Writeup: Putnam 2007 A1

19 Jul, 2025

Problem Statement

Find all values of $\alpha$ for which the curves $y=\alpha x^2+\alpha x+\frac{1}{24}$ and $x=\alpha y^2+\alpha y+\frac{1}{24}$ are tangent to each other.

My Solution:

Let the first curve (quadratic in $x$) be $C_1$, the second be $C_2$. Then note that $C_1$ and $C_2$ are reflections of each other across the line $y=x$. If a point of tangency does not lie on the this line, then it must have a corresponding point of tangency reflected across this line. Note also, that if the two intersect, there must be at least one solution that lies on the line $y=x$. Therefore, we can't have a point of tangency not on the line $y=x$, otherwise we would get $5$ solutions which is impossible for a quartic system. So all points of tangency must lie on the line $y=x$.

Now note that for $C_1$, we have $\frac{dy}{dx}=\alpha (2x+1)$, and for $C_2$, implicitly we have $\frac{dy}{dx}=\frac{1}{\alpha (2y+1)}$. The points of tangency will have the same slope for both curves, and we therefore must have $\alpha (2x+1)=\frac{1}{\alpha (2x+1)}$, or ${\alpha }^2=\frac{1}{(2x+1{)}^2}$. We therefore must have $\alpha =\pm \frac{1}{(2x+1)}$.

Now note that at the point of tangency, we must have $x=\alpha x^2+\alpha x+\frac{1}{24}$. At this point, we substitute both values of $\alpha$ in and solve for $x$. For "positive" $\alpha$, we get $\alpha =\frac{2}{3},\alpha =\frac{3}{2}$. For "negative" $\alpha$, we get $\alpha =\frac{13}{12}\pm \frac{\sqrt{601}}{12}$.