# Puzzler: a squarable region from Leonardo da Vinci

It is famously impossible to square the circle. That is, given a circle, it is impossible, using only a compass and straightedge, to construct a square having the same area as the circle.

I will let you read elsewhere about the exact rules behind compass and straightedge constructions. The punchline is that if you begin with two points 1 unit apart, then you can construct a line segment segment of length ${a}$ if and only if ${a>0}$ can be created from the integers using the operations ${+}$, ${-}$, ${\times}$, and ${\div}$, and by taking square roots.

Thus it is possible to construct a line segment of length ${(1+\sqrt{5})/2}$ (the golden ratio), but it is impossible to construct a segment of length ${\sqrt{2}}$ (hence it is impossible to double the cube).

A circle with radius 1 has area ${\pi}$. A square having this same area would have side-length ${\sqrt{\pi}}$. Because ${\pi}$ is transcendental, this is not a constructible length. Thus it is impossible to square the circle.

Even though the circle is not squarable, some regions with curved boundaries are. For example, in the fifth century BCE Hippocrates of Chios (no, not that Hippocrates) showed that several lunes are squarable.

Let’s give a brief proof that the shaded lune shown below is squarable. Suppose the large circle has radius 1. Then triangle ${ACD}$ has area ${1/2}$ and sector ${ADCE}$ has area ${\pi/4}$. Thus segment ${ACE}$ has area ${\pi/4-1/2}$. The smaller circle has radius ${\sqrt{2}/2}$. So the semicircle ${ACB}$ has area ${\pi(\sqrt{2}/2)^{2}/2=\pi/4}$. It follows that the lune has area ${\pi/4-(\pi/4-1/2)=1/2}$. A square with the same area as the lune would have side-length ${\sqrt{2}/2}$, which is constructible. Thus the lune is squarable. It turns out Leonardo da Vinci played around with squarable figures, and he discovered many beautiful examples. Below I’ve included one of Leonardo’s figures (the on the left). I’ve included the center and right-hand figures to give more information on how Leonardo’s design is created. So here’s the puzzle: show that Leonardo’s figure is squarable.

(Hint: assume that the radius of the large circle is 1. Then find the total area of the shaded regions. Show that the area, and hence the square root of the area, is a constructible number.)

Have fun!

1. In section 53, of PLATO’S “TIMAEUS”, PLATO speaks about the triangular shapes of the Four Elemental Bodies, of their kinds and their combinations : These Bodies are the Fire (Tetrahedron) the Earth (Cube), the Water (Icosahedron), and the Air (Octahedron). These are bodies and have depth. The depth necessarily, contains the flat surface and the perpendicular to this surface is a side of a triangle and all the triangles are generated by two kinds of orthogonal triangles : the “ISOSCELES” Orthogonal and the “SCALΕΝΕ” Orthogonal. From the two kinds of triangles the “Isosceles” Orthogonal has one nature. (i.e. one rectangular angle and two acute angles of 45 degrees), whereas the “scalene” has infinite (i.e. it has one rectangular angle and two acute angles of variable values having, these two acute angles, the sum of 90 degrees). From these infinite natures we choose one triangle “THE MOST BEAUTIFUL”. Let us choose then, two triangles, which are the basis of constructing the Fire and the other Bodies : One of these two is the “ISOSCELES” orthogonal triangle, the other is the “SCALENE” orthogonal triangle, its hypotenuse having a value equal to the “CUBE” of the value of its horizontal smaller side and having its vertical bigger side the value of the “SQUARE” of its smaller horizontal side. The value of the smaller horizontal side is equal to the square root of the GOLDEN NUMBER, the ratio of the sides is equal, again, to the square root of the GOLDEN NUMBER (geometrical ratio) and the Tangent of the angle between the hypotenuse and the smaller horizontal side is also equal to the SQUARE ROOT of the GOLDEN NUMBER (Θ=51 49-38-15-9-17-19-54-37-26-24-0 degrees). By applying THE PYTHAGOREAN THEOREM, on this triangle we obtain a biquadratic ((fourth order)) equation:

(T4-T2-1=0, from, T6 = T4+T2, via Φ2-Φ-1=0)

from which we obtain the size of the small perpendicular (T) as the Square root of the Golden Number (T=SQR(Φ)).

It is concluded here that by “THE MOST BEAUTIFUL TRIANGLE”, PLATO correlates the four elements (UNIFIED THEORY) through the General Analogies of their sides (Fire, Air, Earth and Water), i.e. Fire/Air is equal to Air/Water is equal to Water/Earth, to T, where T is equal to the SQUARE ROOT of the GOLDEN NUMBER.

T = SQR ((SQR.(5) + 1)/2) = 1.27201965

1. …THE PYTHAGOREAN THEOREM, on this triangle we obtain a biquadratic ((fourth order)) equation:

(T4-T2-1=0, from, T6 = T4+T2, via Φ2-Φ-1=0)

THE PYTHAGOREAN THEOREM, on this triangle we obtain a biquadratic ((fourth order)) equation:

(T**4-T**2-1=0, from, T**6 = T**4+T**2, via Φ**2-Φ-1=0)

PANAGIOTIS STEFANIDES

1. ………..On “Quadrature of the Circle” cont.:
In my book :http://www.stefanides.gr/pdf/BOOK%20_GRSOGF.pdf

On Page 44 [Autocad Drawing ] one can see very simply why pi = 3.141592654.. does not Suare the Circle.
Whereas on page 45 it does for π = 4.14460551..

ERRATUM
—————–
π = 4.14460551..
π = 3.14460551..

2. TEXT CONTINUATION
BY PANAGIOTIS STEFANIDES

The values of the sides of this triangle are given by surd numbers, (solution of a fourth degree equation). Reorganizing this triangle, we get another one with the same angle values, which has its bigger vertical side equal to FOUR (4), its smaller horizontal side equal to FOUR divided by the SQUARE ROOT of the GOLDEN NUMBER, and its hypotenuse equal to FOUR multiplied by the SQUARE ROOT of the GOLDEN NUMBER. (Four divided by the SQUARE ROOT of the GOLDEN NUMBER is equal to 3.14460551)

Using this value for Pi 4/1.27201965 [= 3.14460551 ] as the small side of an orthogonal triangle , 4 its bigger vertical and its hypotenuse 5.088078598, inscribing this triangle in a circle of diameter D equal to the hypotenuse then product of

Pi*D = 3.14460551*5.O88078598 = 16 . THIS IS THE CIRCUMFERENCE OF THE CIRCLE= TO THE SQUARE OF THE BIGGER SIDE 4*4= 16 = TO THE IS SQUARE’S PERIMETER.

Multiplying 4*D ==20.35231439 = AREA OG=F CIRCLE =Pi*[D^2]/4 =

=3.14460551*[ 25.88854382]/4=20.35231439.

So we get this Squaring of the Circle relationship, elaborating on PLATO’S TIMAEUS, and more we get it drawn by Ruler and Compass.

Up to now those who had rejected the insoluble quadrature of the circle apparently did not elaborate on this triangle, even though some attempted trying [approximately] usin a similar ,but not the same, and not as fair triangle as this, the MAGIRUS TRIANGLE given to KEPLER