Is the shortest line between two smooth arcs normal to both arcs?

Question

Ok, this seems obvious to me, but how would one prove it?

Let $<f(t),g(t)>$ and $<h(t),p(t)>$ be parametrized arcs in the cartesian plot. If $f,g,h,p$ are all continous and the arcs don't intersect, then there will be a line between the two that will be the shortest distance. Prove this line is normal to both arcs.

Is this proof non trivial? It seems so obvious, but i am not sure how it would be done.

Answer 1

This is not true; consider when the shortest line segment between them contains one of an arc's endpoints.

Answer 2

If you assume that the two curves are defined over an open interval, and there is a shortest segment connecting the two curves, then yes; just differentiate the squared length of the joining segment by one of the two parameters.

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Okay, let me be more explicit.

I assumes that the two smooth curves $\alpha:I\to \mathbb{R}^2, \beta:I\to \mathbb{R}^2$ defined over an open interval $I\subseteq \mathbb{R}$ are given. For the sake of convenience, I write the component as $\alpha(t)=(\alpha_1(t),\alpha_2(t))$ etc. If we assumes that the two curves never meet and there exists a shortest line segment(which is not necessarily unique) joining the two curves, and the endpoints of the shortest segment be $\alpha(t_0)$ and $\beta(s_0)$. We can think of the squared length function $f: (t,s)\mapsto |\alpha(t)-\beta(s)|^2=(\alpha_1(t)-\beta_1(s))^2+(\alpha_2(t)-\beta_2(s))^2$, which is smooth by the assumption. Then as $(t_0,s_0)$ is a minimum point of this function, the derivative of $f$ at $(t_0,s_0)$ should be zero. Thus $2(\alpha_1(t_0)-\beta_2(s_0))\alpha_1'(t_0)+2(\alpha_2(t_0)-\beta_2(s_0))\alpha_2'(t_0)=0$, and similarly for $\beta'(s)$. This is exactly the condition that the segment is normal to the two curves.

Note that the assumption that the curve is defined over an open interval is necessary in this proof because the derivative may not be zero if there is a end point in the domain of the curves.

Answer 3

(The following is more of a comment on the context and intuition behind the result, rather than a formal answer - which has been largely provided by @cjackal.)

Starting with a restatement of the problem: let $C_1, C_2$ be two smooth curves in the $2D$ plane. Assume that, among all segments $AB$ with $A \in C_1, B \in C_2$, there exists a shortest one $PQ$. Further assume that the endpoints of this segment are interior points of the respective curves i.e. $P \in \text{int} \;C_1, Q \in \text{int}\;C_2$. Then $PQ$ is normal to both curves at the points of contact, which is equivalent to the tangents at $P,Q$ respectively to $C_1,C_2$ being parallel.

• The smoothness assumption ensures that normality and tangency are well defined.

• The existence of $P,Q$ is a necessary assumption since a shortest segment does not always exist, even for bounded curves. For example, there is no shortest distance between open segments $(0,0)$ and $(1, 0)$ on the real axis.

• The assumption that $P,Q$ are interior points is required to eliminate cases where the minimum distance would be otherwise achieved at an endpoint of either curve. For example, the shortest distance between two closed segments which are not parallel does exist, but is not normal to at least one of them.
  (As a side note, the interior assumption is slightly more general than the assumption of $C_1,C_2$ being non-self-intersecting and open used in @cjackal's proof, for example it includes the case of two circles, but that proof itself could be adjusted fairly easily to relax those assumptions.)

Now back to the problem itself, draw the tangents at $P,Q$ to the respective curves. Since $P,Q$ are interior points, each has an open vicinity on the respective curve. For small enough such vicinities, they will each be on opposite sides of the respective tangents, and for infinitesimally small such vicinities the tangents virtually approximate the curves around the two points. If the tangents are not parallel, then slightly shifting the points towards the intersection of the two tangents would yield two new points $P',Q'$ with $|P'Q'| \lt |P Q|$ but that would contradict the assumption of $P,Q$ being the shortest distance. Thus the tangents must be parallel, and $P Q$ must be normal to both.