GMAT Club World Cup 2022 (DAY 3): Three tangent circles are inscribed

Question

https://gmatclub.com/forum/download/file.php?id=67254 Three tangent circles are inscribed in a 60° angle, as shown above. If the radius of the blue circle is 27, what is the radius of the yellow circle ? A. 1 B. \sqrt {27} C. 3 D. \sqrt{27} E. 9 1.png

Archit3110 · Answer

joining the circles from the point at 60* we see that its 30-60-90 ∆
and with each circle the distance from the point is 3 times that of the previous circle radius
so middle circle radius is 9 and yellow circle radius is 3
 option C

LeopardLiu · Answer

Answer is C 
For the blue circle, the distance from the point of the angle to the farthest point on the blue circle is 27+27+27 since the radius of the blue circle is 27. Same applies on the middle one. Radius of the middle circle should be 3*r = 27, which is 9.
And same applies on the yellow circle, 3*r=9, then we can get the radius of the yellow circle is 3.

Berdiyor · Answer

The correct answer is C.
The picture shows the detailed solution to the problem.

Iotaa · Answer

Referring to the picture attached, ACF is a 30-60-90 triangle, which makes the sides in the ratio of 2: sqr.3: 1. So AF = 2 * 27 = 54.

Now, triangle ACF & ABE are similar (Angle). So, CF/BE= AF/AE; 27/R = 54/ a+2r+R, So  R= 2r+a  ... (1)

Apply similarity in triangle ABE & smaller triangle, we will get r=a.

We also know AF = 54 = a+2r+2R+27; 27=3r + 6r; 9r=27

Therefore r= 27/9 = 3  (Option C)

NextGenNerd · Answer

Correct answer : choice C

went with an intuitive approach:
So the line from the origin to the tangent of the circle forms 90 degrees
this means we have a 30 60 90 triangle with radius which is given as 27 opposite to the 30 degrees side.
this implies the line from the point to the center of blue circle is 54 by the property of 1 root(3) 2
now 27 is the radius of the circle , so the remaining line becomes 27

applying the same process , the remaining length becomes 9 for the second circle

applying the same process, the radius of the yellow circle becomes 3. hence choice C

KabirKanha · Answer

Three Circles.png We know that if two tangents are drawn from an external point to a circle, the angle bisector of the angle made by these two lines passes through the centre of said circle. In this case, it will pass through the centres of all three circles . Consider the triangle ABC, \sin 30^\circ{} = \frac{AB}{AC} = \frac{1}{2} \Rightarrow \frac{27}{AC} = \frac{1}{2} \Rightarrow AC = 54 Let us denote the radii of the three circles as r_1, r_2, r_3 from smallest to largest. Then we can see by extending the above logic that, 3r_1 = r_2 And 3r_2 = r_3 \Rightarrow r_1= \frac{r_3}{9} = 3 Option C

dsp9846 · Answer

OA - C

as per the figure attached,

let's say smaller circle radius BC = r1
 
triangle ABC is 30-90-60 triangle , so based on the pythagorean theorem AC= 2*BC = 2*r1

Similarly, triangle ADE is also 30-90-60 triangle, so AE = 2* DE = 2*r2

now, AE = AC + CE = 2*r1 + r1 + r2
 so 2*r1 + r1 + r2 = 2*r2 
 r2 = 3*r1

now triangle AFG is also 30-90-60 triangle, so AG = 2*GF = 2*r3

now, AG = AC + CE + FG = 2*r1 + r1 + r2 + r2 + r3 = 3*r1 + 2*r2 + r3
 so 2*r3 = 3*r1 + 2*r2 +r3
 
r3 = 9*r1
 now r3 = 27 given

so r1 = 3

gmatophobia · Answer

Let O be the circle of the blue circle - In the figure shown below Screenshot 2022-07-14 153150.png riangle PQR \angle QRP = 60 \angle RPQ = 90 \angle PQR = 30 riangle OMQ \angle QOM = 60 \angle OMQ = 90 \angle OQM = 30 We know that OP = 27 ; OM = 27 From 30-60-90 triangles OQ = 54; RP = RM = MQ = 27\sqrt{3} Therefore \angle ORM = \angle ORP = 30 Screenshot 2022-07-14 160423.png OX = RX = 27 So, let the radius of the white circle be w 3w = 27 w = 9 Let the radius of the yellow circle be y 3y = 9 y = 3 IMO C

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GMAT Club World Cup 2022 (DAY 3): Three tangent circles are inscribed

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