Question

In a circle with centre $O$, suppose $A,P,B$ are three points on its circumference such that $P$ is the mid-point of minor arc $AB$. Suppose when $\angle AOB=\theta$.
$\frac{area\left(△AOB\right)}{area\left(△APB\right)}=\sqrt{5}+2$,
If $\angle AOB$ is doubled to $2\theta$, then the ratio $\frac{area\left(△AOB\right)}{area\left(△APB\right)}$ is

• A. $\frac{1}{\sqrt{5}}$
• B. $\sqrt{5}-2$
• C. $2\sqrt{3}+3$
• D. $\frac{\sqrt{5}-1}{2}$

Answer 1

The correct option is A $\frac{1}{\sqrt{5}}$

Area $△AOB=\frac{1}{2}ab\sin \theta =\frac{1}{2}{r}^2\sin \theta$

Area$\left(△APB\right)=$ Area$\left(△AOP\right)+$ Area$\left(△POB\right)-$ Area$\left(△AOB\right)$

$=\frac{1}{2}{r}^2\sin \left(\frac{\theta }{2}\right)+\frac{1}{2}{r}^2\sin \left(\frac{\theta }{2}\right)-\frac{1}{2}{r}^2\sin \theta$
$=\frac{1}{2}{r}^2[2\sin \left(\frac{\theta }{2}\right)-\sin \theta ]$
$\frac{Area\left(△AOB\right)}{Area\left(△APB\right)}=\frac{\frac{1}{2}{r}^2\sin \theta }{\frac{1}{2}{r}^2[2\sin \left(\frac{\theta }{2}\right)-\sin \theta ]}=\sqrt{5}+2$
$\Rightarrow \left[\frac{2\sin \left(\frac{\theta }{2}\right)\cos \left(\frac{\theta }{2}\right)}{2\sin \left(\frac{\theta }{2}\right)-2\sin \left(\frac{\theta }{2}\right)\cos \left(\frac{\theta }{2}\right)}\right]=\sqrt{5}+2$
$\Rightarrow \frac{\cos \left(\frac{\theta }{2}\right)}{1-\cos \left(\frac{\theta }{2}\right)}=\sqrt{5}+2$
$\cos \left(\frac{\theta }{2}\right)=(\sqrt{5}+2)[-\cos \left(\frac{\theta }{2}\right)+1]$
$(3+\sqrt{5})\cos \left(\frac{\theta }{2}\right)=(2+\sqrt{5})$
$\cos \left(\frac{\theta }{2}\right)=\frac{2+\sqrt{5}}{(3+\sqrt{5})}\cos \theta =\frac{2+\sqrt{5}}{(7+3\sqrt{5})}$

If $\theta$ is the doubled

$Area\left(△AOB\right)=\frac{r^2}{2}\sin 2\theta Area\left(△APB\right)=Area\left(△AOP\right)+Area\left(△POB\right)-Area\left(△AOB\right)$
$=\frac{1}{2}{r}^2\sin \theta +\frac{1}{2}{r}^2\sin \theta -\frac{1}{2}{r}^2\sin 2\theta$
$\frac{Area\left(△AOB\right)}{Area\left(△APB\right)}=\frac{\frac{1}{2}{r}^2\sin 2\theta }{\frac{1}{2}{r}^2(2\sin \theta -\sin 2\theta )}=\frac{2\sin \theta \cos \theta }{2\sin \theta -2\sin \theta \cos \theta }$
$=\frac{\cos \theta }{1-\cos \theta }=\frac{2+\sqrt{5}}{5+2\sqrt{5}}=\frac{1}{\sqrt{5}}$