Putnam 2007, Problem A-1

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.

Solution:
The only such $\alpha$ are $2/3, 3/2, (13 \pm \sqrt{601})/12$.

**First solution:** Let $C_1$ and $C_2$ be the curves $y = \alpha x^2 + \alpha x + \frac{1}{24}$ and $x = \alpha y^2 + \alpha y + \frac{1}{24}$, respectively, and let $L$ be the line $y = x$. We consider three cases.

If $C_1$ is tangent to $L$, then the point of tangency $(x,x)$ satisfies
$$2\alpha x + \alpha = 1, \quad x = \alpha x^2 + \alpha x + \frac{1}{24};$$
by symmetry, $C_2$ is tangent to $L$ there, so $C_1$ and $C_2$ are tangent. Writing $\alpha = 1/(2x+1)$ in the first equation and substituting into the second, we must have
$$x = \frac{x^2 + x}{2x+1} + \frac{1}{24},$$
which simplifies to $0 = 24x^2 - 2x - 1 = (6x+1)(4x-1)$, or $x \in \{1/4, -1/6\}$. This yields $\alpha = 1/(2x+1) \in \{2/3, 3/2\}$.

If $C_1$ does not intersect $L$, then $C_1$ and $C_2$ are separated by $L$ and so cannot be tangent.

If $C_1$ intersects $L$ in two distinct points $P_1, P_2$, then it is not tangent to $L$ at either point. Suppose at one of these points, say $P_1$, the tangent to $C_1$ is perpendicular to $L$; then by symmetry, the same will be true of $C_2$, so $C_1$ and $C_2$ will be tangent at $P_1$. In this case, the point $P_1 = (x,x)$ satisfies
$$2\alpha x + \alpha = -1, \quad x = \alpha x^2 + \alpha x + \frac{1}{24};$$
writing $\alpha = -1/(2x+1)$ in the first equation and substituting into the second, we have
$$x = -\frac{x^2 + x}{2x+1} + \frac{1}{24},$$
or $x = (-23 \pm \sqrt{601})/72$. This yields $\alpha = -1/(2x+1) = (13 \pm \sqrt{601})/12$.

If instead the tangents to $C_1$ at $P_1, P_2$ are not perpendicular to $L$, then we claim there cannot be any point where $C_1$ and $C_2$ are tangent. Indeed, if we count intersections of $C_1$ and $C_2$ (by using $C_1$ to substitute for $y$ in $C_2$, then solving for $y$), we get at most four solutions counting multiplicity. Two of these are $P_1$ and $P_2$, and any point of tangency counts for two more. However, off of $L$, any point of tangency would have a mirror image which is also a point of tangency, and there cannot be six solutions. Hence we have now found all possible $\alpha$.

**Second solution:** For any nonzero value of $\alpha$, the two conics will intersect in four points in the complex projective plane $\mathbb{P}^2(\mathbb{C})$. To determine the $y$-coordinates of these intersection points, subtract the two equations to obtain
$$(y-x) = \alpha(x-y)(x+y) + \alpha(x-y).$$
Therefore, at a point of intersection we have either $x = y$, or $x = -1/\alpha - (y+1)$. Substituting these two possible linear conditions into the second equation shows that the $y$-coordinate of a point of intersection is a root of either $Q_1(y) = \alpha y^2 + (\alpha-1)y + 1/24$ or $Q_2(y) = \alpha y^2 + (\alpha+1)y + 25/24 + 1/\alpha$.

If two curves are tangent, then the $y$-coordinates of at least two of the intersection points will coincide; the converse is also true because one of the curves is the graph of a function in $x$. The coincidence occurs precisely when either the discriminant of at least one of $Q_1$ or $Q_2$ is zero, or there is a common root of $Q_1$ and $Q_2$. Computing the discriminants of $Q_1$ and $Q_2$ yields (up to constant factors) $f_1(\alpha) = 6\alpha^2 - 13\alpha + 6$ and $f_2(\alpha) = 6\alpha^2 - 13\alpha - 18$, respectively. If on the other hand $Q_1$ and $Q_2$ have a common root, it must be also a root of $Q_2(y) - Q_1(y) = 2y + 1 + 1/\alpha$, yielding $y = -(1+\alpha)/(2\alpha)$ and $0 = Q_1(y) = -f_2(\alpha)/(24\alpha)$.

Thus the values of $\alpha$ for which the two curves are tangent must be contained in the set of zeros of $f_1$ and $f_2$, namely $2/3, 3/2$, and $(13 \pm \sqrt{601})/12$.

**Remark:** The fact that the two conics in $\mathbb{P}^2(\mathbb{C})$ meet in four points, counted with multiplicities, is a special case of Bézout's theorem: two curves in $\mathbb{P}^2(\mathbb{C})$ of degrees $m,n$ and not sharing any common component meet in exactly $mn$ points when counted with multiplicity.

Many solvers were surprised that the proposers chose the parameter $1/24$ to give two rational roots and two nonrational roots. In fact, they had no choice in the matter: attempting to make all four roots rational by replacing $1/24$ by $\beta$ amounts to asking for $\beta^2 + \beta$ and $\beta^2 + \beta + 1$ to be perfect squares. This cannot happen outside of trivial cases ($\beta = 0, -1$) ultimately because the elliptic curve 24A1 (in Cremona's notation) over $\mathbb{Q}$ has rank 0. (Thanks to Noam Elkies for providing this computation.)

However, there are choices that make the radical milder, e.g., $\beta = 1/3$ gives $\beta^2 + \beta = 4/9$ and $\beta^2 + \beta + 1 = 13/9$, while $\beta = 3/5$ gives $\beta^2 + \beta = 24/25$ and $\beta^2 + \beta + 1 = 49/25$.