Difference between revisions of "2007 IMO Problems/Problem 6"
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− | We | + | We will prove the result using the following Lemma, which has an easy proof by induction. |
'''Lemma''' Let <math>S_1 = \{0, 1, \ldots, n_1\}</math>, <math>S_2 = \{0, 1, \ldots, n_2\}</math> and <math>S_3 = \{0, 1, \ldots, n_3\}</math>. If <math>P</math> is a polynomial in <math>\mathbb{R}[x, y, z]</math> that vanishes on all points of the grid <math>S = S_1 \times S_2 \times S_3</math> except at the origin, then <cmath>\deg P \geq n_1 + n_2 + n_3.</cmath> | '''Lemma''' Let <math>S_1 = \{0, 1, \ldots, n_1\}</math>, <math>S_2 = \{0, 1, \ldots, n_2\}</math> and <math>S_3 = \{0, 1, \ldots, n_3\}</math>. If <math>P</math> is a polynomial in <math>\mathbb{R}[x, y, z]</math> that vanishes on all points of the grid <math>S = S_1 \times S_2 \times S_3</math> except at the origin, then <cmath>\deg P \geq n_1 + n_2 + n_3.</cmath> |
Latest revision as of 07:39, 13 April 2015
Problem
Let be a positive integer. Consider as a set of points in three-dimensional space. Determine the smallest possible number of planes, the union of which contain but does not include .
Solution
We will prove the result using the following Lemma, which has an easy proof by induction.
Lemma Let , and . If is a polynomial in that vanishes on all points of the grid except at the origin, then
Proof. We will prove this by induction on . If , then the result follows trivially. Say . WLOG, we can assume that . By polynomial division over we can write Since is a monomial, the remainder must be a constant in , i.e., is a polynomial in two variables , . Pick an element of of the form and substitute it in the equation. Since vanishes on all such points, we get that for all . Let and . For every point in we have where . Therefore, the polynomial vanishes on all points of except the origin. By induction hypothesis, we must have . But, and hence we have .
Now, to solve the problem let be planes that cover all points of except the origin. Since these planes don't pass through origin, each can be written as . Define to be the polynomial . Then vanishes at all points of except at the origin, and hence .
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
2007 IMO (Problems) • Resources | ||
Preceded by Problem 5 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 6 |
All IMO Problems and Solutions |