Difference between revisions of "2006 Romanian NMO Problems/Grade 10/Problem 1"
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==Solution== | ==Solution== | ||
+ | I claim that <math>f(A)=|A|</math>, for all <math>A\in \mathcal P(M)</math>. Clearly this function works; we must now show that it is the only function with the two given properties. We shall do this by constructing a function <math>f</math> that does satisfy both properties, and then showing that this function is in fact <math>f(A)=|A|</math>. | ||
+ | |||
+ | Note that if <math>|A|=1</math>, then <math>f(A)\geq 1</math>, from property (a). Let <math>X=\{x\}</math> and <math>Y=\{x,y\}</math> for some <math>x,y\in M</math>. We then have from property <math>B</math> that <math>f(Y)=f(X)+f(Y\backslash X)</math>. Since <math>|X|=|Y\backslash X|=1</math>, <math>f(Y)\geq 2</math>. This shows that for all <math>Y</math> with <math>|Y|=2</math>, <math>f(Y)\geq 2</math>. | ||
+ | |||
+ | Lemma: <math>f(A)\geq |A|</math> for all <math>A\in \mathcal P(M)</math>. | ||
+ | Proof: We proceed by induction on <math>|A|</math>. Clearly <math>f(\phi)\geq 0</math> and <math>f(|A|)\geq 1</math> if <math>|A|=1</math> from property (a). This is our base case. | ||
+ | Now we do the inductive step. Assume that for all <math>i\leq k\leq n</math> for some positive integer <math>k</math>, <math>f(A)\leq |A|</math> if <math>|A|=i</math>. | ||
+ | Let <math>X</math> and <math>Y</math> be subsets of <math>M</math> such that <math>|X|=1</math>, <math>|Y|=k+1</math> and <math>X</math> is a subset of <math>Y</math>. | ||
+ | We then have from property 2 that <math>f(Y)=f(X)+f(Y\backslash X)</math>. From our inductive hypothesis, this sum is at least <math>k+1</math>. | ||
+ | This completes the inductive step, which completes the proof. | ||
+ | |||
{{solution}} | {{solution}} | ||
==See Also== | ==See Also== | ||
− | *[[2006 Romanian NMO Problems/Grade 9/Problem | + | *[[2006 Romanian NMO Problems/Grade 9/Problem 4 | Previous problem]] |
− | *[[2006 Romanian NMO Problems/Grade | + | *[[2006 Romanian NMO Problems/Grade 10/Problem 2 | Next problem]] |
*[[2006 Romanian NMO Problems]] | *[[2006 Romanian NMO Problems]] | ||
[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] |
Revision as of 15:19, 7 May 2012
Problem
Let be a set composed of
elements and let
be its power set. Find all functions
that have the properties
(a) , for
;
(b) , for all
, where
.
Solution
I claim that , for all
. Clearly this function works; we must now show that it is the only function with the two given properties. We shall do this by constructing a function
that does satisfy both properties, and then showing that this function is in fact
.
Note that if , then
, from property (a). Let
and
for some
. We then have from property
that
. Since
,
. This shows that for all
with
,
.
Lemma:for all
. Proof: We proceed by induction on
. Clearly
and
if
from property (a). This is our base case. Now we do the inductive step. Assume that for all
for some positive integer
,
if
. Let
and
be subsets of
such that
,
and
is a subset of
. We then have from property 2 that
. From our inductive hypothesis, this sum is at least
. This completes the inductive step, which completes the proof.
This problem needs a solution. If you have a solution for it, please help us out by adding it.