
Spanish: 
 
En este video mostramos el ingenioso metodo
que Arquimedes uso para aproximar Pi.
Arquimedes eso poligonos y el teorema de Pitagoras!
Pero, como lo hizo?
 
Recuerdas Pitagoras?
Lo usamos para hallar lados de triangulos rectangulos.
Para ello necesitamos saber los otros dos lados.
Por ejemplo, en este triangulo nos dan
la hipotenusa y uno de los catetos.
Haciendo un simple calculo
sabemos que b es mas o menos 0.866.
Para ello Arquimedes dibujo un circulo de radio 1,
con circumferencia 2Pi.
Arquimedes sabia que si dentro dibujaba un hexagono
podria facilmente calcular el perimetro.
Y es que los lados coinciden con el radio 1.

English: 
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In this video, we are using Archimedes ...
method to find an approximation for the number ...
Pi. He did that by simply using ...
Pythagoras Theorem to approximate the ...
circumference of a circle using the perimeter of a polygon.
Remember Pythagoras Theorem?
To find any missing side in a right-angle triangle,
we need the two other sides. For example,
in this triangle, the hypotenuse is 1 and one ...
of the legs is 0.5. To find the other leg,
we substitute the values and find b by taking the square root!
Now let's see how to approximate Pi.
The circumference of a circle with radius 1 is equal to 2Pi.
Arquimedes used a regular hexagon inscribed in that ...
circle because he knew that this is made of 6 equilateral ...
triangles. In other words,
that side AB equals the radius 1.

Spanish: 
El perimetro 6 es menor que la circumferencia
asi que sabemos que 2Pi es mas grande que 6.
O lo que es lo mismo
Pi es mas grande que 3.
Arquimedes se dio cuenta de que
si escogia un dodecagono
la aproximacion seria mejor.
Lo dificil es encontrar el lado del dodecagono
Lo facil es multiplicarlo por 12.
Esto nos dara una mejor aproximacion del numero Pi.
Como lo hizo?
Arquimedes se dio cuenta de que
la linea que une el centro con el nuevo vertice C
divide el lado del hexagono anterior en dos partes iguales.
Usando Pitagoras pudo hallar la distancia

English: 
So the perimeter is 6.
Since the hexagon is inscribed inside the circle,
the perimeter is smaller than the circumference.
2Pi must be bigger than 6. So Pi must be bigger than 3.
Archimedes realised that by dividing the circle in regular ...
polygons with more sides, he could get a better ...
approximation. And so he did!
Drawing a 12 sided regular polygon inside the circle and ...
multiplying the length of each new side by 12 will give us a ...
better approximation for the circumference.
The first step is to half the previous side AB.
The length is 0.5. We can now use Pythagoras,
since the angle is 90 degrees. The side joining the centre 0 ...
and the midpoint is the square root of the difference of the ...
square of the hypotenuse 1 and the previous length 0.5.

English: 
Since OC equals the radius 1 then the length joining C and ...
the midpoint is 1 take away 0.866.
Using this result and the length 0.5,
we can use Pythagoras again to figure out the hypotenuse ...
AC. The length of the side is 0.518
Since the 12 sided polygon is inscribed inside the circle,
2Pi must be bigger than 12AC.
So Pi must be bigger than 6AC.
Replacing AC wich 0.518, Pi must be bigger than 3.108
Let's carry on…
Again, we half the previous side AC,
which results in 0.259. We can now use Pythagoras ...
twice: firstly inside the right triangle ...
with vertices OAM and secondly inside the left ...
triangle AMD. OM is the square root of the ...

Spanish: 
necesaria para calcular la longitud
del nuevo lado AC.
Para ello uso repitio el calculos usando
ambos triangulos rectangulos.
Arquimedes dio con el lado AC.
Lo multiplico por 12
y pudo aproximar el valor de Pi un poco mejor.
Pi tiene que ser mayor que 6 lados AC
Pi tiene que ser mayor que 3.108.
 
Arquimedes persevero!
Repitio el proceso.
Dividio el lado anterior por dos.

English: 
difference of 1 square and 0.259 squared.
We subtract this from 1, because OD is the radius of ...
our polygon. And this is still 1.
The length DM is 0.034. We use Pythagoras again to ...
figure out the hypotenuse AD. The length of each side of this ...
polygon is 0.261.
Since this is a 24 sided polygon,
2Pi must be bigger than 24AD.
So Pi must be bigger than 12AD.
Replacing AD with 0.261, we know that Pi is bigger than 3.132.
Archimedes carried on with a 48 sided polygon.
He halved the previous side AD.
He used Pythagoras on the right triangle,
to calculate the distance between the centre 0 and the ...
new midpoint: the square root of the ...
difference of the square of the same hypotenuse 1 and the ...
previous length 0.1305. Since OE equals the radius 1 ...
then the length joining E and the new midpoint is 1 take ...
away 0.9914. Using this result and the ...
length AM, we can figure out the ...
hypotenuse AE of the left triangle.

Spanish: 
Uso Pitagoras una
y otra vez!
Hasta dar con el nuevo lado.
 
 
Multiplicandolo por 12.
Pi tiene que ser mayor que 3.132.
 
 

English: 
The length of each side is 0.1307.
Since this is a 48 sided polygon,
2Pi must be bigger than 48AE.
So replacing AE wih 0.1307, we know that Pi is bigger than 3.1368.
Archimedes even used a 96 sided polygon.
He halved the previous side AE.
He used Pythoras to calculate the distance between the ...
centre 0 and the new midpoint:
0.99786. Since OF is the radius 1 then ...
the length FM is 1 take away 0.99786.
Using this result and the length AM,
he figured out the hypotenuse AF:
0.0654. Pi must be bigger than 48AF ...
which is 3.1392.
Arquimedes method returns approximations for pi that ...
converge to the actual value of the number pi.
Very clever if we take into account the simplicity of the Maths behind.
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Spanish: 
Arquimedes repitio con 48 lados!
Y con 96!
 
Los numeros se aproximan a Pi
Todo un genio este Arquimedes!
 
 
