<s>[INST] Finding the largest triangle inscribed in the unit circle Among all triangles inscribed in the unit circle, how can the one with the largest area be found?
 [/INST] Given any triangle $\triangle ABC$ of sides $a,b$ and $c$, let $R$ be its circumradius and $\mathcal{A}$ be its area. We have this interesting identity:
$$4 R \mathcal{A} = abc$$
When $ABC$ is inscribed inside the unit circle, $R = 1$ and by $GM \le AM$, we have
$$\mathcal{A} = \frac14 abc \le \frac14 \left(\frac{a^2+b^2+c^2}{3}\right)^{3/2}$$
Notice 
$$\begin{align}a^2 + b^2 + c^2 
&= |\vec{A} - \vec{B}|^2 + |\vec{B} - \vec{C}|^2 + |\vec{C}-\vec{A}|^2\\
&= 6 - 2\left(\vec{A}\cdot\vec{B} + \vec{B}\cdot\vec{C} + \vec{C}\cdot\vec{A}\right)\\
&= 9 - |\vec{A} + \vec{B} + \vec{C}|^2
\end{align}
$$
This leads to an upper bound for the area
$$\mathcal{A} \le \frac14 \left(\frac{9}{3}\right)^{3/2} = \frac{3\sqrt{3}}{4}$$
Since this upper bound is attained by an equilateral triangle of side $\sqrt{3}$, the maximum area is $\frac{3\sqrt{3}}{4}$.
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