Engineer
POLI USP
carlisp@terra.com.br
copyright reserved by:
record
book 618 381 1,186 265 sheet (National Library)
For high
school students
São
Paulo, September 2013.
Dear
reader:
In
January 2010 I became aware of a problem called doubling the cube. Normally
would not have given him confidence, but that was different given his seniority
and his name consists of only three words: doubling the cube, if we assume that
the preposition + more from the "what" form a word, which seems a
exaggeration. I began to think about the problem without major commitment. Over
time I realized that, besides not know how to solve, I could not put it
down.Become an obsession. I felt like a mouse stuck in a mousetrap. Doubling
cube is to multiply the volume of a cube and determine the two edges of the
product.
Chapter I,
Figures:
This
chapter aims to review the geometry of the middle course, focusing and
amplifying the part that matters to the proof of the theorem of the cube. When
starting this work, the correct would scour the library to search for a needle
in a haystack or the Internet, where I find everything but what I seek. From
this work, giving the reader a bibliography. Certainly, that way, would not
find a proposed solution. The death find me first. It is more convenient and
fast search for information within the head itself that is right there at your
fingertips. The problem with this philosophy is the plagiarism. Put on paper
figures and properties without mentioning the author, whose name unaware. I'd
be reinventing something that has been invented. Would be reinventing the
wheel. I ask the reader, who is young, full of energy, full of the future, I
committed plagiarism check and inform so that I can make retraction.
Figure 1
below is very important and will be used extensively. Nowhere found the name of
the author to be able to mention it. It is very likely to be Pythagoras, but
the story is omitted. In this text, I will call it xis first theorem, referring
to Figure 1, abbreviated: 1TX. To demonstrate this, the hypothesis states that
the right triangles AOC and AOB are similar. The thesis establishes that h 2
= mn. Statement: I will suggest to the reader a demonstration model, step by
step:
1) Check
with a handle either a pair of equal angles, eg the angle OAB, and OCA The
vertex, vertex C are equal because their sides are perpendicular to each other,
two by two.
2) Check
with two handles another pair of equal angles that are OAC and OBA. The reason
is the same of equal Item 1. The angles opposite these sides are called
homologous.
3)
Isolate mentally a triangle, for instance triangle OAB
4) write
a fraction using the measure as the numerator (n) opposite the angle of a
handle side.
5) use as
the denominator the measure of the opposite angle of the two handles (h) side.
the fraction is n / h.
 |
| Figure 1 |
6) do the
exact same thing with the other triangle, the OAC, the new fraction is: h / m.
7) equals
the two fractions: n / h = h / m
Figure 1
and the theorem here called 1TX are placed the way I learned in high school
during the first half of the last century, this course, which at that time had
another name. Unaware of the current pattern. That course of action deserves
repairs. For example, the ends of the segment and C is denoted by m or (CO)? In
a simple case like figure 1 and the theorem 1TX where spare space, we can
squander space, giving two names to the same entity. There may be cases,
however, where space is small there will be congestion. That's not all.
In the
above equation h = m 1/2, the letter m would be the designation of a
segment? If the equation is wrong. There is no square root of a thread. The
equation may be correct if hem are real numbers, ie, measures of the respective
segments in the same unit. This path, however, is full of stones that need to
be removed. Take measures of segments is prohibited.
The
Internet reports that, in geometry, is forbidden the use of graduated scale.
With all these restrictions, no way to attack the theorem of the cube. As you
know, in geometry, not worth miracle. Dear reader, here you can practice
geometry and taking measurements. Figure 2 below is to amend the convention
segments. The letter c (lowercase) is the measure of the segment (OC), better
saying, c is a real number given by the relation c = (OC) / (OR) where (OR) is
a unitary, arbitrary thread, any thread that measures of Fig.
In Figure
2 also put measures a, b to clarify better. We will not do more, it is
understood, from now on, that x (lowercase) is the measure of (OX). Ask noted
that the figure shows three segments focus steps a, b of interest to our work
with each other according to two corresponding segments and all originate from
the point O. The figure is stiff, I mean: all segments are constant.
Here, the
new convention, we can write a2 = bc or a = (bc) 1/2.
Note that it may seem that we are calculating the square root of "a"
of a real number, positive bc. Was it really?
Actually did a drawing, we took measures, did the math and found that the accounts reproduce the measurements. Actually, what we did was draw a square root. This is a mechanism for extracting geometric square root.
Is clear at this point that geometry, with its structure of exact words, with its concepts such as point without dimension, without thick straight segment, primitive notions, postulates, properties, lemmas and theorems, finally, its infrastructure, with all this care that looks more like the description of the ceremonial of a religious cult, is still lacking.
Every new concept requires the invention of a new word that characterizes and makes possible the transmission of knowledge. Missing words in geometry.
Transactions
between segments can not be called product, quotient, root, etc.. here, for
lack of words, we will continue using the old words. In Figure 3 above, the
segment (OA) is constant, (OB) and (OC) are variable. It is a constant figure
product. (OB) is assuming the values (OB1) and (OB2)
and (OC) assuming (OC1) and (OC2).
Whatever
the product, for example: (OB1)(OC1) = (OA)2.
If the segment (OB1) tends to infinity, (OC1) tends to
zero. The figure contains zero and infinity.Arthur Koestler wrote "Zero
and Infinity" (book).
In Figure
4, a factor (B) is constant, the other factor, (OC) and (OA)
are variable. This is the figure constant ratio, for (OB) = (OA)2/(OC)
and (OB) is constant.
Figure 5
below is an example of using a ruler to verify the validity of 1TX. Can be used
a graduated wooden or plastic ruler for scale. It is also interesting to use
computer with autocad. It can also split a segment into equal parts providing a
ruler like that figure where the dotted arc shows that the square root of 4 is
2
In Figure 6 the OCC1C2, rectangle whose area is bc
product was rearranged to form the square A1A2C1C2.
 |
| Figure 6 |
Theorem 2TX.
I do not know the name of the author of the theorem 2TX. This is a generalization of the theorem 1TX. It could be stated thus: a circle, the product of the two separate parts of the same rope by a point O is constant.
Demonstration: in
figure 7 above the angles C1B1C2 and C1B2C2
are equal because they both refer to the same arc C1C2.
The
angles (B2OC1) and (B1OC2) are
equal because they are the opposite vertex.
The
triangles (B2OC1) and B1OC2. So
sides (OC2) and (OC1) opposite these angles are
homologous. Similarly sides (OB2) and (OB1) are
homologous. In the fractions (OB1)/(OC2) and (OB2)/(OC1)
the numerators are homologous, ditto the denominators. The terms of each
fraction are sides of one triangle.
It has
(OB1)/(OC2) = (OB2)/(OC1).
Transforming into product comes (OB1)(OC1) = (OB2)(OC2).
Transforming into product comes (OB1)(OC1) = (OB2)(OC2).
Mr. X2,
whose name I ignore, author of the theorem and the corresponding figure 2TX,
performed an extraordinary feat. The 1TX theorem is a particular case of 2TX,
as we shall see. The theorem 2TX does not contain zero and infinity. In a
single stroke, Mr. X2 eliminated the zero and its inverse. The
inverse of zero, which we call infinite privately, that is not a number, it is
a concept, because the physical horrors and nightmares, they call it
uniqueness.
The most
common case is that of black holes, where everything that does not come out.
Here the mass density is infinite. However, in physics, two united lords
repeated the feat of Mr. X2.
The Mr.
Werner Heisenberg proposed the uncertainty principle which means that it is
impossible to measure two properties of a particle simultaneously. The side
effect of this principle is that empty space is not as empty as it seemed.
There are no floating electric potential, ie, the uncertainty principle ensures
that there the electric field at a point is not zero, even in the absence of
electrical charge next. Also the supposedly empty space is populated by virtual
particles that appear and disappear as the flash of a firefly in dark night.
This means that zero, I mean the empty was removed from the space. The
statement: nature abhors a vacuum is an absolute radicalism.
The transcendental question - why is there anything rather than nothing? Mr. Heisenberg replies, there is everything because nothing is dead. The sr.Hawking, in turn, eliminated the inverse zero. He said the black holes evaporate. Evaporate, means that the infinite density of matter and the consequent infinite gravitational potential at a point does not exist.
In my way of thinking, he eliminated the infinite. The sr. X2, drawing, did the encounter between physics and geometry. In a single stroke, eliminated the two extremes. It is no small thing!
The transcendental question - why is there anything rather than nothing? Mr. Heisenberg replies, there is everything because nothing is dead. The sr.Hawking, in turn, eliminated the inverse zero. He said the black holes evaporate. Evaporate, means that the infinite density of matter and the consequent infinite gravitational potential at a point does not exist.
In my way of thinking, he eliminated the infinite. The sr. X2, drawing, did the encounter between physics and geometry. In a single stroke, eliminated the two extremes. It is no small thing!
The converse of the theorem is false 2TX. There may be a plethora of
competing strings at a point O, such that the product of the two parts of each
string one is constant and all of them may not belong to the same circle.
However the four ends of two ropes determine a circle. It is possible to prove
that the circumference passing through the two ends of the same rope and the
other end should contain the other end of the other.
In fact: two segments are not colinear, (B1C1) and
(B2C2*) which intersect at a point such that the (OB1)(OC1)
= (OB2)(OC2*) (Equation 1) then the four
points B1, C1, B2, C*2 are
cocirculares.
Demonstration: the first three points B1, C1, B2
determine a circle. Suppose the point C2* is outside the circle, as
in Figure 8. Then the segment (OC2*) intersects the circle at a
point C2. By Theorem 2TX should be: (OB1)(OC1)
= (OB2)(OC2) (equation 2).
So the equation is false because 1 (OB2)(OC2*)>
(OB2)(OC2). Note that two data segments
intersect at a point O, which goes to the equation 2, and no other
restrictions, there is a multitude of circles passing through the four points B1,C1,
B2 C2, provided that the angle between they are
variable. See that every arbitrary angle C1OC2
corresponds a circle. Ultimately, for this angle tends to zero, the radius of
the circle tends to infinity.
 |
| Figure 9 |
The
maximum angle of 90 degrees, is obtained when taking the largest segment as
diameter and the smallest is minimal. In this case, the point O is the midpoint
(A1A2) e (OA1) = (OA2).
See
Figure 9. So 2TX By Theorem (OA1)(0A2)
=
(OB1)(OC1) = (OA1)2.
The 1TX
theorem is a particular case of Theorem 2TX.
Demontration:
Figure 10 is self-explanatory.
We have: (OA2) = (OA1) = ((OB1)(OC1))1/2,
(OA) = ((OB)(OC))1/2 and (OB1)(OC1) =
(OB)(OC).
Então (OA1) = (OA2) = (OA).
The two parts of Figure 10 indicate that the circumference of diameter (B1C1),
I indicates it by the symbol C(B1C1) , and similarly all
other is the horizontal projection of a hemispherical surface at which point
the it belongs and the point O is the horizontal projection of it.
The
semicircle BAC is the horizontal projection of a section of the circle diameter
(B1C1) by a vertical plane.
Curves level
The
curves level of a hemispherical surface are concentric circumferences as in
Figure 11 below. The cicunferência radius (OA) is the curve of zero. All steps
between two consecutive curves have the same height, so the horizontal
distances vary with the first, toward the top, very short.
Semiesféricas intersection of surfaces.
In Figure
12 below the rope (B1C1) is the horizontal projection of
the intersection of two surfaces semiesféricas.
Any point
belong to the intersection of the two surfaces, so that the internal curves
appear in the drawing have the same level.
Chapter
II: Theorem hypercube. Particular case, the four-dimensional cube.
I am
addressing the theorem of the hypercube, first, because the theorem of the cube
is very difficult to demonstrate and hypercube, easier, could indicate a
suggestion.
In Figure
13 we have the above segments (B1C1) and (B2C2)
divided into two parts (unequal) through point O.
The
geometric mean (OB1) and (OC1 is (OA1), to (OB2)
and (OC2) is (OA2).
The
geometric mean (OA1) and (OA2) is (OA).
It is
easy to see that it is equation (OA) = ((OA1)(OA2))1/2
= ((OB1)(OC1)(OB2)(OC2))1/4.
 |
| Figure 14 |
In Figure
14 above the segment (B1C1) is divided into 17 equal
parts, taken as units of measurement. The intention is to draw the fourth root
of 16.
The
figure shows that (OA1) = ((OB1)OC1))1/2
and OA1) = (OA2) = 4 units.
Now we
have (OA) = ((OB1)(OA2))1/2, equation 2.
But (OA)
= (OA3) and (OA3) = 2 units.
At first
glance it seems that the findings relating to Figures 13 and 14 are different,
but they are not.
In Figure
14 we also operate with 4 segments: (OB1), (OC1), (OB1),
(OA2).
The Term
(OB1) was used as the unit of measure. Note that, geometrically, we
do not extract the fourth root of a real number but we take the square
root.Therefore solve the problem by extracting twice the square root. Is within
reach of a teenage middle course.
Following
the same recipe we can determine the edge of a hypercube of dimension m since
any factor m equals two.
Then m =
2n with en integer n> 1. The drama is to extract the cube root.
Chapter
III: Theorem of the cube, 1TC.
First
part, 1TC1
The
doubling cube is a special case of the theorem of the cube that is to multiply
the unit volume of a cube of unit edge for any real number and determine the
edge of the product.
In the
case of duplication multiply by two. To clarify, let's multiply the volume
given by 8, as pictured 1FC.
The
advantage of this is that the edge is multiplied by two, integer, making it
simple to check using compass and ruler graduated the example in Figure 6 where
the rectangle OCC1C2 had rearranged his unit squares
forming the larger square A1A2B1B.
In Figure
1FC1, have the unit cube edge. In 1FC2, a parallelepiped rectangle with square
base unit side. The height of the parallelepiped measuring units 8. It is
derived from the left cube that had one of its edges multiplied by 8. In 1FC3,
a cube of edge equal to two units, and therefore cubic volume equal to 8
units.This is derived from the cobblestone had rearranged his eight small unit
cubes. The center to the right was taken a transformation at constant volume.
Seems
thermodynamics. A suggestion for the theorem of the cube, geometric mean of
three segments, is from the general case: x1/2= y1/3 and
consider that, geometrically, we deal with the first member and know not deal
with the second. Demonstrate the theorem of the cube is to solve that equation
geometrically, however that equation we have only the first known member, which
can be put in the form: a= (vz)1/2, where, instead of the respective
steps are (OA), (OV) and (OZ).
Chapter
IV: Theorem of the cube, 2TC.
First
part, 2TC1.
Let's
imagine a circle diameter (BC) and a string (VZ) The point passers by, common
to both. The rope (VZ) revolves around the point O.
Variables
rather take all possible values.
There will be a pair of values such that v = z1/2.
Let c = 1, where c is the measure (OC) 8 = b, where b is the measure (B).
By
Theorem 2TX and Figure 7 we have vvz = 8, v = 2 and z = 4 = 22
because we assume that v = z ½. Substituting is: 23= 8 ou 2 = 81/3.
To solve
the problem we must find the position of the string, BOZ angle that produces
the pair v = 2 and z = 4.
Following
this line of reasoning, let's look at Figure 1FC4 drawn as the above.
Cancel
the portion of the string (OV) because we consider only the z1/2 s
values and adopt the variable s = z ½ The figure above shows how
1FC5 do so by obtaining the point S such that s = z1/2.
Note that
the point S belongs to a new curve must be a circle. Imagine the segment (UZ)
rotating around the point O. (OR) = (OC) keeps constant (OZ) and (OS) vary. The
point S describes a circle, it will be proved later. To draw it we need to
determine two points.
Before
proceeding, I would like the reader to take a look at Figure 14 which has
something in common with figure 1FC5. That one of the factors is constant, the
geometric mean and the other factor are variables. In 1FC5 also the factor (OU)
is constant and taken as unity for convenience. Consider the figure 1FC6, point
A determines the segment (OA), measure "a" is a = b1/2,
where b is the maximum value of z. Therefore "a" is the maximum value
of the root of z. As the diameter (BC) is vertical and point B is above point A
is horizontal to the left. Similarly, when the Z point C through the point C z
= 1, the measure (OU1) is 1 and the point U1 is
horizontal and to the right. Then the S, A and U1 points belong to
the same curve should be a circle. However, the + 1 is the sum of the maximum
and minimum measurement values of the root of z, so it is a measure of the
diameter (U1A) of Fig 1FC6 and 1FC7.
In principle it was not necessary to determine the point S because the diameter is sufficient to determine the circumference. The usefulness of the S point is to see that the circle passes through it. In the figure the dotted 1FC6, center O, passing through S circumference passes near the point V. In this case we have almost the solution. As it is placed in Figure 1FC6, we have to solve by trial, which is not valid. 1FC7 Figure 1 determines the point A by a dotted arc of 90 degrees at the center, such that (OA1) = (OA).
In principle it was not necessary to determine the point S because the diameter is sufficient to determine the circumference. The usefulness of the S point is to see that the circle passes through it. In the figure the dotted 1FC6, center O, passing through S circumference passes near the point V. In this case we have almost the solution. As it is placed in Figure 1FC6, we have to solve by trial, which is not valid. 1FC7 Figure 1 determines the point A by a dotted arc of 90 degrees at the center, such that (OA1) = (OA).
You can see in Figure 1FC6 that segments (OS) and (OV) to
be compared with each other, forming an angle of 90 degrees. Rather than making
comparisons by a plethora of attempts can only do one.
Second
part, 2TC2.
This
second part is to transform the figure as indicated by 2FC1.
To do
this simply take a rotation of 90 degrees to the circle diameter (U1A)
or C(U1A) as the notation adopted, changing the point A through
point A1 using the dotted arc AA1, center O as suggested
by Figure 1FC7.
In the
particular case b = 8, used as an example, we have x = 2 and z = 4, indicated
by the dotted radius equal to 2 and dotted arc of radius 4 circumference, then
1x8 = 2x4, this last relation proved in 2TX. Also we have 2 = 4 ½,
as the TC sends theorem. This means that the equation X ½= Y 1/3,
is solved geometrically.
Note that
X and X1 appear two solutions such that (OX) = (OX1)
because the system of equations to solve is high school. To find a proposal was
necessary to change an equation of the third degree, Y1/3, for a system
of equations of the second.
Part Three - comment.
The third
part here already posted and read by almost 200 visitors in Portuguese is being
rewritten because I'm thinking that it is defective. I apologize to those
readers for the inconvenience.
Part
Three - theorem 3TC
The third
part consists in proving that the curve whose diameter is (AU1),
1FC6 figure is a circle. For this we put 3FC1 and 3FC2 Figures.
Demonstration:
In Figure
below 3FC1 have the semicircle diameter (VY) that determines the point A5.
A C(AA1), radius (OA) is the geometric mean of the segments (OV) and
(OY) whatever the position of the string (VY) as the product of its parts (OV)
and (Y) is constant. The semicircle diameter (UY) determines the point S
 |
| Figure 3FC1 |
In figure
3FC2 , not to congest , we delete the two semicircles , the A5 point and
diameter ( VY ) and added what is found there . In this figure the angle AA2A1
, vertex A2 is straight . The ASU1 angle vertex in S is
right because its sides are parallel to the sides of the angle AA2A1
by construction . We prove that the parallel side passes through the point of
intersection U1 segment (AA1) with C (U3U1)
. For this we must note that C(AA1) and C(U3U1)
are concentric , so (AU3) = (U1A1). We segments
(AA2) and (U3A4) by parallel construction. The
quadrilaterals ASA4U3 and SA4A1U1
are parallelograms. Then (SA4) parallel to (AU3 equals U1A1)
. Therefore, the U1 belongs to the intersection point (AA1)
and C(U3U1). The angular distance between the point S and
the diameter (AU1) is 90 degrees, then S belongs to C(AU1).
Figure 2FC3 below, which shows the doubling cube , the segment (BC) is divided into three equal parts. The segment ( OC ) = ( OU) is the unit of data cube edge . (OB) = 2u is the height of the rectangle parallelepiped volume equal to 2u3. The obtained solution volume equal to the cube of the 2u3, should edge measuring 2 1/3 u. In the figure, point X is located to the right because the left is congested . Unfortunately , the measure ( OX ) is an irrational number and it is not possible to place compass as before . Figure has no beauty.
 |
| Figure 3FC2 |
The
Internet informs the demonstration of a theorem of the 19th century proving
that the doubling cube has no solution with ruler and compass. With my theorem
of the cube, shown with ruler and compass, it creates a paradox. It is possible
to explain, assuming hypothetically that this was demonstrated in a
three-dimensional space. Therefore, it would be correct because in three
dimensions the solution does not exist. The reason would be trivial: we do not
know write or draw in this space. Ali the ruler and compass is useless. There
are other reasons that can be invoked, for example, the geometry prohibits the
use of graduated scale. From the first steps we use the graduated rule, then,
from there, the problem no longer became geometry and physics. If Mr. X19
author's theorem T19, tried the solution using geometry, he could not even
solve. But if the problem has no solution in space, it has a solution in
four-dimensional hyperspace where demonstrated. I did not give this information
before not to confuse the reader's head. The four-dimensional hyperspace can be
decomposed into two plans that are designed on paper. One plan is one that
contains the unit circle, the other contains the C(BC). The two are confused
about the role, then work on a plan, using plane figures. Unable to use spatial
figure, three dimensions, for example, demonstrate the theorem using the design
of a box. To clarify, we give the following Figure 3FC3 which we add the
segment (PN), passing through point O, containing the strings (PQ) of C(U1A1)
and rope (MN) C(BC).
 |
| Figure 3FC3 |
Thus, we
have the following equations: (OP) = ((OC)(ON))1/2 identically (OQ)
= ((OC)(OM))1/2. These relations are valid because, by construction,
any point of C(U1A1) has this property. She is geometric
mean. Then, multiplying both members is: (OP)(OQ) = ((OC)(OM)(OC)(ON))1/2 Rearranging
comes: (OP)(OQ) = ((OC)2(OM)(ON))1/2.
However,
(BC) and (MN) pass through the point O and strings are the same circumference.
Then
(OM)(ON) = (OC)(OB Substituting and rearranging, we have: (OP)(OQ) = ((OC)3(OB))1/2,
(Equation 3).
This
equation shows that the Figure 1FC2 is not a rectangle parallelepiped with a
square base unit side and height equal to 8 but a cubic rectangle
hiperparalelepípedo base edge (OC) unit).
The
height of hiperparalelepípedo (B) measuring unit 8. The cube base looks like a
square because one side is the fourth dimension which can not be drawn. Figure
1FC3 is a hiperparalelepípedo cubic base and edge measuring two units. The
height of this element measures a unit. This is the fourth dimension.
To
finalize this proposal, I must say that most of the time and energy was spent
looking for a volumetric figure that could solve the problem. I went around the
world and found. Always believing that such a figure existed and always
believing in my incompetence in finding. Lately I found a story that makes me
believe that this figure does not exist and I'm not as incompetent as believed.
The article is called "Space is Digital". The author is Mr. Michael
Moyer. I bet he is French or that source because of the surname Moyer. The
central point of this paper is occupied by information science. Parenthetically
put some excerpts: (... information, not matter and energy, is the most basic
unit of existence Information rides on tiny bits, these bits comes the
cosmos.). In addition to praising the information the author strives to explain
how and where the information is logged and that "where" is the
two-dimensional space, ie, it is not a volume but a spherical flat surface etc..
A record of information is called "holographic principle" which is
the record in the spherical surface surrounding a black hole. This surface is
formed by the points of no return.
What is
written is: (This theory, the holographic principle states that when an object
falls into a black hole, the material itself may be lost, but the object
information would somehow printed on the surface around the hole black). I
should explain, reader, that this exaggeration stems from the belief of
scientists in the conservation of information. It can not be destroyed, even
when an object falls into a black hole. As I was given to understand, the
holographic principle is a particular case of what they call the fabric of
space-time, which, incredible as it may seem, is a surface:. (The principle is
more than an accounting trick that involves while the world we see appears to
take place in three dimensions, all the information about it is stored on
surfaces with only two dimensions). It follows from the above that if the information
can not be destroyed, it is eternal. If, hypothetically, what we call soul is
information, the hypothesis is the stumbling block. If it is removed, you, the
reader gains a soul.
Antonio Carli
Engineer POLI USP
carlisp@terra.com.br
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