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

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

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)

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

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

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

this is a 30-60-90 triangle.. sides will be in ratio 1:\(\sqrt{3}\):2
\(r1=x\)
similarly
\(r2=3r1\)
and\(27=3r2=9r1\)

r1 = 3

Please refer to the attachment for constructions on the diagram required for the solution:

Now Triangles GEQ and GFQ are congruent (SSS - Common side, Side equal to radius and 2 tangents from common point to two points on circumference are equal)
Therefore GO line is angular bisector and 60° angle is divided into 2 30° angles.

Also, triangles FQG ~ DPG ~ BOG [all 3 similar triangles as FQ || PD || OB and corresponding angles are equal].
These triangles are all 30-60-90 triangles [Angle QGF=30° and Angle QFG=90° (radius touches tangent perpendicularly]

As shown in image let radius of BLUE circle = R1, WHITE circle=R2 and YELLOW circle=R3

Therefore, 2*R3/R3=(R2+3*R3)/R2 .. [from similarity triangle theorem for triangles FQG and DPG]
OR R2=3R3 .. (1)

Again ,
(9R3+R1)/R1=2R3/R3 .. [from similarity triangle theorem for triangles FQG and BOG]
OR R1=9R3 .. (2)

Therefore since R1=27
R3 (or radius of yellow circle) = 27/9 = 3

ANS C

Solution Image -Answer is C 3

Given: Three tangent circles are inscribed in a 60° angle, as shown above.
Asked: If the radius of the blue circle is 27, what is the radius of the yellow circle ?

Line connecting 0 to centers of the 3 circles makes 30° with line OC.
Angle C1OC = 30°

\(OA = r1/tan30° = r1\sqrt{3}\)
\(OB = r2/tan30° = r2\sqrt{3}\)
\(OC = r3/tan30° = 27\sqrt{3}\)

Since OA is tangent to smallest circle
OAˆ2 = 3r1ˆ2= x (x + 2r1)
xˆ2 + 2r1x - 3rˆ2 = 0
xˆ2 +3r1x - r1x - 3r1ˆ2 = 0
(x + 3r1)(x-r1) = 0
x = r1 since x>0

Since OB is tangent to middle circle
OBˆ2 = 3r2ˆ2 = (x+2r1)(x + 2r1 +2r2) = 3r1(3r1 + 2r2)
r2ˆ2 - 2r1r2 - 3r1ˆ2 = 0
r2ˆ2 - 3r1r2 +r1r2 - 3r1ˆ2 = 0
(r2 - 3r1)(r2+r1) = 0
r2 = 3r1

Since OC is tangent to the largest circle
OCˆ2 = 3r3ˆ2 = 3*27ˆ2 = (x+2r1+2r2)(x+2r1+2r2+2r3) = 9r1(9r1+2*27)
9r1(9r1+54) = 3*27ˆ2
9ˆ2 r1(r1+6) = 3*27ˆ2 = 27*9ˆ2
r1ˆ2 + 6r1 = 27
r1ˆ2 + 6r1 - 27 = 0
r1ˆ2 +9r1 - 3r1 - 27 = 0
(r1 +9)(r1-3) = 0
r1 = 3

The radius of yellow circle is 3 .

IMO C

Let's draw an imaginary line (n) starting from the angle and cutting the centers of the circles up to the blue circle center. Now we have a line that has an inclination of 30 degrees. Next, draw a perpendicular (m) from the center of the circle to the bottom line

Now we have a triangle that is a 30-60-90 degrees combination. Hence we can use the property of the sides to be 1:√3:2

Let's say the radius of the middle circle is r2
Let's say the radius of the yellow circle is r3. We have to find r3

It is given that the perpendicular 'm' is 27. Hence line n becomes 54 by the sides principle. If we remove the radius of the large circle from line 'n' we get the rest to be 27. Hence basis the 30-60-90 degrees combination of triangles we can write
( 27 - r2 ) / r2 = 2:1 => r2 = 9

Now we know the radii of both the blue and middle circle. Hence the remaining length of line 'n' is 54 - 27 - 18 = 9. Hence using the same principle we can write
(9 - r3) / r3 = 2:1 => r3 = 3

IMHO Option C

Please see attached picture...
In the picture, BO = 27, ABO = 90 and OAB = 60/2 = 30 (Line AO will divide the angle A in half)

Triangle ABO is 30-60-90 Triangle, hence, AB = 27\sqrt{3} and AO = 54. (Since, 30:60:90 = 1:\sqrt{3}:2)
Now, PC/BO = AP/AO. Since APC is also 30-60-90 Triangle, AP = 2PC. And AO = AP+PC+BO = 3PC+BO
So, PC/BO = 2PC/(3PC+BO). Solving this, we get PC = 1/3*BO ---> PC = 9

Now, similarly, QD/PC = AQ/AP. Since AQD is also 30-60-90 Triangle, AQ = 2QD. And AP = AQ+QD+PC = 3QD+PC
So, QD/PC = 2QD/(3QD+PC). Solving this, we get QD = 1/3*PC ---> QD = 3

Answer C

The radius of the yellow circle is 3. The answer is C

Answer is C

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Let O be the circle of the blue circle -

In the figure shown below



\(\triangle PQR \)

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

\(\triangle OMQ \)

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

We know that OP = 27 ; OM = 27

From 30-60-90 triangles

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

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



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

The radius of the white circle will be [27]^1/2 = 9

The radius of the yellow circle will be [27]^1/4 = Option B