¿Hay alguna forma de dibujar un ángulo de 1 grado utilizando sólo regla y compás?

Question

Hay formas de dibujar $180^\circ, 90^\circ, 45^\circ, 30^\circ, 60^\circ, \dots$ ángulos.

Pero, ¿hay alguna forma de dibujar un $1^\circ$ ¿Ángulo?

En otras palabras, cómo dividir un círculo en $360$ ¿a partes iguales?

Answer 1

Respuesta corta: No.

La prueba por contradicción

Supongamos que usted podría construir $1°$ en el círculo unidad. Entonces usted puede simplemente añadir que el ángulo a sí mismo un número de veces para terminar con $40°$. Se puede dibujar una línea a través del punto resultante perpendicular a una de las patas del ángulo. Que la línea tendría una distancia desde el origen que es

$$\cos(40°)=\sqrt[3]{\frac{1}{16} i \, \sqrt{3} - \frac{1}{16}} + \frac{1}{4 \, \sqrt[3]{\frac{1}{16} i \, \sqrt{3} - \frac{1}{16}}}\approx0.766$$

Notice the cubic roots there? That's not something you can construct in general. So this is already a strong indication that the number $\cos(40°)$ isn't constructible, which in turn means your $1°$ can't be constructible either.

But perhaps you think I just chose a stupid representation of that number. So I have to rule out that there could be an easier solution. Consider the irreducible polynomial $X^3-3X+1$. The number $x_1=2\cdot\cos(40°)$ is a root of that polynomial (the others are $x_2\approx0.174$ and $x_3\approx-0.940$). But the Galois group of the field extension $\mathbb Q(x_1)/\mathbb Q$ has order $3$. So you can't solve it using square roots, you absolutely need cubic roots.

Construction can only provide square roots. If you were able to construct $1°$, you could go on and construct $x_1$ as outlined. Since $x_1$ isn't constructible, $1°$ can't have been constructible either.

So an angle of $\bf{1°}$ cannot be constructed.

Note that when I wrote about the constructibility of $1°$ above, I meant as an angle, not a number. The length $1°=\pi/180$ is even more obviously not constructible, since it's not even algebraic but transcendental. But that's not what you were asking, and not what I was referring to.

(I'm not too experienced with Galois theory myself. Perhaps there are better ways to put the core of my argument into terms more readily understandable by others. So feel free to edit my post if you understand what I tried to say and know a better way of saying it.)

How I found this

I arrived at this answer with some educated trial and error, and the help of Sage as my computer algebra system. I knew about the concept of constructible numbers, so I knew that I'd have to investigate whether some length derived from that angle would be expressible with only square roots or not. I started by looking at the minimal polynomial of $1°$:

sage: deg = QQbar.zeta(360)
sage: deg.real().minpoly()
x^48 - 12*x^46 + 135/2*x^44 - 473/2*x^42 + 37023/64*x^40 - 33579/32*x^38 + 5995185/4096*x^36 - 6580089/4096*x^34 + 11535777/8192*x^32 - 509423/512*x^30 + 597177831/1048576*x^28 - 277344081/1048576*x^26 + 834808065/8388608*x^24 - 126664335/4194304*x^22 + 1969138215/268435456*x^20 - 378710755/268435456*x^18 + 908975295/4294967296*x^16 - 26063991/1073741824*x^14 + 142445393/68719476736*x^12 - 8735955/68719476736*x^10 + 2925693/549755813888*x^8 - 38697/274877906944*x^6 + 4527/2199023255552*x^4 - 27/2199023255552*x^2 + 1/281474976710656

That zeta(360) is $\exp(i2\pi/360)=\exp(i\cdot1°)$, i.e. a complex number $1°$ along the unit circle in the set $\bar{\mathbb Q}$ of algebraic numbers. So its real and imaginary parts correspond to the cosine and sine of that angle. The minimal polynomial of the real part is the smallest (monic) polynomial (i.e. lowest degree) in $\mathbb Q[X]$ which has this cosine as one of its roots.

In this case that's a fairly complicated polynomial. So I looked for ways to simplify the situation. For example, bisecting an angle is no problem, but doing so numerically likely involves additional square roots, increasing the degree and complexity of the minimal polynomial. So for example $8°$ is already easier.

sage: (deg^8).real().minpoly()
x^12 - 3*x^10 - 1/8*x^9 + 27/8*x^8 + 9/32*x^7 - 7/4*x^6 - 27/128*x^5 + 105/256*x^4 + 31/512*x^3 - 9/256*x^2 - 3/512*x + 1/4096

Powers of a complex number multiply the angle with the exponent, so deg^8 represents $8°$. This computation suppots my heuristic intuition: If I divide $360°$ into fewer parts I stand a better chance of simplifying the situation. So how many parts do I need? As you are probably aware, trisecting arbitrary angles using ruler and compass is impossible. On the other hand, trisecting the full circle into $120°$ parts is really easy. That made me look at the prime factor decomposition of $360=2^3\cdot3^2\cdot5$. Next candidates would be $2^3=9$ leading to $360°/9=40°$ or $5$ leading to $360°/5=72°$. But you may know that regular pentagons are constructible, or read that fact off the quadratic minimal polynomial:

sage: (deg^72).real().minpoly()
x^2 + 1/2*x - 1/4

So that led me to investigating the $40°$ more closely. The cosine (real part) looked easier than the sine (imaginary part), so I stayed with that.

sage: (deg^40).real().minpoly()
x^3 - 3/4*x + 1/8
sage: (deg^40).imag().minpoly()
x^6 - 3/2*x^4 + 9/16*x^2 - 3/64

The monic polynomial with integer coefficients used in my example would have been easy enough to find manually, but in this case (me not seeing the wood for the trees) came from the following computation:

sage: NumberField((deg^40).real().minpoly().numerator(), "y")
Number Field in y with defining polynomial 8*x^3 - 6*x + 1
sage: _.galois_closure("z", map=True)
(Number Field in z with defining polynomial x^3 - 3*x + 1, Ring morphism:
   From: Number Field in y with defining polynomial 8*x^3 - 6*x + 1
   To:   Number Field in z with defining polynomial x^3 - 3*x + 1
   Defn: y |--> 1/2*z)

So this is the source of that factor $2$ in front of the cosine. Then I verified the degree of the Galois group:

sage: _[0].defining_polynomial()
x^3 - 3*x + 1
sage: _.galois_group()
Transitive group number 1 of degree 3

I also had Sage help me with the radical expression above:

sage: (deg^40).real()
0.7660444431189781?
sage: latex(_.radical_expression())
{\left(\frac{1}{16} i \, \sqrt{3} - \frac{1}{16}\right)}^{\frac{1}{3}} + \frac{1}{4 \, {\left(\frac{1}{16} i \, \sqrt{3} - \frac{1}{16}\right)}^{\frac{1}{3}}}

$${\left(\frac{1}{16} i \, \sqrt{3} - \frac{1}{16}\right)}^{\frac{1}{3}} + \frac{1}{4 \, {\left(\frac{1}{16} i \, \sqrt{3} - \frac{1}{16}\right)}^{\frac{1}{3}}}$$

Cross connection

I hadn't looked at this when I first wrote my answer, but the comment by @AakashM points out Can a circle's circumference be divided into arbitrary number of equal parts using straight edge and compass only? That question essentially asks whether one can construct $360°/k$ for arbitrary integers $k$, so you are dealing with the special case $k=360$. The accepted answer there points out that $k=2^\alpha\cdot p_1\cdots p_s$ works if all the $p_s$ are distinct Fermat primes. For $k=360=2^3\cdot3^2\cdot5$ all the prime factors are indded Fermat primes, but the exponent at $3^2$ is a problem, making this non-constructible. In fact $k=2^3\cdot3\cdot5=120$ leads to $3°$ como el entero más pequeño que el número de grados que puede ser construido.

También ver la Wikipedia o MathWorld página en construibles de polígonos regulares.

Answer 2

Así se construye 1 grado con regla y compás (aquí usé una regla de 10 cm de largo, pero cualquiera serviría):

1. Construye un ángulo de 54 grados (no se muestra para mayor claridad. Esto es sólo 45+72/8 grados, y se construye trivialmente.

2. Construye un ángulo de 55 grados:

   • Primero construye un círculo de radio 10 y llama eje x al diámetro horisontal.

   • Ahora construye un ángulo de 15 grados, BAC

   • Coloca la regla a través de C y ajústala de forma que un extremo toque el eje x y el otro extremo el círculo. Marca el punto donde toca el círculo B'

   • Ahora el ángulo BAB' es de 55 grados.

Superponer sobre el ángulo de 54 grados, y restar la diferencia, 1 grado. Ahora puedes construir cualquier grado entero por adición.

¡BOOOM!

Puedes ver rápidamente (geogebra o con calculadora) que esto es correcto hasta 20 dígitos. Dejo la prueba de esto a usted.

Pista: Observa que tanto AB', DB' y AC es 10

construir 55 grados