## IMO longlist 1989 [reformed]

Discussion on International Mathematical Olympiad (IMO)
sourav das
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### IMO longlist 1989 [reformed]

$0<a<1\ constant,\ f:R\mapsto R, \ f(0)=0,\ f(1)=1,\ f(\frac{x+y}{2})=(1-a)f(x) +$ $af(y)$ for $x\leq y$. if $f(x)$ is not equal$\ 0$ for all $x$, prove that $a = \frac{1}{2}$
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*Mahi*
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### Re: IMO longlist 1989 [reformed]

Let $m=2n$ be positive real numbers.
Now,replacing $y$ with $m$ and $x$ with $0$ in the main equation,
$$f(n)=af(m)...(i)$$
And replacing $x$ with $-m$ and $y$ with $0$,
$$f(-n)=(1-a)f(-m)...(ii)$$
Adding $i$ and $ii$
$$f(n)+f(-n)=af(m)+(1-a)f(-m)$$
But again,replacing $x$ with $-m$ and $y$ with $m$,
$$f(0)=af(m)+(1-a)f(-m)=0...(iii)$$
So,$-f(x)=f(-x)$ for all $x \in \mathbb{R}$.
And now we just put this in $(iii)$,
$$af(x)-af(-x)=-f(-x)$$
Or, $af(x)+af(x)=f(x)$
So, $a= \frac 1 2$
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