74th Romanian Mathematical Olympiad

a) Because $f$ is continuous, $F$ is continuous and differentiable, with $F^{\prime }(x)=f(x)>0$, for any $x\in [0,1]$. Hence, $F$ is strictly increasing, and thus injective. Being continuous, $F$ has the intermediate value property, and since $F(0)=0$ and $F(1)=A$, $F$ is surjective. Thus, $F$ is bijective, hence invertible. Also, because $F^{\prime }(x)=f(x)>0$ for any $x\in [0,1]$, $F^{-1}$ is differentiable, with $$(F^{-1}{)}^{\prime }(x)=\frac{1}{f(F^{-1}(x))},\text{for any }x\in [0,A].$$

b) The equality in the statement ${\int }_0^{x}f(t)dt={\int }_{g(x)}^{1}f(t)dt$ becomes $F(x)=F(1)-F(g(x))=A-F(g(x))$, or, equivalently, $F(g(x))=A-F(x)$, for any $x\in [0,1]$. The function $F$ being increasing, it follows that $A-F(x)\in [0,A]$, for any $x\in [0,1]$, so that the function $g:[0,1]\to [0,1]$, defined by $g(x)=F^{-1}(A-F(x))$ for any $x\in [0,1]$, is well defined and unique satisfying (1).

c) Since the functions $F$ and $F^{-1}$ are differentiable, the function $g$ defined above is also differentiable, and $$g^{\prime }(x)=\frac{(A-F(x){)}^{\prime }}{f(F^{-1}(A-F(x)))}=\frac{-f(x)}{f(F^{-1}(A-F(x)))},\text{for any }x\in [0,1].(2)$$ The function $g$ is strictly decreasing, and so has a unique fixed point $c\in [0,1]$. For this we have: $2F(c)=F(c)+F(g(c))=A$, so that $F(c)=\frac{1}{2}\cdot A$ and $c=F^{-1}(\frac{1}{2}\cdot A)$.

Because $g$ is differentiable, the limit $$\underset{x\to c}{\lim }\frac{g(x)-c}{x-c}=\underset{x\to c}{\lim }\frac{g(x)-g(c)}{x-c}=g^{\prime }(c),$$ exists and is equal to $$g^{\prime }(c)=\frac{-f(c)}{f(F^{-1}(A-F(c)))}=\frac{-f(c)}{f(c)}=-1.$$