Let $a, b, c$ be real numbers such that $a \neq 0$. Consider the parabola with equation
$$
y=a x^{2}+b x+c,
$$
and the lines defined by the six equations
$$
\begin{array}{lll}
y=a x+b, & y=b x+c, & y=c x+a, \\
y=b x+a, & y=c x+b, & y=a x+c .
\end{array}
$$

Suppose that the parabola intersects each of these lines in at most one point. Determine the maximum and minimum possible values of $\frac{c}{a}$.
We will show that the maximum possible value of $\frac{c}{a}$ is 5 , and the minimum possible value is 1 .

First note that any linear scaling of the tuple $(a, b, c)$ results in a quadratic which still satisfies the hypothesis of the problem, and also doesn't affect the value of $\frac{c}{a}$. Without loss of generality we may assume that $a=1$, so the quadratic is
$$
y=x^{2}+b x+c \text {. }
$$

First consider the intersection of $y=x^{2}+b x+c$ and $y=x+c$. Solving $x^{2}+b x+c=x+c$ gives us $x=0$ and $x=(1-b)$. These solutions must be equal and thus $b=1$. Now the quadratic must be
$$
y=x^{2}+x+c .
$$

Next consider the intersection of $y=x^{2}+x+c$ and $y=c x+1$. Solving $x^{2}+x+c=c x+1$ leads us to the quadratic $x^{2}+(1-c) x+(c-1)=0$. The discriminant of this quadratic must be nonpositive in order for it to have no more than one real solution, and thus
$$
\begin{aligned}
0 & \geq(1-c)^{2}-4(c-1) \\
& =c^{2}-6 c+5 \\
& =(c-5)(c-1)
\end{aligned}
$$

Hence $1 \leq c \leq 5$, and therefore $1 \leq \frac{c}{a} \leq 5$.
Finally we also need to check that $y=x^{2}+x+1$ and $y=x^{2}+x+5$ both work.
- $y=x^{2}+x+1$ is tangent to $y=x+1$ at $(0,1)$.

Therefore $y=x^{2}+x+1$ works and so the minimal value of $\frac{c}{a}=1$ is possible.
- $y=x^{2}+x+5$ is tangent to $y=x+5$ at $(0,5)$.
- $y=x^{2}+x+5$ is tangent to $y=5 x+1$ at $(2,11)$.
- $y=x^{2}+x+5$ does not intersect $y=x+1$.

Therefore $y=x^{2}+x+5$ works and so the maximal value of $\frac{c}{a}=5$ is possible.