If three quarters of the surface of a sphere is colored black, then an inscribed tetrahedron can be rotated so that all of its vertices are black 
Let $P$ be a tetrahedron inside an sphere such that all of its vertices are on the surface of the sphere. Suppose that three quarters of sphere's surface is colored black. Show that there is a rotation of $P$ such that all of its vertices (when rotated) are black. 

Clarifications from comments: The black region need not be connected, but must be measurable.
 A: The problem as stated is false. However, by slightly adapting the conditions of the problem, we’re able to solve it by using @Hagen von Eitzen’s approach.
To show falsehood, consider a sphere, divide it into octants, and paint all of them black, save for two opposite octants and their borders. By considering its symmetry group, it’s easy to show that any regular tetrahedron on this sphere will necessarily have a white vertex.
Nevertheless, the problem is true if more than $\frac34$ of the sphere is black. We can prove this as follows.
Let $A$, $B$, $C$, $D$ be the vertices of the tetrahedron, and let $R_A$, $R_B$, $R_C$, $R_D$ be the sets of rotations of the sphere that make their respective vertices black. If $\mu(X)$ denotes the measure of a set, recall that $$\mu(X\cap Y)\geq \mu(X)+\mu(Y)-1.$$ Since $\mu(R_A)=\mu(R_B)=\mu(R_C)=\mu(R_D)>\frac34$, applying this equation recursively, we obtain that $$\mu(R_A\cap R_B\cap R_C\cap R_D)>4\cdot\frac34-3=0.$$ In particular, the intersection of these sets must be non-empty, and there must exist a rotation that makes all vertices black, just as we wanted to prove. $\blacksquare$
