A sliding equilateral triangle

The area of an ellipse is $A=\pi\cdot a\cdot b$ where $a$ is the semi-major axis and $b$ is the semi-minor axis. Thus $a$ is the furthest our vertex gets from the origin, while $b$ is the closest it gets to the origin. We will take advantage of the symmetry of the triangle's movement and the fact that there is only one extreme point in each quadrant: the maximum and minimum distances must occur when the other two vertices are equidistant from the origin.

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I will use the above diagram to find $a$ . Using that notation, $a=DC=DE+EC$ . Note that $\triangle ADB$ is a 45-45-90 right triangle with hypotenuse $1$ . Thus its altitude $DE$ will have length $\frac{1}{2}$ . $EC$ is the altitude of a equilateral triangle, and thus has length $\frac{\sqrt{3}}{2}$ .

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I will use the above diagram to find $b$ . Similarly to before, $b=DC=EC-DE$ . Again, $EC=\frac{\sqrt{3}}{2}$ and $DE=\frac{1}{2}$ .

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So the area of the ellipse is $\pi\cdot a\cdot b = \pi\left(\frac{\sqrt{3}}{2} + \frac{1}{2}\right)\left(\frac{\sqrt{3}}{2} - \frac{1}{2}\right) = \pi\left(\left(\frac{\sqrt{3}}{2}\right)^2 - \left(\frac{1}{2}\right)^2\right) = \pi\left(\frac{3}{4} - \frac{1}{4}\right) = \boxed{\dfrac{\pi}{2}}$