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Difference between revisions of "1973 IMO Problems/Problem 5"

(Created page with "<math>G</math> is a set of non-constant functions of the real variable <math>x</math> of the form <cmath>f(x) = ax + b, a \text{ and } b \text{ are real numbers,}</cmath> and <ma...")
 
(→‎Solution)
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==Solution==
 
==Solution==
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First, observe that for each function <math>f(x)=ax+b</math> in <math>G</math>, if <math>a=1</math> then <math>b=0</math>. This is a result of (c); for example, <math>f(x)=x+1</math> could not be in <math>G</math> because it does not have a fixed point. Or if <math>f(x)=x</math>, then every point is a fixed point.
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Also, for each function <math>f(x)=ax+b</math> in <math>G</math>, if <math>a \neq 1</math> then the fixed point of <math>f</math> is where <math>y=ax+b</math> intersects <math>y=x</math>, namely where <math>x=\frac{b}{a-1}</math>.
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Now, take <math>f_1(x)=a_1 x+b_1</math> and <math>f_2(x)=a_2 x+b_2</math>, both in <math>G</math>. By (a), <math>(f_1 \circ f_2)(x)=f_1(f_2(x))=f_1(a_2 x+b_2)=a_1 a_2 x + a_1 b_2 + b_1</math> and <math>(f_2 \circ f_1)(x)=f_2(f_1(x))=f_2(a_1 x+b_1)=a_1 a_2 x + a_2 b_1 + b_2</math> must also both be in <math>G</math>. By (b), <math>(f_1 \circ f_2)^{-1}(x)=\frac{x - a_1 b_2 - b_1}{a_1 a_2}</math> must also be in <math>G</math>. Finally, by (a),
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<math>((f_1 \circ f_2)^{-1} \circ (f_2 \circ f_1))(x)=\frac{a_1 a_2 x + a_2 b_1 + b_2 - a_1 b_2 - b_1}{a_1 a_2}</math>
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<math>= x + \frac{a_2 b_1 + b_2 - a_1 b_2 - b_1}{a_1 a_2}</math>
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must also be in <math>G</math>.
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Using our first observation, <math>\frac{a_2 b_1 + b_2 - a_1 b_2 - b_1}{a_1 a_2}=0</math>. Rearranging, we get <math>\frac{b_1}{a_1-1}=\frac{b_2}{a_2-1}</math>. Therefore, the fixed point of <math>f_1</math> equals the fixed point of <math>f_2</math>. Since we made no assumptions about <math>f_1</math> and <math>f_2</math>, this is true for all <math>f</math> in <math>G</math>.

Revision as of 18:15, 25 October 2014

$G$ is a set of non-constant functions of the real variable $x$ of the form \[f(x) = ax + b, a \text{ and } b \text{ are real numbers,}\] and $G$ has the following properties:

(a) If $f$ and $g$ are in $G$, then $g \circ f$ is in $G$; here $(g \circ f)(x) = g[f(x)]$.

(b) If $f$ is in $G$, then its inverse $f^{-1}$ is in $G$; here the inverse of $f(x) = ax + b$ is $f^{-1}(x) = (x - b) / a$.

(c) For every $f$ in $G$, there exists a real number $x_f$ such that $f(x_f) = x_f$.

Prove that there exists a real number $k$ such that $f(k) = k$ for all $f$ in $G$.

Solution

First, observe that for each function $f(x)=ax+b$ in $G$, if $a=1$ then $b=0$. This is a result of (c); for example, $f(x)=x+1$ could not be in $G$ because it does not have a fixed point. Or if $f(x)=x$, then every point is a fixed point.

Also, for each function $f(x)=ax+b$ in $G$, if $a \neq 1$ then the fixed point of $f$ is where $y=ax+b$ intersects $y=x$, namely where $x=\frac{b}{a-1}$.

Now, take $f_1(x)=a_1 x+b_1$ and $f_2(x)=a_2 x+b_2$, both in $G$. By (a), $(f_1 \circ f_2)(x)=f_1(f_2(x))=f_1(a_2 x+b_2)=a_1 a_2 x + a_1 b_2 + b_1$ and $(f_2 \circ f_1)(x)=f_2(f_1(x))=f_2(a_1 x+b_1)=a_1 a_2 x + a_2 b_1 + b_2$ must also both be in $G$. By (b), $(f_1 \circ f_2)^{-1}(x)=\frac{x - a_1 b_2 - b_1}{a_1 a_2}$ must also be in $G$. Finally, by (a),

$((f_1 \circ f_2)^{-1} \circ (f_2 \circ f_1))(x)=\frac{a_1 a_2 x + a_2 b_1 + b_2 - a_1 b_2 - b_1}{a_1 a_2}$ $= x + \frac{a_2 b_1 + b_2 - a_1 b_2 - b_1}{a_1 a_2}$

must also be in $G$.

Using our first observation, $\frac{a_2 b_1 + b_2 - a_1 b_2 - b_1}{a_1 a_2}=0$. Rearranging, we get $\frac{b_1}{a_1-1}=\frac{b_2}{a_2-1}$. Therefore, the fixed point of $f_1$ equals the fixed point of $f_2$. Since we made no assumptions about $f_1$ and $f_2$, this is true for all $f$ in $G$.

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