It is currently 20 Mar 2018, 14:41

### 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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 44351
Four circles of radius 2 with centers A, B, C and D are arranged symme [#permalink]

### Show Tags

20 Nov 2017, 01:15
1
KUDOS
Expert's post
00:00

Difficulty:

(N/A)

Question Stats:

73% (02:20) correct 27% (02:16) wrong based on 22 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

[Reveal] Spoiler:
Attachment:

2017-11-20_1215.png [ 24.4 KiB | Viewed 843 times ]
[Reveal] Spoiler: OA

_________________
Senior Manager
Joined: 09 Mar 2017
Posts: 355
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

20 Nov 2017, 01:36
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

VP
Joined: 22 May 2016
Posts: 1424
Four circles of radius 2 with centers A, B, C and D are arranged symme [#permalink]

### Show Tags

20 Nov 2017, 11:47
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

[Reveal] Spoiler:
Attachment:
The attachment 2017-11-20_1215.png is no longer available

Attachment:

rrrrrrr.png [ 30.32 KiB | Viewed 548 times ]

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

Find the length of the square's diagonal, then the length of the square's side.

From the square's side, subtract two radii of length 2: that is the diameter of small circle.

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

Length of square's diagonal AC (see diagram)
Diagonal length = (radius of circle A) + (radius of middle circle * 2) + (radius of circle C)
Diagonal length = (2 + 4 + 2) = 8

Side length, $$s$$, of the square with diagonal $$d$$?

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

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

Side of square and diameter of small circle
$$s$$ = (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 = $$(s - 4) = (4\sqrt{2} - 4)$$

Diameter of small circle = (side length) - 4

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

Radius of small circle = $$\frac{1}{2}$$diameter, so

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.

_________________

At the still point, there the dance is. -- T.S. Eliot
Formerly genxer123

Manager
Joined: 03 May 2017
Posts: 113
Four circles of radius 2 with centers A, B, C and D are arranged symme [#permalink]

### Show Tags

20 Nov 2017, 12:49
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

[Reveal] Spoiler:
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.
Four circles of radius 2 with centers A, B, C and D are arranged symme   [#permalink] 20 Nov 2017, 12:49
Display posts from previous: Sort by