An arc AC of a circle subtends a right angle at the centre O.

Question

An arc $AC$ of a circle subtends a right angle at the centre $O$. The point $B$ on arc $AC$ divides it such that $\dfrac{	ext{arc }AB}{	ext{arc }BC} = \dfrac15$. If $\vec{OC} = \alpha\vec{OA}+\beta\vec{OB}$, then $\alpha+\sqrt2(\sqrt3-1)\beta$ is equal to:

Accepted Answer

Since $\angle AOC=90^\circ$ and $\dfrac{\text{arc }AB}{\text{arc }BC}=\dfrac15$: $\angle AOB=15^\circ$, $\angle BOC=75^\circ$.

Treating $\vec{OA},\vec{OB},\vec{OC}$ as unit vectors: $\vec c = \alpha\vec a+\beta\vec b$. Dotting with $\vec a$ (using $\vec a\cdot\vec c=\cos90^\circ=0$, $\vec a\cdot\vec b=\cos15^\circ$):
$$0 = \alpha+\beta\cos15^\circ \quad (1)$$
Dotting with $\vec b$ (using $\vec b\cdot\vec c=\cos75^\circ$):
$$\cos75^\circ = \alpha\cos15^\circ+\beta \quad (2)$$

From (1): $\alpha=-\beta\cos15^\circ$. Substituting into (2): $\cos75^\circ = \beta(1-\cos^215^\circ)=\beta\sin^215^\circ$, so $\beta = \dfrac{\cos75^\circ}{\sin^215^\circ}=\dfrac1{\sin15^\circ} = \dfrac{2\sqrt2}{\sqrt3-1}$.

$\alpha = -\dfrac{\cos15^\circ}{\sin15^\circ} = -\dfrac{\sqrt3+1}{\sqrt3-1}$.

$$\alpha+\sqrt2(\sqrt3-1)\beta = -\frac{\sqrt3+1}{\sqrt3-1} + \sqrt2(\sqrt3-1)\cdot\frac{2\sqrt2}{\sqrt3-1} = -\frac{\sqrt3+1}{\sqrt3-1}+4$$

Rationalizing $\dfrac{\sqrt3+1}{\sqrt3-1} = \dfrac{(\sqrt3+1)^2}2 = \dfrac{4+2\sqrt3}2=2+\sqrt3$:

$$= -(2+\sqrt3)+4 = 2-\sqrt3$$

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