SOLUTION: 4 congruent circles, each of which is tangent externally to 2 of the other circles, are circumscribed by a square of area 16. A small circle is then placed in the center so that it

Since the area of the square is 16 square units, each side of the 
square is the square root of 16 or 4 units.

This green line: 



is also 4 units and goes across both circles on the left.
so each of the four larger circles has radius 1.

Now let's draw a diagonal of the square and letter some points:




ABC is a right triangle,

AB=1, BC=1, so by the Pythagorean theorem AC = 

and since GH = AC, GH = 

CD and FG are both radii of the larger circles so they are 1 each.

Triangle AIH is a right triangle, AI = 4, IH = 4, so by the
Pythagorean theorem, diagonal AH = 

We add up the parts of the diagonal AH and equate the sum to  

AC + CD + DF + FG + GH =  

Substituting the values for the parts that we know the lengths of:

+1+DF+1+ = 

Now we solve for DF

+2+DF = 

DF =  

DF is the diameter of small circle E, so its radius

is  or  



Edwin