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Bunuel
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MayankSingh
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R AND R creates aright angle triangle with thr ratio R:R: R*ROOT 2
So the Side of the square is about 20 (a bit less)*root 2= 20*1.4=a bit less than 28

a bit less than 28 *4= 110
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Visualization is important here. Let's bring the square into a single Circle.

Diameter of circle: 2*19.5 = 39
Diagonal of Square: \(\sqrt{2}\)*a (a is side of square)
Here
Diagonal = Diameter
\(\sqrt{2}\)*a = 39
a =39/\(\sqrt{2}\)

Perimeter of Square : 4*39/\(\sqrt{2}\) = 110

110
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Connect the centre of all the circles.
Lets suppose A, B,C,D be the centre of the circle passses through the vertices of squares M, N, P &Q
M = vertices between the lines AB
N= vertices between the lines BC
P = vertices between the lines CD
Q= vertices between the lines DA
Then,
AB = AM + MB = 19.5+19.5 = 39
BC = BN + NC = 19.5+ 19.5 = 39

In Triangle MBN,

MN^2 = MB^2 + BN^2 = (19.5)^2 + (19.5)^2
MN = sqrt[ (19.5)^2 + (19.5)2]
MN = sqrt[ 2*19.5*19.5]
MN = 19.5 * 1.4

Perimeter of square = 4* side of square = 4*MN = 4* 19.5*1.4 = 4*195*14/10*10 = 780*14/100
= 110cm

Answer : D
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MayankSingh
IMO D

Refer attached Image.
Let ABCD be the square & O1, O2, O3, O4 be the center of the circles.
Let join the center of the circles to each other & opposite vertices of the sqaure.

Now, diagonal AC & BD are acting as tangent to the circle at point of contact.
And, Tangent is perpendicular to the radius at point of contact .

So, AC perpendicular to O1O2 & BD AC perpendicular to O2O3
In Quadrilateral AEBO2, Angle AO2B = 90


so AB = Sqrt (R^2+R^2) = 19.5 Sqrt 2
Perimeter = 4x19.5 Sqrt2 = 110


Can you please explain why the diagonal of the square will be a tangent to the circle? the question only mentions that the two circles touch at the vertice of the square
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