Belorusija 2012

Since the graphs of $f(x)$ and $h(x)$ has exactly one common point, the equation $a{x}^2+bx+c=cx+b$, i.e. the equation $$a{x}^2-(c-b)x+(c-b)=0,(1)$$ has exactly one root. So, its discriminant is equal to $0$, i.e. $(c-b{)}^2-4a(c-b)=0\iff (c-b)(c-b-4a)=0$. By condition, $c\ne b$, i.e. $c-b\ne 0$, so we have $c-b-4a=0$ or $$c-b=4a.(2)$$

From (1) and (2) it follows $a{x}^2-4ax+4a=0$, whence $x^2-4x+4=0\iff (x-2{)}^2=0\iff x=2$. Thus $x=2$ is an abscissa of the common point of all three graphs.

Since the graphs of $g(x)$ and $h(x)$ has also exactly one common point, the equation $\frac{cx+b}{cx+a}=cx+b$, i.e. the equation $cx+b=(cx+b)(cx+a)$ has exactly one root. Therefore, $x=-\frac{b}{c}=-\frac{1-a}{c}$. But $x=2$, so $$-\frac{b}{c}=2,\text{and}2=\frac{1-a}{c}.(3)-(4)$$ From (2), (3) and (4) we obtain $a=\frac{3}{11}$, $b=-\frac{8}{11}$ and $c=\frac{4}{11}$. It is easy to see that for these values of $a$, $b$ and $c$ the graphs of all three functions have exactly one common point.