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.
Bibliography.
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