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Missouri State University's Advaned Problem February 2013
Vinent Pantaloni (Orléans, Frane)
Problem : A solid is formed by placing congruent regular square pyramids on each face of a cube as shown in the figure below.
• What is the maximum value of V/S, where V is the volume and S the surface area of the solid ?
• Can one fill space with congruent copies of the solid that attains this maximum ? Source : Hungarian National Olympiad
Solution :
Here’s a figure showing a square face ABCD of the cube and one triangle∆BCE. We are looking for the heighthof the pyramid. I will writesfor the side of the cube and afor the altitude EF of∆BEC. The volume of one pyramid isVP =s2×h3. The volume of the entire polyhedron isV =s3+ 6VP, so :
V =s3+ 2s2h=s3
1 +2h s
(1) The surface area S is 24 times that of ∆BCE so S = 12sa. Using the Pythagorean theorem to solve forawe finda=q
h2+s42 = 12p
s2+ (2h)2so :
S = 6sp
s2+ (2h)2= 6s2 s
1 + 2h
s 2
(2)
a h
s
b
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Hence the ratioV /Scan be calculated : V
S = s3 1 + 2sh 6s2
q
1 + 2sh2 Cancelings2 and settingx=2h
s we have : V S = s
6× 1 +x
√1 +x2
We now have to study the function f defined for every positive x by f(x) = 1 +x
√1 +x2. Its derivative is f′(x) = 1−x
(1 +x2)3/2 which is positive before 1 and negative after, so f reaches a maximum when x= 1. That means that the polyhedron that has the greatest ratioV /S is the one for which 2h/s= 1, i.e. h = s/2. Inspecting the figure, it means that ∆GF E is a right isosceles triangle. In other words,
∠EF G = 45◦. The same situation appears on the other faces of the cube so ∆BEC and the other triangle that shares the edge BC will lie in the same plane, forming a rhombus face. The polyhedron we are looking for is therefore made of 12 congruent rhombic faces (one for each edge of the cube). It is usually called the rhombic dodecahedron. And yes, it can tesselate space, every honey bee knows that.
Bees use this shape (half of it actually) to join the opposite hexagonal prisms in which they put the honey. Using the construction we have just made of the rhombic dodecahedron we can easily prove this polyhedron tesselates space and also give the coordinates of its vertices by choosing the center of the cube as the origin and the axis parallel to the sides of the cube, we get fors= 2:
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•Eight vertices of the cube(±1;±1;±1)
•Six vertices as the tops of the pyramids :(±2; 0; 0),(0;±2; 0),(0; 0;±2).
In order to prove that it is possible to fill space with congruent copies of our rhombic dodecahedron we will consider our construction. First imagine that we fill space with cubes alternating red and blue, like a 3D chessboard. See figure below, on the left hand side.
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Notice that the pyramidal hats we put on the faces of the cube to make the rhombic dodecahedron can all six be put back inside the cube since their height is one half of a cube’s side. The summits of all six pyramids meeting at the center of the cube. These six pyramids do fill the inside of all the blue cubes. Now let’s imagine that we slice all the blue cubes into six of these pyramids (you only need four cuts to do so). They become the hats of the adjacent red cubes, turning them into rhombic dodecahedra therefore filling space.
For some DIY, you can make a rhombic dodecahedron by a paper folding method known as modular origami. Since it has 12 faces, you can print a calendar on it.http://www.ii.uib.no/~arntzen/kalender/
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