Worker bees secrete beeswax to build hives, which are used as storerooms for the queen bee to lay eggs, brood and store honey and pollen. Seen from the front, the hive is composed of many hexagonal hollow columnar storage rooms. As shown in Figure 8, of course, it is best for readers to have the experience of seeing the hive on the spot.
Figure VIII
Seen from the whole three-dimensional beehive, there are storage rooms on the left and right (or front and back) sides. Its cross section is shown in fig. 9. Figure 10 is a cylindrical storage room, and its bottom is composed of three congruent rhombic faces ASBR, ASCQ and PBSC.
Figure 9
Figure X
Humans were surprised by the observation of honeycomb structure, and then put forward two mathematical problems:
㈠
Why a regular hexagon?
㈡
Why is the bottom edge composed of three congruent rhombic faces?
Let's discuss these two issues.
The first problem involves the old isoperimetric problem: how to enclose a closed field with fixed-length line segments on the plane to maximize its area?
This problem is also called Dido problem. In ancient Greek legend, Princess Dido (the queen who founded Carthage) intuitively put forward the correct answer: circle. However, it was not until the nineteenth century, more than two thousand years later, that this point was truly and strictly proved through the study of variational methods.
For the isoperimetric problem, the ancient Greek mathematician Zhi Nuo Dorothy (about 180 BC) has proved the following results:
(i) Among all N-polygons, the regular N-polygon has the largest area, and the more sides, the larger the area;
(ii) The area of a circle is larger than that of any regular polygon.
On the other hand, the ancient Egyptians already knew that there were only three ways to pave the floor with regular polygons of the same shape and size, as shown in figure 1 1.
Figure Xi
That is, we can only use regular triangles, squares and regular hexagons, and nothing else. This is a simple inference. The sum of the three internal angles of a triangle is 180.
Bees secrete beeswax to build nests. From the cross section, this is equivalent to using a fixed amount of wax to circle the largest area. This is an isoperimetric problem. According to Zenodorus' results, there are only three styles of floor, so bees have only three choices: regular triangle, square and regular hexagon, and bees instinctively choose the best regular hexagon. In other words, bees act according to the most economical principle.
Pappus, a geometer from Alexandria, published eight volumes of mathematical anthology in about 300 AD, in which the fifth volume discussed isoperimetric problems and honeycomb structures. He especially praised the ability of bees to "reason by instinct" and their inherent "certain geometric foresight".
Secondly, we discuss the second problem of honeycomb, that is, the geometric structure at the bottom of each cell. This question is more difficult.
We observe the storage room of a beehive, which is a hollow regular hexagonal cylinder with three rhombic faces at the bottom, and these three faces intersect at the center vertex S at the bottom (see figure 12). Let's review a period of history first.
Figure XII
17 12, the astronomer G.F. Maraldi of Paris observatory measured the angle of the diamond, and the results were 70 32' and109 28', as shown in figure 12. Mallardi inquired about nature on the spot and thought that bees followed two principles in building nests: simplicity and mathematical beauty.
Mallardi's results attracted the interest of the famous French naturalist Reaumur, who speculated that bees must have chosen these two angles for a reason, perhaps to minimize the surface area under a fixed volume, that is, to make a storage room with the least beeswax and the largest volume. Therefore, Reaumur asked the young Swiss mathematician Samuel K.? Nigel, the following questions:
Given a regular hexagonal column, the bottom is composed of three congruent diamonds. What should I do to save the most materials?
Reaumur didn't tell Connich that the problem was caused by urticaria.
Wait until k? Nigel sent the calculation results of 70 34 "and109 26" to Reaumur, and Reaumur told K? Nig's measurement results of honeycomb and Maraldi. They were shocked that the difference between theory and measurement was only 2 ". k？ Nigel's result supports Reaumur's guess that bees act according to the "most economical principle". k？ Nig solved the above extreme value problem by differential method. He said, "bees solve problems beyond the ability of classical geometry and must use Newton and Leibniz's calculus." However, a generation of scholar Fontainelle (Permanent Secretary of French Academy of Sciences) made a famous judgment in 1739. He denied that bees have wisdom and thought that bees only used advanced mathematics "unwittingly" according to the instructions of nature and the creator.
About k? Nig's 2-point difference was later recalculated by Cramer, Boskovich, Maclaurin and others, and it was found that bees were right, but where was K wrong? Nig, and k Nig's small mistake is that when calculating $\sqrt{2}$, the numerical table used printed out an wrong number.
Let's settle Reaumur versus K? Extreme value problem proposed by nig.
Consider the regular hexagonal cylinder in figure 13, and cut out three equal tetrahedrons ABFM, CDBO and EDFN at a, c and e, as shown in figure 14, making it figure 15. Three planes BFM, BDO and DFN extend and intersect at vertex P, as shown in figure 16. From figure 13 to figure 16, the cutting volume is exactly equal to the filling volume. Therefore, the volumes of Figure 13 and Figure 16 are equal, but their surface areas are not equal.
Figure XIII
So the original extreme value problem is equivalent to finding the minimum surface area under a fixed volume. The surface of a storage room in the hive (Figure 16) is composed of six trapezoids (BMGH et al. ) and three diamonds. In figure 14, let AB=a, BH=h, AM=x(x is a variable), then the diagonal of rhombic PBMF can be obtained by cosine law and Pythagorean theorem.
\ begin { display math } BF = \ sqrt { 3 } a \ quad，\ quad mp=2\sqrt{x^2+\frac{a^2}{4}} \ end { display math }
At present, the area of each rhombus is $ \ sqrt {3} a \ cdot2 \ sqrt {x 2+\ frac {a 2} {4}} $ and the area of each trapezoid is $ah-\frac{ 1}{2}ax$, so the total surface area of a storage room is.
\ begin { display math } a(x)= 3\sqrt{3}a\sqrt{x^2+\frac{a^2}{4}}+3a(2h-x)\ end { display math }(5)
Let A'(x)=0 by differential method.
\begin{displaymath}3\sqrt{3}ax\cdot\frac{ 1}{\sqrt{x^2+\frac{a^2}{4}}}-3a=0\end{displaymath}
solve
\ begin { display math } x = \ frac { \ sqrt { 2 } } { 4 } a \ end { display math }(6)
Using the second differential, it is easy to verify that $x=\frac{\sqrt{2}a}{4}$ is indeed a minimal point. Under $ ＄x = \ frac {\ sqrt {2} a} {4}, further make the acute angle of the diamond $ \ angle PBM = \ theta $,and then
\ begin { display math } \ tan(\ frac { 1 } { 2 } \ theta)= \ frac { \ sqrt { 2 } } { 2 } \ end { display math }
therefore
\ begin { display math } \ tan \ theta & amp； = & amp2 \ sqrt { 2 } \ end { display math }(7)
\ begin { displaymath } & amp& amp\ theta \ quad \ mbox { { \ font family { CWM 2 } \ font series { m } \ selec......{ CWM 1 } \ font series { m } \ select font \ char 107 } } \ quad 70^\circ 32 ' \ end { display math }
Exercise: In Figure 16, let α represent the included angle between diagonal PO and central axis PQ, and try to prove that the total surface area of a storage room is
\ begin { display math } a(\alpha)=6ha+\frac{3}{2}a^2(\frac{\sqrt{3}}{\sin\alpha}-\cot\alpha)\ end { display math }(8)
Re-solving $A'(\alpha)=0$ gives
\ begin { display math } \ cos \ alpha = \ frac { 1 }......family { CWM 1 } \ font series { m } \ select font \ char 107 } } \ quad 0.57735 \ end { display math }(9)
So $ \ alpha = 54 \ about 44' $
Note: We can also derive Equation (9) by using Equation (6) and then matching with Figure 16.
It is not satisfactory to deal with an elementary extreme value problem by differential method. So some people, such as Maclaurin( 1743) and L 'Huilier (178 1), began to seek elementary and simple algebraic and geometric solutions.
(A) the configuration method of algebra
We notice that all the above solutions have nothing to do with a and h, so we might as well assume a= 1 from the beginning. Then equation (5) becomes
\begin{displaymath}a(x)=3\sqrt{3}\sqrt{x^2+\frac{ 1}{4}}+6h-3x\end{displaymath}
Since 6h is a constant, it is only necessary.
\begin{displaymath}f(x)=\frac{3\sqrt{3}}{2}\sqrt{ 1+4x^2}-3x\end{displaymath}
Minimum value. manufacture
\ begin {eqn array *} y& =& \ frac {3 \ sqrt {3}} {2} \ sqrt {1+4x 2}-3x \ y+3x sum; = & amp\frac{3\sqrt{3}}{2}\sqrt{ 1+4x^2} \ end { eqn array * }
Square on both sides and simplify.
\ begin { display math } y^2-\frac{27}{4} & amp； = & amp 18x^2-6xy { display math }( 10)
Correct formula, simplify the complex.
\begin{displaymath}3y^2-\frac{27}{2}=(6x-y)^2\geq0\end{displaymath}
Therefore, when y=6x, the minimum value of y is $y=\frac{3\sqrt{3}}{2}$ Therefore,
\ begin { display math } x = \ frac { y } { 6 } = \ frac { \ sqrt { 2 } } { 4 } \ end { display math }
Get the same answer as equation (6) (a= 1).
(B) the discriminant method of quadratic equation
From (10)
\ begin { display math } 18x^2-6xy-(y^2-\frac{27}{4})=0 \ end { display math }( 1 1)
As a quadratic equation of X. Since X is always a real number, the discriminant of (1 1) formula.
\begin{displaymath}\delta=36y^2+4\times 18 \times(y^2-\frac{27}{4})\geq0\end{displaymath}
Tidy it up and simplify it
\begin{displaymath}y^2\geq\frac{9}{2}\end{displaymath}
So the minimum value of y is $\frac{3\sqrt{2}}{2}$, which is obtained by substituting $y=\frac{3\sqrt{2}}{2}$ into the formula (1 1).
\ begin { display math } x = \ frac { \ sqrt { 2 } } { 4 } \ end { display math }
Darwin praised the beehive as "the most amazing achievement in the known instinctive structure" He added: "To surpass such a perfect structure, natural selection is impossible, because as far as we can see, no matter from the labor force or the use of beeswax, the hive conforms to the most economical principle and is absolutely perfect. 」
In nature, besides bees following the "minimum principle", there are water drops on lotus leaves, sidewalks on campus lawns, the shortest path principle of light in Heron, Fermat's shortest time principle and so on. This makes us guess that nature operates according to some kind of "minimum principle".
/kloc-in the 7th century, Leibniz philosophically demonstrated that "this is the best in all possible worlds". Physicists finally discovered the "minimum action principle" of dynamics in 18 and 19 centuries, which became the most beautiful achievement in mathematical science.