I have seen a claim that for the sofa problem, an upper bound for the area of a sofa is $2 \sqrt 2$, and that this can be proved by a "simple" argument. But I can't find a proof. What that argument?

(The articles and papers I found all seem to point back to On the enfeeblement of mathematical skills by modern mathematics and by similar soft intellectual trash in schools and universities, but that paper gives this bound as a problem, and I can't find the solution.)


1 Answer 1


It does not seem useful to me to just repeat the solution in e.g. Ian Stewart's book Another Fine Math You've Got Me Into... as linked from the Wikipedia Moving sofa problem page, so I'll just describe how I understood it. (Note: This is take 2. Edited on 2017-12-12 to fix the +2 to -2 in the formula, and to expand a bit on how the areas are calculated.)

Consider the situation when the sofa is rotated 45 degrees with respect to the corner. In order to fit into the straight part of the corridor, the sofa must lie within the shaded area (corresponding to a straight corridor of width 1, denoted by dashed lines) in the illustrations below.

We can parametrize the different possibilities by the distance $h$ between the outer corner and the part that fits into the straight corridor.

There are two basic cases:

  1. The sofa fits within the corner triangle, $0 \le h \lt \sqrt{2} - 1$: First case Note that the area of both triangles is $(2 h + 2)(h + 1)/2 = h^2 + 2 h + 1$, the area of the upper, white triangle is $2hh/2 = h^2$. In this case, the area is $$A_1(h) = h^2 + 2 h + 1 - h^2 = 2 h + 1$$ which reaches its maximum $2 \sqrt{2} - 1 \approx 1.828427$ at the upper limit, $h = \sqrt{2} - 1$.

  2. The sofa does not fit into the corner triangle; $\sqrt{2} - 1 \le h \le \sqrt{2}$: Second case Note that the area of the triangle above the blue line is $2\sqrt{2}\sqrt{2}/2 = 2$, the area of the white triangle is $2 h h / 2 = h^2$, and the area of each of the parallelograms below the blue line is $\sqrt{2}(1 - (\sqrt{2} - h)) = \sqrt{2} h + \sqrt{2} - 2$. In this case, the shaded area, the area the sofa can possibly occupy, is $$A_2(h) = (2 - h^2) + 2 (\sqrt{2} h + \sqrt{2} - 2) = 2\sqrt{2} - 2 + 2\sqrt{2} h - h^2$$ which reaches its maximum, $2\sqrt{2} \approx 2.828427$, at the upper end of the range, $h = \sqrt{2}$.

Note that $h \lt 0$ makes no physical sense (we'd just be rejecting part of the available width for the sofa for zero gain), and that $h \gt \sqrt{2}$ means the sofa would split into two in the cornering.

In short, the maximum area the sofa can occupy, as a function of $h$, is $$A(h) = \begin{cases} 0, & h \lt -1 \quad \text{(no sofa at all)} \\ (h + 1)^2, & -1 \le h \lt 0 \\ 2 h + 1, & 0 \le h \lt \sqrt{2} - 1 \\ 2\sqrt{2} - 2 + 2\sqrt{2} h - h^2, & \sqrt{2} - 1 \le h \le \sqrt{2} \\ 2 \sqrt{2}, & h \gt \sqrt{2} \quad \text{ but split in two } \end{cases}$$ which reaches a maximum at $h = \sqrt{2}$ (still connected by an infinitesimally thin part at the middle), with area $2 \sqrt{2}$: Maximum area case at h = sqrt(2)

Indeed, all sofa variants, convex and concave, and even cars (that turn both ways), must reside within the shaded area -- just with different values of $0 \le h \le \sqrt{2}$. If they do not, they either do not fit into the straight corridor, or they do not fit into the corner at a 45 degree angle.

This means the maximum area for the moving sofa (and cars) cannot exceed $2\sqrt{2}$.

(Note that because we only consider the non-rotated and rotated-by-45-degrees cases, this does not prove that there is a sofa (or car) with area $2\sqrt{2}$ that can turn the corner. We only proved that no sofa (or car) cannot be larger than $2\sqrt{2}$ in area, because it would be impossible for a sofa (or car) to fit into the corridor when rotated by 45 degrees.)

  • $\begingroup$ This only works if you assume the sofa is convex. The sofa pictured on Wikipedia is longer than $2\sqrt 2$. $\endgroup$
    – dankness
    Commented Jul 9, 2016 at 20:11
  • $\begingroup$ @dankness: Darn, you're right. Lemme rewrite the answer from another point. Needs an illustration, though. $\endgroup$ Commented Jul 9, 2016 at 22:15
  • $\begingroup$ @dankness: Rewrite done. This is no longer an original answer, however; it is basically just a rewording from the Stewart book. $\endgroup$ Commented Jul 9, 2016 at 23:28

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