Let $f(p)$ be the number of common tangent lines of two parabolas $x^2=2y$ and $\left(y+\dfrac{1}{2}\right)^2 = 4px.$

Find a real number $k$ that satisfies $\displaystyle \lim_{p\to k^{+}}f(p) > f(k).$

- It is expected that you solve this in 10 minutes.

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Let's slowly decrease $p$ when it's a positive number.

We need to have a discontinuity in $f(p)$ where, right past a certain value it should output a value higher than before.

That certain value is obviously when the two parabolas are tangent to each other.

However note that decreasing $p$ means "expanding" the parabola.

Then, after being tangent, the function value is smaller, because now the two parabolas intersect without being tangent to each other.

Since this is impossible, let's slowly decrease $p$ when it's a negative number.

Again we need to have a discontinuity in $f(p)$ where, right past a certain value it should output a value higher than before, that certain value is blablabla.

Note that decreasing $p$ now means "shrinking" the parabola. (Because the

absolute valueof $p$ increases.)So, after being tangent, the function value is larger, because now the two parabolas don't intersect with each other.

Hence the $k$ we're looking for is the value of $p<0$ such that the two parabolas are tangent to each other.

Let the point of tangency be $(\alpha,~\beta).$

Then, from $\alpha^2=2\beta$ and $\left(\beta+\dfrac{1}{2}\right)^2=4k\alpha,$ we have $(\alpha^2+1)^2 = 16k\alpha.$

Nextly, the parabola $x^2=2y$ satisfies $\dfrac{dy}{dx} = x$ and the parabola $\left(y+\dfrac{1}{2}\right)^2=4kx$ satisfies $\dfrac{dy}{dx}=\dfrac{4k}{2y+1}.$

So, $\alpha=\dfrac{4k}{2\beta+1}=\dfrac{4k}{\alpha^2+1}$ so that $4k=\alpha(\alpha^2+1).$

$(\alpha^2+1)^2 = 16k\alpha = 4\alpha^2(\alpha^2+1) \\ \alpha^2+1 = 4\alpha^2 \\ \alpha = -\dfrac{1}{\sqrt{3}}~(\because\,\alpha<0)$

Hence $k=\dfrac{\alpha(\alpha^2+1)}{4}=\boxed{-\dfrac{\sqrt{3}}{9}}.$