\[
\begin{aligned} & {\left[\begin{array}{l}\text { Law of Sines }\end{array}\right]} \\ & \text { [Inscribed angle subtended by the diameter] }\end{aligned}
\]

In triangle $ABC$, point $D$ lies on side $BC$, and point $O$ lies on segment $AD$. It is known that points $C, D$, and $O$ lie on a circle, the center of which is on side $AC$, $4AC = 3\sqrt{2}AB$, angle $DAC$ is twice angle $BAD$, and angle $OCA$ is half of angle $OCB$. Find the cosine of angle $ABC$.
Denote $\angle A C O=\alpha, \angle D C O=2 \alpha, \angle B A D=\beta, \angle C A D=2 \beta$. Express $\beta$ in terms of $\alpha$ and apply the Law of Sines to triangle $A B C$.

## Solution

Let

$$
\angle A C O=\alpha, \angle D C O=2 \alpha, \angle B A D=\beta, \angle C A D=2 \beta
$$

Let $C E$ be the diameter of the circle passing through points $C, D$, and $O$. Then

$$
\angle O D E=\angle O C E=\alpha, \angle C D E=90^{\circ}, \angle D E C=90^{\circ}-3 \alpha
$$

Since $D E C$ is the exterior angle of triangle $A D E$, we have

$$
\angle D E C=\angle D A E+\angle A D E=\angle D A E+\angle O C E, \text { or } 90^{\circ}-3 \alpha=\alpha+2 \beta,
$$

from which $\beta=45^{\circ}-2 \alpha$. Therefore,

$$
\begin{aligned}
\angle A B C & =180^{\circ}-\angle A C B-\angle B A C=180^{\circ}-3 \alpha-3 \beta= \\
& =180^{\circ}-3 \alpha-135^{\circ}+6 \alpha=45^{\circ}+3 \alpha
\end{aligned}
$$

By the Law of Sines,

$$
\frac{\sin \angle A B C}{\sin \angle A C B}=\frac{A C}{A B} \text {, or } \frac{\sin \left(45^{\circ}+3 \alpha\right)}{\sin 3 \alpha}=\frac{3 \sqrt{2}}{4} \text {. }
$$

After obvious transformations, we get

$$
3 \sqrt{2} \sin 3 \alpha=2 \sqrt{2}(\sin 3 \alpha+\cos 3 \alpha), \sin 3 \alpha=2 \cos 3 \alpha, \operatorname{tg} 3 \alpha=2
$$

from which we find that

$$
\cos ^{2} 3 \alpha=\frac{1}{1+\operatorname{tg}^{2} 3 \alpha}=\frac{1}{5}
$$

Since $\operatorname{tg} 3 \alpha=2$, we have $3 \alpha<90^{\circ}$, so

$$
\cos 3 \alpha=\frac{1}{\sqrt{5}}, \sin 3 \alpha=\frac{2}{\sqrt{5}}
$$

Therefore,

$$
\begin{aligned}
\cos \angle A B C= & \cos \left(45^{\circ}+3 \alpha\right)=\cos 45^{\circ} \cos 3 \alpha-\sin 45^{\circ} \sin 3 \alpha= \\
& =\frac{\sqrt{2}}{2}\left(\frac{1}{\sqrt{5}}-\frac{2}{\sqrt{5}}\right)=-\frac{1}{\sqrt{10}}
\end{aligned}
$$

## Answer

$-\frac{1}{\sqrt{10}}$