[ Law of Cosines [Angles subtended by equal arcs and equal chords]

A square is inscribed in a circle of radius 5. A point on the circle is marked, the distance from which to one of the vertices of the square is 6. Find the distances from this point to the other three vertices of the square.

#
Apply the Law of Cosines.

## Solution

Let the distance from point $M$, lying on the circumcircle of square $A B C D$, to vertex $A$ be 6. Since the side of the square is $5 \sqrt{2} > 6$, point $M$ lies on the smaller arc $A B$ (or $A D$). Then

$$
\angle A M C = \angle A B C = 90^{\circ}, \angle D M C = \angle D B C = 45^{\circ} = \angle C A B = \angle C M B.
$$

Let $M B = x, M C = y, M D = z$. Then

$$
y = M C = \sqrt{A C^{2} - A M^{2}} = \sqrt{10^{2} - 6^{2}} = 8
$$

Applying the Law of Cosines to triangle $D M C$, we get the equation

$$
D C^{2} = M D^{2} + M C^{2} - 2 M D \cdot M C \cdot \cos \angle D M C, \text{ or } 50 = z^{2} + 8^{2} - 2 \cdot 8 z \cdot \frac{\sqrt{2}}{2},
$$

where $z > 6$. From this, we find that $z = 7 \sqrt{2}$.

Applying the Law of Cosines to triangle $B M C$ similarly, we find that $x = \sqrt{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_ae1b0d3bbbf9aa5c38e8g-37.jpg?height=760&width=778&top_left_y=546&top_left_x=640)

## Answer

$8, \sqrt{2}, 7 \sqrt{2}$.