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.
