[Area of a triangle (through two sides and the angle between them)] Inscribed angle is half of the central angle ]  [Cauchy's inequality

A unit square $A B C D$ is inscribed in a circle, and a point $M$ is chosen on this circle.

What is the maximum value that the product $M A \cdot M B \cdot M C \cdot M D$ can take?
Let $M$ not coincide with any of the vertices of the square (otherwise, the value of the product is zero and cannot be the greatest) and lie on the arc $BC$, $O$ - the center of the circumscribed circle (see the figure). Note that $\angle A M B = 1 / 2$

![](https://cdn.mathpix.com/cropped/2024_05_06_3b3613da627e706c6878g-11.jpg?height=86&width=1978&top_left_y=777&top_left_x=14)
$45^{\circ}$, so $M A \cdot M B \cdot M C \cdot M D = 8 S_{A M B} \cdot S_{C M D}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_3b3613da627e706c6878g-11.jpg?height=440&width=418&top_left_y=906&top_left_x=820)

Let $x$ be the distance from point $M$ to the line $AB$, then the distance from $M$ to the line $CD$ is $1 - x$.

Thus, $S_{A M B} = 1 / 2 \cdot A B \cdot x = x / 2$, and

$S_{C M D} = 1 / 2 \cdot C D \cdot (1 - x) = 1 / 2 \cdot (1 - x)$. Therefore, $M A \cdot M B \cdot M C \cdot M D = 2 x (1 - x)$. Since $0 < x < 1$, the obtained expression reaches its maximum value at $x = 1 / 2$. This value is $1 / 2$.

## Answer

0.5.