Inellipse

The Steiner inellipse has the maximum area of any inellipse of a triangle.

According to Marden's theorem , If the vertices of a triangle $z_1,z_2,z_3$ are the complex roots of a cubic polynomial $f(x)$ , then the foci of the Steiner inellipse are the roots of the derivative $f'(x)$ .

In this case letting $z_1=0, z_2=1, z_3=\sqrt{3}i$

We have $f(x)=x(x-1)(x-\sqrt{3}i)$

Letting $f'(x)=0$ , we get $3x^2-2(1+\sqrt{3}i)x+\sqrt{3}i=0$

Now, the distance between the foci of the ellipse is equal to the absolute value of the difference between the roots of the above equation.

Which is $\left|\dfrac{2}{3}\sqrt{-2-\sqrt{3}i}\right|=\boxed{\dfrac{2}{3}\sqrt[4]{7}}$

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• Click here for the proof that the Steiner inellipse has the greatest area of all inellipses of a triangle.

• Click here for the proof of Marden's theorem.

A Steiner inellipse has the maximum area of any inellipse of a triangle, and for a triangle with sides $a$ , $b$ , and $c$ it has semi-major and semi-minor axes $\frac{1}{6}\sqrt{a^2 + b^2 + c^2 \pm 2Z}$ where $Z = \sqrt{a^4 + b^4 + c^4 - a^2b^2 - b^2c^2 - c^2a^2}$ . Therefore, for a triangle with sides $a = 1$ , $b = \sqrt{3}$ , and $c = 2$ , the semi-major and semi-minor axes $a_e$ and $b_e$ are $\frac{1}{6}\sqrt{8 \pm 2\sqrt{7}}$ .

The distance $d$ between the foci of an ellipse is twice its linear eccentricity $c_e$ , and $c_e = \sqrt{a_e^2 - b_e^2}$ . Therefore, $d = 2c_e$ $=$ $2\sqrt{a_e^2 - b_e^2}$ $=$ $2\sqrt{(\frac{1}{6}\sqrt{8 + 2\sqrt{7}})^2 - (\frac{1}{6}\sqrt{8 - 2\sqrt{7}})^2}$ $=$ $\boxed{\frac{2}{3}\sqrt[4]{7}}$ .

Since solution does not compile in this space, I have to give it below.
The given formulas are from link below.
link text

What is the area of this Steiner inellipse ?

It is pi/6