In a $\triangle ABC$, $\angle A = 30^\circ, BC = 13.$

Question

In a $\triangle ABC$, $\angle A = 30^\circ, BC = 13.$ Given two circles $\gamma_1,\gamma_2$ with radius $r_1,r_2$ respectively, contain $A$ and touch the side $BC$ at $B$ and $C$ respectively. Find $r_1r_2$.

What I Tried:- First, I am having a bit trouble understanding the question. I couldn't get the part that if both the circles contain point $A$ in their circumference, or contain point $A$ just inside them. Can someone just resolve this confusion and post a correct picture of this?

Next, by having confusion on the circles. I have little idea to proceed. I was only able to figure out that by applying Cosine Rule on $\triangle ABC$ gives me :-

$$\rightarrow 13^2 = b^2 + c^2 - \sqrt{3}bc$$

Other than this I have no idea.

Can someone help me? Thank You.

Edit :- I drew a figure of the question, can someone confirm if this is correct or not?

This Question is from LIMITS $2020$ Objective Paper (Conducted by ISI Students)

Answer 1

The point $A$ can be is located anywhere on an upper arc of the circle centered at $O$: $\triangle OCB$ is equilateral. For chosen point $A$ the center $O_1$ is located at the intersection of the perpendicular bisector of $AC$ and the perpendicular to $BC$ through $C$, the other center $O_2$ is found similarly.

Prove that $r_1r_2=|BC|^2$.

Answer 2

Your diagram could be correct. It all depends on those words "touch" and "contain". If we assume the word "touch" means "tangential" and that "contain" means "at the circumference" then an approximate diagram is below (obtained using geogebra). The trouble is that there are two possible places for A (by moving the centres of the circles it may be possible get 30 at the second intersection, but perhaps then the angle A is not "contained" by the circles?).