A sliding equilateral triangle - 2

By symmetry, the ellipse is centered at the intersection of the two orange lines, and the maximum and minimum distances from this center (the semi-major axis $a$ and and the semi-minor axis $b$ ) occur when the two sliding vertices of the equilateral triangle are equal distances away from this center.

The semi-major axis $a$ can be found by finding $AO$ in the diagram below, when $OB = OC$ and $\angle BOC = 135°$ .

Since $\triangle BOC$ is an isosceles triangle, $\angle OBC = \frac{180° - 135°}{2} = 22.5°$ . Since BD is half a unit length, BD = \frac{1}{2}. Solving right triangle $\triangle BDO$ gives $DO = \frac{1}{2} \tan 22.5° = \frac{\sqrt{2} - 1}{2}$ . $AD$ is the height of a unit equilateral triangle, so $AD = \frac{\sqrt{3}}{2}$ . Therefore, $a = AO = DO + AD = \frac{\sqrt{2} - 1 + \sqrt{3}}{2}$ .

The semi-minor axis $b$ can be found by finding $A'O$ in the diagram below, when $OB' = OC'$ and $\angle B'OC' = 45°$ .

Since $\triangle B'OC'$ is an isosceles triangle, $\angle OB'C' = \frac{180° - 45°}{2} = 67.5°$ . Since B'D' is half a unit length, B'D' = \frac{1}{2}. Solving right triangle $\triangle B'D'O$ gives $D'O = \frac{1}{2} \tan 67.5° = \frac{\sqrt{2} + 1}{2}$ . $A'D'$ is the height of a unit equilateral triangle, so $A'D' = \frac{\sqrt{3}}{2}$ . Therefore, $b = A'O = D'O' - A'D' = \frac{\sqrt{2} + 1 - \sqrt{3}}{2}$ .

Therefore, the area of the ellipse is $A = \pi ab = \pi (\frac{\sqrt{2} - 1 + \sqrt{3}}{2})(\frac{\sqrt{2} + 1 - \sqrt{3}}{2}) = \boxed{\frac{(\sqrt{3} - 1) \pi}{2}}$ .