Problem of the Month
Problem 2: November 2023

This month’s problem is based on the following general question: If you draw a "loop" in the Cartesian plane, is it always possible to find four points on that loop that are the vertices of a square? For example, the diagram below has a loop (the solid line) and a square drawn (the dashed line) with its four vertices on the loop.

Although it is a bit informal, it should be sufficient to think of a "loop" as a curve that you could draw by starting your pencil somewhere on a page and moving the pencil around the page eventually ending up where it started. Such a loop could be "smooth" (like a circle), "jagged" (like a polygon), or some combination of the two.

1. In each of parts (i) through (v), find four points on the loop that are the vertices of a square.

   1. the circle with equation $x^2+y^2=1$

   2. the ellipse with equation ${\displaystyle \frac{x^2}{a^2}}+{\displaystyle \frac{y^2}{b^2}}=1$ where $a$ and $b$ are fixed positive real numbers

   3. the polygon with vertices $(1,0)$, $(\frac{1}{2},\frac{\sqrt{3}}{2})$, $(-\frac{1}{2},\frac{\sqrt{3}}{2})$, $(-1,0)$, $(-\frac{1}{2},-\frac{\sqrt{3}}{2})$, and $(\frac{1}{2},-\frac{\sqrt{3}}{2})$

   4. the boundary of the region enclosed by the parabola with equation $y=-{\displaystyle \frac{1}{2}}{x}^2+{\displaystyle \frac{1}{6}}x+{\displaystyle \frac{16}{9}}$ and the line with equation $y=x$

   5. the boundary of the region enclosed by the parabolas with equations $y=x^2+{\displaystyle \frac{2}{3}}x-{\displaystyle \frac{4}{3}}$ and $y=-x^2+{\displaystyle \frac{2}{3}}x+{\displaystyle \frac{4}{3}}$

2. Show that for every acute triangle there are exactly three squares whose vertices all lie on the perimeter of the triangle.