Difference between revisions of "1992 IMO Problems/Problem 5"
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Latest revision as of 23:43, 16 November 2023
Problem
Let be a finite set of points in three-dimensional space. Let ,,, be the sets consisting of the orthogonal projections of the points of onto the -plane, -plane, -plane, respectively. Prove that
where denotes the number of elements in the finite set . (Note: The orthogonal projection of a point onto a plane is the foot of the perpendicular from that point to the plane)
Solution
Let be planes with index such that that are parallel to the -plane that contain multiple points of on those planes such that all points of are distributed throughout all planes according to their -coordinates in common.
Let be the number of unique projected points from each to the -plane
Let be the number of unique projected points from each to the -plane
This provides the following: [Equation 1]
We also know that [Equation 2]
Since be the number of unique projected points from each to the -plane,
if we add them together it will give us the total points projected onto the -plane.
Therefore, [Equation 3]
likewise, [Equation 4]
We also know that the total number of elements of each is less or equal to the total number of elements in
That is, [Equation 5]
Multiplying [Equation 1] by [Equation 5] we get:
Therefore,
Adding all we get:
[Equation 6]
Substituting [Equation 2] into [Equation 6] we get:
Since, ,
Then,
[Equation 7]
Substituting [Equation 3] and [Equation 4] into [Equation 7] we get:
~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 |