Cho tam giác OAB đều cạnh a. Trên đường thẳng d qua O

* Hướng dẫn giải

Đáp án B

Ta có

$\begin{array}{l}AF\perp OB,AF\perp MO\\ \Rightarrow AF\perp \left(MOB\right)\\ \Rightarrow AF\perp MB\end{array}$

Mà  $MB\perp AE$nên $MB\perp \left(AEF\right)\Rightarrow MB\perp EF$  .

Suy ra $\Delta MOB\backsim \Delta MEN$  , mà $\Delta MEN\backsim \Delta FON$ nên $\Delta MOB\backsim \Delta FON$ . Khi đó $\frac{OB}{OM}=\frac{ON}{OF}$$\Rightarrow ON=\frac{OB.OF}{OM}$$=\frac{a.\frac{a}{2}}{x}=\frac{a^2}{2x}$  .

Từ

$\begin{array}{l}{V}_{ABMN}=V_{M.OAB}+V_{N.OAB}\\ =\frac{1}{3}.S_{\Delta OAB}.\left(OM+ON\right)\\ =\frac{1}{3}.\frac{a^2\sqrt{3}}{4}.\left(x+\frac{a^2}{2x}\right)\end{array}$

$\Rightarrow V_{ABMN}=\frac{a^2\sqrt{3}}{12}\left(x+\frac{a^2}{2x}\right)$$\ge \frac{a^2\sqrt{3}}{12}.2\sqrt{x.\frac{a^2}{2x}}$$=\frac{a^2\sqrt{3}}{12}.\sqrt{2}a$$=\frac{a^3\sqrt{6}}{12}$

Dấu “=” xảy ra

$\begin{array}{l}\iff x=\frac{a^2}{2x}\\ \iff 2x^2=a^2\\ \iff x=\frac{a\sqrt{2}}{2}.\end{array}$