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Difference between revisions of "2007 IMO Problems/Problem 6"
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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>.  
 
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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==Solution2==
 
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let <math>x_n</math> be ways of doing so, we see easily <math>x_n</math>=<math>x_{n-1}</math> +3 hence <math>x_n</math> = 3n +c put n=1, c=0 hence answer is 3n. ~SANSGANKRSNGUPTA
 
  
  

Latest revision as of 12:38, 23 March 2024

Problem

Let $n$ be a positive integer. Consider \[S=\{(x,y,z)~:~x,y,z\in \{0,1,\ldots,n \},~x+y+z>0\}\] as a set of $(n+1)^3-1$ points in three-dimensional space. Determine the smallest possible number of planes, the union of which contain $S$ but does not include $(0,0,0)$.

Solution

We will prove the result using the following Lemma, which has an easy proof by induction.

Lemma Let $S_1 = \{0, 1, \ldots, n_1\}$, $S_2 = \{0, 1, \ldots, n_2\}$ and $S_3 = \{0, 1, \ldots, n_3\}$. If $P$ is a polynomial in $\mathbb{R}[x, y, z]$ that vanishes on all points of the grid $S = S_1 \times S_2 \times S_3$ except at the origin, then \[\deg P \geq n_1 + n_2 + n_3.\]

Proof. We will prove this by induction on $n = n_1 + n_2 + n_3$. If $n = 0$, then the result follows trivially. Say $n > 0$. WLOG, we can assume that $n_1 > 0$. By polynomial division over $\mathbb{R}[y, z][x]$ we can write \[P = (x - n_1)Q + R.\] Since $x-n_1$ is a monomial, the remainder $R$ must be a constant in $\mathbb{R}[y, z]$, i.e., $R$ is a polynomial in two variables $y$, $z$. Pick an element of $S$ of the form $(n_1, y, z)$ and substitute it in the equation. Since $P$ vanishes on all such points, we get that $R(y, z) = 0$ for all $(y, z) \in S_2 \times S_3$. Let $S_1' = \{0, 1, \ldots, n_1 - 1\}$ and $S' = S_1' \times S_2 \times S_3$. For every point $(x, y, z)$ in $S'$ we have \[P(x, y, z) = (x - n_1)Q(x, y, z) + R(y, z) = (x-n_1)Q(x, y, z),\] where $x - n_1 \neq 0$. Therefore, the polynomial $Q$ vanishes on all points of $S'$ except the origin. By induction hypothesis, we must have $\deg Q \geq n_1 - 1 + n_2 + n_3$. But, $\deg P \geq \deg Q + 1$ and hence we have $\deg P \geq n_1 + n_2 + n_3$.

Now, to solve the problem let $H_1, \ldots, H_m$ be $m$ planes that cover all points of $S$ except the origin. Since these planes don't pass through origin, each $H_i$ can be written as $a_i x + b_i y + c_i z = 1$. Define $P$ to be the polynomial $\prod (a_ix + b_iy + c_iz - 1)$. Then $P$ vanishes at all points of $S$ except at the origin, and hence $\deg P = m \geq 3n$.


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

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