= Solution
Let $M=M_1+M_2$ and $\alpha=M_2/M$, with $0<\alpha<1$. In the <centre of mass> frame the two primary distances are $r_1=\alpha R$ and $r_2=(1-\alpha)R$. Applying <Newton's law of universal gravitation> and circular <centripetal acceleration> to the first primary gives $M_1\omega^2\alpha R=GM_1M_2/R^2$, and hence
$$
\boxed{\omega^2=\frac{GM}{R^3}.}
$$
The same result follows from the second primary or from the relative two-body <equation of motion>.
A rotating-frame equilibrium has zero rotating-frame <velocity>, so its <Coriolis force> vanishes. Above the orbital plane both primaries have downward gravitational components; below it both have upward components. The centrifugal <force> has no $z$ component. More explicitly,
$$
a_z=-Gz\left(\frac{M_1}{|\mathbf r-\mathbf r_1|^3}+\frac{M_2}{|\mathbf r-\mathbf r_2|^3}\right),
$$
which vanishes only at $z=0$ for finite nonsingular points. Thus \b[all equilibria lie in the orbital plane].
The <equation of motion in a rotating frame> for the dwarf is
$$
m\ddot{\mathbf r}=-\nabla U-2m\boldsymbol\omega\mathbin{\times}\dot{\mathbf r}-m\boldsymbol\omega\mathbin{\times}(\boldsymbol\omega\mathbin{\times}\mathbf r),\qquad U=-\frac{GmM_1}{|\mathbf r-\mathbf r_1|}-\frac{GmM_2}{|\mathbf r-\mathbf r_2|}.
$$
On the orbital plane the centrifugal contribution is $m\omega^2\mathbf r$. Thus the <rotating-frame effective potential> is $E=U-m\omega^2(x^2+y^2)/2$; equilibria are exactly its stationary points. The <Coriolis force> affects motion about those equilibria but not their positions. This is the <circular restricted three-body problem>, with the dwarf treated as a test <mass>.
\b[The printed energy unit has the wrong dimensions.] The correct total <energy> unit is $mGM/R$, or $GM/R$ for specific <energy>; $GM/R^2$ is an <acceleration>. Dividing $E$ by $mGM/R$ and measuring lengths in units of $R$ gives the dimensionless <effective potential>
$$
\mathcal E(x,y)=-\frac{x^2+y^2}{2}-\frac{1-\alpha}{\sqrt{(x+\alpha)^2+y^2}}-\frac{\alpha}{\sqrt{(x-1+\alpha)^2+y^2}}.
$$
The primary coordinates are $(-\alpha,0)$ and $(1-\alpha,0)$. Restricting to $y=0$ gives $F(x)=\mathcal E(x,0)$ and
$$
F'(x)=-x+\frac{(1-\alpha)(x+\alpha)}{|x+\alpha|^3}+\frac{\alpha(x-1+\alpha)}{|x-1+\alpha|^3},\qquad F''(x)=-1-\frac{2(1-\alpha)}{|x+\alpha|^3}-\frac{2\alpha}{|x-1+\alpha|^3}<0.
$$
The absolute values are essential; dropping them would reverse the attraction on the left of a primary. On each of the intervals separated by the two primaries, $F'$ decreases strictly from $+\infty$ to $-\infty$. There is exactly one <Collinear Lagrange point> on each interval, a maximum along the $x$ direction.
For the <inner Lagrange point>, write $x=1-\alpha-\delta$ with $\delta>0$. For the <L2 Lagrange point>, write $x=1-\alpha+\delta$. The respective equations are
$$
0=-(1-\alpha-\delta)+\frac{1-\alpha}{(1-\delta)^2}-\frac{\alpha}{\delta^2}=3\delta-\frac{\alpha}{\delta^2}+O(\delta^2+\alpha\delta),
$$
$$
0=-(1-\alpha+\delta)+\frac{1-\alpha}{(1+\delta)^2}+\frac{\alpha}{\delta^2}=-3\delta+\frac{\alpha}{\delta^2}+O(\delta^2+\alpha\delta).
$$
Balancing the leading terms gives $\delta=(\alpha/3)^{1/3}+O(\alpha^{2/3})$. Therefore the <small-mass-ratio collinear Lagrange points> obey
$$
\boxed{x_{L_1}=1-(\alpha/3)^{1/3}+O(\alpha^{2/3}),\qquad x_{L_2}=1+(\alpha/3)^{1/3}+O(\alpha^{2/3}).}
$$
The omitted $-\alpha$ displacement of the secondary is smaller than the stated error. The leading distance from the secondary is the dimensionless <Hill radius>.
For the <L3 Lagrange point> put $x=-1+c\alpha+O(\alpha^2)$. It lies to the left of both primaries, so
$$
F'=-x-\frac{1-\alpha}{(x+\alpha)^2}-\frac{\alpha}{(x-1+\alpha)^2}.
$$
Expanding the two positive inverse-square factors gives $(1-\alpha)/(x+\alpha)^2=1+(2c+1)\alpha+O(\alpha^2)$ and $\alpha/(x-1+\alpha)^2=\alpha/4+O(\alpha^2)$. Thus $F'=-(3c+5/4)\alpha+O(\alpha^2)$, so
$$
\boxed{x_{L_3}=-1-\frac5{12}\alpha+O(\alpha^2).}
$$
All three have $y=z=0$. The small-$\alpha$ equalities in the paper are asymptotic expressions, not exact coordinates for finite mass ratio.
For the remaining <Lagrange points>, introduce plane <polar coordinates> about the first primary: $x=-\alpha+s\cos\theta$, $y=s\sin\theta$. Its distance to the second primary is $d=(s^2+1-2s\cos\theta)^{1/2}$, and
$$
\mathcal E=-\frac12(s^2-2\alpha s\cos\theta+\alpha^2)-\frac{1-\alpha}{s}-\frac\alpha d.
$$
The angular stationary condition is $\partial_\theta\mathcal E=\alpha s\sin\theta(d^{-3}-1)=0$. For an off-axis point $s>0$ and $\sin\theta\ne0$, so $d=1$. The radial stationary condition becomes
$$
\partial_s\mathcal E=-s+\alpha\cos\theta+\frac{1-\alpha}{s^2}+\frac{\alpha(s-\cos\theta)}{d^3}=(1-\alpha)(s^{-2}-s)=0.
$$
Hence $s=1$, and $d=1$ implies $\cos\theta=1/2$. The <Triangular Lagrange points> are exactly
$$
\boxed{L_4=(\tfrac12-\alpha,\tfrac{\sqrt3}{2},0),\qquad L_5=(\tfrac12-\alpha,-\tfrac{\sqrt3}{2},0).}
$$
Both triangles are equilateral. The collinear and off-axis calculations together exhaust the stationary points.
For the requested sketch, $\mathcal E$ tends to $-\infty$ at either primary and at large in-plane radius. Each <Collinear Lagrange point> is a saddle of the planar <effective potential>: its $x$ curvature is negative and its $y$ curvature is positive. To see the latter, let $d_1=|x+\alpha|$, $d_2=|x-1+\alpha|$ and $D=(1-\alpha)/d_1^3+\alpha/d_2^3$. At an inner point both distances are below one, so $D>1$. At an exterior stationary point, use barycentric positions $a_1=-\alpha$, $a_2=1-\alpha$ and weights $1-\alpha,\alpha$: the equation implies $x(D-1)=\sum w_i a_i/|x-a_i|^3$. To the right, the weighted sum is positive because the positive primary is closer; to the left it is negative because the negative primary is closer. Thus $D>1$ there as well, and $\mathcal E_{yy}=D-1>0$.
At either <Triangular Lagrange point>, the planar <Hessian> is
$$
D^2\mathcal E=-3\begin{pmatrix}1/4&\pm\sqrt3(1-2\alpha)/4\\\pm\sqrt3(1-2\alpha)/4&3/4\end{pmatrix}.
$$
Its trace is $-3$ and determinant $27\alpha(1-\alpha)/4>0$, so these are local maxima. Their height is $\mathcal E=-3/2+\alpha(1-\alpha)/2$. The sketch below uses a finite mass ratio to separate the five points clearly; its singular wells are clipped only for display. These curvature labels concern the <effective potential>, not a stability test that ignores the <Coriolis force>.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2012/iii/paper-59-effective-potential.png]
{title=Rotating binary effective-potential surface and contours at mass fraction 0.10, with all five Lagrange points marked}
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