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] |
− | + | We also know that <math>|S|=\sum_{i=1}^{n}|Z_{i}|\;</math> [Equation 2] | |
− | |||
− | We also know that | ||
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− | <math>|S|=\sum_{i=1}^{n}|Z_{i}|\;</math> [Equation 2] | ||
Since <math>a_{i}</math> be the number of unique projected points from each <math>Z_{i}</math> to the <math>yz</math>-plane, | Since <math>a_{i}</math> be the number of unique projected points from each <math>Z_{i}</math> to the <math>yz</math>-plane, | ||
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if we add them together it will give us the total points projected onto 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, | + | Therefore, <math>|S_{x}|=\sum_{i=1}^{n}a_{i}\;</math> [Equation 3] |
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− | <math>|S_{x}|=\sum_{i=1}^{n}a_{i}\;</math> [Equation 3] | ||
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− | |||
− | <math>|S_{y}|=\sum_{i=1}^{n}b_{i}\;</math> [Equation 4] | + | likewise, <math>|S_{y}|=\sum_{i=1}^{n}b_{i}\;</math> [Equation 4] |
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> | 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, | + | That is, <math>|Z_{i}| \le |S_{z}|\;</math> [Equation 5] |
− | |||
− | <math>|Z_{i}| \le |S_{z}|\;</math> [Equation 5] | ||
Multiplying [Equation 1] by [Equation 5] we get: | Multiplying [Equation 1] by [Equation 5] we get: | ||
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<math>|Z_{i}|^{2} \le a_{i}b_{i}|S_{z}|</math> | <math>|Z_{i}|^{2} \le a_{i}b_{i}|S_{z}|</math> | ||
− | Therefore, | + | Therefore, <math>|Z_{i}| \le \sqrt{a_{i}b_{i}}\sqrt{|S_{z}|}</math> |
− | |||
− | <math>|Z_{i}| \le \sqrt{a_{i}b_{i}}\sqrt{|S_{z}|}</math> | ||
Adding all <math>|Z_{i}|</math> we get: | Adding all <math>|Z_{i}|</math> we get: | ||
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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: | ||
− | <math>|S|^{2} \le |S_{x}|</math> | + | <math>|S|^{2} \le |S_{x}| \cdot |S_{y}| \cdot |S_{z}|</math> |
+ | ~Tomas Diaz. orders@tomasdiaz.com | ||
+ | {{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 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 |