When the absurd of the absurd is real.

In fact, * ^{i}*√

*i*=

*e*

^{π/2}= 4.81…

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# Category: Mathematics

## Imaginary Root of An Imaginary Number

## Geometry & Electronics

## Paradoxical Elastic Squares

## Parasitic Number

## Right Triangle with Rational Sides

## Brahmagupta’s Theorem

## Amazing Roman Rock-crystal Icosahedron Die

## Perfect “Square” Circle

## Sum of Consecutive Cubes (Visual Proof)

## Edible Geometry.

Geometric shapes are not limited only to the figurative aspect, they can also play active roles, for instance, serving in microelectronics to build operational printed circuits such as: small inductors (magnified, fig. a below), resistors (fig. b) and capacitors (fig. c). (image taken from my book “*Almanach du Mathématicien en Herbe*“)

A math-magic article I wrote for the German magazine Zeit Wissen: with the 13 triangular and square pieces (fig. 1) it is possible to form two large squares shown in fig. 2. Though the second large square has an extra piece the dimensions of the squares seem to be the same! Can you explain why this is possible?

This puzzle is available as greeting cards from my online store.

Useless, yet intriguing arithmetical fact… Multiplying this large number by 2, the rightmost digit 2 seems to pop to the front.

Such numbers are called “**parasitic numbers**“, read more: https://en.wikipedia.org/wiki/Parasitic_number

The simplest right triangle with rational sides (the longest side has a denominator of 45 digits!) and area 157, was found by Don Zagier in 1993.

If a **cyclic quadrilateral **( = with vertices lying on a common circle) has diagonals which are perpendicular, then the perpendicular to a side from the point of intersection of the diagonals will bisect the opposite side (AF = FD).

Here is an intriguing Roman crystal 20-sided die (icosahedron), used in fortune-telling, ca. 1st century AD.

Continue reading “Amazing Roman Rock-crystal Icosahedron Die”

Numbers 1 to 32 are placed along the circumference of a circle without repeating any number and still the sum of any two adjacent numbers in this circle is a perfect square!

The sum of the first *n* cubes is the square of the *n*th triangular number:

1^{3} + 2^{3} + 3^{3} + 4^{3} + 5^{3} + . . . + *n*^{3} = (1 + 2 + 3 + 4 + 5 + . . . + *n*)^{2}.

The tutorial of the day: Turn a carrot into a cube-octahedron!

More edible geometry from: https://www.archimedes-lab.org/pastashape.html