1992 IMO Problems/Problem 5

Problem

Let $S$ be a finite set of points in three-dimensional space. Let $S_{x}$ , $S_{y}$ , $S_{z}$ , be the sets consisting of the orthogonal projections of the points of $S$ onto the $yz$ -plane, $zx$ -plane, $xy$ -plane, respectively. Prove that

$|S|^{2} \le |S_{x}| \cdot |S_{y}| \cdot |S_{z}|,$

where $|A|$ denotes the number of elements in the finite set $|A|$ . (Note: The orthogonal projection of a point onto a plane is the foot of the perpendicular from that point to the plane)

Solution

Let $Z_{i}$ be planes with index $i$ such that $1 \le i \le n$ that are parallel to the $xy$ -plane that contain multiple points of $S$ on those planes such that all points of $S$ are distributed throughout all planes $Z_{i}$ according to their $z$ -coordinates in common.

Let $a_{i}$ be the number of unique projected points from each $Z_{i}$ to the $yz$ -plane

Let $b_{i}$ be the number of unique projected points from each $Z_{i}$ to the $xz$ -plane

This provides the following:

$|Z_{i}| \le a_{i}b_{i}\;$ [Equation 1]

We also know that

$|S|=\sum_{i=1}^{n}|Z_{i}|\;$ [Equation 2]

Since $a_{i}$ be the number of unique projected points from each $Z_{i}$ to the $yz$ -plane,

if we add them together it will give us the total points projected onto the $yz$ -plane.

Therefore,

$|S_{x}|=\sum_{i=1}^{n}a_{i}\;$ [Equation 3]

likewise,

$|S_{y}|=\sum_{i=1}^{n}b_{i}\;$ [Equation 4]

We also know that the total number of elements of each $Z_{i}$ is less or equal to the total number of elements in $S_{z}$

That is,

$|Z_{i}| \le |S_{z}|\;$ [Equation 5]

Multiplying [Equation 1] by [Equation 5] we get:

$|Z_{i}|^{2} \le a_{i}b_{i}|S_{z}|$

Therefore,

$|Z_{i}| \le \sqrt{a_{i}b_{i}}\sqrt{|S_{z}|}$

Adding all $|Z_{i}|$ we get:

$\sum_{i=1}^{n}|Z_{i}| \le \sqrt{|S_{z}|}\sum_{i=1}^{n}\sqrt{a_{i}b_{i}}\;$ [Equation 6]

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

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1992 IMO Problems/Problem 5

Problem

Solution