# Background

A student asked me to prove that the regular tetrahedron is the minimum energy geometry available to describe the locations of the electron domains in three dimensions – the prediction of valence shell electron pair repulsion (VSEPR) theory for a central atom with 4 mutually repelling electron domains. With the tetrahedron demonstrated, it could then be shown using vector multiplication that the angles between all substituent atoms is $109.5^{\circ}$.

# Some ideas

I can think of a few ways to start this proof:

• As mentioned above, we could define a regular tetrahedron and work backwards to see if the distance between all vertices is equal and maximized. Start with the vertices: A(0, 0, 0), B(1, 1, 0), C(1, 0, 1) and D(0, 1, 1), and a central point O($\frac{1}{2}$, $\frac{1}{2}$, $\frac{1}{2}$). With these, we can get vectors that point to each vertex. Call them $\vec{r_1}$ through $\vec{r_4}$. We remember that we can calculate the angle between any two vectors using the geometric and algebraic definitions of the dot product. I won't bore you with the details of this calculation but if anyone wishes to see it, I would be happy to share it. If you do it you will see that all the angles are equal and satisfy: $\cos^{-1}\left(-\frac{1}{3}\right) = \theta_i \approx 109.47122^{\circ}$. From here we would need to calculate the six distances bewteen the four vertices to show that the distances are all equal.

• Using multivariable calculus, we could treat this as an optimization problem subject to the constraint matrix, $g$; the sum square distance function, $f$; and use the optimization formalism described by Lagrange. This results in a system of 13 unknowns – the 4 points with three coordinates each plus $\lambda$ – and 13 equations – the four constraint equations plus the nine partial derivative equations from the Lagrange method:

$$g = 1 = \left\{\begin{array}{c}g_1(x_1,y_1,z_1) = {x_1}^2 + {y_1}^2 + {z_1}^2\\ g_2(x_2,y_2,z_2) = {x_2}^2 + {y_2}^2 + {z_2}^2\\ g_3(x_3,y_3,z_3) = {x_3}^2 + {y_3}^2 + {z_3}^2\\ g_2(x_4,y_4,z_4) = {x_4}^2 + {y_4}^2 + {z_4}^2\\ \end{array}\right.$$ $$f = \mathop{\sum_{i=1}^{3}\sum_{j=1}^{3}}_{i<j}\left[\left({x_i}-{x_j}\right)^2+\left({y_i}-{y_j}\right)^2+\left({z_i}-{x_j}\right)^2\right]$$ $$\nabla f=-\lambda \nabla g$$

• An approach similar to the one outlined above could be used with spherical coordinates instead. This version reduces the set of variables from 13 to 8 (the four sets of two angles for each point $\theta$'s and $\phi$'s; note the spherical radius could be set to 1 or any other bond distance, $a$, so is not a variable). These could be further reduced using symmetry arguments. For example, if we take the first vertex at the point $(0, 0, 1)$, we know based on symmetry that if we look down the z axis at the xy-plane, the angle between the three other vectors must have an angle of $120^{\circ}$ with respect to one another. Let one of the other vertices have the point $(x, 0, -\sqrt{1 - x^2})$ – satisfying the radius of 1 unit constraint we have imposed. Now based on symmetry, we know that the other two points must be $(x\cos(\frac{2\pi}{3}),x\sin(\frac{2\pi}{3}),-\sqrt{1-x^2})$ and $(x\cos(\frac{4\pi}{3}),x\sin(\frac{4\pi}{3}),-\sqrt{1-x^2})$ using spherical symmetry. From here, we can write a formula for the total sum square distance between all the points and set the derivative with respect to $x$ to zero to find the maximum sum square distance. Notice we have reduced the problem with these symmetry arguments from a multivariable calculus problem to a calculus BC problem. Alternatively, we could use some other guess and check method to solve for $x$. Note, we should find the same answer using this method as the others above (and below). 