Other correct solutions were sent in by Nathan Pauli. Further solutions were submitted by alumni Ray Kremer, Brian Laughlin ('81), and from Sudipta Das, Mike Young, Philippe Fondanaiche, Bryan Fluhrer, Al Zimmermann.  There were a great many incorrect solutions submitted.

First of all, notice that we need consider only packages whose length equals one of the other two dimensions, since longer packages will have a smaller girth.  So consider a package whose dimensions are a, a, b with a + 2(a + b) = 3a + b = 130, and b £ a.  Hence b = (130 - 2a)/3. There are two basic ways in which a package can be pushed through a square window -- with sides parallel to the sides of the window, or with sides parallel to the diagonal of the window.  In the first case, the length of one side of the square is a, and the diagonal has length d = Ö2 a.  In the second case, the perimeter of the box is twice the diagonal of the square, so d = a + b = (130 + b)/3.  In the first case, the diameter decreases as b increases, while in the second case the diameter increases with increasing b.  We will find our desired result when these two expressions for the diameter are equal.  Solving gives b = (130 (Ö2 - 1))/(1 + 2 Ö2 ) and the length of the side of the window would be a = (130 (2Ö2 -1)/7 = 33.9565...

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