Square on a Parabola

Let $(p, p^2)$ be the coordinates of the leftmost vertex of the square, $(q, q^2)$ be the coordinates of the topmost vertex of the square, and $(r, r^2)$ be the coordinates of the bottommost vertex of the square, as shown below:

Then the slope between $p$ and $q$ is $m = m_1 = \frac{p^2 - q^2}{p - q} = p + q$ , and the slope between $p$ and $r$ is $m_2 = \frac{p^2 - r^2}{p - r} = p + r$ . These slopes are perpendicular, so $m_1m_2 = -1$ . Also, the sides $s$ of the square are congruent, so the area $A$ is $A = s^2 = (p - q)^2 + (p^2 - q^2)^2 = (p - r)^2 + (p^2 - r^2)^2$ . Solving these equations in terms of $m$ for $m > 0$ gives us $p = \frac{m^3 + 1}{2m(m - 1)}$ , $q = \frac{m^3-2m^2-1}{2m(m-1)}$ , $r = -\frac{m^3+2m-1}{2m(m-1)}$ , and $A = \frac{(m^2+1)^3}{m^2(m-1)^2}$ for $m \neq 1$ . (When $m < 0$ these equations make the fourth vertex inside the parabola.)

For $m > 0$ and $m \neq 1$ , $\frac{dA}{dm} = \frac{2m(m+1)(m^2-3m+1)}{m^4(m-1)^3}$ and $\frac{d^2A}{dm^2} > 0$ , so the minimum area occurs when $m^2 - 3m + 1 = 0$ , or at $m = \frac{3 \pm \sqrt{5}}{2}$ (which in terms of the golden ratio $\phi = \frac{1 + \sqrt{5}}{2}$ , $m = \phi^2$ and $m = \phi^2 - 3$ ). Both of these values of $m$ give symmetrical coordinates for the square and an area of $A = \boxed{27}$ .

I’ve used a slightly different parametrization. Every square in the plane has coordinates $(a+u,b+v),(a-u,b-v),(a-v,b+u),(a+v,b-u)$ . The square is centered at $(a,b)$ and its half diagonal is $(u,v)$ .

Having three vertices on the parabola provides three equations, and they can be transformed into

$u=\frac{4a^2+1}{4a^2-1}, v=2a u$

with the proviso that $a>1/2$ .

We wish to minimize $u^2+v^2$ , which is

$u^2+v^2=\frac{(4a^2+1)^3}{(4a^2-1)^2}$

And the minimum (in the allowed domain) is $\frac{27}{2}$ , corresponding to a square of area 27.

If the diametrically opposite corners of a square are $(a,a^2)$ and $(b,b^2)$ , with $a \neq b$ , then the other two corners have coordinates $\left(\tfrac12(a+b+a^2-b^2),\tfrac12(b-a+a^2+b^2)\right) \hspace{2cm}\left(\tfrac12(a+b+b^2-a^2),\tfrac12(a-b+a^2+b^2)\right)$ For the first of these to lie on the parabola, we require that $0 \; = \; (a+b + a^2 - b^2)^2 - 2(b-a + a^2 + b^2)$ which simplifies to give $a - b \; = \; \frac{2(1+u^2)}{1-u^2} \hspace{1.5cm} a + b \; = \; u$ where $u^2 \neq 1$ . For the second of these points to lie outside the parabola, we require that $0 \; < \; (a + b + b^2 - a^2)^2 - 2(a -b + a^2 + b^2)$ which simplifies to give $-\frac{8(1+u^2)^2}{1-u^2} \; > \; 0$ and hence $u^2 > 1$ . The area of the square is $\tfrac12\big((a-b)^2 + (a^2-b^2)^2\big) \; = \; \frac{2(1+u^2)^3}{(1-u^2)^2}$ Thus we need to minimize $A(t) \; = \; \frac{2(1+t)^3}{(1-t)^2} \hspace{2cm} t > 1$ Since $\frac{dA}{dt} \; = \; \frac{2(1+t)^2(5-t)}{(1-t)^3}$ we see that $A$ is minimized when $t=5$ , giving a minimum area of $\boxed{27}$ .