Old Putnam problem that I could use some help on

Question

I've been beginning to practice for the Putnam this year, and I came across a problem from a few years back that I could use some help on. I want to 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.

Hence, I believe that what I am looking for is all of the $\alpha$s such that the defined curves intersect at one point and only one point (based on the tangential aspect of this problem). Say this point is $(x_1,y_1)$ we need to show that this point uniquely satisfies $y_1 = \alpha x_1^2 + \alpha x_1 + \frac{1}{24}$ and $x_1 = \alpha y_1^2 + \alpha y_1 + \frac{1}{24}.$ What I had imagined that we would need to do is plug one equation into the other, and find the $\alpha$s such that the given equation has only one root (although, this is primarily just a hypothesis of what may work). So far, I set $$y_1 = \alpha (\alpha y_1^2 + \alpha y_1 + \frac{1}{24})^2 + \alpha (\alpha y_1^2 + \alpha y_1 + \frac{1}{24}) + \frac{1}{24}$$ Although I am having some difficulties figuring out what may be the best way to manner to simplify this equation, given that it will end up having one side with a polynomial of degree $4$ (which seems somewhat unweildy). Any recommendations?

Answer 1

$$C_1:y=\alpha x^2+\alpha x+\frac1{24}$$ $$C_2:x=\alpha y^2+\alpha y+\frac1{24}$$ $$L:y=x$$ $x$ and $y$ are swapped between the two equations, so $C_1$ is the reflection of $C_2$ across $L$ and vice versa.

There can be at most four intersections with multiplicity (IM) between $C_1$ and $C_2$. If $C_1$ does not cross $L$, there are no intersections/tangencies at all; if it does, there cannot be a distinct pair of tangencies across $L$ since that would make for more than four IMs. Hence all tangencies must lie on $L$.

Furthermore, the tangent directions at tangencies must be parallel (A) or perpendicular (B) to $L$. For case (A): $$x=\alpha x^2+\alpha x+\frac1{24}$$ $$\alpha x^2+(\alpha-1)x+\frac1{24}=0$$ If the discriminant of this polynomial is zero, $C_1$ will be tangent to $L$ and thus to $C_2$: $$(\alpha-1)^2-4\alpha\cdot\frac1{24}=0$$ $$\alpha^2-2\alpha+1-\frac16\alpha=0$$ $$\alpha^2-\frac{13}6\alpha+1=0$$ Solving for $\alpha$ we get $$\alpha=\frac23\text{ or }\alpha=\frac32$$ For case (B), if the derivative of $C_1$ where it meets $L$ is $-1$, it will also be tangent to $C_2$ (on the same side) by symmetry: $$\frac{dy}{dx}=2\alpha x+\alpha=-1$$ $$x=\frac{-\alpha-1}{2\alpha}$$ $$\alpha x^2+(\alpha-1)x+\frac1{24}=0\to\alpha\left(\frac{-\alpha-1}{2\alpha}\right)^2+(\alpha-1)\frac{-\alpha-1}{2\alpha}+\frac1{24}=0$$ $$\frac{(\alpha+1)^2}{4\alpha^2}-(\alpha-1)\frac{\alpha+1}{2\alpha^2}+\frac1{24\alpha}=0$$ $$(\alpha+1)^2-2(\alpha-1)(\alpha+1)+\frac16\alpha=0$$ $$\alpha^2+2\alpha+1-2\alpha^2+2+\frac16\alpha=0$$ $$-\alpha^2+\frac{13}6\alpha+3=0$$ $$6\alpha^2-13\alpha-18=0$$ Solving for $\alpha$ we get $$\alpha=\frac{13\pm\sqrt{601}}{12}$$

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In conclusion, the values of $\alpha$ where both curves are tangent to each other are $$\alpha=\frac23,\frac32,\frac{13\pm\sqrt{601}}{12}.$$

Answer 2

You might consider the difference of the two functions giving the curves. At a point of intersection $(x,y)$, $$ x-y = ay^2 - ax^2 +ay - ax. $$ See if that helps.

Answer 3

Extended hint (too long for a comment).

Curves $y = P(x)$ and $x = P(y)$ are symmetric with respect to $y = x$. They can have at most $4$ intersection points (counting multiplicities) because of the degrees of the polynomials involved.

• If one of them doesn't intersect $y = x$ then neither will the other, because each one will be strictly contained in either half plane separated by $y = x$, So this case can be ignored.

• If one of them does intersect $y = x$ then that would be an intersection point with its mirror image, and there can be no tangency points outside $y = x$, because its symmetric with respect to $y = x$ would also be a tangency point, and that would give a total of $1+2+2=5 \gt 4$ intersection points (again, counting multiplicities).

Therefore, any points of tangency must be on the axis of symmetry $y = x$, which gives:

$$ (1) \;\;\;\;\;\;\;\; P(x) = x \;\;\iff\;\; \alpha x^2 + \alpha x + \frac{1}{24}= x $$

Furthermore, the tangents must be the same, but the only lines invariant to the symmetry across $y = x$ are $y = x$ itself and the normals $y = -x + y_0$. Therefore:

$$ (2) \;\;\;\;\;\;\;\; P'(x) = \pm 1 \;\;\iff\;\; 2 \alpha x + \alpha = \pm 1 $$

Eliminating $x$ between $(1)$ and $(2)$ results in two quadratics in $\alpha$, which happen to have real and distinct roots each, so in the end there are $4$ solutions for $\alpha$.