How to argue that a moving point in 2D crosses a moving line segment? I have three smooth functions $a, b, z: \mathbb R\to\mathbb R^2$, such that for all $t$, the points $a(t)$, $b(t)$, and $z(t)$ are all different. I want to prove that if $z$ touches the line segment between $a$ and $b$, it must immediately cross onto a particular side of the line segment.
More precisely, let $h(t)$ be the signed distance between $z(t)$ and the oriented line between $a(t)$ and $b(t)$. I want to prove that if $z(0)$ lies on the line segment between $a(0)$ and $b(0)$, then for every $\epsilon>0$, there exists a $t\in(-\epsilon,\epsilon)$ such that $h(t)>0$. For example, if I could prove that either $h'(0)\neq0$ or $h''(0)>0$, I'd be done.

*

*Of course, the claim is trivially false if $a$, $b$, and $z$ are not constrained further. For a counterexample, we could take three constant functions at collinear points!

In my case, the curves aren't given by nice formulas, but I do have some control over them. In particular, I think I can argue that the second derivative $\frac{d^2}{dt^2}(z-a)$ or $\frac{d^2}{dt^2}(z-b)$ is "large" for $t$ near $0$, and pointing in the right direction that would make the claim true if $a$ and $b$ were constant functions. I can also prove that, although $a-b$ is not constant, its first derivative $\frac{d}{dt}(a-b)$ is small for $t$ near $0$. I can also control the logarithmic derivative $\frac{1}{|a-b|}\frac{d}{dt}(a-b)$ and the second derivative $\frac{d^2}{dt^2}(a-b)$, if necessary. I may also be able to argue that the projection of $\frac{d}{dt}z$ in the direction of $a-b$ is large, or perhaps control some third derivatives...
Is there some nice identity or inequality, involving a small set of easy derivatives that I've hinted at, that can help me prove what I want? Perhaps a standard identity for $h''$, or some surrogate function with a nicer second derivative? Did I skip the textbook that tells me how to do such things?
(For context, I'm trying to reason about meridian slices of the "regular" embedding of a "fat" $p$-adic solenoid, where $a$ and $b$ are extreme points of a slice at longitude $t$.)
 A: Instead of signed distance it's probably easier to analyze
$$f(t) = (a[t]-b[t]) \times (z[t]-b[t])$$
where $\times$ is the "2D cross product" $v\times w = v_x w_y - v_y w_x.$ Since $f(t)$ has the same sign as $h(t)$, you want to show that $f'(0)\neq 0$.
Since you say that $|a'-b'|$ is small, you'll also need to be able to bound $(a-b)\times (z'-b')$ away from zero; note that geometrically, this is simply saying that the velocity of $z$ in the direction perpendicular to the line $ab$ is large.
As you say, it's easy to create counterexamples to the general claim that the point always crosses the line at the time of intersection, even if you restrict to non-constant polynomials for $a,b,$ and $z$:

So you will need some fairly strong extra guarantees about the motion of the line and point to make the claim true.
A: (Following up on the above, with details that wouldn't fit in a comment)
Slightly adjusting user7530's answer, this is working for me:
$$
f(t) = \big(a(t)-b(t)\big) \times \big((z(t)-\alpha a(t) - \beta b(t)\big)
$$
where the constants $\alpha+\beta=1$ in $\alpha,\beta\in(0,1)$ are chosen so that $\alpha a(0) + \beta b(0) = z(0)$. Then we get:
$$
f'(t) = \big(a'(t)-b'(t)\big) \times \big((z(t)-\alpha a(t) - \beta b(t)\big) \\
+ \big(a(t)-b(t)\big) \times \big((z'(t)-\alpha a'(t) - \beta b'(t)\big)
$$
$$
f''(t) = \big(a''(t)-b''(t)\big) \times \big((z(t)-\alpha a(t) - \beta b(t)\big) \\
+ 2 \big(a'(t)-b'(t)\big) \times \big((z'(t)-\alpha a'(t) - \beta b'(t)\big) \\
+ \big(a(t)-b(t)\big) \times \big((z''(t)-\alpha a''(t) - \beta b''(t)\big)
$$
which simplifies at $t=0$ to:
$$
f'(0) = (a-b) \times (z'-\alpha a' - \beta b')
$$
$$
f''(0) = 2 (a'-b') \times (z'-\alpha a' - \beta b')
+ (a-b) \times (z''-\alpha a'' - \beta b'')
$$
Now if $f'(0)=0$, then that tells us that $a-b$ and $z'-\alpha a' - \beta b'$ are parallel. And that information can help us (at least, it helps in my application) to argue that $f''(0)$ has the desired sign. Of the two terms in $f''(0)$, we either need to argue that:

*

*they both have the desired sign, or that

*one of them is large in the desired direction, and the other is small.

(I originally thought that, by making $|a'-b'|$ small, I could take strategy 2. I now understand my application a little better, and I'll probably go with strategy 1. But that's outside the scope of this question.)
