Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution)
Line 18: Line 18:
  
 
<math>|Z_{i}| \le a_{i}b_{i}</math>
 
<math>|Z_{i}| \le a_{i}b_{i}</math>
 +
 +
We also know that <math>|S|=\sum_{i=1}^{n}|Z_{i}|</math>
  
  

Revision as of 13:14, 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 $|S|=\sum_{i=1}^{n}|Z_{i}|$



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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1992_IMO_Problems/Problem_5&oldid=202643"