Solution to Putnam 2007 A-1 
2007 A-1: Find the 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: Notice that the second equation is obtained from the first by swapping $x$ and $y$. This happens when you reflect the graph of the first equation across the $x=y$ line. Hence, the two curves will intersect only once, and therefore be tangent to each other at that point of intersection, if each curve individually also intersects the $y=x$ line only once. Hence $x=\alpha x^2 + \alpha x + \frac{1}{24}$, on solving which we get $\alpha=\frac{43}{12},-\frac{17}{12}$. 
This is not the answer or solution given on Kedlaya's putnam page. Where am I going wrong?

Let $(x,y)=(a,b)$ be the point of tangency between the two curves.  Thus,
$$b=\alpha a^2+\alpha a+\frac{1}{24}\text{ and }a=\alpha b^2+\alpha b+\frac{1}{24}\,.\tag{*}$$
We claim that $a=b$.  To show this, suppose that $a\neq b$.  By symmetry, $(x,y)=(b,a)$ is also a point of tangency of the two parabolas.  By Bézout's Theorem, there are at most $4=2\cdot2$ common points of the two parabolas (which are algebraic curves of degree $2$), where points of tangency are counted with multiplicities greater than $1$.  Thus, we have two distinct common points $(x,y)=(a,b)$ and $(x,y)=(b,a)$, each with multiplicity at least $2$.  Consequently, there are no other common points.  However, since $a\neq b$ and a parabolic curve is continuous, both parabolas must intersect the line $y=x$.  By symmetry, this intersection point (indeed, if one intersection along the line $y=x$ existed, then there had to be at least $2$) is another common point of the parabolas, which is absurd.  Therefore, $a=b$ must hold.
The slopes at $(x,y)=(a,b)$ of the two parabolas must be equal.  For the  graph $y=\alpha x^2+\alpha x+\dfrac{1}{24}$, the slope is $2\alpha a+\alpha$.  For the other graph $x=\alpha y^2+\alpha y+\dfrac{1}{24}$, the slope is $\dfrac{1}{2\alpha b+\alpha}$ (whence $\alpha\neq 0$).  That is,
$$\alpha^2\,(2a+1)(2b+1)=1\,.\tag{#}$$
We obtain $(2a+1)^2=(2a+1)(2b+1)=\dfrac{1}{\alpha^2}$.  Thus,
$$2a+1=\pm\frac{1}{\alpha}\,,\text{ or }a=b=-\frac{1}{2}\pm\frac{1}{2\alpha}\,.$$
Now, plug this result into (*), we have
$$-\frac{1}{2}+\frac{s}{2\alpha} =\alpha \,\left(-\frac{1}{2}+\frac{s}{2\alpha}\right)^2+\alpha\,\left(-\frac{1}{2}+\frac{s}{2\alpha}\right)+\frac{1}{24}\,,$$
where $s=\pm1$.  Thus,
$$\alpha^2-\frac{13}{6}\alpha+2s-1=0\,.$$
If $s=+1$, then 
$$\alpha^2-\frac{13}{6}\alpha+1=\left(\alpha-\frac{3}{2}\right)\left(\alpha-\frac{2}{3}\right)\,,\text{ so }\alpha\in\left\{\dfrac{2}{3},\dfrac{3}{2}\right\}\,.$$
If $s=-1$, then
$$\alpha^2-\frac{13}{6}\alpha-3=0\,,$$
that is, $$\alpha=\frac{13\pm\sqrt{601}}{12}\,.$$
We have found four values of $\alpha$: $\dfrac{3}{2}$, $\dfrac{2}{3}$, $\dfrac{13-\sqrt{601}}{12}$, and $\dfrac{13+\sqrt{601}}{12}$.  Clearly, these values work.
