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

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Substituting [Equation 3] and [Equation 4] into [Equation 7] we get:
 
Substituting [Equation 3] and [Equation 4] into [Equation 7] we get:
  
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<math>|S|^{2} \le |S_{x}|</math>
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<math>|S|^{2} \le |S_{x}| \cdot |S_{y}| \cdot |S_{z}|</math>
  
  

Revision as of 13:45, 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:

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

Substituting [Equation 2] into [Equation 6] we get:

$|S| \le \sqrt{|S_{z}|}\sum_{i=1}^{n}\sqrt{a_{i}b_{i}}$

$|S|^{2} \le |S_{z}| \left( \sum_{i=1}^{n}\sqrt{a_{i}b_{i}} \right)^{2}$

Since, $\sum_{i=1}^{n}\sqrt{a_{i}b_{i}} \le \sqrt{\sum_{i=1}^{n}a_{i}}\sqrt{\sum_{i=1}^{n}b_{i}}$,

Then, $|S|^{2} \le |S_{z}| \left( \sqrt{\sum_{i=1}^{n}a_{i}}\sqrt{\sum_{i=1}^{n}b_{i}} \right)^{2}$

$|S|^{2} \le |S_{z}| \left( \sum_{i=1}^{n}a_{i}\right)\left( \sum_{i=1}^{n}b_{i}\right)\;$ [Equation 7]

Substituting [Equation 3] and [Equation 4] into [Equation 7] we get:

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



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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