A unit sphere is on a level surface in the rain. Find the largest of each of the following shapes that can be placed under the sphere and not get rained on:

a) Sphere
b) Cube
c) Regular tetrahedron

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:

d² = 1 – (1 – 2a)²           (2)

Eliminating d from (1) and (2) gives:

1 + 3a² – 4a*sqrt(1 – a²) = 4a – 4a²

7a² – 4a + 1 = 4a*sqrt(1 – a²)

Squaring and simplifying gives:   65a⁴ – 56a³ + 14a² - 8a + 1 = 0

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:

            (sqrt(1 – b²) – (b*sqrt3)/3)² + (1 – 2*b*(sqrt(2/3))² = 1

Squaring and simplifying:    2(sqrt3)b² – 4(sqrt2)b +sqrt3 = 2b*sqrt(1 – b²)

Squaring and simplifying:  16b⁴ – 16(sqrt6)b³ +40b² – 8(sqrt6)b + 3 = 0

Factorising:        (4b² – 2(sqrt6)b + 1)(4b² – 2(sqrt6)b + 3) = 0

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 I’ve found that gives a smaller maximum edge:

viz:    (1 + sqrt2 – 2^3/4)/sqrt3  =  0.422863...

Posted by Harry on 2012-05-02 22:24:12