Angle between lines joining tetrahedron center to vertices

The required tetrahedral angle is $\arccos\left(-\frac13\right)\approx109.5^\circ$. You can use the law of cosines to show this... or more transparently, you can exploit the fact that a tetrahedron is easily embedded inside a cube:

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I suppose now's as good a time as any to post the synthetic proof.

One can use the Pythagorean theorem to show that a square with unit edge length has a diagonal of length $\sqrt 2$. The Pythagorean theorem can be used again to show that a right triangle with leg lengths $1$ and $\sqrt 2$ will have a hypotenuse of length $\sqrt 3$ (corresponding to the triangle formed by an edge, a face diagonal, and a cube diagonal). We know that the diagonals of a rectangle bisect each other; this can be used to show that the diagonals of a cube bisect each other. From this, we find that the side lengths of the (isosceles!) triangle formed by two half-diagonals of the cube (corresponding to two of the arms of your caltrops) and a face diagonal are $\frac{\sqrt 3}{2}$, $\frac{\sqrt 3}{2}$, and $\sqrt{2}$. From the law of cosines, we have

$$2=\frac34+\frac34-2\frac34\cos\theta$$

where $\theta$ is the obtuse angle whose measure we are seeking. Algebraic manipulation yields $\cos\,\theta=-\frac13$.

One way is to write the vertices as vectors $a,b,c,d$ with norm $\|\cdot\|=1$. Then $a+b+c+d=0$. But

$$ 0=\|a+b+c+d\|^2=4+2{4 \choose 2}\cos\theta,$$

so $\theta = \arccos(-1/3)$.

(Assuming the tetrahedron is supposed to be regular) Take the tetrahedron with vertices $(1,1,1), (1,-1,-1), (-1,1,-1), (-1,-1,1)$, which has centre at the origin, and use the dot product formula:

$a\cdot b = |a| |b|\cos\theta$

which gives $\cos\theta=-\frac13$