2009 February « Konstantin’s Weblog

Archive for February, 2009

Verschlüsseln macht verdächtig

February 27, 2009

Udo Vetter’s law blog cites under the title Verschlüsseln macht verdächtig from an interrogation transcript. [German]

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The “Broken Windows” Theory of Crimefighting

February 27, 2009

Bruce Schneier’s blog post The “Broken Windows” Theory of Crimefighting quotes the article Breakthrough on ‘broken windows’ from The Boston Globe. Interesting information on what is effective — and what not.

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Question 4 from IMO 2008

February 24, 2009

Problem: Find all functions $f: \mathbb{R}^{+} \to \mathbb{R}^{+}$ , s.t.

$\displaystyle \frac{f(w)^2 + f(x)^2}{f(y^2)+f(z^2)} = \frac{w^2+x^2}{y^2+z^2} \ \ \ \ \ (1)$

whenever $wx=yz$ . (Author: Hojoo Lee, South Korea)

Remark: Obvious solutions are $\mathop{id}: x \mapsto x$ and $\mathop{inv}: x \mapsto 1/x$ . We show, that those are also the only solutions.

Fact 1: $f(1) = 1$ .

Proof: Let $w=x=y=z=1$ .

Fact 2: $f(x^2) = f(x)^2$ .

Proof: Let $w=y=1$ and $x=z$ .

Using Fact 2, we can rewrite $(1)$ as

$\displaystyle \frac{f(w)+f(x)}{w+x}=\frac{f(y)+f(z)}{y+z} \ \ \ \ \ (2)$

whenever $wx=yz$ .

Fact 3: For every $x \in \mathbb{R}^{+}$

$\displaystyle f(x) = x \text{ or } f(x) = 1/x$ .

Proof: Let $w=x, y=1 \text{ and } z = x^2$ . Then $(2)$ implies

$\displaystyle \frac{2f(x)}{2x} = \frac{1+f(x)^2}{1+x^2}$

which is a quadratic equation for $f(x)$ with the two solutions $x$ and $1/x$ .

Fact 4 (

Non-mixing Lemma): If $f(x) = x$ for some $x \neq 1$ , then for all $x \in \mathbb{R}^{+}$ .

Proof: Assume $f(x) = x$ , for some $x \neq 1$ , but $f(w) \neq w$ for some $w \neq 1$ . By Fact 3, we know that in this case $f(w) = 1/w$ . We will show, that now the two possibilities for $f(xw)$ , i.e. $f(xw) = xw$ and $f(xw) = 1/(xw)$ both lead to contradictions.