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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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20 Nov 2017, 00:15
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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

Attachment:

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

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

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20 Nov 2017, 00:36
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--
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Posts: 3847
Four circles of radius 2 with centers A, B, C and D are arranged symme  [#permalink]

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20 Nov 2017, 10:47
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
$$s^2+s^2=d^2$$

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

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

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

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

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20 Nov 2017, 11:49
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

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

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14 Aug 2019, 01:45
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?
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Four circles of radius 2 with centers A, B, C and D are arranged symme  [#permalink]

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15 Aug 2019, 17:29
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

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

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15 Aug 2019, 21:16
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
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Four circles of radius 2 with centers A, B, C and D are arranged symme   [#permalink] 15 Aug 2019, 21:16