Leningrad 1973 Problem

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Leningrad 1973 Problem

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EQSon

#1 h Mar 23, 2011, 6:47 PM • 2 Y

Y by Adventure10, Mango247

Three vertexes of a square are given. Each step, we can add symmetry of a point from another point to our collection. Is it possible to add that square's fourth point to our collection?

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#2 h Mar 23, 2011, 8:52 PM • 2 Y

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EQSon wrote:

Three vertexes of a square are given. Each step, we can add symmetry of a point from another point to our collection. Is it possible to add that square's fourth point to our collection?

Let the three vertices be $(0,0)$ , $(0,1)$ and $(1,0)$ . Obviously, any reflection only yields lattice points.
Now, if two lattice points $(a_1,a_2)$ and $(b_1,b_2)$ are symmetrical w.r.t. a third one, $a_1+a_2$ and $b_1+b_2$ must be even. But $(1,1)$ has both coordinates odd, so if it has been added at some point, it is the reflection of another point with both odd coordinates. As there is no such point at the beginning, we can never obtain one (and it is easy to see that all others can be obtained).

This post has been edited 1 time. Last edited by spanferkel, Mar 23, 2011, 8:53 PM

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#3 h Mar 23, 2011, 8:53 PM • 2 Y

Y by Adventure10, Mango247

Take the first three points as $(0,0),(0,1),(1,0)$ . Taking symmetries about one of these points has one property, namely that the product of the coordinates of any resulting point is even.
Now, taking symmetries of a point $(a,b)$ with property $ab$ even about a point $(r,s)$ takes us to a point with coordinates $(2r-a,2s-b)$ , and $(2r-a)(2s-b)$ is even because $ab$ is. Therefore, all points we can reach have the product of coordinates even. The fourth point of the square has coordinates $(1,1)$ and the product of the coordinates is odd, therefore we cannot add the fourth point of the square to our collection.

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