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 $\angle AOB=\theta$,
$\frac{\text{area}\left(△AOB\right)}{\text{area}\left(△APB\right)}=\sqrt{5}+2,$
If $\angle AOB$ is doubled to $2\theta$, then the ratio is $\frac{\text{area}\left(△AOB\right)}{\text{area}\left(△APB\right)}$ 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 $\frac{1}{\sqrt{5}}$


Let the radius of the circle be '$r$'
$\text{area}\left(△AOB\right)=\frac{1}{2}\times r\times r\sin \theta =\frac{1}{2}{r}^2\sin \theta$
Now,
$\text{area}\left(△APB\right)=\text{area}\left(△OAP\right)+\text{area}\left(△OBP\right)-\text{area}\left(△AOB\right)\Rightarrow \text{area}\left(△APB\right)=2\left(\frac{1}{2}{r}^2\sin \frac{\theta }{2}\right)-\frac{1}{2}{r}^2\sin \theta \Rightarrow \text{area}\left(△APB\right)=r^2\sin \frac{\theta }{2}(1-\cos \frac{\theta }{2})$

We know that $\frac{\text{area}\left(△AOB\right)}{\text{area}\left(△APB\right)}=\frac{\frac{1}{2}{r}^2\sin \theta }{r^2\sin \frac{\theta }{2}(1-\cos \frac{\theta }{2})}\Rightarrow \frac{\text{area}\left(△AOB\right)}{\text{area}\left(△APB\right)}=\frac{\cos \frac{\theta }{2}}{1-\cos \frac{\theta }{2}}\Rightarrow \sqrt{5}+2=\frac{\cos \frac{\theta }{2}}{1-\cos \frac{\theta }{2}}\Rightarrow \frac{1}{\sqrt{5}+2}=\frac{1}{\cos \frac{\theta }{2}}-1\Rightarrow \sqrt{5}-2=\frac{1}{\cos \frac{\theta }{2}}-1\Rightarrow \cos \frac{\theta }{2}=\frac{1}{\sqrt{5}-1}=\frac{\sqrt{5}+1}{4}$

$\therefore \cos \theta =2{\cos }^2\frac{\theta }{2}-1=\frac{5+1+2\sqrt{5}}{8}-1=\frac{\sqrt{5}-1}{4}$
When the angle $\theta$ is doubled then,
$\frac{\text{area}\left(△AOB\right)}{\text{area}\left(△APB\right)}=\frac{\cos \theta }{1-\cos \theta }=\frac{\frac{\sqrt{5}-1}{4}}{1-\frac{\sqrt{5}-1}{4}}=\frac{\sqrt{5}-1}{5-\sqrt{5}}=\frac{1}{\sqrt{5}}$