China Mathematical Competition

Since $f(x-4)=f(2-x)$ for $x\in \mathbb{R}$, it is known that the quadratic function $f(x)$ has $x=-1$ as its axis of symmetry. By condition (3), we know that $f(x)$ opens upward, that is, $a>0$. Hence $$f(x)=a(x+1{)}^2(a>0).$$ By condition (1), we get $f(1)\ge 1$ and by (2), $f(1)\le {\left(\frac{1+1}{2}\right)}^2=1$. It follows that $f(1)=1$, i.e. $a(1+1{)}^2=1$. So $a=\frac{1}{4}$. Thereby, $f(x)=\frac{1}{4}(x+1{)}^2$.

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Since the graph of the parabola $f(x)=\frac{1}{4}(x+1{)}^2$ opens upward, and a graph of $y=f(x+t)$ can be obtained by translating that of $f(x)$ by $t$ units. If we want the graph of $y=f(x+t)$ to lie under the graph of $y=x$ when $x\in [1,m]$, and $m$ to be maximal, then $1$ and $m$ should be two roots of an equation with respect to $x$. $$\frac{1}{4}(x+t+1{)}^2=x.\stackrel{◯}{1}$$ Substituting $x=1$ into (1), we get $t=0$ or $t=-4$. When $t=0$, substituting it into (1), we get $x_1=x_2=1$ (in contradiction with $m>1$). When $t=-4$, substituting it into (1), we get $x_1=1$, and $x_2=9$; and so $m=9$. Moreover, when $t=-4$, for any $x\in [1,9]$, we have always $$\begin{gathered} (x-1)(x-9)\le 0 \\ \iff \frac{1}{4}(x-4+1{)}^2\le x, \end{gathered}$$ that is $$f(x-4)\le x.$$ Therefore, the maximum value of $m$ is $9$.