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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.

enter image description here

Jyotishraj Thoudam
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  • Am I to understand that the 60 degree angle is formed by a diameter and a chord? Otherwise the problem seems underdetermined. – law-of-fives May 18 '17 at 19:14
  • No. The 60 degree angle is formed a chord on the circle, which is not drawn on the image. The radius of the bigger circle is given to be of length $1$ – Jyotishraj Thoudam May 18 '17 at 19:16
  • There’s not enough information given to solve this. There needs to be some other constraint on the chord, since for some 60° chords there is no such inscribed circle in the first place. – amd May 18 '17 at 19:34
  • how is this unsolvable? could you please show the conditions. please help – Jyotishraj Thoudam May 18 '17 at 19:36
  • The problem says, "angle of 60 is subtended by a [single] chord." To draw the angle in your diagram, you have two chords. You may wish to research what "subtended" means in this context, because I don't think your diagram accurately reflects the problem. – scott May 18 '17 at 20:09
  • I edited the question – Jyotishraj Thoudam May 18 '17 at 20:18
  • @scott I edited the question. please check? – Jyotishraj Thoudam May 18 '17 at 20:24
  • $\angle{APB}=60°$ determines the length of the chord $AB$, but it doesn’t determine the location of $P$, which can be anywhere on the major arc $\overparen{AB}$. The size of the inscribed circle will vary with the location of this point. – amd May 19 '17 at 02:47

2 Answers2

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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?

![enter image description here

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 $\frac {R_1}{R_2}=\frac 32$

fleablood
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  • fleablood, but what if 2R1 is not so. 2 R1 comes only when we assume the line joining P and your triangle mets both circle at the point of tangency. which may not be true in every case – Jyotishraj Thoudam May 18 '17 at 20:52
  • If the angle is 60 then yes it does have to happen. – fleablood May 18 '17 at 20:55
  • But why 2R1? It's not true if the triangle doesn't meet at z – Jyotishraj Thoudam May 18 '17 at 20:57
  • Consider this. Draw the radii of r_2. one goes to the point label x, one to the point labelled y, and one to the point where the two circles are tangent. center to x is r_2, angle center to vertex to x is 30. and so it is a 30- 60-90. So center to vertex is twice center to x which is r1. so center of small circle to vertex is 2r_2. so the diameter of the big circle is 2r_1 = 3 r_2. – fleablood May 18 '17 at 20:59
  • The triangle will meet at z because it is equilateral. But ignore the triangle. We actually don't need it. Just use the triangles small center-x-vertex. It has sides of length r_2, 2r_2 and root(2)r_2. So from z to center to vertex is r_2 + 2r_2 = 3r_2. – fleablood May 18 '17 at 21:02
  • I'm not able to get it. Could you please present it in more accessible format? – Jyotishraj Thoudam May 18 '17 at 21:04
  • Look at the picture. – fleablood May 18 '17 at 21:16
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    Aren’t you assuming that $\angle P$ is bisected by a diameter? Otherwise, there might not even be an inscribed circle, let alone one for which the circumscribed triangle is equilateral? – amd May 19 '17 at 02:45
  • Pz doesn't bisect the angle P let alone be the diameter – Jyotishraj Thoudam May 19 '17 at 05:49
  • @amd please help? – Jyotishraj Thoudam May 20 '17 at 20:09
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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.

Diagram 1 Diagram 2

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.

amd
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