The paradox with Fibonacci numbers. Effective trick and its solution

EFFECTIVE FOCUSES AND THEIR CLUES
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The paradox with Fibonacci numbers. Focus Secret

Spectacular tricks and their clues

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Focus Description:

The lengths of the sides of the four parts that make up the figures (Fig. 1 and 2) are members of the Fibonacci series, that is, a series of numbers starting with two units: 1, 1, each of which, starting from the third, is the sum of the two previous ones. Our row looks like 1, 1, 2, 3, 5, 8, 13, 21, 34, ...

Focus Paradox with Fibonacci numbers
Ris.1

Focus Paradox with Fibonacci numbers
Fig. 2

The arrangement of the parts into which the square was cut, in the form of a rectangle, illustrates one of the properties of the Fibonacci series, namely the following: when squaring any member of this series, the product of two adjacent members of the series plus or minus one is obtained. In our example, the side of the square is 8, and the area is 64. The 5 in the Fibonacci series is located between 13 and 5. Since the numbers 13 and 65 become the lengths of the sides of the rectangle, its area should be equal to XNUMX, which gives an increase in area by one unit.

Thanks to this property of the series, it is possible to construct a square whose side is any Fibonacci number greater than one, and then cut it in accordance with the two preceding numbers of this series.

If, for example, we take a square of 13 x 13 units, then its three sides should be divided into segments of length 5 and 8 units, and then cut, as shown in Fig. 2. The area of this square is 169 square units. The sides of the rectangle formed by the parts of the squares will be 21 and 8, giving an area of 168 square units. Here, due to the overlapping of parts along the diagonal, one square unit is not added, but lost.

If we take a square with a side of 5, then there will also be a loss of one square unit. It is also possible to formulate a general rule: taking for the side of the square some number from the "first" subsequence of the Fibonacci numbers (3, 8, ...) located through one and composing a rectangle from the parts of this square, we get along its diagonal a gap and as a consequence of the apparent increase in area by one unit. Taking some number from the "second" subsequence (2, 5, 13, ...) as the side of the square, we get overlapping areas along the diagonal of the rectangle and the loss of one square unit of area.

The further we move along the Fibonacci series, the less noticeable the overlaps or gaps become. And vice versa, the lower we go down the row, the more significant they become. You can build a paradox even on a square with a side of two units. But then there is such an obvious overlap in the 3x1 rectangle that the effect of the paradox is completely lost.

Using other Fibonacci series for paradox, you can get: countless options. So, for example, squares based on a row of 2, 4, 6, 10, 16, 26, etc. result in an area loss or gain of 4 square units. The magnitude of these losses or gains can be found by calculating for a given series the difference between the square of any of its terms and the product of its two adjacent terms on the left and right. Row 3,4,7, I, 18,29, etc. gives a gain or loss of five square units.

T. de Moulidar gave a drawing of a square based on the series 1, 4, 5, 9, 14, etc. The side of this square is taken equal to 9, and after converting it into a rectangle, 11 square units are lost. The row 2, 5, 7, 12, 19, ... also gives a loss or gain of 11 square units. In both cases, the overlaps (or gaps) along the diagonal are so large that they can be seen immediately.

Denoting any three consecutive Fibonacci numbers by A, B and C, and by X - loss or gain in area, we get the following two formulas:

A+B=C

B2=AC±X.

If we substitute for X the desired gain or loss, and for B the number that is taken as the length of the side of the square, then we can construct a quadratic equation from which two other Fibonacci numbers can be found, although these, of course, will not necessarily be rational numbers. It turns out, for example, that by dividing a square into figures with rational side lengths, one cannot obtain an increase or loss of two or three square units. With the help of irrational numbers, this can of course be achieved. Thus, the Fibonacci series √2, 2√2, 3√2, 5√ ... gives an increase or loss of two square units, and the series √3, 2√3, 3√3, 5√3, ... results in to a gain or loss of three square units.

Author: M.Gardner

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