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

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==Solution== | ==Solution== | ||

− | {{solution}} | + | 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) & \\ |

+ | + 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}</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. | ||

== Resources == | == Resources == |

## Latest revision as of 23:12, 17 July 2009

## Problem

Let denote the set of all ordered triples of nonnegative integers. Find all functions such that

## Solution

We can see that for and for satisfies the equation. Suppose there exists another solution . Let . Plugging in we see that satisfies the relationship , so that each value of is equal to 6 points around it with an equal sum . This implies that for fixed , is constant. Furthermore, some values of are always zero; for example, by the problem statement, and similarly, , so . Thus, must be identically zero, so is the only function satisfying this equation.