Difference between revisions of "2001 IMO Shortlist Problems/A1"
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==Problem== | ==Problem== | ||
− | {{ | + | 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 |
+ | <center><math>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}</math></center> | ||
==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 == | ||
+ | |||
+ | * [[2001 IMO Shortlist Problems]] | ||
+ | * [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17447 Discussion on AOPS/MathLinks] | ||
+ | |||
+ | [[Category:Olympiad Algebra Problems]] |
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
![$f(p,q,r) = \begin{cases} 0 & \text{if} \; pqr = 0, \\ 1 + \tfrac{1}{6}\{f(p + 1,q - 1,r) + f(p - 1,q + 1,r) & \\ + f(p - 1,q,r + 1) + f(p + 1,q,r - 1) & \\ + f(p,q + 1,r - 1) + f(p,q - 1,r + 1)\} & \text{otherwise.} \end{cases}$](http://latex.artofproblemsolving.com/d/2/d/d2d4e8080418f5bb567971c0ee129fd02c7c7b13.png)
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.