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.
In each of parts (i) through (v), find four points on the loop that are the vertices of a square.
the circle with equation \(x^2 + y^2 = 1\)
the ellipse with equation \(\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1\) where \(a\) and \(b\) are fixed positive real numbers
the polygon with vertices \((1,0)\), \(\left(\frac{1}{2},\frac{\sqrt 3}{2}\right)\), \(\left(-\frac{1}{2},\frac{\sqrt 3}{2}\right)\), \((-1,0)\), \(\left(-\frac{1}{2},-\frac{\sqrt 3}{2}\right)\), and \(\left(\frac{1}{2},-\frac{\sqrt 3}{2}\right)\)
the boundary of the region enclosed by the parabola with equation \(y = -\dfrac{1}{2}x^2 +\dfrac{1}{6}x+\dfrac{16}{9}\) and the line with equation \(y = x\)
the boundary of the region enclosed by the parabolas with equations \(y = x^2 + \dfrac{2}{3}x - \dfrac{4}{3}\) and \(y = -x^2 + \dfrac{2}{3} x + \dfrac{4}{3}\)
Show that for every acute triangle there are exactly three squares whose vertices all lie on the perimeter of the triangle.