A Triangle Balances on a Circle

Let $R$ and $r$ be the radius of the large and small circle, respectively.

$\sin~30=\dfrac{x}{R}$ $\implies$ $\dfrac{1}{2}=\dfrac{x}{R}$ $\implies$ $x=\dfrac{R}{2}$

Then,

$R=x+2r=\dfrac{R}{2}+2r$

$R-\dfrac{R}{2}=2r$ $\implies$ $\dfrac{R}{2}=2r$ $\implies$ $R=4r$ $\implies$ $\dfrac{R}{r}=\dfrac{4}{1}$

Square both sides to get the ratio of the areas, we have

$\dfrac{R^2}{r^2}=\dfrac{16}{1}$

The desired answer is $\boxed{16}$ .

A non trigonometric solution:

Hence, the answer is 16.

Let PQR be the inscribed triangle and O be the centre of the circle. Now draw PA perpendicular to QR which passes through O and join OR. Let AB be the diameter of small circle with center C. Let r be the radius of the circle. Now, ORA=30°. sin30°=OA/OR or 1/2=OA/r Therefore, OA=r/2 Now, AB=OB-OA=r - r/2 = r/2. Radius CA=AB/2=r/4 Therefore, (area of big circle)/(area of small circle)=πr²/(πr²/16)=16. Therefore, a:1=16:1. So, a=16.

The key is to notice that the perpendicular bisector of the triangle is $\frac {3}{2}r$ where r is the radius of the smaller circle.

(Centroid divides median into segments of length 2:1)

It follows easily that the ratio is $4^2=\boxed {16}$ .