Circles in a Square

Date: 09/15/2001 at 14:00:48
From: Ash Thotambilu
Subject: Circles and squares

A circle of radius 1 is inside a square whose side has length 2. Show 
that the area of the largest circle that can be inscribed between the 
circle and the square is (pi(17 - 12sqrt(2))).

Date: 09/15/2001 at 15:03:25
From: Doctor Jubal
Subject: Re: Circles and squares

Hi Ash,

Thanks for writing Dr. Math.

The little circle is inscribed in the corner of the square so that it
touches the big circle and both sides of the square. Its diameter lies
along one of the square's diagonals.

Draw in that diagonal of the square, and label the following points:

A: the center of the big circle
B: where the diagonal intersects both the big and the little circles.  
   The two circles are tangent at this point
C: the center of the little circle
D: where the diagonal intersects the little circle, opposite B
E: the corner of the square that the little circle is inscribed in

  

A,B,C,D,E should lie on the diagonal in that order.

The square has side length 2, and the diagonal of a square is sqrt(2) 
times longer than its side. (You can prove this to yourself with the 
Pythagorean theorem. Feel free to write back if you have trouble.)  
So, the square's diagonal has length 2*sqrt(2). AE is half the 
square's diagonal, so AE has length sqrt(2).

AB is the big circle's radius, with length 1. Comparing AB to AE, we 
can arrive at the following theorem:

  When a circle is inscribed inside a right angle,

   distance(center of circle to vertex of angle)     AE
  ----------------------------------------------- = ---- = sqrt(2)
                 radius of circle                    AB

Now, the smaller circle is also a circle inscribed inside a right 
angle, so we can use this theorem to write

  CE / CD = sqrt(2)

CD is just the radius of the little circle (I'll call this r from here 
on out), and we can find CE by noting
 
  CE = BE - BC

But BC has length r, and BE is AE - AB = sqrt(2) - 1.  So, we can 
write

  CE = sqrt(2) - 1 - r
  
   sqrt(2) - 1 - r
  ----------------- = sqrt(2)
          r 

I'll leave it to you to solve for r and use r to find the area of the 
circle (A = pi*r^2), but if you have any further trouble, feel free to 
write back. 

- Doctor Jubal, The Math Forum
  http://mathforum.org/dr.math/