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Difference between revisions of "2001 IMO Shortlist Problems/A1"

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==Problem==
 
==Problem==
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{{problem}}
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Let <math>T</math> denote the set of all ordered triples <math>(p,q,r)</math> of nonnegative integers. Find all functions <math>f:T \rightarrow \mathbb{R}</math> such that
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<center><math>f(p,q,r) = \begin{cases} 0 & \text{if} \; pqr = 0, \\
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1 + \tfrac{1}{6}\{f(p + 1,q - 1,r) + f(p - 1,q + 1,r) & \\
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+ f(p - 1,q,r + 1) + f(p + 1,q,r - 1) & \\
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+ f(p,q + 1,r - 1) + f(p,q - 1,r + 1)\} & \text{otherwise.} \end{cases}</math></center>
  
 
==Solution==
 
==Solution==
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{{solution}}
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We can see that <math>h(p,q,r)=\frac{3pqr}{p+q+r}</math> for <math>pqr\neq0</math> and <math>h(p,q,r)=0</math> for <math>pqr=0</math> satisfies the equation. Suppose there exists another solution <math>f(p,q,r)</math>. Let <math>g(p,q,r)=f(p,q,r)-h(p,q,r)</math>. Plugging in <math>f=g+h,</math> we see that <math>g</math> satisfies the relationship <math>g(p,q,r)=\begin{cases} \tfrac{1}{6}\{g(p + 1,q - 1,r) + g(p - 1,q + 1,r) & \\
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+ g(p - 1,q,r + 1) + g(p + 1,q,r - 1) & \\
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+ g(p,q + 1,r - 1) + g(p,q - 1,r + 1)\}\end{cases}</math>, so that each value of <math>g</math> is equal to 6 points around it with an equal sum <math>p+q+r</math>. This implies that for fixed <math>p+q+r</math>, <math>g(p,q,r)</math> is constant. Furthermore, some values of <math>g</math> are always zero; for example, <math>f(p,2,0)=0</math> by the problem statement, and similarly, <math>h(p,2,0)=0</math>, so <math>g(p,2,0)=0-0=0</math>. Thus, <math>g</math> must be identically zero, so <math>h</math> is the only function satisfying this equation.
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== Resources ==
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* [[2001 IMO Shortlist Problems]]
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* [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17447 Discussion on AOPS/MathLinks]
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[[Category:Olympiad Algebra Problems]]

Latest revision as of 22:12, 17 July 2009

Problem

Let $T$ denote the set of all ordered triples $(p,q,r)$ of nonnegative integers. Find all functions $f:T \rightarrow \mathbb{R}$ such that

$f(p,q,r) = \begin{cases} 0 & \text{if} \; pqr = 0, \\ 1 + \tfrac{1}{6}\{f(p + 1,q - 1,r) + f(p - 1,q + 1,r) & \\ + f(p - 1,q,r + 1) + f(p + 1,q,r - 1) & \\ + f(p,q + 1,r - 1) + f(p,q - 1,r + 1)\} & \text{otherwise.} \end{cases}$

Solution

We can see that $h(p,q,r)=\frac{3pqr}{p+q+r}$ for $pqr\neq0$ and $h(p,q,r)=0$ for $pqr=0$ satisfies the equation. Suppose there exists another solution $f(p,q,r)$. Let $g(p,q,r)=f(p,q,r)-h(p,q,r)$. Plugging in $f=g+h,$ we see that $g$ satisfies the relationship $g(p,q,r)=\begin{cases} \tfrac{1}{6}\{g(p + 1,q - 1,r) + g(p - 1,q + 1,r) & \\ + g(p - 1,q,r + 1) + g(p + 1,q,r - 1) & \\ + g(p,q + 1,r - 1) + g(p,q - 1,r + 1)\}\end{cases}$, so that each value of $g$ is equal to 6 points around it with an equal sum $p+q+r$. This implies that for fixed $p+q+r$, $g(p,q,r)$ is constant. Furthermore, some values of $g$ are always zero; for example, $f(p,2,0)=0$ by the problem statement, and similarly, $h(p,2,0)=0$, so $g(p,2,0)=0-0=0$. Thus, $g$ must be identically zero, so $h$ is the only function satisfying this equation.

Resources

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