypotheses: ABC is an equilateral triangle; B', C' the reflections of B, C wrt A; B" =
reflection of B' wrt C; U = reflection of B wrt the line C'B". U is not a vertice of the grid generated by ABC.

[SCL, 12/4/01]: Okay, I see the error of my ways. But, even a mistake can lead to profit.
Here's why I came across the example:
If p,q,r are three distinct points with the angle pqr of measure \alpha which is not a rational multiple of \pi, then we can proceed as follows:
Reflection in pq, then in pr, is a rotation by angle 2\alpha. Therefore, repeated images p under this rotation are dense on the circle with center q and radius qp. Let T be the set of points of S on this circle.
Now, for any line m (even if m doesn't have points of S) we can find a pair of points x,y of T which are nearly diametrical (as close as you like, without being a diameter), with the angle between xy and m as small as you like. (Of course, I'm just taking the diameter d of the circle which is parallel to m, and picking two points of T very close to the two ends of the diameter d (and on the same side of d). If both ends of d are in T, make sure you pick one of x or y not to be an end of d.
Now, reflect T in the line xy. This "thickens" circle in the direction perpendicular to d.and we can continue the process (and do it in both directions) until we have a band across the plane. It won't have straight edges, but we could, say, take half of the length of d and make sure that we cover a straight-edged band of the plane with width d/2. It is then an easy process to translate this band in the perpendicular direction.
My problem was in not seeing what to do if \alpha was a rational multiple of \pi. And the example gives me an idea. (I haven't worked out the details yet).
If k\alpha is the first multiple of 2\pi. Then images of p will only give k points, p_1, p_2, p_3, ..., p_k, on the circle C of radius qp centered at q. We can reflect q in the line p_ip_{i+1} to get a point q_i. If we then reflect p_{i+1} in the line p_iq_i, it will fall inside the angle p_{i-1}qp_i. If we can show that it is not on the line qq_{i-1}, one of the angles p_iqr_{i+1} or p_{i-1}qr_{i+1} will be less than half of angle p_iqp_{i+1}, and we get a finer set of points on the circle C. Then, we're done.
Maybe the best way is then reflect the point p_i in the line q_ir_{i+1} to get a point s_i. If s_i is not on qp_i, the angle p_iqs_i is less than half angle p_iqp_{i+1}. And, then only way this can happen is if k=6. We get the grid I mentioned. However, it is now obvious that if one goes far enough from q there is a point t with the angle p_iqt as small as one may desire.
I think we're got all the pieces.

[JPE, 12/5/01]: Yes, you have. I've proceeded in a slightly different way to avoid the distinction between rational or not multiples of /pi and the problem due to the fact that we can start from an equilateral triangle. It was easier for me because I knew that the examiner helped the candidate and asked him to prove that it was possible to find a rectangle with vertices in S. Starting from this one, the idea is to find arbitrary small rectangles with vertices in S.
1) Let ABC a non degenerated triangle with vertices in S. Then we can find a non degenerated rectangle with vertices in S and included in a disk with radius twice the length of the smallest side of ABC.
Proof : Suppose that a is the smallest side and a<c (If ABC is equilateral, take the reflection of B wrt AC instead of A).
D = foot of the B-altitude of ABC
A' = reflection of A wrt BD
M = reflection of C wrt A'B
Using the reflections of M wrt AC and BD, we get a rectangle MNPQ with center D and DM <= 2a because BM = a and BD = a sin C
2) For any integer n, consider Q_n on the half-line PQ such as PQ_n = n PQ and M_n = reflection of M wrt NQ_n. By 1) applied to the triangle MNM_n, we can construct a rectangle R_n with vertices in S and included in a disk with radius 2 MM_n. Clearly MM_n -> 0 when n->infi and the vertices of the grid generated by R_n lie in S.


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