Zero force for three charges in a triangle Following is a problem from the book, Electricity and Magnetism by Purcell and Morin 3e.

Two positive ions and one negative ion are fixed at the vertices of an
equilateral triangle. Where can a fourth ion be placed, along the
symmetry axis of the setup, so that the force on it will be zero? Is
there more one such place? You will need to solve something
numerically.

Please don't just post a solution. I already have it. I just don't get how they arrived at it. I have already tried to use columbs law to add the forces to zero and solve for the distance required for the fourth ion. I keep getting a different answer and setup from the solution.
Any explanation would be helpup. I don't even care about a full solution. I just need the proper setup and how it was derived.
 A: You need a diagram like this:

Note that since two of the charges have a positive charge and one has a negative charge, the point where the forces cancel out is on the line of symmetry, and outside of the triangle.
By symmetry, the horizontal components of force cancel. The vertical components sum to zero; this means that
$$0 = \frac{Q}{\left(d\sqrt{3}/2+h\right)^2} - 2\cdot \frac{Q}{\left(\frac{h}{\sin\alpha}\right)^2}\sin\alpha$$
A bit of rearranging turns this into an expression in $h$ and $\sin^3\alpha$:
$$\sin^3\alpha = \frac{h^2}{2(d\sqrt{3}/2+h)^2}$$
Noting that $\sin\alpha = \frac{h}{\sqrt{h^2 + d^2/4}}$ this reduces to an (ugly) expression in h, which can be solved numerically.
Does this help? Note that if you make $h$ go negative, the two green forces start pointing up, and you would find another solution above the negative charge (where the red force points down). Since the way I formulated the force only involves $h^2$, you need to do a bit of work to find the other solution. I leave that up to you.
