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by shankar.ashwin » Wed Sep 21, 2011 7:13 am
In a circular paper of radius 5, four congruent circles are drawn such that no two circles intersect each other. What is the maximum possible radius of the inner circles?

A) 5(sqrt(2)-1)
B) 2(2 - sqrt3)
C) 2sqrt2 - 1
D) 2.5
E) sqrt3/2 (2 - sqrt2)
Last edited by shankar.ashwin on Wed Sep 21, 2011 8:08 am, edited 1 time in total.
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by gmatclubmember » Wed Sep 21, 2011 7:45 am
shankar.ashwin wrote:In a circular paper of radius 5, four congruent circles are drawn such that no two circles intersect each other. What is the maximum possible radius of the inner circles?

A) 5(2- sqrt2)
B) 2(2 - sqrt3)
C) 2sqrt2 - 1
D) 2.5
E) sqrt3/2 (2 - sqrt2)
Consider the radius of one of the congruent circles which is inscribed in one of the 4 sectors (1/4th of circle) of circle is 'x'.
then the x+sqrt(2)x=5 => which gives x=5(sqrt(2)-1), which I dont see among the answers :(

Could you please confirm the answer choices ??

Cheers
Ami/-
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by cavaliar » Wed Sep 21, 2011 7:51 am
I agree with you ami---- > answer is

5 /(sqrt2 + 1) or 5*(sqrt2 - 1)
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by shankar.ashwin » Wed Sep 21, 2011 8:09 am
Oops sorry guys... typo.. Apologies.

Corrected it now. And gmatclubmember- Bang on!!
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by ashutoshkumar7 » Wed Sep 21, 2011 10:32 am
Ami,
How did u reach to this equation?

x+sqrt(2)x=5
Thanks,
Ashutosh
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by gmatclubmember » Wed Sep 21, 2011 10:43 am
ashutoshkumar7 wrote:Ami,
How did u reach to this equation?

x+sqrt(2)x=5
Ashu for your eyes only :) (Sorry 007)
Please see the attached figure:
lets consider the radius of smaller circle as 'a'.
then OI = sqrt(IC^2 + OC^2) = a * sqrt2
OE = 5 = EI + OI = a + a*sqrt(2)
a=5/(sqrt2+1).

Cheers
Ami/-
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by gmatclubmember » Wed Sep 21, 2011 10:44 am
Image

Forgot to attach the image earlier :(((
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by ashutoshkumar7 » Wed Sep 21, 2011 11:02 am
Thanks Amy.

I didn't realize that the the diagonals of the square are equal :)
Thanks,
Ashutosh
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