The vertices of a tetrahedron correspond to four alternating corners of a cube. By using analytical geometry, demonstrate that the angle made by connecting two of the vertices to a point at the center of the cube is \(109.5^{\circ},\) the characteristic angle for tetrahedral molecules.

Let's assign coordinates to the vertices of the cube, considering the cube's edge of length 1. We place the origin of the coordinate system at one vertex of the cube, with three edges coming out along the positive x, y, and z axes: \( A(0,0,0) \), \( B(1,0,0) \), \( C(0,1,0) \), \( D(0,0,1) \), \( E(1,1,0) \), \( F(1,0,1) \), \( G(0,1,1) \), and \( H(1,1,1) \). Now we choose four alternating corners of the cube to be the vertices of the tetrahedron, for instance, A, D, E, and G.

To find the center of the cube, we will average the coordinates of two opposite vertices, for example, the vertices A and H: \[ O = (\frac{A_x + H_x}{2},\frac{A_y + H_y}{2},\frac{A_z + H_z}{2}) \] Which simplifies to \( O = (0.5,0.5,0.5) \).

We will find the vectors by subtracting the center coordinates O from the coordinates of the tetrahedron vertices A, D, and E: \[ \vec{OA} = A - O = (0,0,0) - (0.5,0.5,0.5) = (-0.5,-0.5,-0.5) \] \[ \vec{OD} = D - O = (0,0,1) - (0.5,0.5,0.5) = (-0.5,-0.5,0.5) \] \[ \vec{OE} = E - O = (1,1,0) - (0.5,0.5,0.5) = (0.5,0.5,-0.5) \]

We will use the formula with the dot product and vectors' length to find the angle \(\theta\) between the vectors \(\vec{OA}\) and \(\vec{OD}\) first: \[ cos(\theta) = \frac{\vec{OA}\cdot\vec{OD}}{\|\vec{OA}\|\|\vec{OD}\|} \] We first calculate the dot product and the length of vectors: \[ \vec{OA}\cdot\vec{OD} = (-0.5)(-0.5) + (-0.5)(-0.5) + (-0.5)(0.5) = 0.75, \] \[ \|\vec{OA}\| = \sqrt{(-0.5)^2 + (-0.5)^2 + (-0.5)^2} = \frac{\sqrt{3}}{2}, \] \[ \|\vec{OD}\| = \|\vec{OA}\| = \frac{\sqrt{3}}{2}\] Plug these into the formula: \[ cos(\theta) = \frac{0.75}{(\frac{\sqrt{3}}{2})(\frac{\sqrt{3}}{2})} = \frac{0.75}{\frac{3}{4}} = 1 - \frac{1}{3} =\frac{2}{3} \] Now, calculate \(\theta\): \[ \theta = arccos(\frac{2}{3}) \approx 109.5^{\circ} \] Since all vertices of the tetrahedron are symmetric, we can conclude that the angle between any of the vectors connecting the center of the cube to its vertices will be approximately \(109.5^{\circ}\), which is the characteristic angle for tetrahedral molecules.