Hong Kong Preliminary Selection Contest

Since $A$ and $B$ both lie on the straight line $y=x+k$, we may let $A=(\alpha ,\alpha +k)$ and $B=(\beta ,\beta +k)$. Combining the equations of the straight line and the parabola, we get $x^2+x+k-1=0$. Its two roots are $\alpha$ and $\beta$ since the two graphs meet at $A$ and $B$. Hence we have $\alpha +\beta =-1$ and $\alpha \beta =k-1$.

It follows that $A{B}^2=2(\alpha -\beta {)}^2=2(\alpha +\beta {)}^2-8\alpha \beta =10-8k$. The height $h$ from $C$ to $AB$ is $\frac{|1+k|}{\sqrt{2}}$, so the area of $△ABC$ is $$\begin{gathered} \frac{AB\cdot h}{2}=\frac{1}{2}\sqrt{(5-4k)(1+k{)}^2} \\ =\frac{1}{2}\sqrt{2\left(\frac{5}{2}-2k\right)(1+k{)}^2}. \end{gathered}$$ Finally, by the AM-GM inequality, we have $$\left(\frac{5}{2}-2k\right)(1+k{)}^2\le {\left[\frac{\left(\frac{5}{2}-2k\right)+(1+k)+(1+k)}{3}\right]}^3=\frac{27}{8}.$$ Equality holds when $\frac{5}{2}-2k=1+k$, or $k=\frac{1}{2}$. The greatest possible area of $△ABC$ is thus $\frac{1}{2}\sqrt{2\cdot \frac{27}{8}}=\frac{3\sqrt{3}}{4}$.