Sam Loyd

Sam Loyd (1841-1911) is one of the World’s greatest known puzzle inventors, chess puzzle creators, and recreational mathematicians. He shares this accolade with other notables such as Martin Gardner, Henry Ernest Dudeney, and Boris Kordemsky.

Probably the most well-known book by Sam Loyd is the 1914 compilation: Cyclopedia of 5,000 puzzles tricks and conundrums (with answers). This is considered to be the most fabulous and exciting collection of puzzles ever assembled in one volume. Reprints are still available today.

The following is one of the puzzles from this book (page 245), called The Three Napkins.

Paraphrasing, you are given three unit-sized napkins that you can place onto a square tabletop in any orientation. You can fold the napkins, let them droop over the edge, or overlap them. The only restriction is that you cannot cut them. The question is to determine the largest size square table that you can cover (meaning at least one napkin is over every point of the tabletop).

Sam’s solution, in the answer section of the book, is a little sparse, and provides the solution, but no derivation. His solution states that three 12” napkins will cover a 15¼” table. This represents a ratio of napkin to table edge length of 1:1.271 Let’s see if we can derive and prove this answer …

The solution involves placing one napkin in a corner, flush with the edge of the table, and then placing the two others, at a bias, rotated to just cover the gaps. At the limit these two biased triangles form two triangles which will droop down the sides (If you imagine rotating the napkin more clockwise, then a gap would appear on the left edge of the smaller triangle, similarly for a rotation counter-clockwise and the right edge. A larger table would produce gaps on both sides).

Through symmetry, if we look at just one of these napkins we can see that the two drooping triangles are similar. We’ll define the length of the edge of a napkin to be one unit, and the difference between the side of a napkin and the side of the table to be g units.

Through Pythagoras, we can determine all the lengths of the sides of the triangles. The larger triangle has a hypotenuse of (1+g), and a long side of 1, so the remaining side is length:

The hypotenuse of the smaller triangle is just g, and from simple subtraction, the length of longer side is:

Because the triangles are similar, the ratio of the longer side over their hypotenuses is the same.

To simplify the math, let’s define t to be the length of the edge of the table. This is equal to (1+g)

We end up with a quartic, but if we make another substitution that u = t², then we get the following quadratic which we can solve with the standard formula.

Taking the real solution we find out that the maximum table length is ≈ 1.272 times the unit length of a napkin. This corresponds very well with Loyd’s answer of 1.271 (If the napkins were exactly 12” per side, the largest table would be 15.264” which is, as the original puzzle states, is just greater than 15¼”).

If this answer seems familiar, it's all related to the Golden Ratio, and you can read a little more about it in an article a couple of years ago about folding a sheet of paper to get the shortest crease.

Here is a little app that allows you to move napkins around (red, green, blue). At the bottom are buttons to allow you to change the size of the table relative to the napkins (white). Drag the center of each napkin to move it, and the corners to rotate it.