Fractions - camels - parquets

30

FRACTIONS - CAMELS -
PARQUETS

Candidate of Physical and Mathematical Sciences
V. L. GUTENMACHER

It often happens that the same mathematical problem naturally arises in modern situations - from puzzles to applied calculations. Today we are almost talking about one of these tasks.

Arithmetic problem

Find all quadruples of natural numbers (p, q, r, s) for which the equality on the right is

$$\frac{1}{p} + \frac{1}{q} + \frac{1}{r} + \frac{1}{s} = 1.$$

We will assume that $p \leq q \leq r \leq s$. Then the smallest number - $p$ - is not more than 4 (otherwise $\frac{1}{p} + \frac{1}{q} + \frac{1}{r} + \frac{1}{s} < 1$) and, of course, not less than 2. Therefore, we need to consider three cases: $p=2$, $p=3$, $p=4$.

Let $p=2$. Then $\frac{1}{q} + \frac{1}{r} + \frac{1}{s} = \frac{1}{2}$. Let’s estimate the smallest number again - $q$. It cannot be more than 6 (otherwise $\frac{1}{q} + \frac{1}{r} + \frac{1}{s} < \frac{1}{2}$) and must be at least 3.

Now into the equality $\frac{1}{q} + \frac{1}{r} + \frac{1}{s} = \frac{1}{2}$ we substitute four possible

32

be equal to $\frac{1}{2}, \frac{1}{3}, \frac{1}{7}$ respectively*). The judge adds one of his camels (it turns out 42) and separates 21, 14, 6 camels for the brothers respectively (41 in total), and takes his camel back.

Practical problem

The simplest of the “regular” parquets is a division of the plane into squares (Fig. 1). It is interesting to find out how many such parquets there are: each vertex of the parquet is adjacent to four regular polygons and all vertices of the device are the same (the latter means that the parquet can be moved so that any
*) The first four stands out among all the others in that it contains the most number - 42, with 42 = 2·3·7. Among the triplets of numbers whose reciprocal sum is 1, the beginning of our quadruple has the same property: (2, 3, 7). It turns out that among the paterok there will be (2, 3, 7, 43, 1806), where 1806 = 2·3·7·43; among sixes - (2, 3, 7, 43, 1807, 3263443), and then it will be the same. This is a very difficult theorem, which our older readers encountered in the 7th issue of «Kvant» for 1987 in the article “Partitioning of Units”.

If its given vertex goes to any other given vertex and all lines coincide). This is a completely practical task.

We know that the sum of the angles of a regular $n$-gon is $180^\circ(n-2)$, and its one angle is equal to

$$\frac{180^\circ(n-2)}{n} = 180^\circ - \frac{360^\circ}{n}.$$

Let the corners of four regular polygons be combined at the vertex of the parquet: $p$-gon, $q$-gon, $r$-gon and $s$-gon. The sum of these four angles should be $360^\circ$. Let’s write this condition:

$$180^\circ - \frac{360^\circ}{p} + 180^\circ - \frac{360^\circ}{q} + 180^\circ - \frac{360^\circ}{r} + 180^\circ - \frac{360^\circ}{s} = 360^\circ.$$

It is easy to understand that this equality leads to the familiar correspondence

$$\frac{1}{p} + \frac{1}{q} + \frac{1}{r} + \frac{1}{s} = 1.$$

33

If we assume that $p \leq q \leq r \leq s$, then we know all four of these - there are 14 of them. Since we are talking about polygons, we need to discard those four where $p=2$. There will be four four left:

(4,4,4,4), (3,4,4,6), (3,3,6,6), (3,3,4,12).

The first quadrangle corresponds to a parquet of identical squares (each vertex is adjacent to 4 regular quadrangles - see figure 1).

The second four (3,4,4,6) represents two possibilities for constructing the top (Fig. 2, $a$, $\delta$), but for the correct parquet it is possible to complete only the parquet at figure 2, $a$ - it turns out figure 2, $\delta$.

The third four (3,3,6,6) also corresponds to two polygon arrangements:

Lightning rod, politics and…
hats

(For the beginning, see p. 13)

Czech communications officer Prokop Diviš according to the rules of the unofficial tipula “magus of electricity” (magus electricus). In 1754, he built the first lightning rod in Europe, which he invented independently of Franklin. But let’s fast forward a few years ago.

In 1750, Viennese professor Joseph Franz decided to show Divish his electrical experience. Franz had a static electricity generator typical of that time - a doctorable glass ball with a conductor (outlet) in the form of a metal tube or rod suspended on silk cords. To test any electrical experiment, it was necessary to rotate a glass ball, which was then electrified by friction and transmitted to the parquet.

Divish came to Franz in a wig, as fashion demanded. But the wig had a secret: it was covered all over with scraps of wire. It has long been known that the absence of metal “takes away electricity” (this is one of the phenomena on which the action of a lightning rod is based). By laying his head on the conductor, Divish thereby diverted the electric train. Probably, Divish experienced unpleasant mistakes, sending a couple into the ground through his body, but why not endure it for the sake of a good joke!

ki! Franz was unsafe: he could not get a charge on the conductor. He never managed to surprise the “magician of electricity.”

If Divish had carried out borrowing from wire cuttings (in a hidden way, of course), he would have been spared from embarrassing mistakes. An idea of ​​this kind arose in the second half of the 18th century. We are talking about a lightning rod hat. figure (see page 2 of the cover) shows Parisian models with personal lightning protection in the form of a middle chain that goes down to the toes from a metal rod mounted on the hat. In the 18th century this was no joke.

L. N. Krizhanovsky