What is the length of the sides of the largest square which can be inscribed in an annulus given the radii, $r_1$ and $r_2$?

The circles are concentric and the square is completely outside the inner circle (the center point of one side is the tangent point) and completely inside the outer one (two corners lie on it).

Partial ASCII diagram: o[])

  • $\begingroup$ If I have understood the question correctly, a square can never be inscribed in an annulus in the way you describe, except in the special case where $r_2=r_1\sqrt2\,$. Am I missing something? $\endgroup$
    – David
    Jan 22, 2014 at 2:41
  • 3
    $\begingroup$ @David, I think he means the square and its interior lie in the annulus. He doesn't intend for the inner circle to lie within the square. $\endgroup$
    – MPW
    Jan 22, 2014 at 3:02
  • $\begingroup$ @David: MPW is correct, the inner circle and the square do not intersect. $\endgroup$ Jan 22, 2014 at 3:07
  • $\begingroup$ Ah... got it. Thanks. In that case, not too hard, answer coming up ;-) $\endgroup$
    – David
    Jan 22, 2014 at 3:11

1 Answer 1


I can't draw diagrams in my posts so please draw one for yourself. If the square has side $x$, one side is tangent to the inner circle and the opposite two corners lie on the outer circle, then by Pythagoras we have $$(r_1+x)^2+\Bigl(\frac{x}{2}\Bigr)^2=r_2^2\ .$$ This simplifies to $$\frac{5}{4}x^2+2r_1x+(r_1^2-r_2^2)=0\ ,$$ and since the constant term is negative there will be a unique positive solution for $x$.

  • $\begingroup$ So $x = 2/5 (-2 r_1+\sqrt(-r_1^2+5 r_2^2))$ $\endgroup$ Jan 22, 2014 at 3:28
  • $\begingroup$ @Dennis Correct. TeX suggestion: try using curly brackets, \sqrt{....} to get a "long" square root sign. $\endgroup$
    – David
    Jan 22, 2014 at 3:41
  • $\begingroup$ I don't think your formula is correct. For $r_1 = 0$ and $r_2 = 1$ then $x$ should be $0.707$ and I get $0.894$ $\endgroup$ Jan 22, 2014 at 3:48
  • $\begingroup$ @Dennis, I don't think that's right. For $x=0.707$, the diagonal of the square would have length $1$. That is, the radius of the outer circle would be the diagonal of the square. However in fact - draw a diagram - this radius should be the line from one corner to the middle of the opposite side. But maybe I have misunderstood your original question. $\endgroup$
    – David
    Jan 22, 2014 at 3:58
  • $\begingroup$ Oh, you're right. I was visualizing that case incorrectly. $\endgroup$ Jan 22, 2014 at 4:07

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