Question: My textbook says: Suppose that $E$ is a convex region in the plane bounded by a curve $C$. [Where a convex region is defined as for all $x, y \in E, sx + ty \in E$ where $0 \le s, t \le 1$ and $s + t = 1$.] Show that $C$ has a tangent line except at a countable number of points.

So my thinking is roughly that points without tangent lines look like sharp corners with some angle $\theta < 180^\circ$ and so if $\theta\_m$ is the largest $\theta$ in $C$ then the most corners you can pack into $C$ is the regular $n$-gon with $n=\frac{2}{180-\theta\_m}$. Except that seems to suggest that $C$ must have a tangent line except at finitely many points whereas the question clearly asks about countably many. So what's the pathological convex curve with infinitely many discontinuities that thwarts my proof?

Answer: 
Line segments from $(n,n^2)$ to $((n+1),(n+1)^2)$ for each integer $n$ you should draw. A nice and simple convex region you will get.
Here's a compact example. Start with a circle of radius $1$. Cut off a minor segment with a chord of length $1$, and close the resulting major segment. From this major segment, cut off a minor segment with an adjacent chord of length $\frac12$, and close the rest. Continue this process, cutting off minor segments with adjacent chords successively of length $\frac14,\frac18,...,1/2^n,...$. The resulting figure is convex with no defined tangent at the infinitely many corners between the successive chords.