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Re: GMAT Club World Cup 2022 (DAY 3): Three tangent circles are inscribed [#permalink]
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The correct answer is C.
The picture shows the detailed solution to the problem.
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Re: GMAT Club World Cup 2022 (DAY 3): Three tangent circles are inscribed [#permalink]
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Bunuel wrote:

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. \(\sqrt[4]{27}\)

C. \(3\)

D. \(\sqrt{27}\)

E. \(9\)


 


This question was provided by GMAT Club
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Attachment:
The attachment 1.png is no longer available



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)
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Re: GMAT Club World Cup 2022 (DAY 3): Three tangent circles are inscribed [#permalink]
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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
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Re: GMAT Club World Cup 2022 (DAY 3): Three tangent circles are inscribed [#permalink]
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Attachment:
Three Circles.png
Three Circles.png [ 28.39 KiB | Viewed 2745 times ]

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
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Re: GMAT Club World Cup 2022 (DAY 3): Three tangent circles are inscribed [#permalink]
2
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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
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Re: GMAT Club World Cup 2022 (DAY 3): Three tangent circles are inscribed [#permalink]
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Let O be the circle of the blue circle -

In the figure shown below

Attachment:
Screenshot 2022-07-14 153150.png
Screenshot 2022-07-14 153150.png [ 54.28 KiB | Viewed 2453 times ]


\(\triangle PQR \)

\(\angle QRP = 60\)
\(\angle RPQ = 90\)
\(\angle PQR = 30\)

\(\triangle OMQ \)

\(\angle QOM = 60\)
\(\angle OMQ = 90\)
\(\angle OQM = 30\)

We know that OP = 27 ; OM = 27

From 30-60-90 triangles

OQ = 54; RP = RM = MQ = \(27\sqrt{3}\)

Therefore \(\angle ORM = \angle ORP = 30\)

Attachment:
Screenshot 2022-07-14 160423.png
Screenshot 2022-07-14 160423.png [ 61.71 KiB | Viewed 2458 times ]


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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Re: GMAT Club World Cup 2022 (DAY 3): Three tangent circles are inscribed [#permalink]
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