Difference between revisions of "2007 IMO Problems/Problem 6"
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==Solution== | ==Solution== | ||
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+ | We will prove the result using the following Lemma, which has an easy proof by induction. | ||
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+ | '''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> | ||
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+ | ''Proof.'' We will prove this by induction on <math>n = n_1 + n_2 + n_3</math>. If <math>n = 0</math>, then the result follows trivially. Say <math>n > 0</math>. | ||
+ | WLOG, we can assume that <math>n_1 > 0</math>. | ||
+ | By polynomial division over <math>\mathbb{R}[y, z][x]</math> we can write <cmath>P = (x - n_1)Q + R.</cmath> | ||
+ | Since <math>x-n_1</math> is a monomial, the remainder <math>R</math> must be a constant in <math>\mathbb{R}[y, z]</math>, i.e., <math>R</math> is a polynomial in two variables <math>y</math>, <math>z</math>. | ||
+ | Pick an element of <math>S</math> of the form <math>(n_1, y, z)</math> and substitute it in the equation. Since <math>P</math> vanishes on all such points, we get that <math>R(y, z) = 0</math> for all <math>(y, z) \in S_2 \times S_3</math>. | ||
+ | Let <math>S_1' = \{0, 1, \ldots, n_1 - 1\}</math> and <math>S' = S_1' \times S_2 \times S_3</math>. | ||
+ | For every point <math>(x, y, z)</math> in <math>S'</math> we have <cmath>P(x, y, z) = (x - n_1)Q(x, y, z) + R(y, z) = (x-n_1)Q(x, y, z),</cmath> where <math>x - n_1 \neq 0</math>. | ||
+ | Therefore, the polynomial <math>Q</math> vanishes on all points of <math>S'</math> except the origin. By induction hypothesis, we must have <math>\deg Q \geq n_1 - 1 + n_2 + n_3</math>. But, <math>\deg P \geq \deg Q + 1</math> and hence we have <math>\deg P \geq n_1 + n_2 + n_3</math>. | ||
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+ | Now, to solve the problem let <math>H_1, \ldots, H_m</math> be <math>m</math> planes that cover all points of <math>S</math> except the origin. Since these planes don't pass through origin, each <math>H_i</math> can be written as <math>a_i x + b_i y + c_i z = 1</math>. Define <math>P</math> to be the polynomial <math>\prod (a_ix + b_iy + c_iz - 1)</math>. Then <math>P</math> vanishes at all points of <math>S</math> except at the origin, and hence <math>\deg P = m \geq 3n</math>. | ||
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{{alternate solutions}} | {{alternate solutions}} |
Latest revision as of 06:39, 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 | ||
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