Let S be a finite set of points in three-dimensional space. Let Sx, Sy, Sz be the sets consisting of the orthogonal projections of the points of S onto the yz-plane, zx-plane, xy-plane respectively. Prove that:
|S|2 <= |Sx| |Sy| |Sz|, where |A| denotes the number of points in the set A.
[The orthogonal projection of a point onto a plane is the foot of the perpendicular from the point to the plane.]
Solution
by Gerhard Wöginger
Induction on the number of different z-coordinates in S.
For 1, it is sufficient to note that S = Sz and |S| ≤ |Sx| |Sy| (at most |Sx| points of S project onto each of the points of Sy).
In the general case, take a horizontal (constant z) plane dividing S into two non-empty parts T and U. Clearly, |S| = |T| + |U|, |Sx| = |Tx| + |Ux|, and |Sy| = |Ty| + |Uy|.
By induction, |S| = |T| + |U| ≤ (|Tx| |Ty| |Tz|)1/2 + (|Ux| |Uy| |Uz|)1/2. But |Tz|, |Uz| ≤ |Sz|, and for any positive a, b, c, d we have (a b)1/2 + (c d)1/2 ≤ ( (a + c) (b + d) )1/2 (square!).
Hence |S| ≤ |Sz|1/2( ( |Tx| + |Ux| ) ( |Ty| + |Uy| ) )1/2 = ( |Sx| |Sy| |Sz| ) 1/2.
Solutions are also available in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
2 Sep 1999