Virtual Math Tutor: Math problem #84

## Wednesday, January 26, 2011

### Math problem #84

Math problem: The so-called Cassini's paradox can be explained as follows. If we cut a chess board (or, - lest destructively, - an 8x8 square piece of graph paper) into two equal trapeziums and two equal right angled triangles, like this:

the four pieces can be rearranged so as to form a 5x13 rectangle:

Note that the total area of the square piece is 8x8 = 64, while the area of the rectangle is 5x13 = 65. Does it mean we have just proven that 64 = 65?
Use the same four pieces of the 8x8 square to "prove" that 64 = 63.

Discussion and hints: Recall that the above arrangement is a consequence of Cassini's identity:

$F_{n-1}F_{n+1} - F^2_n = (-1)^n,$

where $F_{n-1}, F_{n}, F_{n+1}$ are Fibonacci numbers:

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ... (generated by the recurrence relation $F_n = F_{n-1} + F_{n-2}, \, F_0 = 0, F_1 = 1.)$

Note that 5,8, and 13 are three consecutive  Fibonacci numbers, as such satisfying Cassini's identity above. This explains Cassini's paradox. Indeed, in view of the above we have

$5\cdot13 \mbox{(the area of the rectangle)} - 8\cdot 8 \mbox{(the area of the square)} =(-1)^6 = 1.$

Pretty neat, eh? We note, however, that 63 cannot be represented as a product of two Fibonacci numbers. Therefore, the resulting geometric figure of the total area of 63 square units should be something other than a rectangle. As for two areas above, - they are different indeed, and the difference is precisely one. More specifically, the diagonal of the rectangle is so "thick" that its area is exactly equal to one.

Cassini's paradox (identity) is named after Giovanni Domenico Cassini (1625 – 1712), an Italian/French mathematician, astronomer, engineer, and astrologer.

Solution