Show that at some time the hour hand of the first clock points to the tip of the hour hand of the second clock Consider two round clocks of different sizes lying on a table:

As shown on the picture, the clocks can be oriented differently but they are both set to the same time.
The problem is to show that at some time of the day, the hour hand of the first clock points to the tip of the hour hand of the second clock.
Implicit assumptions:


*

*The hands rotate continuously

*The clocks don't overlap


My try: This seems like an application of the intermediate value theorem, but I find it very hard to make my argument rigourous. 
Intuitively, if we extend the hour hand of the first clock infinitely, it swaps the entire second clock in some portion of the day that last less than 6 hours. During that time, surely it must meet the tip of the hour hand of the second clock. But how to prove this rigourously.
 A: This assume that the clocks don't overlap.
Let $O$ denote the cneter of the first  clock 
If the clocks don't overlap, than by convexity of the second clock, there exists a line $\ell$ through $O$ such that the second clock is entirely in one of the half planes defined by it. 
Fix a system of coordinates with origin $O$ at the centre of the first clock, and with $\ell$ as $y$-axis and with the second clock in the negative $x$ half plane.
Let $a(t)$ denote the position vector of the tip of the first clock's hour hand at time $t$, similarly $b(t)$ for the second clock.
Let $n(t)$ denote the normal vector to $a(t)$ in clockwise direction, or more formally just let $n(t)=a(t+3\,\mathrm h)$.
Now define the function $f(t)=\langle n(t),b(t)\rangle$.
We know that $a(t)$ and $n(t)$ describe a circle around $O$, especially that there are times $t_+$ and $t_-$ where $n(t_+)$ is on the positive $x$-axis and $n(t_-)$ is on the negative $x$-axis, and these times are uniquely determined up to multiples of twelve hours. Then clearly $f(t_+)<0$ and $f(t_-)>0$. By the 12 hour periodicity, we may assume wlog. that $t_-<t_+<t_-+12\,\mathrm h$.
As $f$ is continuous ($a,n,b$ are continuous in $t$ by assumption and the scalar product is continuous), we conclude that $f$ has a zero $\tau_1$ in $(t_-,t_+)$ and a zero $\tau_2$ in $(t_+,t_-+12\,\mathrm h)$. At a zero of $f$, $a(t)$ must either point to $b(t)$ or away from it. As $\tau_1,\tau_2$ are separated by $t_+$, $a(\tau_1$ and $a(\tau_2)$ are in different half planes, at the one belonging to negative $x$ values we must have that $a(\tau)$ points to $b(\tau)$.

Note that the claim indeed becomes false if we allow the clocks to overlap. Foe example, put the in the same position but slightly rotated against each other.

I think a more general result (with Dali-like molten clocks) holds, bu there are a number of subtleties involved:
Conjecture. Let $a,b\colon S^1\to\mathbb C$ be continuous with nonintersecting images, $a[S^1]\cap b[S^1]=\emptyset$. Assume that $\mathbb C\setminus (a[S^1]\cup b[S^2])$ has among others two components $U_\infty$ and $U_0$ where $U_\infty$ is not bounded and $\partial U_\infty$ intersects both $a[S^1]$ and $b[S^1]$, and $U_0$ is bounded and $\partial U_0$ intersects $a[S^1]$.
Assume  $0\in U_0$. Then there exists $t\in S^1$ such that $b(t)=\lambda\cdot a(t)$ for some $\lambda>1$.
A: Here is a more general statement which implies yours as a special case. Inspired by Hagen von Eitzen's ending remark, it could be interpreted as a generalization of the original question with "Dali-like molten clocks".
Proposition: Let $\alpha,\beta:S^1\to \Bbb R^2-\{0\}$ be two continuous maps that are not homotopic. Then, there exists $z_0\in S^1$ such that $\alpha(z_0)$ and $\beta(z_0)$ are collinear.
Proof: Let $f,g:S^1\to S^1$ be the radial projections of $\alpha$ and $\beta$ on $S^1$. Then, $f\simeq \alpha$, $g\simeq \beta$, and $\alpha(z_0),\beta(z_0)$ are collinear iff $f(z_0)=g(z_0)$. Thus, by contrapositive, it suffices to show that if $f,g:S^1\to S^1$ are continuous maps such that $f(z)\ne g(z)$ for all $z\in S^1$ then $f\simeq g$. View $S^1$ as a subset of $\Bbb C$ and let $\theta:S^1-\{1\}\to\Bbb R$ be an angle coordinate (see here). Then $f(z)/g(z)\in S^1-\{1\}$ for all $z\in S^1$ so we have a continuous map $\varphi:S^1\to \Bbb R$, $\varphi(z)=\theta(f(z)/g(z))$ satisfying $f(z)=g(z)e^{i\varphi(z)}$. Now we get a homotopy between $f$ and $g$ by
$$H:S^1\times I\to S^1,\quad H(z,t)=g(z)e^{i\varphi(z)(1-t)}.$$
$$\tag{Q.E.D.}$$

To connect with the original question, the first clock is centred at the origin, $\alpha$ is the tip of the first hour-hand and $\beta$ the tip of the second hour-hand.
A: Let the tip of the hour-hand on the big click simply be a point on the plane. 
Let the small hour-hand represent a line originating from another point on the plane (the center of the small clock). That line originates at an angle. 
Angles are continuous, so it follows that for every single point on the plane, you can reach it with a straight line from the center of the small clock. 
(The mental picture you should have is a single point with an arbitrarily long line extending from it, representing the small clock and hand. Over to the top-right, you have a circle, representing the possible positions of the tip of the hour hand of the big clock.)
Then you want to show that you can sweep that line in one clockwise revolution around the center of the small clock, and will certainly hit the hour-hand point of the big clock.
And that is clear: the point of the hour-hand on the big clock is continuously moving on a circle. We can use the line to bisect the circle into two parts: the part it has gone over (A) and the part it has not gone over (B). At every time, the point will be either in A or in B. 
We start out with A being empty and B being all the points. Later, the line will have swept the entire circle, so B will be an empty set and A will contain all the points. That means that the point of the hour hand will move from set B to set A at some point. And the only way this move is possible is if it is 'swept' by the line, i.e. if it intersects the line, i.e. if it is pointed to by the small hour hand.
