I believe that each sheltered shape needs to be inside and touching the wall of a hollow (imaginary) vertical cylinder that escribes the unit sphere and defines the dry region; and also touching the underside of the unit sphere.
(a) Sphere with radius r The line joining the centres of the two spheres passes through the intersection of the cylinder and base plane at an angle of 45 degrees and forms a triangle which yields:(1 + r + r*sqrt2))2 = 2
From which,r = (sqrt2 1)/(sqrt2 + 1)=3 2*sqrt2=0.1715728... (b) Cube with side 2a Two vertical edges of the cube will touch the cylinder wall, so that the distance, d, of its inner vertical face from the axis of the cylinder is given by:
d = sqrt(1 a2) 2a(1)
Working in the vertical plane through the centre of sphere and cube, lines through their point of contact form a triangle which yields:
Sadly, no easy factors, but a numerical solution gives 2 real roots, one of which relates to our problem:
a = 0.1438558... therefore the cube has sides of length 0.287712...
(c) Tetrahedron with edges of length 2b Arrange** the tetrahedron so that two of its vertices touch the cylinder wall where it meets the base plane; and the apex of the tetrahedron touches the unit sphere. The outer edge which joins the vertices that touch the cylinder is at a distance of
sqrt(1 b2) from the axis of the cylinder and, since the altitude of each
face of the tetrahedron is b*sqrt3, it follows that the distance from the apex
to the axis of the cylinder is sqrt(1 b2) b*(sqrt3)/3.
The height of the apex above the base plane is 2*b*sqrt(2/3) and, by working in a vertical plane, the line from the apex to the centre of the sphere can be used as a hypotenuse to derive:
Only the first factor has real roots, only one of which relates to our problem:
b = (sqrt2)(sqrt3 1)/4=0.258819...
So the tetrahedron has edges of length 0.517638... _________ ** NB the tetrahedron could be turned through 180 degrees so that a vertex touches the cylinder, but Ive found that gives a smaller maximum edge: