B2. On the circle $\mathcal{K}$, points $A$ and $B$ are such that the length of the arc $\widehat{A B}$ is equal to $\frac{1}{3}$ of the circumference of the circle $\mathcal{K}$. Point $C$ lies on the arc $\widehat{A B}$, and point $D$ lies on the chord $A B$. Calculate the area of triangle $A C B$, given that $|A D|=2$, $|B D|=1$, and $|C D|=\sqrt{2}$.
B2.

![](https://cdn.mathpix.com/cropped/2024_06_07_e347356273e0c1397da7g-25.jpg?height=617&width=646&top_left_y=214&top_left_x=725)

Let $S$ be the center of the circle $\mathcal{K}$. Since the length of the arc $\widehat{A B}$ is $\frac{1}{3}$ of the circumference of the circle $\mathcal{K}$, we have $\angle A S B = \frac{2 \pi}{3}$ or $\angle B S A = \frac{4 \pi}{3}$. By the theorem of the inscribed and central angles, $\angle B C A = \frac{1}{2} \angle B S A = \frac{2 \pi}{3}$. Let $|B C| = a$ and $|A C| = b$, and $\angle A D C = \alpha$. By the cosine rule for triangles $A C B$, $A C D$, and $C B D$, we have

\[
\begin{aligned}
9 & = a^{2} + b^{2} - 2 a b \cos \frac{2 \pi}{3} \\
b^{2} & = 4 + 2 - 2 \cdot 2 \cdot \sqrt{2} \cos \alpha \\
a^{2} & = 1 + 2 - 2 \cdot 1 \cdot \sqrt{2} \cos (\pi - \alpha)
\end{aligned}
\]

Considering the properties $\cos \frac{2 \pi}{3} = -\frac{1}{2}$ and $\cos (\pi - \alpha) = -\cos \alpha$, the equations simplify to

\[
\begin{aligned}
9 & = a^{2} + b^{2} + a b \\
b^{2} & = 6 - 4 \sqrt{2} \cos \alpha \\
a^{2} & = 3 + 2 \sqrt{2} \cos \alpha
\end{aligned}
\]

If we add twice the third equation to the second equation, we get $2 a^{2} + b^{2} = 12$. From the first equation, we can express $a^{2} + b^{2} = 9 - a b$. Now we combine these two equations. Subtracting the first equation from the second gives $a^{2} = 3 + a b$, and subtracting twice the second equation from the first gives $b^{2} = 6 - 2 a b = 2(3 - a b)$. Multiplying the derived equations, we get

\[
a^{2} b^{2} = 2(3 - a b)(3 + a b) = 2(9 - a^{2} b^{2}) = 18 - 2 a^{2} b^{2}
\]

From this, we express $a^{2} b^{2} = 6$ or $a b = \sqrt{6}$. The area of triangle $A C B$ is therefore

\[
p = \frac{1}{2} a b \sin \frac{2 \pi}{3} = \frac{1}{2} \cdot \sqrt{6} \cdot \frac{\sqrt{3}}{2} = \frac{3 \sqrt{2}}{4}
\]

2nd method. Since $|S A| = |S B|$ and $\angle A S B = \frac{2 \pi}{3}$, triangle $A B S$ is composed of two halves of an equilateral triangle. Since $|A B| = 3$, it follows that $|S A| = |S B| = \sqrt{3}$.

Using the cosine rule in triangle $D B S$, we calculate

\[
|D S| = \sqrt{|B D|^{2} + |B S|^{2} - 2 |B D| \cdot |B S| \cdot \cos \angle S B D} = \sqrt{1 + 3 - 2 \cdot 1 \cdot \sqrt{3} \cdot \frac{\sqrt{3}}{2}} = 1
\]

From this, it follows that $\angle B D S = \frac{2 \pi}{3}$, since we have just shown that triangle $B D S$ is isosceles with vertex $D$.

Now we observe that $|S D|^{2} + |C D|^{2} = 1 + 2 = 3 = |S C|^{2}$, which means that $C D S$ is a right triangle with a right angle at vertex $D$. It follows that $\angle A D C = \frac{\pi}{6}$, and using the cosine rule, we calculate

\[
|A C| = \sqrt{|A D|^{2} + |C D|^{2} - 2 |A D| \cdot |C D| \cdot \cos \frac{\pi}{6}} = \sqrt{4 + 2 - 2 \cdot 2 \sqrt{2} \cdot \frac{\sqrt{3}}{2}} = \sqrt{6 - 2 \sqrt{6}}
\]

and

\[
|B C| = \sqrt{|B D|^{2} + |C D|^{2} + 2 |A D| \cdot |C D| \cos \frac{\pi}{6}} = \sqrt{1 + 2 + 2 \cdot \sqrt{2} \cdot \frac{\sqrt{3}}{2}} = \sqrt{3 + \sqrt{6}}
\]

It follows that the area of triangle $A B C$ is

\[
\frac{1}{2} |A C| \cdot |B C| \cdot \sin \angle B C A = \frac{1}{2} \sqrt{(6 - 2 \sqrt{6})(3 + \sqrt{6})} \cdot \frac{\sqrt{3}}{2} = \frac{1}{2} \sqrt{6} \cdot \frac{\sqrt{3}}{2} = \frac{3 \sqrt{2}}{4}
\]

1st METHOD AND METHODS SIMILAR TO THE 1st.
Calculation of $\angle B C A = \frac{\pi}{3}$ or $|S A| = \sqrt{3}$ ..... 1 point.
Correct writing of one cosine rule ..... 1 point.
Writing a system of $n$ equations with $n$ unknowns (for some $n$) that includes the lengths $|A C|$ and $|B C|$ ..... 1 point.
Converting the system to one equation in one unknown ..... 2 points.
Solving the obtained equation ..... 1 point.
Calculating the area using the obtained solution ..... 1 point.
2nd METHOD.

Calculation of $|S A| = \sqrt{3}$

1 point.
Calculation of $|S D| = 1$ ..... 2 points.
Proof that $\angle S D C$ is a right angle ..... 2 points.
Calculation of the lengths of segments $A C$ and $B C$ or the height of triangle $A B C$ on side $A B$ ..... 1 point.
Calculation of the area of triangle $A B C$ ..... 1 point.