Shape of an elastic steel ruler bent between 3 pins An thin elastic steel ruler is bent around 3 pins. It's not attached to the pins and can slide freely (not attached to the pins).
Let's assume that the pin coordinates are $(0,0)$, $(x,y)$, $(1,0)$
What formula can describe the shape of the ruler?
**What are the formulas for the reaction forces of the pins?

 A: I am afraid that the question as it is is ill-posed and has no unique solution (I am not sure what a solution is actually). For instance, one  needs to know what the Lagrangian is (i.e., the elastic potential) as it depends on cross-section/thickness, ... Also, what is the length of the ruler, at which points should it meet the pins and what are the boundary conditions.
So for the sake of giving an answer I made up a couple of hypotheses that suit me: the cows are spherical and live in a vacuum the ruler is infinitely thin and the pins are point-like. The ruler is also inextensible of length $L$ and has a bending energy proportional to the square of its curvature:
$$
E \propto \int_0^L \kappa^2(s) \mathrm d s
$$
where $s$ is the coordinate along the ruler. If $x$ is the position of particle $s$ then
$$
\kappa^2 = \langle\ddot x,\ddot x\rangle = \dot\theta^2
$$
with $\dot x = (\cos \theta,\sin \theta)$ since under the inextensibility constraint $\langle \dot x,\dot x\rangle = 1$. Minimizing $E$ yields the Euler-Lagrange equation (EDIT: the following equations were corrected):
$$
2\ddot \theta + \lambda_1 \cos\theta + \lambda_2 \sin\theta = 0
$$
where $\lambda_{1,2}$ are Lagrange multipliers. This can be integrated once after multiplication by $\dot \theta$. All in all, the equation of the ruler between two pins is given implicitly by
$$
\int_0^s \frac{\mathrm d\theta}{\sqrt{c-\lambda_1\sin\theta+\lambda_2\cos\theta}} = s.
$$
Therein, $c$ is an integration constant.
Once we have $\theta$ (somehow numerically), we need to integrate once more to get $x$. As for the constants $(c,\lambda_{1,2})$, we need to ensure smoothness across pins and to know at which $s_i$ the ruler goes through the pin at $x_i$. Note that this system may not admit a solution if the positions $x_i$ are too far apart so that inextensibility cannot be ensured.
EDIT:
A couple more details. The total length of the ruler does matter. Here is an extreme scenario. If the length of the ruler is infinite, there exists a sequence of configurations parametrized with $R$ such that the bending energy goes to $0$ as $R$ goes to infinity. In that case, a solution that minimizes bending energy does not exist. The buckled shape in the figure should be familiar. It would help a lot to say for instance that the ruler goes through the first pin at $s_0=0$ and through the third one at $s_3$ = L. As for $s_2$, it can be determined by minimizing bending energy over the reduced set of candidate solutions. But even in these cases, I am not sure that buckling can be completely precluded.
