Question

If a rectangle is inscribed in an equilateral triangle of side length \(2\sqrt2\) as shown in the figure, then the square of the largest area of such a rectangle is _____________.

Hint:

Let the rectangle's height be \(y\); its width is the side minus \(\frac{2y}{\sqrt3}\). Maximise the area.

Answer 1

Correct Answer: 3

Side \(s=2\sqrt2\). The base angles are \(60^\circ\), so at height \(y\) the width of the triangle is \(s-\frac{2y}{\sqrt3}\), which is also the width of the rectangle with top side at height \(y\).
Area \(A(y)=y\left(2\sqrt2-\frac{2y}{\sqrt3}\right)\). \(A'(y)=2\sqrt2-\frac{4y}{\sqrt3}=0\Rightarrow y=\frac{\sqrt6}{2}\) (which is half the height \(\sqrt6\) of the triangle).
\(A_{max}=\frac{\sqrt6}{2}\left(2\sqrt2-\sqrt2\right)=\frac{\sqrt6}{2}\cdot\sqrt2=\sqrt3\).
\(A_{max}^2=3\).
Answer: 3