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?

$$(\alpha - 1)^2 - \frac{\alpha}6=0$$
$$\alpha^2-2\alpha-\frac{\alpha}6+1=0$$
$$6\alpha^2-12\alpha - \alpha + 6 = 0$$
$$6\alpha^2-13\alpha+6=0$$
$$(2\alpha-3)(3\alpha-2)=0$$
Also, here is the visualization for the other solutions.

