 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Spherical shelter (Posted on 2012-04-27) 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

 No Solution Yet Submitted by Jer No Rating Comments: ( Back to comment list | You must be logged in to post comments.) Solution Comment 2 of 2 | 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 � 23/4)/sqrt3  =  0.422863...

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

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