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Difference between revisions of "1992 IMO Problems/Problem 5"

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if we add them together it will give us the total points projected onto the <math>yz</math>-plane.
 
if we add them together it will give us the total points projected onto the <math>yz</math>-plane.
  
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This will be the value of all the elements of <math>S_{x}</math>
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This will be the value of all the elements of <math>S_{z}</math>
 +
 
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That is, <math>|Z_{i}| \le |S_{z}|</math>
  
 
Therefore,
 
Therefore,

Revision as of 13:21, 12 November 2023

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:

$|Z_{i}| \le a_{i}b_{i}$

We also know that

$|S|=\sum_{i=1}^{n}|Z_{i}|$

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.

This will be the value of all the elements of $S_{z}$

That is, $|Z_{i}| \le |S_{z}|$

Therefore,

$|S_{x}|=\sum_{i=1}^{n}a_{i}$

likewise,

$|S_{y}|=\sum_{i=1}^{n}b_{i}$

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




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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