Heights of two points on a circle Given a circle of arbitrary radius centred on C, let each point on the circle have an associated height which varies continuously and smoothly as you go around the circle. How can I prove or disprove that there will always be at least one pair of diametrically opposite points (i.e. there is a straight line which joins them going through C) which are at the same height? And if true, is there anything special about that height?

Suppose the height of each point on $C$ is $f(\theta)$ where $\theta$ runs from 0 to $2\pi$. Consider any two diametrically opposite points on $C$, $x$ and $x+\pi$, and their associated heights $f(x)$ and $f(x+\pi)$. Let $h(x)=f(x+\pi)-f(x)$ and vary $x$ from 0 to $\pi$; during this movement $h(x)$ will start out at one value and end at its negative. Since $f(x)$ is continuous, so is $h(x)$; since $h(0)=-h(\pi)$, by the intermediate value theorem there is some $0\le x<\pi$ where $h(x)=0$, i.e. the two diametrically opposite points have the same height.

However, there is nothing special about this height other than the antipodal property.