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Difference between revisions of "2006 Romanian NMO Problems/Grade 10/Problem 1"

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==Solution==
 
==Solution==
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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>.
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Lemma: <math>f(A)\geq |A|</math> for all <math>A\in \mathcal P(M)</math>.
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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>.
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This completes the inductive step, which completes the proof.
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{{solution}}
 
{{solution}}
  
 
==See Also==
 
==See Also==
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*[[2006 Romanian NMO Problems/Grade 9/Problem 1 | Previous problem]]
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*[[2006 Romanian NMO Problems/Grade 9/Problem 4 | Previous problem]]
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*[[2006 Romanian NMO Problems/Grade 9/Problem 3 | Next problem]]
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*[[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 $M$ be a set composed of $n$ elements and let $\mathcal P (M)$ be its power set. Find all functions $f : \mathcal P (M) \to \{ 0,1,2,\ldots,n \}$ that have the properties

(a) $f(A) \neq 0$, for $A \neq \phi$;

(b) $f \left( A \cup B \right) = f \left( A \cap B \right) + f \left( A \Delta B \right)$, for all $A,B \in \mathcal P (M)$, where $A \Delta B = \left( A \cup B \right) \backslash \left( A \cap B \right)$.

Solution

I claim that $f(A)=|A|$, for all $A\in \mathcal P(M)$. 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 $f$ that does satisfy both properties, and then showing that this function is in fact $f(A)=|A|$.

Note that if $|A|=1$, then $f(A)\geq 1$, from property (a). Let $X=\{x\}$ and $Y=\{x,y\}$ for some $x,y\in M$. We then have from property $B$ that $f(Y)=f(X)+f(Y\backslash X)$. Since $|X|=|Y\backslash X|=1$, $f(Y)\geq 2$. This shows that for all $Y$ with $|Y|=2$, $f(Y)\geq 2$. 

Lemma: $f(A)\geq |A|$ for all $A\in \mathcal P(M)$.
Proof: We proceed by induction on $|A|$. Clearly $f(\phi)\geq 0$ and $f(|A|)\geq 1$ if $|A|=1$ from property (a). This is our base case.
Now we do the inductive step. Assume that for all $i\leq k\leq n$ for some positive integer $k$, $f(A)\leq |A|$ if $|A|=i$.
Let $X$ and $Y$ be subsets of $M$ such that $|X|=1$, $|Y|=k+1$ and $X$ is a subset of $Y$.
We then have from property 2 that $f(Y)=f(X)+f(Y\backslash X)$. From our inductive hypothesis, this sum is at least $k+1$.
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.

See Also

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