<s>[INST] 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?
 [/INST] Let $f(t)=a t^2+b t+c$, then
\begin{align}
  y &= f(x) \\
  x &= f(y) \\
  &= f(f(x)) \\
  0 &= f(f(x))-x \\
  &= ay^2+by+c-x \\
  &= ay^2 \color{red}{-ax^2}+by \color{blue}{-bx}
     \color{red}{+ax^2} \color{blue}{+bx}+c-x \\
  &= a(y^2-x^2)+b(y-x)+y-x \\
  &= (y-x)[a(y+x)+b+1] \\
  &= [ax^2+(b-1)x+c][a^2x^2+a(b+1)x+ac+b+1]
\end{align}
For tangency,
$$(b-1)^2=4ac \quad \text{or} \quad (b+1)^2=4(ac+b+1)$$
Plugging $a=b=\alpha$ and $c=\dfrac{1}{24}$ gives
$$(\alpha-1)^2=\frac{\alpha}{6} \quad \text{or} \quad
(\alpha+1)^2=4\left( \frac{\alpha}{24}+\alpha+1 \right)$$
On solving,
$$\alpha=\frac{2}{3} \: ,  \frac{3}{2} \: , \frac{13 \pm \sqrt{601}}{12}$$
</s>