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1
CIRCLING THE SQUARE WITH
STRAIGHTEDGE & COMPASS IN
EUCLIDEAN GEOMETRY
by Tran Dinh Son, UK,
E-mail: trandinhson2010@gmail.com
16 JANUARY 2024.
ABSTRACT
There are three classical problems remaining from ancient Greek
mathematics which are extremely influential in the development of
Geometry. They are Trisecting An Angle, Squaring The Circle, and
Doubling The Cube problems. I solve the Squaring The Circle problem,
of which paper is published in the International Journal Of Mathematics
Trends And Technology (Volume 69, June 2023). Upstream from this
method of exact “Squaring The Circle”, we can deduce,
conversely/inversely, to get a new Mathematical challenge "CIRCLING
THE SQUARE" with a straightedge & a compass in Euclidean Geometry.
This study idea came from my exact solution “Squaring The Circle by
Straightedge & compass in Euclidean Geometry”, published by IJMTT in
June 2023 at https://ijmttjournal.org/Volume-69/Issue-6/IJMTT-
V69I6P506.pdf for this ancient Greek Geometry problem. In this
research, I adopt the ANALYSIS method to prove the process of solving
this new challenge problem, which has not existed in the Mathematics field
till today. The process is an inverse/converse solution solving the ancient
Greek Geometry challenge problem of “Squaring The Circle”, using a
straightedge & a compass. I hereby commit that this is my own personal
research project.
Keywords
circling the square; circulating square; circle mature of square; make
square circled; find circle area same as square; make a square rounded
2
1. INTRODUCTION
‘Doubling a cube’, ‘trisecting an angle’, and ‘squaring the circle’ by
straightedge & compass, are the problems first proposed in Greek
mathematics, which were extremely influential in the development of
Geometry. The history of the “Squaring The Circle” problem dates back
millennia to around 450 B.C. (nearly 2,500 years), according to Quanta, a
science and mathematics magazine. Mathematician Anaxagoras of
Clazomenae was imprisoned for radical ideas about the sun, and while in
prison, he worked on the now-iconic problem involving a compass and
straightedge [1].
The present article studies what has become the most famous for these
problems, namely the problem of squaring the circle or the quadrature of
the circle, as it is sometimes called [2]. One of the fascinating aspects of
this problem is that it has been of interest throughout the history of
mathematics. From the oldest mathematical documents known up to today,
the problems and related problems concerning π have interested both
professional and non-professional mathematicians. In geometry,
“straightedge and compass” construction is also known as Euclidean
construction or classical construction [3].
Despite the proof of the impossibility of "squaring the circle," the problem
has continued to capture the imaginations of mathematicians and the
general public alike, and it remains an important topic in the history and
philosophy of mathematics. In 2022, I solved the “Trisecting An Angle”
problem with straightedge and compass [4], and published it in the IJMTT
journal as a counter-proof to the Wantzel, L. proof in 1837 [5].
3
Although the above ancient Greek Maths Challenges are closely linked, I
chose to solve the “Squaring The Circle” problem as my second research
study after solving the “Trisecting An Angle” exactly and successfully.
They were then published in the IJMTT, [4] and [6].
In June 2023 I solved exactly and accurately the ancient Greek problem
that has challenged mathematicians for over 2500 years - "Squaring The
Circle" with a straightedge & compass. Then, the International Journal
Of Mathematics Trends And Technology (IJMTT) published this paper
on 23 June 2023 [6]. After the paper was published, an idea derived from
the solved result to create a new mathematical challenge, which had not
existed before. The idea is as follows:
"If we can square a given circle then how about to circle a given square,
inversely/conversely". The following diagram illustrates this concept.
“Squaring The Circle” “ “Circling The Square”
(inversely/conversely from one another)
In seeking the solutions to many mathematical problems, geometers
developed a special technique, which they called “analysis”. They
assumed that the problem had been solved, and then, by investigating the
properties of this solution, worked back to find an equivalent problem that
could be solved based on the givens. To obtain a formally correct solution
to the original problem, the geometers reversed the procedure. First,
the data were used to solve the equivalent problem derived in the analysis,
and from the solution obtained, the original problem was solved. In
contrast to this analysis, this reversed procedure is called “synthesis”. I
adopted the technique “ANALYSIS” to solve accurately the “Squaring
The Circle” problem with only a straightedge & compass [24]. I also used
the technique “ANALYSIS” to solve the current challenge problem
“CIRCLING THE SQUARE”, derived from my “Squaring The Circle”
solution mentioned above. I hereby commit that this is my own personal
research project.
4
It is not saying that a circle of equal area to a square does not exist. If the
square has an area equal to A, then a circle with a radius r =
has the
same area. Moreover, it is not saying that it is impossible, since it is
possible, under the restriction of using only a straightedge and a
compass [2].
This thesis paper includes a new geometrical shape defined as a “Conical-
Arc,” a proof of the intersection of an arbitrarily given square and a circle
with the same area as the given square. In addition, proofs of some
theorems show that the intersection of the square and the resulting circle is
a regular octagon inscribed in the given square.
Lao Tzu (Author of Tao Te Ching):
“The great Tao is simple, very simple!”
(Lão T - c Kinh: o thì gin d, rt gin d !).
2. PROPOSITION
2.1 Definition 1: “Conical-Arc” shape
Given a circle (O, r) and an angle
with its vertex outside the
circle such that the bisector of the angle passes through the centre O
5
of the circle, then the special shape formed by the 2 sides of the angle
and arc
can be called a Conical-Arc (in Figure 1 below, the red
shape ADE is a Conical-Arc). If
is a right angle then the shape
ADE is called a Right-Conical-Arc.
2.2 Theorem 1:
If there is a circle (O, r) of area
r² = a², of which the centre
coincides with the centre of a given square ABCD (yellow colour in
Figure 2 below), side a and area a²,
then
a. The circumference of the circle (O, r) intersects the square
ABCD at 8 points a, b, c, d, e, f, g & h.
b. Square ABCD and circle (O, r) make 4 equal circle segments,
attached to 4 sides AB, BC, CD & DA of the given square
ABCD and located outside this square (in Figure 2 below).The
result of this shows the areas of the 4 Right-Conical-Arcs are
equal.
6
c. The areas of the four Right-Conical-Arcs formed by the four
right angles
,
,
and the four circle arcs ha, bc, de & fg
of the circle (O, r) are equal to the areas of the four circle
segments mentioned in section b. above.
PROOF:
a. Consider circle (O, OA), which is the circumscribed circle of
the given square ABCD (orange colour in Figure 2 below).
This circle occupied an area larger than area a² of the given
square ABCD because area
of the circle is larger than area
a² of the square.
Then, consider the inscribed circle (blue colour in Figure 2
below) of the square, which has an area
less than the area a²
of the given square ABCD. And then,
< a² <
(1)
Thus, by (1) circumference of the concentric circle (O, r)
with area a² is located in between the circumscribed circle
(O, OA) and the inscribed circle (blue colour) of the square
and intersects the 4 sides of the concentric square at 8 points
a, b, c, d, e, f, g, h, as required (Figure 2 below).
7
Figure 2: Four small segments (bordered by yellow line and black
arc) locate above side AB, below side CD and left & right of sides AC
& BD of the square ABCD.
b. In section a., the circle (O, r) cuts four sides AB, BC, CD, and
DA of the given square at a, b, c, d, e, f, g, and h (Figure 3
below). Therefore, the intersection of the square and the black
circle (O, r) shows us four circular segments, described in
Figure 3 below, by the four regions limited between arcs ab, cd,
ef, and gh and the four line segments (red colour in Figure 3
below).
Because the circle (O, r) and square ABCD are
concentric, the distances from the four line segments ab, cd, ef,
and gh to the centre O are equal. This result shows that the
above four circular segments are equal (Figure 3).
8
This concentric property of the circle (O, r) and the
square ABCD results in the four areas of the four Right-Conical-
Arcs Aah, Bbc, Cde, and Dgf being equal (Figure 3 below).
AB = a
Diameter of the yellow circle = a
2
Radius of the black circle (O, r), r = a
Figure 3: Four equal areas of the 4 circle segments, described by the
4 regions limited in between the arcs ab, cd, ef & gh (black colour)
and the 4 line segments (red colour). And also 4 equal areas of the 4
Right-Conical-Arcs Aah, Bbc, Cde & Dgf.
c. Aim to prove the areas of the four Right-Conical-Arcs formed
by 4 right angles
,
,
and 4 arcs of the circle (O, r) are
equal to areas of the 4 segments in section b. above.
9
Assume there exists a circle (O, r) that is concentric to the given
square ABCD with side a, and has the same area as the area of the
square a² = r². Then, in section b., four circle segments, which are
outside the square, are equal (Figure 4 below). These four circular
segments were excluded from the intersection area of the square and
circle (O, r). By the constraint of the assumption above {the area of
the square a² = the area of the circle r²},
either the intersection has to include these 4 circle segments to
be equal to the area
r² of the circle (O, r),
or the intersection has to include these 4 Right-Conical-Arcs to be
equal to the area a² of the square.
Therefore, the areas of the four equal Right-Conical-Arcs Aah, Bbc,
Cde & Dgf formed by the four right angles
,
,
and the areas
of the four equal circle segments are equal (Figure 4 below).
Diameter of the yellow circle = a
2
Radius of the black circle (O, r), r = a
Figure 4: The intersection area of the square and the circle (O, r)
needs EITHER the 4 areas of the 4 circle segments ab, cd, ef & gh to
be equal to the area
r² of the circle (O, r) OR the 4 areas of the 4
right Conical-Arcs Aah, Bbc, Cde &Dfg to be equal to the area of the
given square ABCD.
10
2.3 Theorem 2: “ANALYSIS THEOREM”
Given a square ABCD with area a². If there exists a circle
(O, r) with area
r² = a² (assumed), which is concentric with
the square ABCD, then 4 sides of the square ABCD are
overlapped 4 non-consecutive sides of a regular octagon
abcdefgh, which is inscribed in the circle (O, r).
PROOF:
Assuming that there exists a circle (O, r) of area a² – black colour in
Figure 5 below, which is concentric with the given square ABCD of
area a², then by b. in section 2.2 of Theorem 1, the four circle
segments, formed by the circle (O, r) at four sides of ABCD, are
equal.
Figure 5: abcdefgh (red and black colours) is the regular Octagon
inscribed in Circle (O, r).
Consider the area of the Conical-Arc Aah in Figure 5 above (as
defined in Definition 1) of vertex A of the square ABCD. From the
11
expression {area a² of ABCD = area
r² of the circle (O, r)}, we
obtain the following similar/equivalent expressions:
{the area of the Conical-Arc Aha = the area of the
circle segment ab (red colour)} (2)
{the area of the Conical-Arc Bbc = the area of the circle
segment cd} (3)
{the area of the Conical-Arc Dde = the area of the
circle segment ef} (4)
{the area of the Conical-Arc Cfg = the area of the circle
segment gh} (5)
Note that all expressions (2), (3), (4) & (5) above are illustrated in
Figure 5 above and Figure 6 below.
Figure 6: The inscribed squares ABCD (red and black colours) &
EFGH (blue colour) of the circle (O, R).
Let the given square ABCD, the circumscribed circle (O, R) – green
colour - of this square ABCD, and the assumed circle (O, r) of area a²
(black colour) all be concentric. Then, the extended arc chord ah of
12
(O, r), in Figure 6, meets the circumscribed circle (O, R) – marked
green dash in Figure 6 above – at E and H. Connect the diameter of
(O, R) which gets through E and O. From E, draw a symmetric chord
to EH that meets the green dashed circle (O, R) at F. In the special
octagon abcdefgh, inscribed in the given circle (O, r) with four
equal and parallel side pairs, section c. of Theorem 1 shows that EF is
the symmetric chord of EH through the symmetric EO-axis (green
colour). From section c. of Theorem 1 above, the distances between O
and the two chords ha & bc are the same, and this equality shows that
chord EF (blue colour in Figure 6 above) in the green dashed circle
(O, R) overlaps chord bc of the given circle (O, r). Similarly, chord
FG in the green dashed circle (O, R) also overlaps chord de (Figure 6
above) of the given black circle (O, r). By Section c. of Theorem 1,
FG // EH, then chords EF and GH of the green dashed circle (O, R)
are equal and parallel. This implies
EF = FG = GH = HE (6)
and
EFGH (blue) is the inscribed square of the circle (O, R) (7).
Then, (6) and (7) show that the areas of the two squares ABCD
(black) & EFGH (blue) are the same and equal to a².
The locations of the eight sides of the equal squares ABCD and
EFGH above show 8 chords ab, bc, cd, de, ef, fg, gh & ha of the
given circle (O, r) are equal. Therefore, these eight equal chords show
the shape abcdefgh is a regular octagon that inscribes in the
assumed circle (O, r), as required.
Note that this proof also shows that the regular octagon abcdefgh in
Figure 6 above has 4 sides overlapping the given square ABCD.
2.4 Theorem 3: The regular octagon abcdefgh,
mentioned in the above Theorem 2 has an inscribed circle
(O,
) that is also the inscribed circle of the given square
ABCD.
13
PROOF:
According to Theorem 2 above, the octagon abcdefgh has four sides
attached to the given square ABCD (Figure 7 below). From the eight
equal distances between the centre O and eight sides of the regular
octagon abcdefgh, the inscribed circle (O,
) of this octagon (red
colour in Figure 7 below) is also the inscribed circle of the given
square ABCD. Thus, this circle had a radius of
(Figure 7).
Figure 7: The inscribed circle (O,
) red colour - of the given
squares ABCD (red and black colours).
14
2.5 Theorem 4: “
The circle (O, r) of area a², equalling the area a² of the given
square ABCD of side a, and mentioned in the Analysis
Theorem, creates the inscribed circle (O,
) of the square.
This circle (O,
) is called the RULER
THE SQUARE problem.
PROOF:
According to Theorems 2 and 3 in Sections 2.3 & 2.4, the blue circle
(O,
) in Figure 8 below is the inscribed circle in both the given
square ABCD and the regular octagon abcdefgh. This octagon is also
an inscribed regular octagon of the resulting circle (O, r) of area r² =
a² and r = a
.
Note that the resulting circle (O, r) of area
r² = a² equalling
the area of the given square ABCD is unconstructed by
straightedge & compass because its radius is the irrational
number r = a
.
However, this RULER THEOREM shows us a RULER of
problem, which can be
used to construct the resulting circle (O, r = a
) with a
straightedge & compass, as this
THE SQUARE problem is proved a circle (O,
) having
radius
problem.
15
Figure 8: The blue circle (O,
) that inscribes in both the given
squares ABCD (red and black colours) and the octagon abcdefgh
(red colour) is the RULER OF THE CIRCLING THE SQUARE
problem, mentioned in Theorem 4 above.
2.6 Theorem 5: SOLUTION FOR
It is possible to construct a circle, which has the same area
as the area a² of a given square ABCD (side a) with a
straightedge and a compass.
16
PROOF:
Given a square ABCD of side a, a ℝ and area a², then by the
Definition 1 and the Theorems 1, 2, 3 & 4 above, we have the Ruler
of the “Circling The Square” problem, which is the inscribed circle
(O,
) of both the square and the inscribed regular octagon of
the square. This circle (O,
) can be constructed using a straightedge
and a compass.
Meanwhile, the circumscribed regular octagon abcdefgh of the circle
(O,
) is also constructed by a straightedge and compass (Figure 9
below). Finally, the circumscribed circle (O, r) with area r² = a² and r
= a
of the octagon is constructed by a straightedge and compass, as
its radius r is the constructive distance from the vertex a/b/c/d/e/f/g or
h to the centre O (Figure 9).
2.7 Construction Solution:
From the above sections 2.1, 2.2, 2.3, 2.4, 2.5, and 2.6, the
construction of the exact solution for the “CIRCLING THE
SQUARE” problem with a straightedge and compass is as follows:
a. For the given square ABCD side a, use a straightedge and a
compass to draw two line segments that divide the square
into four equal parts. Then, draw its inscribed blue circle (O,
) and its diagonals AC and BD (the 1st image in Figure 9
below).
17
Figure 9: The image illustrates the constructive solution of
the problem.
b. Use a straightedge and compass to draw the regular octagon
abcdefgh, which circumscribes the circle (O,
) by NOTE that
this octagon has 4 sides perpendicular to the 2 diagonals
of square ABCD and the other 4 sides overlaps the 4
sides of ABCD. Then, in the octagon, use the distance
from any of its vertexes a, b, c, d, e, f, g and h to the
centre O to construct the circle (O, r), area r² = a², and r =
a
(the 2nd image of Figure 9 above).
18
This circle (O, r), area
r² = a² and r = a
, constructed with
only straightedge & compass, is the solution for the
3. DISCUSSION AND CONCLUSON
Can mathematicians use a compass and a straightedge to construct
a circle having an area equal to a given square exactly/accurately?
This question is the same as for constructing the mentioned circle or
finding an accurate solution for the new challenge Mathematics
problem “CIRCLING THE SQUARE”. Surprisingly, only I, myself,
was still working on this question because the challenge problem had
just arisen contemporarily when I ended up my original research
article
, published by IJMTT in June 2023 [6].
In 2017, Andras Máthé and Oleg Pikhurko of the University of
Warwick and Jonathan Noel of the University of Victoria were the
latest authors who joined this ancient tradition challenge (Squaring
The Circle problem). These authors showed how a circle can be
squared by cutting it into pieces that can be visualized and drawn.
This result builds on a rich history. Mathematicians named this
method “the equidecomposition” but it is also theoretical proof that
the problem can be solved (without a straightedge & compass) by
cutting the circle into pieces and rearranging it into a square and none
knows the number of pieces. Nevertheless, no computers existed in
the ancient Greek era [25].
In June 2023, I solved exactly and accurately the ancient Greek
problem that has challenged mathematicians for over 2500 years -
"Squaring The Circle" with a straightedge & compass.
19
Then, the International Journal Of Mathematics Trends And
Technology (IJMTT) published this paper on 23 June 2023 [6].
After the paper was published, an idea derived from the solution to
create a new mathematical challenge, which had not existed before.
The idea is as follows:
"If we can square a given circle then how about to circle a given
square, inversely/conversely". The following diagram illustrates this
concept idea:
“Squaring The Circle” “Circling The Square”
(inversely/conversely from one another)
The “Analysis” method is applied correctly to Geometry to complete
this research study to gain an exact/accurate solution to this new
challenge “CIRCLING THE SQUARE” problem in Mathematics.
The results of my independent research show that the correct answer
(O, r), constructed by compass and straightedge, has the exact area a²
=
r², therefore if the given square to circle is a unit square a² = 1,
then in terms of geometry, can be constructive/expressed by a circle
with area
r² = 1. This circle comes from the RULER of the
problem yielded by the given unit
square. Subsequently, the exact geometric length of =
was
determined. In practice, if the International Bureau of Weights and
Measures (BIPM), the International System of Units, or any accurate
laser measurement is used to measure the arithmetic value r of the
answer circle (O, r), we can use this r to measure as accurately as
possible to obtain the arithmetic value of =
.
The above arithmetic value of
could be the nearest arithmetical
value of the irrational number
ever seen.
20
My construction method is quite different from approximation
and is based on using a straightedge and compass within
secondary Geometry so that any secondary student can solve
the problem for any given square. Moreover, this method
shows that the value r =
can be expressed accurately, and
the value =
or =
, a ℝ can also be expressed
accurately in terms of Geometry. This Geometrical expression
of the irrational number could be an interesting field for
mathematicians in the 21st century. In other words, algebraic
geometry can express exactly any irrational number k , k
ℝ.
In addition, this research result can be used for further research
in the “SPHERING THE CUBE” challenge, with only “a
straightedge & a compass” in Euclidean Geometry.
Furthermore, the research also opens some new challenge
problems which can be “CIRCLING THE EQUILATERAL
TRIANGLE”, “CIRCLING THE REGULAR PENTAGON”,
“CIRCLING THE REGULAR HEXAGON”, “CIRCLING THE
REGULAR OCTAGON” etc …, using a straightedge and
compass.
4. Conflicts of Interest
The author declares that there is no conflict of interest regarding
the publication of this paper.
5. Funding Statement
No funding from any financial bodies for this research.
6. Acknowledgements
21
I gratefully acknowledge the constructive suggestions of the friends
and reviewers who participated in the evaluation of the developed
theorems.
7. References
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