Putnam 1946 
Topic: Geometry Algebra

Problem: Let $A$, $B$ be variable points on a parabola $P$, such that the tangents at $A$ and $B$ are perpendicular to each other. Show that the locus of the centroid of the triangle formed by $A$, $B$ and the vertex of $P$ is a parabola $P_1$. Apply the same process to $P_1$, obtaining a parabola $P_2$, and repeat the process, obtaining altogether the sequence of parabolas $P$, $P_1$, $P_2$, \dots, $P_n$. If the equation of $P$ is $y^2 = mx$, find the equation of $P_n$.

Solution: Since $P$ is a parabola with equation $y^2 = mx$, any point of $P$ has coordinates of the form $(mt^2, mt)$ for some real $t$, and conversely, every such point is on $P$. The slope of the line tangent to $P$ at $(mt^2, mt)$ is $1/2t$.

Let $A$ and $B$ be the points $(ms^2, ms)$ and $(mt^2, mt)$, respectively. The tangents to $P$ at $A$ and $B$ are perpendicular if and only if $(1/2s)(1/2t) = -1$, i.e., if and only if
\[(1)\quad st = -\frac{1}{4}.\]

The centroid of the points $A$, $B$, and the vertex $(0, 0)$ of $P$ is
\[(2)\quad \left( \frac{1}{3}m(s^2 + t^2), \frac{1}{3}m(s + t) \right)\]
and this centroid lies on a new parabola
\[(3)\quad y^2 = \frac{1}{3}m\left(x - \frac{m}{6}\right)\]
if the perpendicularity condition (1) is satisfied.

Conversely, any point $(x, y)$ on the parabola (3) has the form (2), since the equations
\[\frac{1}{3}m(s + t) = y\]
\[st = -\frac{1}{4}\]
can always be solved to give real $s$ and $t$. Indeed, $s$ and $t$ are zeros of $S^2 - (3y/m)S - \frac{1}{4}$, which has positive discriminant. Hence the locus in question is the entire parabola $P_1$ given by (3).

Now $P_1$ is obtained from $P$ by changing the constant from $m$ to $m/3$, and displacing the vertex to the right by $m/6$. Consequently, $P_2$ can be obtained from $P_1$ by changing the constant from $m/3$ to $(m/3)/3$ and displacing the vertex $(m/3)/6$ further to the right. The equation of $P_2$ is therefore
\[y^2 = \frac{1}{9}m\left(x - \frac{1}{6}m - \frac{1}{18}m\right).\]

Continuing this reasoning, we see that $P_n$ has the equation
\[y^2 = \frac{1}{3^n}m\left(x - \frac{m}{6} - \frac{1}{6\cdot3}m - \frac{1}{6\cdot3^2}m - \cdots - \frac{1}{6\cdot3^{n-1}}m\right)\]
\[= \boxed{\frac{1}{3^n}m\left(x - \frac{m}{4}\left(1 - \frac{1}{3^n}\right)\right)}.\]