This week’s installment of
mathematical content will be on the inscribed square problem. This problem is
also referred to as the square peg problem, and Toeplitz’ conjecture. This
problem though simple, remains unsolved still to today. There is not a
million-dollar reward for this problem, but in exchange for the lack of
monetary incentive the problem is easy to understand. There have not been
a large number of developments in this problem, so the history is brief. In
order to compensate a rectangular proof will be mentioned. To format this post,
the first part will be the brief history; second will be the problem, and
finally the rectangular proof.
It is called the Toeplitz
conjecture, because Jewish mathematician, Otto Toeplitz, created it. Otto was
born in 1881, and much like his father, and grandfather, he wanted to work in
mathematics. He received his doctorate in 1905, and in 1906 started working at
Gottingen University. At the time, Gottingen was one of the most prestigious
universities in mathematical development. He worked with mathematicians such as
Felix Klein, and David Hilbert. During his time there he worked on linear, and
quadric functions in high special dimensions. Then in the year 1911 he proposed
his conjecture on an inscribed square within a curve.
His conjecture is that on any curve
there can be found four points that can construct a square. For example, a
circle is probably the easiest to visualize a square inscribed within. In fact,
if the square is rotated it can be seen that there can be an infinite number of
squares found within a circle. However, this does not only apply to circles, the conjecture says it applies to all Jordan curves. A
Jordan curve being a closed curve not having an intersection of any kind. For
example, oval shapes, kidney bean shapes, and heart shapes would all be Jordan
curves. However, self-intersecting shapes such as a figure eight, or infinity
symbol would not be Jordan curves. This also includes all irregular shape
variants that do not intersect. This conjecture was purposed a hundred years
ago, and still it remains unsolved today. There has not been any progress on
the square problem, but it has been proven that this problem can be solved with
triangles, and also can be solved with rectangles.
As is the case for most
mathematical problems, the solutions tend to be more complicated than the
questions. This is the case with the rectangular proof of this problem. Just to
be clear, this is not a proof of the inscribed square, but rather an inscribed
rectangle. The approach to solving this is to look at a property of rectangles.
A rectangle has four points, and two pairs of diagonal lines can be draw within
a rectangle. It is important to recognize that the two diagonals have the same
length. In fact, if two line segments share a midpoint, and are the same
length, a rectangle can be formed from those four corresponding points. So
rather than proving that a rectangle can be found, it is easier to prove that
two line segments, that share a midpoint, and have the same length, can be
found in any Jordan curve. The proof involves mapping the curve in three-dimensional
space. Imagine a closed Jordan curve on a horizontal plane, then draw a line
connecting two points on the curve. Take the distance between the two points
and graph it vertically on the z axis, at the midpoint. Continually doing this over every pair
of points on the curve generates a three-dimensional model of the curve. This
being hard to visualize, a circle would map to a nice dome shape. At this point
it needs to be made clear that the two line segments that are being looked at, need to be different line segments. This is to make sure that a rectangle would
be formed, and not just a line. In order to do this the points on the curve
must be unordered pairs. This just means that it removes redundancy from our
problem. This next part will not be fully explained, but it can be proven. Any
closed loop with unordered point pairs maps to a Mobius strip. The outer edge
of the Mobius strip, corresponds to the points of the outer edge of the Jordan
curve. In order to transform the Mobius strip into the three-dimension curve
generated at the midpoints, it requires it to intersect itself. Meaning that
there are two pairs of point on the outside edge that correspond in distance
and midpoint.
The last part was not easily
understandable, with just words. Here is a link to a video that gives a nice
visualization to this problem. https://www.youtube.com/watch?v=AmgkSdhK4K8
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