The Inscribed Squares Problem

THEOREM E: Let J be any simple closed curve, T any triangle, and let V be the set of points on J that are vertex to a triangle inscribed in J that is similar to T. Then the set V is dense in J (meaning that there are points of V arbitrarily close to any given point of J).

Proof: Let A, B, and C be the vertices of T. Let P be a point on J. Define a function

f: R2 -----> R2
by the following rule: Then f amounts to a rotation about P followed by a dilation (stretch or shrink) with center P.

The image f(J) of the curve J will itself be a curve passing through the point P. If J is reasonably well-behaved at P these curves will cross each other there (see the first diagram).

But if f(J) crosses J at P, then part of f(J) lies inside of J while part lies outside of J. So, f(J) must meet J at another point somewhere. This means there is a point f(R) on f(J) that is also on J. But then the points P, R, and f(R) all lie on J and form a triangle similar to T. (See the second diagram.)

So, any point P that is nice enough to make J and f(J) cross will be a vertex to an inscribed triangle similar to T. It takes some topological arguing (the details of which we won't give here), but it can be proved that the candidates for such a well-behaved P are common enough to be dense in the curve J -- that is, there is always such a well-behaved point as close as you want to any point on J.

Note: Mark Meyerson has proved a remarkable fact that strengthens this theorem in the case that T is an equilateral triangle:

This is the best that can be said -- for if J itself is a triangle with two vertices measuring less than 60 degrees then neither of those vertices can be vertex to an equilateral triangle inscribed in J.

It seems likely that something like Meyerson's Theorem is actually true for all triangles. This would strenthen Theorem E considerably!