[ [ Law of Sines

In a circle, two perpendicular and intersecting chords $A B$ and $C D$ are drawn. It is known that $A B=B C=$ $C D$. Determine which is larger: the area of the circle or the area of the square with side $A B$.

#
Find the angles of triangle $ABC$.

## Solution

Let $M$ be the point of intersection of chords $AB$ and $CD$, $O$ be the center of the circle, $R$ be its radius, and $P$ and $Q$ be the projections of point $O$ onto $AB$ and $CD$ respectively.

Then $MP = QO = PO = MQ$. Therefore, $MB = CM$. Hence, $\angle CBA = \frac{\pi}{4}$, and since $AB = BC$, triangle $ABC$ is isosceles. Therefore,

$$
\angle ACB = \frac{\pi - \frac{\pi}{4}}{2} = \frac{3\pi}{8}
$$

Then

$$
AB = 2R \sin \angle ACB = 2R \sin \frac{3\pi}{8}
$$

and the area of the square with side $AB$ is

$$
\begin{aligned}
AB^2 & = 4R^2 \sin^2 \frac{3\pi}{8} = 2R^2 \left(1 - \cos \frac{3\pi}{4}\right) = \\
& = 2R^2 \left(1 + \frac{\sqrt{2}}{2}\right) = R^2 (2 + \sqrt{2})
\end{aligned}
$$

The area of the circle is $\pi R^2$.

Since $2 + \sqrt{2} > 3.4 > \pi$, the area of the square is larger.

![](https://cdn.mathpix.com/cropped/2024_05_06_df2bbd76ac4dd2f0f20fg-16.jpg?height=492&width=486&top_left_y=1272&top_left_x=795)

## Answer

The area of the square.