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Difference between revisions of "2006 Romanian NMO Problems/Grade 10/Problem 1"
 
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==Solution==
 
==Solution==
 +
===Solution 1===
 
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 proving that any such function <math>f</math> that satisfies both properties must be <math>f(A)=|A|</math>.
 
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 proving that any such function <math>f</math> that satisfies both properties must be <math>f(A)=|A|</math>.
  
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Note that <math>f(a_{i}) \ge 1</math> for any <math>i</math>, from property (a). Hence <cmath>f(A_{0}) = 0 < f(A_{1}) < \dotsb < f(A_{n-1}) < f(A_{n}) \le n \;,</cmath> which implies <math>f(A_{i}) = i</math> for all <math>i</math>.
 
Note that <math>f(a_{i}) \ge 1</math> for any <math>i</math>, from property (a). Hence <cmath>f(A_{0}) = 0 < f(A_{1}) < \dotsb < f(A_{n-1}) < f(A_{n}) \le n \;,</cmath> which implies <math>f(A_{i}) = i</math> for all <math>i</math>.
 +
===Solution 2===
 +
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>.
 +
 +
We have <math>f(\emptyset) = f(\emptyset) + f(\emptyset)</math>, hence <math>f(\emptyset) = 0</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>.
 +
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\subset 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.
 +
 +
I shall now prove that <math>f(A)=|A|</math> for all <math>A\in \mathcal P(M)</math>.
 +
 +
Note that our lemma implies that <math>f(M)=n</math>. Consider a set <math>A\subset M</math> with cardinality <math>k</math>. We then have that <math>A \cup M=A</math> and <math> A \Delta M=M\backslash A</math>. We then have that <math>f(A)\geq k</math> and <math>f(M\backslash A)\geq n-k</math>, so their sum (which equals <math>f(M)</math> from property (b)) is at least <math>k+n-k=n</math>. This then implies that <math>f(A)=k</math> and <math>f(M\backslash A)=n-k</math> for all <math>A</math>. We defined <math>k</math> to be <math>|A|</math>, so <math>f(A)=|A|</math>. This completes the proof.
  
 
==See Also==
 
==See Also==
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*[[2006 Romanian NMO Problems/Grade 10/Problem 2 | Next problem]]
 
*[[2006 Romanian NMO Problems/Grade 10/Problem 2 | Next problem]]
 
*[[2006 Romanian NMO Problems]]
 
*[[2006 Romanian NMO Problems]]
−
[[Category:Olympiad Geometry Problems]]
+
[[Category:Olympiad Combinatorics Problems]]

Latest revision as of 10:53, 8 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

Solution 1

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 proving that any such function $f$ that satisfies both properties must be $f(A)=|A|$.

We have $f(\emptyset) = f(\emptyset) + f(\emptyset)$, hence $f(\emptyset) = 0$. Also, if $A \subset B \subset M$, then $f(B) = f(A) + f(B-A)$ since $B = B \cup A$, $A = B \cap A$, and $B-A = B\Delta A$.

Let \[A_{0} = \emptyset \subsetneq A_{1} \subsetneq \dotsb \subsetneq A_{n-1} \subsetneq A_{n} = M\] be an increasing sequence of subsets of $M$, such that $A_{i} = A_{i-1} \cup \{a_{i}\}$ for all $i=1,\dotsc,n$. Then $|A_{i}| = i$ for all $i$.

Note that $f(a_{i}) \ge 1$ for any $i$, from property (a). Hence \[f(A_{0}) = 0 < f(A_{1}) < \dotsb < f(A_{n-1}) < f(A_{n}) \le n \;,\] which implies $f(A_{i}) = i$ for all $i$.

Solution 2

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 $f(A)=|A|$.

We have $f(\emptyset) = f(\emptyset) + f(\emptyset)$, hence $f(\emptyset) = 0$. 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 $X\subset Y$.
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.

I shall now prove that $f(A)=|A|$ for all $A\in \mathcal P(M)$.

Note that our lemma implies that $f(M)=n$. Consider a set $A\subset M$ with cardinality $k$. We then have that $A \cup M=A$ and $A \Delta M=M\backslash A$. We then have that $f(A)\geq k$ and $f(M\backslash A)\geq n-k$, so their sum (which equals $f(M)$ from property (b)) is at least $k+n-k=n$. This then implies that $f(A)=k$ and $f(M\backslash A)=n-k$ for all $A$. We defined $k$ to be $|A|$, so $f(A)=|A|$. This completes the proof.

See Also

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