Area of the triangle generated by the hands of a clock
This is an addendum to Oscar's answer. Just doing some math.

Let t = 0 be 12 o'clock, when all hands are coinciding and the area is 0. We need to calculate the area as a function of t. Let s, m, h be the angles of second, minute and hour with respect to the trigonometric reference. Then
$$s=\frac{\pi}2-\frac{\pi}{30}t\\ m=\frac{\pi}2-\frac{\pi}{1800}t\\ h=\frac{\pi}2-\frac{\pi}{21600}t$$
and let S, M, H be the lengths of clock's hands. Then the area is
$$A=\frac12\left|SM\sin\alpha+MH\sin\beta+HS\sin\gamma\right|$$
where
$$\alpha=s-m,\;\beta=m-h,\;\gamma=h-s$$
Therefore we have
$$\alpha=-\frac{59}{60}\cdot\frac{\pi}{30}t\\ \beta=-\frac{11}{720}\cdot\frac{\pi}{30}t\\ \gamma=\frac{719}{720}\cdot\frac{\pi}{30}t$$
where t is a positive integer. As you see, doing some boring math can sometimes reduce the amount of codes needed for a job.

P.S. and here is a plot of A vs. t. It is periodic indeed.