Ratio of Radii of two circles

Question

The question is this

" Inside a circle of unit radius, an angle of $60$ is formed by the minor arc. A circle is drawn touching the two arms of the chords $PA$ and $PB$ and also the bigger circle tangentially. Find the ratio of the radius of the two circles." i.e.,${\frac{R1}{R2}}=?$

I tried to argue with the power of a point, couldn't get through. I'm missing something to solve the problem.

Answer 1

Let's think this out. $60$ degrees always makes me think of equilateral triangles. If we have a circle inside an equilateral triangle, how does the radius of the circle relate to the sides of an equilateral triangle?

Extend the sides labelled $x$ and $y$ past the circle so that the form an equilateral triangle with the small circle inscribed in it. $2R_1 = $ the height of this triangle. And height of the triangle = $R_2 + $ distance from center of the small circle to the vertex of the triangle.

Now the center of the small circle to the side of the triangle to the vertex of the angle form a 30-60 - 90 triangle so the measures are: radius of small circle = $R_2$, half the side of the triangle = $\sqrt{3}R_2$ and the distance from the center of the small circle to the vertex is $2R_2$.

So $2R_1 = R_2 + 2R_2 = 3R_2$. So $\frac {R_1}{R_2}=\frac 32$

Answer 2

Without additional information, this problem can’t be solved. The points $A$ and $B$ are fixed by the condition that $\angle{APB}$ measures 60°. Unfortunately, this is true for every point $P$ on the major arc. The radius of the inscribed circle will vary with choice of $P$, as demonstrated in the two diagrams below.

One could certainly come up with a formula for the ratio of the radii as a function of $P$, but I suspect that’s not what the problem is asking. Choosing $P$ so that $\triangle{APB}$ is isosceles is a reasonable guess, and that’s covered by fleablood’s answer.