This is a solution to the unit fraction chord length puzzle.
A brief description:
Any continuous function $f:[0,1] \to \mathbb{R}$ such that $f(0) = f(1)$, has a chord of length $\frac{1}{n}$ where $n$ is a positive integer.
Any other chord length (non unit fraction lengths) have functions which don't have those chord lengths.
Solution
Let $g(x) = f(x) - f\left(x + \frac{1}{n}\right)$.
Then we have that
$$g(0)+ g(1/n) + \dots + g((n-1)/n) = f(0) - f(1) = 0 $$
Thus if $g(k/n) \ne 0$, then there are some $i, j$ such that $g(i/n) \lt 0$ and $g(j/n) \gt 0$, and thus by the mean value theorem, some point $c$ where $g(c) = 0$.
For the other part, given any non unit fraction $r$, the function
$$ f_r(x) = \sin^2\left(\frac{\pi x}{r}\right) - x \sin^2\left(\frac{\pi}{r}\right)$$
is a counter-example.
This result is apparently called the Universal Chord Theorem.
A brief description:
Any continuous function $f:[0,1] \to \mathbb{R}$ such that $f(0) = f(1)$, has a chord of length $\frac{1}{n}$ where $n$ is a positive integer.
Any other chord length (non unit fraction lengths) have functions which don't have those chord lengths.
Solution
Let $g(x) = f(x) - f\left(x + \frac{1}{n}\right)$.
Then we have that
$$g(0)+ g(1/n) + \dots + g((n-1)/n) = f(0) - f(1) = 0 $$
Thus if $g(k/n) \ne 0$, then there are some $i, j$ such that $g(i/n) \lt 0$ and $g(j/n) \gt 0$, and thus by the mean value theorem, some point $c$ where $g(c) = 0$.
For the other part, given any non unit fraction $r$, the function
$$ f_r(x) = \sin^2\left(\frac{\pi x}{r}\right) - x \sin^2\left(\frac{\pi}{r}\right)$$
is a counter-example.
This result is apparently called the Universal Chord Theorem.
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