How do I recover a convex region from the set of its tangent planes?

I believe the answer to be the parametric curve $(x(\theta), y(\theta))$ given by $$x(\theta) = M(\theta)cos(\theta) - M^{'}(\theta) sin(\theta)$$ $$y(\theta) = M(\theta)sin(\theta) + M^{'}(\theta) cos(\theta)$$

I've tried computing a proof but in the end I've gotten it with an informal sketch.

Here is my elaboration of the suggestions given in the comments and answers.

$\newcommand{\ntheta}{\hat n_\theta}$Let $A\subseteq \mathbb R^n$ be our convex closed (finite) region. What we know about this region is that, for all $\mathbf x\in A$ and direction $\ntheta\in S^{n-1}$, there is some $M(\theta)$ such that $$\langle \ntheta,\mathbf x\rangle= M(\theta)\equiv\min\{C\ge0 : \forall \mathbf y\in A,\,\, \langle \ntheta,\mathbf y\rangle\le C\}.\tag A$$ In words, $M(\theta)$ is gives the location of the plane normal to $\ntheta$ that touches $A$.

The problem is to go from the knowledge of $M(\theta)$ in (A), to a full characterisation of the region $A$.

The main observation is that we can think of $F(\theta)\equiv M(\theta)-\langle\ntheta,\mathbf x\rangle=0$ as the constraint characterising $A$, and therefore observe that $F'(\theta)$ gives, for every $\theta$, the set of points orthogonal to $A$ at the point $\mathbf x$ such that $\langle \ntheta,\mathbf x\rangle= M(\theta)$. To recover $A$, we therefore just need to find the intersection between these points and the surface itself, that is, solve the system: $$\begin{cases} \partial_\theta[M(\theta)-\langle\ntheta,\mathbf x\rangle]=0, \\ M(\theta) = \langle\ntheta,\mathbf x\rangle. \end{cases}\tag B$$ It's worth noting that this will not have a solution for every $\theta$. Indeed, while $M(\theta)$ is always well-defined, there can in general be angles that do not correspond to any point of $\partial A$ (think for example what happens if $A$ is a triangle, in which case there are only three values of $\theta$ corresponding to planes tangent to $A$).

Here are a few examples of how this works:

Toy example with misplaced circle

Consider the region $A$ defined algebraically as $(x-x_0)^2+(y-y_0)^2=r^2$. For any direction $\theta$, we can easily find that $M(\theta)=\cos\theta x_0 + \sin\theta y_0 + r$. In this case (B) thus reads $$\begin{cases} -\sin\theta (x_0-x) + \cos\theta (y_0-y) = 0, \\ \cos\theta (x_0-x) + \sin\theta (y_0-y) + r= 0, \end{cases}$$ we thus find from the first equation $\theta=\theta(x,y)$ to be given by $\tan\theta = (y-y_0)/(x-x_0)$, which then using the second equation gives back the equation of the circle. The $\cos\theta=0$ case is also handled trivially.

More contrived example with discontinuities

Suppose now $A$ is the convex hull of the graph of $y=x^2$ in the domain $x\in[-1,1]$. In this case, $M(\theta)$ is given by solving $$M(\theta) = \max_{x\in[-1,1]} [x(\cos\theta + \sin\theta\, x)].$$ This is a bit tricky because the maximum is achieved sometimes at the boundary and sometimes in the inside of the region. The local stationary points are given by solving $\cos\theta+2x\sin\theta=0$, and thus correspond to the points $x=-\frac12 \cot\theta$, whenever this is inside $[-1,1]$. One then has to check for each $\theta$ what the maximum is, which is a tedious but straightforward procedure, and finally leads to the result (as given by Mathematica, with a few simplifications):

$$M(\theta) = \begin{cases} 1 & \theta\in\{0,\pi/2,\pi\} \\ -\frac{1}{4} \cos (\theta ) \cot (\theta ) & \cos (\theta )>2 \sin (\theta )\land 2 \sin (\theta )+\cos (\theta )\leq 0 \\ \sin (\theta )-\cos (\theta ) & (\cos (\theta )\leq 2 \sin (\theta )\land \sin (\theta )<0)\lor (\sin (\theta )>0\land \cos (\theta )<0) \\ \sin (\theta )+\cos (\theta ) & (\sin (\theta )>0\land \cos (\theta )>0)\lor (\sin (\theta )<0\land 2 \sin (\theta )+\cos (\theta )>0). \end{cases}$$

which looks like the following when plotted for $\theta\in[0,2\pi]$:

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The corresponding characterisation of the region can be represented as follows:

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Now the question is how to retrieve the original region from this. There are essentially two regions to consider: the smooth lower part of $A$, and the discontinuous upper regions.

The lower part corresponds to the angles such that $-2\arctan(2+\sqrt5)\le\theta\le 2\arctan(2-\sqrt5)$, for which $M(\theta)=-\frac14 \cos\theta\cot\theta$. Here, system (A) becomes, using the simplified notation $\cos\theta\equiv c, \sin\theta\equiv s$, $$\begin{cases} c x + s y = -\frac14 c^2/s, \\ -s x + c y = \frac{c}{4}(1+1/s^2). \end{cases}$$ Solving this system we get $$\begin{cases} x = \frac14 c/s[ -c^2 -s^2 - 1] = -\frac12 (c/s), \\ y = \frac14[ -c^2 + c^2(1+1/s^2) ] = \frac14 (c/s)^2, \end{cases}$$ which parametrises $y=x^2$ in the considered $\theta$ domain, as expected.

Consider now the set of angles in which $M(\theta)=\cos\theta+\sin\theta$. Here, the system becomes $$\begin{cases} c x + sy = c + s, \\ \partial_\theta[c x + s y - (c+s)] = -s x + c y + s - c = 0, \end{cases}$$ which solves to $x=y=1$ for all $\theta$s, again as expected from the figure. The angles with $M(\theta)=-\cos\theta+\sin\theta$ are handled similarly, and give the solution $y=-x=1$.

The final case to handle is $\theta=\pi/2$. This case is actually toughest, because $M'(\theta)$ is discontinuous at this point, and so we cannot use the usual system. This failure is most likely due to the non-injectivity of the mapping between $\mathbf x$ and corresponding angle $\theta$ (that is, the flat part of the region). A way around this is to simply connect the solutions found for the angles $\theta\neq\pi/2$, that is, $x=y=1$ and $y=-x=1$, with a straight segment (although in fairness I'm not sure how to formalise why this works).

Triangle

Another interesting example is one with a region that is the convex hull of a finite number of points. So let's consider a function $M(\theta)$ of the form $$M(\theta) = \begin{cases} \cos\theta, &\theta\in[-\pi/2,\pi/4], \\ \sin\theta, &\theta\in[\pi/4,\pi], \\ 0, &\theta\in[-\pi,-\pi/2]. \end{cases}$$ We have three regions to consider. For $\theta\in[-\pi/2,\pi/4]$, we have $$\begin{cases} -s x + c y = -s, \\ c x + s y = c \end{cases}$$ which solves to $x=1$ and $y=0$. For $\theta\in[\pi/4,\pi]$, we have $$\begin{cases} -s x + c y = c, \\ c x + s y = s, \end{cases}$$ which gives $x=0$ and $y=1$. Finally, for $\theta\in[-\pi,-\pi/2]$, we have $$\begin{cases} -s x + c y = 0, \\ c x + s y = 0, \end{cases}$$ and thus $x=y=0$.

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This tells us that the region must then be the convex closure of the three points $(1,0), (0,1), (0,0)$, that is, a right triangle (which is of course what I used to find $M(\theta)$ in the first place).