Solution to Putnam 2007 A-1

Question

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?

Answer 1

$$(\alpha - 1)^2 - \frac{\alpha}6=0$$

$$\alpha^2-2\alpha-\frac{\alpha}6+1=0$$

$$6\alpha^2-12\alpha - \alpha + 6 = 0$$

$$6\alpha^2-13\alpha+6=0$$

$$(2\alpha-3)(3\alpha-2)=0$$

Also, here is the visualization for the other solutions.

Answer 2

As pointed out by me in the comments, and Siong in his answer, you haven't correctly calculated which $\alpha$ makes the discriminant equal to $0$.

As for the other two solutions given in the answer key you linked, you lost them when you made a false assumption about when the two curves can be tangent. See the pictures below for each of the two given values of $\alpha$:

As you can see, the curves can be tangent to each other even though they're not tangent to the line $x = y$, because at that point they're both perpendicular to that line.

This is even directly adressed in the given answer key, as it explicitly describes the solutions in each of the three cases given by each of the parabolas intersect the line 1) once, 2) no times, and 3) two times.

Answer 3

You are on the right lines! Just check your solution: solving $$(\alpha-1)^2 - \frac{\alpha}6=0$$

should get

$$\alpha = \frac{2}{3}, \frac{3}{2} $$

You're missing solutions though. You have looked for where both curves are tangent to $y=x$. But think about the image: if both curves are perpendicular to $y=x$ at the same point, then they will be tangent to each other! You can't easily use the discriminant in this case. Instead, differentiate $y=\alpha x^2 + \alpha x + \frac{1}{24}$ and set this gradient equal to $-1$ to find $-1=2 \alpha x + \alpha$. Solve this along with $x=\alpha x^2 + \alpha x + \frac{1}{24}$ to find your other solutions.

Answer 4

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.

Answer 5

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,

$$\alpha=\frac{2}{3} \: , \frac{3}{2} \: , \frac{13 \pm \sqrt{601}}{12}$$