Maximum Triangle Side

Let $ABC$ be the triangle in question and $x$ be the length of $BC$ . We place the triangle on a coordinate system so that $B\left( -\dfrac{x}{2},0 \right)$ and $C\left( \dfrac{x}{2},0 \right)$ . Then, vertex $A$ lies on an ellipse $C$ with points $B$ and $C$ being its foci and the length of its major semi-axis being $\alpha =\dfrac{1}{2}\left( AB+AC \right)=\dfrac{1}{2}\left( 12-x \right)=6-\dfrac{x}{2}$ The length of the minor semi-axis of the ellipse is
$\beta =\sqrt{{{a}^{2}}-{{\left( \dfrac{x}{2} \right)}^{2}}}=\sqrt{{{\left( 6-\dfrac{x}{2} \right)}^{2}}-\dfrac{{{x}^{2}}}{4}}=\sqrt{36-6x}$ For the height $h=AD$ of $\triangle ABC$ we have $\left[ ABC \right]=6\Rightarrow \frac{1}{2}xh=6\Rightarrow h=\frac{12}{x}$ Hence, vertex $A$ lies on the line $l$ with equation $y=\dfrac{12}{x}$ , ( $x$ is constant).

We conclude that $\triangle ABC$ exists iff the ellipse and the line do intersect, i.e when $\dfrac{12}{x}\le \beta \Rightarrow \dfrac{12}{x}\le \sqrt{36-6{{x}^{2}}}\overset{x>0}{\mathop{\Leftrightarrow }}\,\dfrac{144}{{{x}^{2}}}\le 36-6{{x}^{2}}\Leftrightarrow {{x}^{3}}-6x^2+24\le 0$ We can easily find that function $f\left( t \right)={{t}^{3}}-6t^2+24$ , $t>0$ , is non negative in an interval $\left[ {{x}_{1}},{{x}_{2}} \right]\subseteq \left( 0,6 \right)$ , with ${{x}_{1}}$ and ${{x}_{2}}$ being its zeros. Hence, the maximum value for the side $x$ is ${{x}_{2}}$ , the greatest root of $f$ .

For the answer, $f\left( 2 \right)={{2}^{3}}-6\times {{2}^{2}}+24=\boxed{8}$ .