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

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Let <math>b_{i}</math> be the number of unique projected points from each <math>Z_{i}</math> to the <math>xz</math>-plane
 
Let <math>b_{i}</math> be the number of unique projected points from each <math>Z_{i}</math> to the <math>xz</math>-plane
  
−
This provides the following:
+
This provides the following: <math>|Z_{i}| \le a_{i}b_{i}\;</math> [Equation 1]
  
−
<math>|Z_{i}| \le a_{i}b_{i}</math>
+
We also know that <math>|S|=\sum_{i=1}^{n}|Z_{i}|\;</math> [Equation 2]
  
−
We also know that
+
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 projected onto the <math>yz</math>-plane.
 +
 
 +
Therefore, <math>|S_{x}|=\sum_{i=1}^{n}a_{i}\;</math> [Equation 3]
  
−
<math>|S|=\sum_{i=1}^{n}|Z_{i}|</math>
+
likewise, <math>|S_{y}|=\sum_{i=1}^{n}b_{i}\;</math> [Equation 4]
  
−
Since <math>a_{i}</math> be the number of unique projected points from each <math>Z_{i}</math> to the <math>yz</math>-plane,
+
We also know that the total number of elements of each <math>Z_{i}</math> is less or equal to the total number of elements in <math>S_{z}</math>
 +
 
 +
That is, <math>|Z_{i}| \le |S_{z}|\;</math> [Equation 5]
 +
 
 +
Multiplying [Equation 1] by [Equation 5] we get:
 +
 
 +
<math>|Z_{i}|^{2} \le a_{i}b_{i}|S_{z}|</math>
  
−
if we add them together it will give us the total points projected onto the <math>yz</math>-plane.
+
Therefore, <math>|Z_{i}| \le \sqrt{a_{i}b_{i}}\sqrt{|S_{z}|}</math>
  
−
This will be the value of all the elements of <math>S_{x}</math>
+
Adding all <math>|Z_{i}|</math> we get:
  
−
Therefore,
+
<math>\sum_{i=1}^{n}|Z_{i}| \le \sqrt{|S_{z}|}\sum_{i=1}^{n}\sqrt{a_{i}b_{i}}\;</math>[Equation 6]
  
−
<math>|S_{x}|=\sum_{i=1}^{n}|a_{i}|</math>
+
Substituting [Equation 2] into [Equation 6] we get:
  
−
likewise,
+
<math>|S| \le \sqrt{|S_{z}|}\sum_{i=1}^{n}\sqrt{a_{i}b_{i}}</math>
  
−
<math>|S_{y}|=\sum_{i=1}^{n}|b_{i}|</math>
+
<math>|S|^{2} \le |S_{z}| \left( \sum_{i=1}^{n}\sqrt{a_{i}b_{i}} \right)^{2}</math>
  
 +
Since, <math>\sum_{i=1}^{n}\sqrt{a_{i}b_{i}} \le \sqrt{\sum_{i=1}^{n}a_{i}}\sqrt{\sum_{i=1}^{n}b_{i}}</math>,
  
 +
Then, <math>|S|^{2} \le |S_{z}| \left( \sqrt{\sum_{i=1}^{n}a_{i}}\sqrt{\sum_{i=1}^{n}b_{i}} \right)^{2}</math>
  
 +
<math>|S|^{2} \le |S_{z}| \left( \sum_{i=1}^{n}a_{i}\right)\left( \sum_{i=1}^{n}b_{i}\right)\;</math> [Equation 7]
  
 +
Substituting [Equation 3] and [Equation 4] into [Equation 7] we get:
  
 +
<math>|S|^{2} \le |S_{x}| \cdot |S_{y}| \cdot |S_{z}|</math>
  
 +
~Tomas Diaz. orders@tomasdiaz.com
  
 
{{alternate solutions}}
 
{{alternate solutions}}
 +
 +
==See Also==
 +
 +
{{IMO box|year=1992|num-b=4|num-a=6}}
 +
[[Category:Olympiad Geometry Problems]]
 +
[[Category:3D Geometry Problems]]

Latest revision as of 23:43, 16 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:

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

~Tomas Diaz. orders@tomasdiaz.com

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

See Also

1992 IMO (Problems) • Resources
Preceded by
Problem 4 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 6
All IMO Problems and Solutions
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