Curve with all its tangent lines intersecting in one point $\implies$ zero curvature How can one show that a regular curve in $\mathbb R^n$ with all its tangent lines intersecting in one point has zero curvature (and therefore is a line)?
 A: Hint: Write that single point (which you might as well assume is the origin) as a linear combination 
$$
X(t) + b(t) T(t) + c(t) N(t) + d(t) B (t)
$$
using the Frenete-Serret frame $(T,N,B)$ of the curve. What do you know about $b, c, d$? Now differentiate like mad. 
(I haven't actually carried out this process except briefly in my head, but it's almost the universal solution to problems like this.) 
I'm assuming the curve is regular enough (order 2?) that it actually HAS a Frenet-Serret frame, of course, and that you CAN differentiate.  
Post-comment addition
Following up on Ted's remark, you don't really need the Frenet frame. 
Let's call the curve $t \mapsto X(t)$, and the point lying on all tangents $A$. Define $Y(t) = X(t) - A$. Then $Y$ is a translated version of $X$ with all tangents passing through $0$. So for every $t$, 
$$
Y(t) = a(t) Y'(t)
$$
for some number $a(t)$. Next, since $X$ (and hence $Y$) is a regular curve, there's a function $s \mapsto h(s)$ with the property that 
$$
\| \frac{d}{ds} Y(h(s)) \| = 1
$$
for all $t$, i.e., by travelling along $Y$ at a possibly altered rate, we can have the tangent vector have length $1$. Define
$$
Z(s) = Y(h(s))
$$
Then we can write
$$
Z(s) = u(s) Z'(s)
$$
where just as before, $u$ is a scalar-valued function. 
Differentiating, we get
$$
Z'(s) = u'(s) Z'(s) + u(s) Z''(s) 
$$
Now since $Z'(s)$ has constant length $1$, we have
$$
Z'(s) \cdot Z'(s) = 1 \\
Z''(s) \cdot Z'(s) + Z'(s) \cdot Z''(s) = 0\\
2 Z'(z) \cdot Z''(s) = 0
$$
i.e., $Z''$ is perpendicular to $Z'$. So if we take the dot product of the formula for $Z'$ with $Z'$, we get
\begin{align}
Z'(s) \cdot Z'(s) &= Z'(s) \cdot \left( u'(s) Z'(s) + u(s) Z''(s) \right)\\
Z'(s) \cdot Z'(s) &= u'(s) Z'(s) \cdot Z'(s) + u(s) Z'(s) \cdot Z''(s) \\
1 &= u'(s) 1 + 0
\end{align}
From this, we conclude that $u'(s) = 1$, so $u(s) = s + C$. 
But we also get something else: by projecting 
$$
Z'(s) = u'(s) Z'(s) + u(s) Z''(s) 
$$
onto the plane perpendicular to $Z'(s)$, via the projection 
$$
P(v) = v - (v \cdot Z'(s)) Z'(s)
$$
(you can check for yourself that every vector in the image of $P$ is perpendicular to $Z'(s)$, and that $P$ sends $Z'(s)$ itself to the zero vector, and that $P^2 = P$, so that $P$ really is the projection I claim!), we get
$$
0 = 0 + u(s) P(Z''(s))
$$
But since $Z''(s)$ is already perpendicular to $Z'(s)$, we have $P(Z''(s)) = Z''(s)$, hence
$$
0 = 0 + u(s) Z''(s)\\
0 = (s + C) Z''(s).
$$
Hence for every value of $s$ except $s = -C$, we have $Z''(s) = 0$, which makes the curvature of $Z$ be zero everywhere except perhaps at $s = -C$. But if we assume that $X$ is at least twice continuously differentiable, then the curvature is a continuous function of position, and a continuous function that's zero everywhere except at one point must be zero everywhere. 

Sadly, in this proof I'm assuming not just that the original curve is differentiable with nonzero tangent at every point, but that it is twice differentiable. Then again, without that assumption, most definitions of curvature don't even make sense, so it's pretty reasonable. But I'm also assuming that the curvature itself is continuous to fill in the final point $s = -C$. 
I suspect that there's a pretty simply proof that says "away from $s = -C$, the curve lies on a line. So overall, the curve could consist of two distinct rays, one for $s < -C$, another for $s > -C$. But the only way for such a curve to have a derivative at $s = C$ is for the two rays to share the same start point, and for their directions to be equal, hence the curve consists of a single straight line." I could probably turn that into a proof, but I need to go build a cabinet instead. 
