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Logic
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An
escalator can never break: it can only become stairs. You should
never see an "Escalator Temporarily Out Of Order" sign,
just "Escalator Temporarily Stairs. Sorry for the convenience"
 Mitch Hedberg

•••
Always
Look on the Bright Side of Life
(see
the video!)
Some
things in life are bad,
They can really make you mad,
Other things just make you swear and curse,
When you're chewing life's gristle,
Don't grumble,
Give a whistle
And this'll help things turn out for the best.
And...
Always
look on the bright side of life.
[whistle]
Always look on the light side of life.
[whistle]
If
life seems jolly rotten,
There's something you've forgotten,
And that's to laugh and smile and dance and sing.
When you're feeling in the dumps,
Don't be silly chumps.
Just purse your lips and whistle.
That's the thing.
And...
Always
look on the bright side of life.
[whistle]
Always look on the right side of life,
[whistle]
For
life is quite absurd
And death's the final word.
You must always face the curtain with a bow.
Forget about your sin.
Give the audience a grin.
Enjoy it. It's your last chance, anyhow.
So,...
Always
look on the bright side of death,
[whistle]
Just before you draw your terminal breath.
[whistle]
Life's
a piece of ****,
When you look at it.
Life's a laugh and death's a joke it's true.
You'll see it's all a show.
Keep 'em laughing as you go.
Just remember that the last laugh is on you.
And...
Always
look on the bright side of life.
Always look on the right side of life.
[whistle]
Always look on the bright side of life!
[whistle]
Always look on the bright side of life!
[whistle]
Always look on the bright side of life!
[whistle]
Repeat to fade...
 Monty
Pytons, 'Brian's Life'
(see the video!)



Previous
Puzzles of the Month + Solutions


MayJune
2008, Puzzle nr 117 
Back to PuzzleoftheMonth
page  Home 
Puzzle
# 117 Difficulty
level: ,
general math knowledge.
The
Geometry of the Bees...
The
shape of the wax
cells is such that two opposing honeycomb layers nest into each other, with
each facet of the closed ends being shared by opposing cells, and with the open
ends facing outward, as illustrated in fig. 1 above. Each cell is actually a
rhombic decahedron, that is an hexagonal prism having three rhombi at its closed
end (as shown in fig. 2 above). In short, each cell is a tensided structure
with one side open. Mathematicians made extensive studies of the isoperimetric
properties of the honeycomb cells and believed them to be the most efficient
design possible. If the faces of every cell contain the LARGEST possible volume
with the LEAST possible surface, what is the value of the angle alpha?
The faces of the hexagonal prism are each 1unit wide...
Key
words: isoperimetric, rhombic decahedron,
rhombus, hexagonal prism.
Related
puzzles:
Square
vs rectangle problem.
The
best cardboard
box.
 La
geometria delle api
Le celle
di cera di un favo sono disposte come illustrato
in fig. 1 qui sopra, con le aperture verso l’esterno.
Ogni cella è, di fatto, un decaedro rombico,
cioè un prisma cavo a sei facce con fondo
cuspidato terminato da tre rombi uguali. Prova
a determinare l'inclinazione dell’angolo
minore dei rombi (angolo alfa nella fig. 2) in
modo che la cella contenga il volume maggiore possibile
con la minore superficie possibile. Le facce del
prisma esagonale sono larghe 1 unità.
Parole
chiave: isoperimetrico, decaedro rombico.
Suggerisci un'altra
soluzione Chiudi
 Abeilles
géomètres
Les alvéoles d’un
rayon de miel sont disposées comme illustré en
fig. 1 cidessus, avec les ouvertures vers l’extérieur.
Chaque alvéole est en fait un décahèdre
rhombique, c’estàdire un prisme à section
hexagonale avec un fond formé de 3 rhombes égaux.
Essaie de déterminer la valeur de l’angle
alpha (fig. 2), de sorte que l’alvéole
contienne le plus grand volume possible avec la
plus petite surface possible. Les faces du prisme
hexagonal ont chacune une largeur de 1 unité.
Mots
clés: isopérimétrique,
décahèdre rhombique.
Propose une
autre solution Fermer

Source
of the puzzle:
©G. Sarcone, "Il Calcolo Differenziale ed Integrale", by Ing.
Gustavo Bessière, Hoepli
Publications, page 114. 


a) l is
the hypotenuse of the right triangle MNP shown
in fig. 3 above, then:
l^{2} =
(2x)^{2} + 1^{2} and
l = √(4x^{2} +
1)
b)
The diagonal AB of the rhombus at the
closed end of the cell does not depend upon its
spatial angulation (the axis AB remains
the same even when the spatial angle changes, fig.
4) but upon the sides of the hexagonal prism and
the radius r that circumscribes the hexagonal
section. Thus the length of AB is:
2√[1^{2}  (1/2)^{2}] = √(4
 4/4) = √3
c)
So, the area of each of the 3 rhombi at the closed
end of the cell is:
dl/2
or
√3 √(4x^{2} + 1)/2
d)
The area of each lateral trapezoid of the prism
is:
1/2[h + (h  x)] · 1 unit, or
(2h  x)/2 units
e)
Therefore, the whole surface of the cell is:
S = 3[√3 √(4x^{2} +
1)/2] + 6(2h  x)/2 units
Reducing to
S = 3/2[√3 √(4x^{2} +
1)] + 6h  3x units
f)
To know where the variable x peaks we take the derivative of S at
the point 0:
S' = 3/2[8√3 / 2√(4x^{2} +
1)]  3 = 0
so
6x(√3)
/ √(4x^{2} + 1) = 3
and
finally
x^{2} = 1/8
g)
If we replace x^{2} in the first equation
a) with this last value, we obtain:
l = √(4x^{2} +
1) = √[4(1/8) + 1] = √1.5
The
ratio of both diagonals of the rhombus gives us
the tangent of the semiangle α of
the rhombus:
l/d =
tg α/2 = √1.5/√3
Thus,
the angle α of the rhombus
at the closed end of the cell is:
2 arctg √1.5/√3 = 70.52877937.... ≈ 70° 31"
(arc = Inverse
trigonometric functions)
The
5 Winners of the Puzzle of the Month are:
No winners... Yes, the puzzle was
a little bit harder.
Sorry!

© 2004 G.
Sarcone, www.archimedeslab.org
You can reuse content from Archimedes’ Lab
on the ONLY condition that you provide credit to the
authors (© G.
Sarcone and/or M.J.
Waeber) and a link back to our site. You CANNOT
reproduce the content of this page for commercial
purposes.

You're
encouraged to expand and/or improve this article. Send
your comments, feedback or suggestions to Gianni
A. Sarcone. Thanks! 

More
Math Facts behind the puzzle
Curiously,
a 3dimensional honeycomb partition is not optimal!
In 1965, the Hungarian mathematician László Fejes
Tóth discovered that a cell end composed
of two hexagons and two smaller rhombi (see fig.
opposite), instead of three rhombi, would actually
be 0.035% (or approximately 1 part per 2850) more
efficient. But, frankly, this difference is too minute
to measure on an actual honeycomb, and irrelevant
to the hive economy in terms of efficient use of
wax, and because the honeycomb walls have a definite
thickness, it is not clear that Tóth's structure
would indeed be an improvement... In that respect,
the honeycomb is more like a wet foam than a dry
foam. Several years ago, the physicist D.
Weaire and his colleague R. Phelan undertook
to construct twolayer foams with equalsized bubbles,
and they found that the dry foams did take on Tóth's
pattern. But when they gradually added liquid to
thicken the bubble walls, something 'quite dramatic'
happened: the structure suddenly switched over to
the threerhombus configuration of a honeycomb (it
seems, then, that the bees got it right after all!).
The switch also occurs in the reverse direction as
liquid is removed.


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