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

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<math>|Z_{i}| \le a_{i}b_{i}</math>
 
<math>|Z_{i}| \le a_{i}b_{i}</math>
  
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We also know that <math>|S|=\sum_{i=1}^{n}|Z_{i}|</math>
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We also know that  
 +
 
 +
<math>|S|=\sum_{i=1}^{n}|Z_{i}|</math>
 +
 
 +
Since <math>a_{i}</math> be the number of unique projected points from each <math>Z_{i}</math> to the <math>yz</math>-plane, if we add them together it will give us the total points on the <math>yz</math>-plane which will be <math>S_{x}</math>
 +
 
  
  

Revision as of 13:17, 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 on the $yz$-plane which will be $S_{x}$



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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