GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 25 May 2020, 02:05 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # Four circles of radius 2 with centers A, B, C and D are arranged symme

Author Message
TAGS:

### Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 64078
Four circles of radius 2 with centers A, B, C and D are arranged symme  [#permalink]

### Show Tags

2 00:00

Difficulty:

(N/A)

Question Stats: 72% (02:38) correct 28% (02:39) wrong based on 64 sessions

### HideShow timer Statistics Four circles of radius 2 with centers A, B, C and D are arranged symmetrically around another circle of radius 2, and four smaller equal circles with centers E, F, G and H each touch three of the larger circles as shown in the figure above. What is the radius of one of the smaller circles?

(A) √2 – 2
(B) √2 – 1
(C) 2√2 – 2
(D) 1
(E) 3√2 – 1

Attachment: 2017-11-20_1215.png [ 24.4 KiB | Viewed 1741 times ]

_________________
Senior Manager  P
Joined: 09 Mar 2017
Posts: 476
Location: India
Concentration: Marketing, Organizational Behavior
WE: Information Technology (Computer Software)
Four circles of radius 2 with centers A, B, C and D are arranged symme  [#permalink]

### Show Tags

1
1
IMO C

let the radius of small circle be x. Then the side of the square ABCD would be 4 +2x. Let this be called S for now
AC (diagonal of ABCD) is 2+4+2 = 8
applying PGT
$$2(S^2) = 8^2$$
$$S^2 = 32$$
$$(4+2x)^2 = 32$$
$$x^2 +4x - 4 = 0$$
need to use Determinant here for there are no direct factors
$$(-4 + \sqrt{32})/2$$ or
$$(-4 - \sqrt{32})/2$$

solving further gets us to$$(2\sqrt{2}) -2) or ( -2\sqrt{2} -2)$$
options given has only the former of the two.
--C--
_________________
------------------------------
"Trust the timing of your life"
Hit Kudus if this has helped you get closer to your goal, and also to assist others save time. Tq Senior SC Moderator V
Joined: 22 May 2016
Posts: 3847
Four circles of radius 2 with centers A, B, C and D are arranged symme  [#permalink]

### Show Tags

2
1
Bunuel wrote: Four circles of radius 2 with centers A, B, C and D are arranged symmetrically around another circle of radius 2, and four smaller equal circles with centers E, F, G and H each touch three of the larger circles as shown in the figure above. What is the radius of one of the smaller circles?

(A) √2 – 2
(B) √2 – 1
(C) 2√2 – 2
(D) 1
(E) 3√2 – 1

Attachment:
The attachment 2017-11-20_1215.png is no longer available

Attachment: rrrrrrr.png [ 30.32 KiB | Viewed 1401 times ]

All large circles have equal radii, $$r$$ = 2. All small circles have equal radii.*

Find the length of the square's diagonal.
From the length of the diagonal, find the length of the side of the square**:
$$s\sqrt{2}=d$$ and
$$s=\frac{d}{\sqrt{2}}$$

From the square's side, subtract the radii length of the two large circles: that is the diameter of small circle.

Divide the result by 2 to get the radius of small circle.

(1) Length of square's diagonal AC (see diagram)

The diagonal of the square runs only through large circles.
Given: $$r =2$$

Diagonal length = $$(r) + (2r) + (r) = 4r$$
Diagonal length = $$(2 + 4 + 2) = 8$$

(2) Calculate the side length, $$s,$$ from the diagonal, $$d$$

$$s = \frac{d}{\sqrt{2}}$$

$$s = \frac{8}{\sqrt{2}} = (\frac{8}{\sqrt{2}} * \frac{\sqrt{2}}{\sqrt{2}})=4\sqrt{2}$$

(3) Find diameter of small circle from the length of the side of the square
From above, $$s = 4\sqrt{2}$$

Diameter of small circle = (side length) - 4
Because
$$s$$ = (radius of circle B) + (diameter of small circle F) + (radius of circle C) (see diagram, BC)
$$s$$ = (2) + (diameter of small circle) + (2)
$$s$$ = (diameter of small circle) + 4
diameter of small circle = $$(s - 4) = (4\sqrt{2} - 4)$$

(4) Radius of small circle is half of that diameter:

Radius of small circle = $$\frac{4\sqrt{2} - 4}{2}$$

Radius of small circle = $$2\sqrt{2} - 2$$

*All the circles are tangent. Their common tangency points are perpendicular to their radii, thus all radii lie on the same line(s).
The radii can be summed to find diagonal and side of square.

** Derived from properties of squares/isosceles right triangles
$$s^2+s^2=d^2$$

$$2s^2 = d^2$$

$$\sqrt{2}\sqrt{s^2} = \sqrt{d^2}$$

$$\sqrt{2}s=d$$
$$s = \frac{d}{\sqrt{2}}$$

_________________
Visit SC Butler, here! Get two SC questions to practice, whose links you can find by date.

Our lives begin to end the day we become silent about things that matter. -- Dr. Martin Luther King, Jr.
Manager  B
Joined: 03 May 2017
Posts: 80
Four circles of radius 2 with centers A, B, C and D are arranged symme  [#permalink]

### Show Tags

Bunuel wrote: Four circles of radius 2 with centers A, B, C and D are arranged symmetrically around another circle of radius 2, and four smaller equal circles with centers E, F, G and H each touch three of the larger circles as shown in the figure above. What is the radius of one of the smaller circles?

(A) √2 – 2
(B) √2 – 1
(C) 2√2 – 2
(D) 1
(E) 3√2 – 1

Attachment:
2017-11-20_1215.png

The diagonal of the square is $$4r = 8$$.

Half of the square forms an isosceles triangle with sides $$a$$, where $$2(a^2)= 8^2$$. Thus a = $$4 \sqrt{2}$$.

The diameter of the small circle is $$a-2r$$ which is

$$4 \sqrt{2} - 4$$

The radius equals $$0.5 *$$ $$4 \sqrt{2} - 4$$ = $$2 \sqrt{2} - 2$$

Option C is correct.
Manager  S
Joined: 20 Apr 2019
Posts: 112
Re: Four circles of radius 2 with centers A, B, C and D are arranged symme  [#permalink]

### Show Tags

x = radius of the small circle
side of the square = s = 4+2x
diagonal of a square = √2 * s = 8
s = 8/√2
4 + 2x = 8/√2
2x = 8/√2 -4
x = 8/2√2 - 2
x = 4√2 - 2
Although I classified x as the radius of the small circle, my result requires me to divide 4 again by 2. Can someone tell me why? What am I doing wrong?
Senior SC Moderator V
Joined: 22 May 2016
Posts: 3847
Four circles of radius 2 with centers A, B, C and D are arranged symme  [#permalink]

### Show Tags

1
Luca1111111111111 wrote:
x = radius of the small circle
side of the square = s = 4+2x
diagonal of a square = √2 * s = 8
s = 8/√2
4 + 2x = 8/√2
2x = 8/√2 -4
x = 8/2√2 - 2
x = 4√2 - 2
Although I classified x as the radius of the small circle, my result requires me to divide 4 again by 2. Can someone tell me why? What am I doing wrong?

Hi Luca1111111111111 , nothing is wrong with your method. Your arithmetic is not correct. Whoops. I numbered your steps.

s = 8/√2
1) 4 + 2x = 8/√2
2) 2x = 8/√2 -4
3) x = 8/2√2 - 2
4) x = 4√2 - 2

You wrote that 8/2√2 in #3 =>
4√2 in #4. Not correct.

Rationalized fraction in #3:

$$(\frac{8}{2\sqrt{2}}* \frac{\sqrt{2}}{\sqrt{2}})=(\frac{8\sqrt{2}}{2\sqrt{2}*\sqrt{2}})=$$

$$\frac{8\sqrt{2}}{4}=2\sqrt{2}$$

Correct: 8/2√2 in #3 =>
2√2 in #4

4) x = 2√2 - 2

Your method is fine. _________________
Visit SC Butler, here! Get two SC questions to practice, whose links you can find by date.

Our lives begin to end the day we become silent about things that matter. -- Dr. Martin Luther King, Jr.
CEO  V
Joined: 03 Jun 2019
Posts: 2889
Location: India
GMAT 1: 690 Q50 V34 WE: Engineering (Transportation)
Four circles of radius 2 with centers A, B, C and D are arranged symme  [#permalink]

### Show Tags

Bunuel wrote: Four circles of radius 2 with centers A, B, C and D are arranged symmetrically around another circle of radius 2, and four smaller equal circles with centers E, F, G and H each touch three of the larger circles as shown in the figure above. What is the radius of one of the smaller circles?

(A) √2 – 2
(B) √2 – 1
(C) 2√2 – 2
(D) 1
(E) 3√2 – 1

Attachment:
2017-11-20_1215.png

Given: Four circles of radius 2 with centers A, B, C and D are arranged symmetrically around another circle of radius 2, and four smaller equal circles with centers E, F, G and H each touch three of the larger circles as shown in the figure above.

Let the radius of one of the smaller circles be x

AC = 2 + 2*2 +2 = 8
AB = $$4\sqrt{2}$$

AB = $$2*2 + 2 x = 4\sqrt{2}$$
$$x = 2\sqrt{2} - 2$$

IMO C
_________________
Kinshook Chaturvedi
Email: kinshook.chaturvedi@gmail.com Four circles of radius 2 with centers A, B, C and D are arranged symme   [#permalink] 15 Aug 2019, 21:16

# Four circles of radius 2 with centers A, B, C and D are arranged symme  