In search of the shortcut

Let $|\overline {BC}|=a, |\overline {AC}|=a\sqrt 2, \angle {ABC}=α$ . Then $\cos α=\dfrac{4-a^2}{4a}\implies \sin α=\dfrac{\sqrt {24a^2-a^4-16}}{4a}=\dfrac {\sqrt {128-(a^2-12)^2}}{4a}\leq \dfrac{2\sqrt 2}{a}$ . Hence, area of $\triangle {ABC}$ is $\dfrac{1}{2}\times 2\times a\sin α=a\sin α\leq 2\sqrt 2$ . So the required answer is $(2\sqrt 2)^2=\boxed 8$ .

Let $BC = x$ , so that the three sides are $a = x$ , $b = \sqrt{2}x$ , and $c = 2$ .

By one form of Heron's formula ,

$S = \frac{1}{4}\sqrt{2(a^2b^2 + a^2c^2 + b^2c^2) - (a^4 + b^4 + c^4)}$

$S = \frac{1}{4}\sqrt{2(2x^4 + 4x^2 + 8x^2) - (x^4 + 4x^4 + 16)}$

$S = \frac{1}{4}\sqrt{-x^4 + 24x^2 - 16}$

$S = \frac{1}{4}\sqrt{128 - (x^2 - 12)^2}$

Therefore, $S$ is a maximum when $x^2 - 12 = 0$ , which makes $S = \frac{1}{4}\sqrt{128 - 0} = 2\sqrt{2}$ , and $S^2 = \boxed{8}$ .

In the cartesian coordinate system, let $A = (1,0)\,,\,B = (-1,0)$ with $AB = 2$ . Now finding the locus of $C$ such that $\large \frac{AC}{BC} =$ $\sqrt{2}$ or $\large \frac{AC^2}{BC^2}$ $= 2$

Let $C$ be at $(h,k)\,,\,$ then

$BC^2 = (h+1)^2 + k^2 = h^2 + 2h + 1 + k^2$ and

$AC^2 = (h-1)^2 + k^2 = h^2 - 2h + 1 + k^2$

Now $AC^2 = 2BC^2 \newline \Rightarrow h^2 - 2h + 1 + k^2 = 2h^2 + 4h + 2 + 2k^2 \newline \Rightarrow h^2 + k^2 + 6h + 1 = 0 \newline \Rightarrow (h+3)^2 + k^2 = 8$

This is a circle with center $(-3,0)$ and radius $2\sqrt{2}$ . So the vertex $C$ must lie on this circle except the points where circle intersects x-axis.

Let $p$ be the height of $\triangle ABC$ with respect to base $AB$ . Then area of $\triangle ABC$ , $S = \frac{1}{2}\cdot AB \cdot p = p\quad [\because AB = 2]$ . Thus we have to maximize $p$ .

The height of vertex $C$ from the line $AB$ is same as the height of the point on circle from the x-axis. The highest point on the circle is one which is just pependicular to x-axis from the center of the circle with the height of $2\sqrt{2} \newline$ Thus $S = p = 2\sqrt{2} \newline \Rightarrow S^2 = 8$