Given a circle $(O)$ with radius $R$, starting from point $A$ on the circumference, we measure on the circle the $60^{\circ}$ arc $\widehat{A B}$, the $90^{\circ}$ arc $\widehat{B C}$, and the $120^{\circ}$ arc $\widehat{C D}$.

$1^{0}$. Calculate the sides of the quadrilateral $A B C D$ as functions of $R$!

$2^{0}$. Show that the diagonals of this quadrilateral are perpendicular to each other! What are the lengths of the segments of the diagonals?

$3^{0}$. Calculate the area of the quadrilateral!
$1^{0}$. The fourth side of the quadrilateral, $D A$, is a chord corresponding to an arc of

$$
360^{\circ}-\left(60^{\circ}+90^{\circ}+120^{\circ}\right)=90^{\circ} \text { -arc }
$$

The quadrilateral is a symmetric trapezoid because $B C=D A$ and $A B \| C D$.

Already now, $A B$ is the side of a regular hexagon $=R$.

$B C=A D$ is the side of a square inscribed in the circle $=R \sqrt{2}$.

$C D$ is the side of an equilateral triangle inscribed in the circle $=R \sqrt{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_b3a5259fa85b02a27e5ag-1.jpg?height=418&width=417&top_left_y=485&top_left_x=846)

$2^{0}$. The measure of the angle formed by the diagonals is half the sum of the arcs they intercept:

$$
\frac{1}{2}(\widehat{A B}+\widehat{C D})=\frac{1}{2} \cdot 180^{\circ}=90^{\circ}=\frac{1}{2}(\widehat{B C}+\widehat{C D})
$$

The diagonals of the symmetric trapezoid are equal: $A C=B D$. Let their intersection point be $K$. The $\Delta C K D$ is isosceles because $C D K \angle \equiv C D B \angle=\frac{1}{2} \widehat{B C}=45^{\circ}$.

Therefore,

$$
D K=C K=\frac{C D}{\sqrt{2}}=R \sqrt{\frac{3}{2}}
$$

The $\Delta A K D$ is a right triangle in which

$$
A D K \angle=\frac{1}{2} \widehat{A B}=30^{\circ} \text { and thus } K A=\frac{1}{2} A D=\frac{1}{2} R \sqrt{2} \text {. }
$$

Similarly, it can be seen that $B K=K A$.

$3^{0} \mathrm{~A}$ The area of the quadrilateral

$$
\begin{gathered}
\frac{1}{2} C A \cdot B K+\frac{1}{2} C A \cdot D K=\frac{1}{2} C A(B K+K D)=\frac{1}{2} C A \cdot B D=\frac{1}{2} \overline{C A}^{2} \\
=\frac{1}{2}\left(R \sqrt{\frac{3}{2}}+\frac{1}{2} R \sqrt{2}\right)^{2}=\frac{R^{2}}{4}(\sqrt{3}+1)^{2}=R^{2} \frac{2+\sqrt{3}}{2}
\end{gathered}
$$

András Deák (Érseki g. V. o. Bp. II.).

Note. $\overline{C A} \cdot \overline{B D}=\overline{C A}^{2}$ can also be calculated based on Ptolemy's theorem.