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

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

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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
2017-11-20_1215.png [ 24.4 KiB | Viewed 725 times ]
[Reveal] Spoiler: OA

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

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New post 20 Nov 2017, 00:36
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--
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Four circles of radius 2 with centers A, B, C and D are arranged symme [#permalink]

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New post 20 Nov 2017, 10:47
Bunuel wrote:
Image
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
rrrrrrr.png [ 30.32 KiB | Viewed 442 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 = \((s - 4) = (4\sqrt{2} - 4)\)

Small circle's radius
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 = \(2\sqrt{2} - 2\)

Answer C

*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.

Kudos [?]: 403 [0], given: 645

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

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New post 20 Nov 2017, 11:49
Bunuel wrote:
Image
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.

Kudos [?]: 19 [0], given: 14

Four circles of radius 2 with centers A, B, C and D are arranged symme   [#permalink] 20 Nov 2017, 11:49
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Four circles of radius 2 with centers A, B, C and D are arranged symme

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