Key Concepts and Formulas

• Position Vectors on a Circle: For points $A, B, C$ on a circle with center $O$ and radius $R$, their position vectors $\overrightarrow{OA}, \overrightarrow{OB}, \overrightarrow{OC}$ satisfy $|\overrightarrow{OA}| = |\overrightarrow{OB}| = |\overrightarrow{OC}| = R$.
• Dot Product: $\vec{u} \cdot \vec{v} = |\vec{u}| |\vec{v}| \cos \theta$, where $\theta$ is the angle between $\vec{u}$ and $\vec{v}$.
• Arc Length and Central Angle: The ratio of arc lengths is equal to the ratio of their subtended central angles: $\frac{\text{arc length}_1}{\text{arc length}_2} = \frac{\theta_1}{\theta_2}$.
• Trigonometric Identities: $\cos(A-B) = \cos A \cos B + \sin A \sin B$, $\sin(A-B) = \sin A \cos B - \cos A \sin B$, $\cos(90^\circ - \theta) = \sin \theta$, $\sin(90^\circ - \theta) = \cos \theta$, $\sin^2 \theta + \cos^2 \theta = 1$.

Step-by-Step Solution

Step 1: Determine the Central Angles

The arc $AC$ subtends a right angle at the center $O$, so $\angle AOC = 90^\circ$. The point $B$ divides the arc $AC$ such that $\frac{\text{length of arc } AB}{\text{length of arc } BC}=\frac{1}{5}$. This implies the ratio of the central angles subtended by these arcs is also $\frac{1}{5}$. Let $\angle AOB = \theta_1$ and $\angle BOC = \theta_2$. We have $\frac{\theta_1}{\theta_2} = \frac{1}{5}$, which means $\theta_2 = 5\theta_1$. Also, $\theta_1 + \theta_2 = \angle AOC = 90^\circ$. Substituting $\theta_2 = 5\theta_1$ into the sum: $\theta_1 + 5\theta_1 = 90^\circ \implies 6\theta_1 = 90^\circ \implies \theta_1 = 15^\circ$. Then, $\theta_2 = 5 \times 15^\circ = 75^\circ$. So, the angles are $\angle AOB = 15^\circ$, $\angle BOC = 75^\circ$, and $\angle AOC = 90^\circ$.

Step 2: Set up Vector Equations using Dot Products

We are given the vector equation $\overrightarrow{OC}=\alpha \overrightarrow{OA}+\beta \overrightarrow{OB}$. Let $R$ be the radius of the circle, so $|\overrightarrow{OA}| = |\overrightarrow{OB}| = |\overrightarrow{OC}| = R$. To find $\alpha$ and $\beta$, we take the dot product of the given equation with $\overrightarrow{OA}$ and $\overrightarrow{OB}$.

• Dot product with $\overrightarrow{OA}$: $\overrightarrow{OC} \cdot \overrightarrow{OA} = (\alpha \overrightarrow{OA}+\beta \overrightarrow{OB}) \cdot \overrightarrow{OA}$ $|\overrightarrow{OC}||\overrightarrow{OA}| \cos(\angle AOC) = \alpha |\overrightarrow{OA}|^2 + \beta \overrightarrow{OB} \cdot \overrightarrow{OA}$ $R \cdot R \cos(90^\circ) = \alpha R^2 + \beta |\overrightarrow{OB}||\overrightarrow{OA}| \cos(\angle AOB)$ Since $\cos(90^\circ) = 0$: $0 = \alpha R^2 + \beta R^2 \cos(15^\circ)$ Dividing by $R^2$ (since $R \neq 0$): $\alpha + \beta \cos(15^\circ) = 0 \quad (1)$

• Dot product with $\overrightarrow{OB}$: $\overrightarrow{OC} \cdot \overrightarrow{OB} = (\alpha \overrightarrow{OA}+\beta \overrightarrow{OB}) \cdot \overrightarrow{OB}$ $|\overrightarrow{OC}||\overrightarrow{OB}| \cos(\angle BOC) = \alpha \overrightarrow{OA} \cdot \overrightarrow{OB} + \beta |\overrightarrow{OB}|^2$ $R \cdot R \cos(75^\circ) = \alpha |\overrightarrow{OA}||\overrightarrow{OB}| \cos(\angle AOB) + \beta R^2$ $R^2 \cos(75^\circ) = \alpha R^2 \cos(15^\circ) + \beta R^2$ Dividing by $R^2$: $\cos(75^\circ) = \alpha \cos(15^\circ) + \beta \quad (2)$

Step 3: Solve for $\alpha$ and $\beta$

From equation (1), we express $\alpha$ in terms of $\beta$: $\alpha = -\beta \cos(15^\circ)$ Substitute this into equation (2): $\cos(75^\circ) = (-\beta \cos(15^\circ)) \cos(15^\circ) + \beta$ $\cos(75^\circ) = \beta (1 - \cos^2(15^\circ))$ Using the identity $1 - \cos^2 \theta = \sin^2 \theta$: $\cos(75^\circ) = \beta \sin^2(15^\circ)$ We know that $\cos(75^\circ) = \cos(90^\circ - 15^\circ) = \sin(15^\circ)$. So, the equation becomes: $\sin(15^\circ) = \beta \sin^2(15^\circ)$ Since $\sin(15^\circ) \neq 0$, we can divide by $\sin(15^\circ)$: $\beta = \frac{1}{\sin(15^\circ)}$ We calculate $\sin(15^\circ) = \sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6}-\sqrt{2}}{4}$. $\beta = \frac{1}{(\sqrt{6}-\sqrt{2})/4} = \frac{4}{\sqrt{6}-\sqrt{2}}$ To rationalize the denominator: $\beta = \frac{4(\sqrt{6}+\sqrt{2})}{(\sqrt{6}-\sqrt{2})(\sqrt{6}+\sqrt{2})} = \frac{4(\sqrt{6}+\sqrt{2})}{6-2} = \frac{4(\sqrt{6}+\sqrt{2})}{4} = \sqrt{6}+\sqrt{2}$ Now, we find $\alpha$ using $\alpha = -\beta \cos(15^\circ)$. We calculate $\cos(15^\circ) = \cos(45^\circ - 30^\circ) = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6}+\sqrt{2}}{4}$. $\alpha = -(\sqrt{6}+\sqrt{2}) \left(\frac{\sqrt{6}+\sqrt{2}}{4}\right) = -\frac{(\sqrt{6}+\sqrt{2})^2}{4}$ $\alpha = -\frac{6 + 2 + 2\sqrt{12}}{4} = -\frac{8 + 4\sqrt{3}}{4} = -(2+\sqrt{3}) = -2-\sqrt{3}$

Step 4: Evaluate the Expression

We need to compute $\alpha+\sqrt{2}(\sqrt{3}-1) \beta$. Substitute the values of $\alpha$ and $\beta$: $\alpha+\sqrt{2}(\sqrt{3}-1) \beta = (-2-\sqrt{3}) + \sqrt{2}(\sqrt{3}-1)(\sqrt{6}+\sqrt{2})$ Let's simplify the term $\sqrt{2}(\sqrt{3}-1)(\sqrt{6}+\sqrt{2})$: $\sqrt{2}(\sqrt{3}-1)(\sqrt{6}+\sqrt{2}) = (\sqrt{6}-\sqrt{2})(\sqrt{6}+\sqrt{2})$ This is a difference of squares: $(a-b)(a+b) = a^2-b^2$. $= (\sqrt{6})^2 - (\sqrt{2})^2 = 6 - 2 = 4$ Now substitute this back into the full expression: $\alpha+\sqrt{2}(\sqrt{3}-1) \beta = (-2-\sqrt{3}) + 4 = 2-\sqrt{3}$

Common Mistakes & Tips

• Angle Calculation: Ensure the division of the $90^\circ$ angle is done correctly based on the arc length ratio.
• Trigonometric Values: Double-check the values of $\sin 15^\circ$ and $\cos 15^\circ$, as errors here propagate.
• Algebraic Manipulation: Be meticulous with algebraic steps, especially when expanding squares and rationalizing denominators.
• Dot Product Application: The dot product method is standard for finding coefficients in vector linear combinations when the vectors have known relationships (like being on a circle).

Summary

The problem involves expressing $\overrightarrow{OC}$ as a linear combination of $\overrightarrow{OA}$ and $\overrightarrow{OB}$ where $A, B, C$ lie on a circle. We first determined the central angles subtended by the arcs using the given arc length ratio. Then, we used the dot product property of vectors to set up a system of two linear equations for the coefficients $\alpha$ and $\beta$. Solving these equations yielded the values of $\alpha$ and $\beta$, which were then substituted into the expression to be evaluated. The final result is $2-\sqrt{3}$.

The final answer is $\boxed{2-\sqrt{3}}$.