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Geometrical Enigmas (education for secondary school students)


This chapter is intended for students attending secondary schools in the most simplest way only with a compass and unmarked straightedge, a purely geometric way, step by step and all this with the purpose of eliminating the many successions of delusions that have for centuries been circulating as enigmatically impossible constructions, giving rise to countless stories, discussions particularly among circles that compile textbooks for primary and secondary schools! This is shameful and incredible. The situation gave birth to dogmas of an alchemic and esoteric nature. The blame is not upon the professors who impart geometry to the students, but upon the pundits who write the books that the students learn from, and their books stick to conclusions of the Middle Ages and uphold theories of the “impossible” that seek to prove how certain solutions are not possible, instead of getting down to work and finding solutions! They certainly find any mention of sacred geometry (read: universal) bothersome because they consider themselves as creative individuals, and all others do not exist! Isn’t that somewhat strange! How is it then that we can not (read: don’t know how), even nowadays, to solve ordinary geometrical problems (if Archimedes were alive today he would comprehend them in a matter of seconds, but in his time he didn’t have the possibilities that we now have). Indeed, we don’t have any exceptional possibilities, but the principle is that in the language of geometry a hundredth degree or a millimeter may here and there cause errors due to the usage of manual drawing. Here we will abide by two entities since the third entity (angle trisection) is nonessential in the sense of ancient architecture construction, but the two other ones are important (the perimeter and area of a circle and the lessening and doubling of a cube). Don’t ask yourself why this is so, and estimates based on guesswork are simply useless. Only the One who gave mankind our moral guidelines 3300 years ago on Mount Sinai knows why. Therefore let us merely focus our attention on pure geometry with the compass and unmarked straightedge, using numbers only for the sake of verification.

* * *


The first error occurs so soon because the circle of a radius (select a measure into the compass) is not depicted in full circles with the same radius, which would otherwise aid in dividing the circles into 6 and 12 parts. I see no reason why the so-called “floral pattern” with its six petals should be a difficulty? What are the divisible circles separated into 6 parts? They are a basic natural code. It is therefore good to always be present – and visible.

* * *


Then why would straight line bisectors be a difficulty when they are also useful, even though we will only use the four opposite intersections of the circle. They are likewise of no trouble since their aid is indispensable in many respects (whoever finds this a nuisance will never be able to carry out what we have in mind).

* * *


For if we divide the circle from the intersection points with the same radius, before that we obtained new bisectors and thus we divided the circle into 24 parts. But we don’t need so many and only use a part of them, with our initial aim in mind – the perimeter and area of the circle.

* * *


We boldfaced this “four-leaf” a little. The other two have been left out since the one now depicted by semicircles of same radius will suffice, as emitted from the 4 opposite poles of the circle with which we started and still wish to find its perimeter and area.

* * *


Thereby we are able to outline the circumscribed square or quadrilateral circle which will likewise be helpful regarding the perimeter and circle area or in other words their diagonal lines. I need one more element.

* * *


The hexagonal star-shaped polygon circle divided oppositely into 6 parts, their lengths 3 diameters to the circumference. Why do we need them? First of all, because our goal is the length of the of the circle’s arc.

* * *


Now we will think a little about number Π (Pi). Its magnitude is Pi = 3.14 (28571) and reduced somewhat lately (adjusting it to something interchangeable, as you will see); to formula 2rΠ or regular d (diametar – one) times 3.14(28571′) = 3 diameters and remainder 14(28571′) and this remainder is (decimally) 1:7=0.1428571 thus exactly one seventh or 3 wholes and one seventh of a diameter. Hence, Pi is a regular perimeter or circumference (not of Greece but of ancient Egypt, since the city of Pi and many other urban centers of Egypt had alphabets before their emergence in Greece). Thus we must find the one-seventh. Numerally, Archimedes was right. The perimeter is 22 sevenths. Of what? Of the diameter of the circle.

* * *


Now our aim is to find the seventh (the heptagon is also sometimes announced an “impossible construction” only with a compass and unmarked straightedge, i.e. only numerically, but isn’t it a part just like the nonagon?). Strange! Then number 3 would also be impossible since one whole divided by 3 = 0.3333””, and even 8; 1÷8 = 0.125. However that’s why we have the hexagonal star-shaped polygon (besides other options). Isn’t it seven parts of a circle; the one-half side of the triangle of a circle’s hexagonal star-shaped polygon.

* * *


And to spare you from being bothered by this, simply verify it. Take 3 diameters, and divide 1 diameter by seven parts the way you were taught in grade school. To the one part add the length of three diameters and then with bisectors divide all into four parts. One is the fourth side of a quadrilateral circumscribed circle and it is equivalent to this one. This is the first division into seven circles from four opposite poles.

* * *


Now it is easy to construct the square (quadrilateral) circumference, namely square Pi – number is the perimeter circle = 3.1428571” or 22 sevenths. It belongs to the lineage of sevenths, its edge is thus 5.5 sevenths (4 x 5.5 = 22) and (22 ÷ 3.1428571 = read as 3 diameters and one-seventh, and it is inscribed as the circle area. (put in measures when using the formula and you will see that it is so).

* * *


Now the drawing is easy once we know the measure of the diameter and that the circle’s circumscribed square is one whole (seven sevenths and the circumference edge is 5.5 sevenths and from this follows = diameter raised to the second power divided by 7 times 5.5 – circle area = four-sided polygon. It doesn’t matter if you use the old or the new formula. Funny that it never came to anybody’s mind that the circle area is in the lineage of the quadrilateral?

* * *


It’s quite simple. To this very day there is a lot of superficial philosophizing on what is Pi, therefore prior to leveling the two rectangles used in the calculation of 7 and 5.5 if we use 14 then we also use 11; if 28 then also 22, and so on. Their relationship is 7:5,5 = 1.272727 and that number times Pi (number 3.1428571’) = 4 (diameters) and so it is easy to depict the circumference (Pi) whereas the other number carries the number called PiT (a duplicated Pi depicted as a doubled door).

* * *


Nothing to it, we just level two circle areas, two rectangles, and that is the same as if we extracted the square root, which can also be numerically verified by the “doubled door” d2 divided by 7 times 5.5 = circle area (rectangle) and divided once more by 7 x 5.5 = area of circumscribed square and the root of the first sum = edge of square circle area and root of second area = edge of circumference (you will note that the geometrical presentation is much easier to follow).

* * *


We simply boiled this down to all the participating elements in their full shapes, but in the absence of circles (circumscribed) so as to have a simplified appearance. That would be that! Whoever has any doubts can take measurements, and nowadays they can be precise – through application of computer simulation, the most up-to-date technology of our epoch. But let’s go on to our second “enigma”, the system of duplication and diminution using the same principle. In the end we will learn why.

* * *


Since the cube is a three-dimensional object, a cardinal mistake already came about in elementary school. Drawing is offhanded, instead of employing the given (assigned) side as a radius in the same way as we went forward with the circumference and circle area. Therefore, the given length is the radius of the circle naturally divided by the circles of the same radius with their six parts – the inscribed hexagon of the circle. The intersections of the divisible circles are monitoring points of proper drawing.

* * *


If we link the opposite poles we get a three-dimensional shape, a cube with its 12 edges where of the first topmost and last lowest angle of its 8 centers we see only 7 angles of the three-dimensional object. Wouldn’t the sole use of a compass (by entering the scope of the given length) and an unmarked or marked straightedge be easier for elementary school pupils?

* * *


Since the three-dimensional object (cube) has 6 equal sides (in construction-work terminology: top view, front view, and side view are the same) it will suffice to draw one square in the manner demonstrated by means of the properly drawn circle – dividing the circle and system into 12 parts.

* * *


Here we have the aid of the division of the arc of the circle into 12 parts. We will use only 8 of the 12 divided poles of the arc of the circumscribed circle of the hexagon. Hence they are the top view, front view and side view, the square edges of the cube’s given lengths (radiuses). Thus the significant role is the circle’s arc. Outside of it every other arc ((every other larger radius) is an enlargement, and inside of every other arc a diminution (smaller radius) of the cube. This is logical. But our other enigma is the duplication of the cube.

* * *


It has long been known that the doubled cube is numerically solvable as the side length of the cube times cube root of 2 or or side length of cube times 1.259921. Now we can do likewise when we know that the radius is the side length of the cube, write the formula = (navesti formulu). Draw it. We will apply the construction-builder’s system and the diagonal straight lines of the cube square will help (if we drew all of them that would be a division of the circle into 24 parts, but for the cube we need only one square.

* * *


If anyone thinks the duplicated star-shaped hexagon polygon could be used for the front view, he will be wrong. It is intended solely for division of the cube into parts. In this case it is simple. We don’t even need the radius of the circle (formula) save for the help of the inscribed hexagon circle and diagonal straight lines of the quadrilateral edges of the assigned cube (read: top view, front view or side view – no matter). Every builder knows what to do. If he has one edge (quadrilateral) of a doubled cube, he has it all. However, our goal is to go forward with the duplications: to duplicate the doubled cube.

* * *


Here we could use the Pi number or its square because its edges (it is smaller) are 7 one-hundredths smaller than the edges of the duplicated doubled cube. Therefore we will employ (one) natural square of a properly drawn division of a circle with its radius, and a straight line division into 12 and partially into 24 segments. All this is done only with a compass and unmarked straightedge.

* * *


It is no wonder that the Pi number and radius or the edges of the duplicated doubled cube have recently changed since they belong to the lineage of number seven. The difference is 7 one-hundredths, but duplications have their own regularities, but that is another subject. We will stick to our theme. We have duplicated the doubled cube. Now we will triplicate it. This procedure is the simplest.

* * *


In the case of threefold duplication there is no “philosophizing”. It is the length of the diameter or the circumscribed square of the circle. The arcs of the circle are not utilized although they incorporate a sequence of data, but we will adhere to the shortest route of duplications.

* * *


Thus we see that the system is steadily repetitive and so we will use the “experience” from the first duplication and our fourth duplication is also simple. However, for the last two of this duplication system we will utilize the “floral four-leaf pattern” of the inscribed circumscribed circle of a doubled hexagon or 2r scope or length of the edge of the cube.

* * *


But not to complicate matters, we will simplify them by a method which I will show you after and which is usable for all duplications and diminutions not only of this system of doubled diameter. We circumscribe the whole system with its circle and from its four opposite poles with the same radius (with semicircles) we draw a “four-leaf”.

* * *


Actually it is a descriptive circumscribed square or cube of two radiuses. It will appear as the sixth duplication. Many of you wonder why not inscribe a cubu within a cube? Simply because it would bring an erroneous perception of a three-dimensional object, but we will approximately be able to do so even today – by means of computer animation. But the task of this enigma is to solve it only with a compass and an unmarked straightedge, with an equal increment on all sides.

* * *


Thus, enough has been said from a builder’s aspect (which any genuine contractor will confirm, or finally any designer who makes modellings of his buildings based on front view ratios). Therefore, we see that 12 divisional straight lines cut the circumscribed square system. We link them with line segments. They cut the straight line division into four, creating intersections.

* * *


We connect the intersections with line segments. This is a front view of the fifth increment of the cube. Its sides are the magnitude of the radius and the sixth increment is the enlargement of the circumscribed square system. This cognition brings us to the simplest ways of decreasing and increasing, and it goes much further because the system was used in ancient times as a natural code, but we won’t go into for what purpose and meaning.

* * *


Simply try to single out this beginning by yourselves. But bear in mind: the cube of the assigned sides (radiuses) and its front view.

* * *


Let’s begin with the front view and front view of the complete system.

* * *


We now simply follow how the rectangle of the four opposite poles opens its “turn” or lineage of duplication or diminution. The system of simple geometry only with a compass and unmarked straightedge is infallible (the dimunution of the cube will be foretold by a dotted line so you can comprehend the system).

* * *


Note that the specificity of the “lineage” manifests itself like an echo. Clearly a very simple procedure that is valid indefinitely not only in the doubled diametrical system but also in the threefold, fourfold, etc., hence we have the first diminution of a square side of a cube.

* * *


And this (dotted line) is the proclamation of the second diminution of a side of the assigned line segment of the cube when once divided by the root of 2 (the third) or by 1.259921 and then divided once again with the obtained result. You are certainly wondering how far this diminution can go? All this surely has a meaning and is related to the ancient constructions, but that is not our theme and we will stick to our system of “as up above, so it is down below”, or as many increases the same amount of decreases.

* * *


This (diminution) is somewhat harder due to the initially small radius (read: magnitude of the assigned line segment edge of the cube), and once more because of our A4 paper format, but regardless of this the system is exact, hence the second diminution.

* * *


And now with the third diminution we pass on to the simple procedure because the third diminution is the decrease of half of the radius of the assigned radius. So we will let it stay up to the end without a comment. We have simply proclaimed our objective: to reach the last, positive sixth diminution, namely the cube edges, the fourths edges that were assigned in the beginning.

* * *


We cannot go further on. It would take us over into the negative sphere of the third root of two, respectively to the decimalization of the decimal (I’m joking), but for the sake of truth all of that would lead to some other “unknown”, but that wouldn’t be for the secondary school student level. This here is just the final picture of the duplication and diminution of an assigned cube.

* * * *


So what must we remember: whatever we know now, whatever we used to know, and what was learned now.
The cube is a solid shape, three-dimensional with its 6 equal quadrilateral surfaces, its 8 angles and 12 edges. Well, we know that. A long time ago somebody made up a story about it that turned it into a so-called enigma; how and why I don’t know but I am surprised because there were many in the course of the centuries who knew but kept it for themselves (or for some other reason) regardless of the method of solving it geometrically; and since the legend (or two of them originated in Greece) was of a construction-builder nature it was necessary to apply constructor methods. Therefore, the front view, which customarily consist of the front view and the front and back sides views, and since the cube has all 6 of its equal surfaces then it is logical by means of a front view or by side views to find some way of geometrically drawing one of the quadrilateral surfaces and the surfaces of their equal duplication on all sides. The first geometrical steps immediately proved to be inadequate because the teaching of geometry is superficial and fails to give results of the kind that would emerge if they were properly drawn instead of being reduced in scope and thereafter equilibrated even today be means of certain strange “acrobatics” that do more to confuse the students than to teach them. But, be it as it may, this has been established as an “enigma” of a geometrical constructor-worker’s nature and that it is solvable through a formula of line segments (read: radius) times third root of two. Basically it was necessary to find the simplest method of duplication, not only whatever kind but one of a system that is universal, in accompaniment of a method for diminution (dividing the radius with third root of two. I can thank the Good Lord for giving me a “pathfinder” spirit that is, it seems, fairly patient with me because it was indeed not easy to discover the simplest method in only seven days, and that spirit always call forth to mind the words of the crucified on which my Christian civilization is founded: “ye who seek, ye shall find”. But when I consider that the enigma has been seeking a depicted solution for who knows how long, what then are a mere seven days! It is true that I sent a somewhat different and more complex solution to relevant worldwide institutions of mathematics inviting their critical reviews, but there were no positive or negative responses, and it seems that I have finally comprehended that this geometry and its results are designated for you, students who will be the men of the future. That is why I have taken this small amount of exertion. Until recently constructor methods demanded the creation of models, but in you youthful world this is also becoming changing via computer animation. If you apply measures you will easily confirm what has been said in this chapter, while I will remain in this geometry only with a compass and unmarked straightedge.

* * * *


I know this will pass
Daybreaks stay mute
As stone
Like this country
This rocky native soil
Lone reminiscences stay
Keepsakes remain
Their tale of whispers lone
Whispery dunes of stone.

All this will pass
Though nothing else will
Nothing different
Nothing new.

All this once was
Long ago
Do worlds go
Through antiquity
Like winds
Like generations
And only
Keepsakes remain
About which they whisper
Like stone
Covered by the foam
Of waves
And then everything remains
A rocklike story
Everything .
(Translation and poetic rendition by Slobodan Drenovac)

My dear friends and readers of these pages, as many of you know I’m a poet, I have decided to end every chapter with one of my poems. This is one of the latest verses of more than 350 thousand so far written during the past 10 years (I doubt whether I will ever have the financial means to publish all of them.) I hope you will understand. Unfortunately (from the history that we are familiar with) that seems to be the story of mankind.

Hr – Rijeka, July 1, 2014
Author: Tomo Periša
Webmaster: SLIM
English: S.F. Drenovac


One Response to “Geometrical Enigmas (education for secondary school students)”

  1. Ricardo says:

    Im gonna tell you again Mr TOmo. I made a simple equation that it belongs to ALL THINGS. Einstein forgot 3 symbols in his E=mc2 🙂 this equation is not a multiplication or a division, its both. Like any other relation.

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