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GIZE, STONEHENGE, CHICHEN ITZA, ETC,ETC: ¿PORQ. "GRAN PIRAMIDE" ESTA DISEÑADA EN FUNCION AL CUBO Q CONTIENE LA TIERRA?
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La razon de la circunferencia y el diametro es obvio que es el numero PI.
22/7=DIA DE MARIA LA MAGDALENA=3.14 =PI
SABIDURIA=SABADO=LETRA S=$
EN EL CIRCULO SE CONFUNDEN EL PRINCIPIO Y EL FIN- HERACLITO
y la Omega, principio y fin, dice el Señor, el que es y que era y que ha de venir, el Todopoderoso.
16. Apocalipsis 1:11 que decía: Yo soy el ALFA y la Omega, el primero y el último. Escribe en un libro lo que ves, y envíalo a las siete iglesias que están en Asia: a Efeso, Esmirna, Pérgamo, Tiatira, Sardis, Filadelfia y Laodicea.
17. Apocalipsis 2:27 y las regirá con vara de hierro, y serán quebradas como vaso de ALFArero; como yo también la he recibido de mi Padre;
18. Apocalipsis 21:6 Y me dijo: Hecho está. Yo soy el ALFA y la Omega, el principio y el fin. Al que tuviere sed, yo le daré gratuitamente de la fuente del agua de la vida.
19. Apocalipsis 22:13 Yo soy el ALFA y la Omega, el principio y el fin, el primero y el último.
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168. Juan 16:21 La mujer cuando da a luz, tiene dolor, porque ha llegado su hora; pero después que ha dado a luz un niño, ya no se acuerda de la angustia, por el gozo de que haya nacido un hombre en el mundo.
24X7=168
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Message 694 of 694 on the subject |
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3 Phi and Triangles
3.1 Phi and the Equilateral Triangle
 Chris and Penny at Regina University's Math Central (Canada) show how we can use any circle to construct on it a hexagon and an equilateral triangle. Joining three pairs of points then reveals a line and its golden section point as follows:
- On any circle (centre O), construct the 6 equally spaced points A, B, C, D, E and F on its circumference without altering your compasses, so they are the same distance apart as the radius of the circle. ABCDEF forms a regular hexagon.
- Choose every other point to make an equilateral triangle ACE.
- On two of the sides of that triangle (AE and AC), mark their mid-points P and Q by joining the centre O to two of the unused points of the hexagon (F and B).
- The line PQ is then extended to meet the circle at point R.
Q is the golden section point of the line PR.
 Q is a gold point of PRThe proof of this is left to you because it is a nice exercise either using coordinate geometry and the equation of the circle and the line PQ to find their point of intersection or else using plane geometry to find the lengths PR and QR.The diagram on the left has many golden sections and yet contains only equilateral triangles. Can you make your own design based on this principle?
Chris and Penny's page shows how to continue using your compasses to make a pentagon with QR as one side.
- Equilateral Triangles and the Golden ratio J F Rigby, Mathematical Gazette vol 72 (1988), pages 27-30.
  Earlier we saw that the 36°-72°-72° triangle shown here as ABC occurs in both the pentagram and the decagon.Its sides are in the golden ratio (here P is actually Phi) and therefore we have lots of true golden ratios in the pentagram star on the left.But in the diagram of the pentagram-in-a-pentagon on the left, we not only have the tall 36-72-72 triangles, there is a flatter on too. What about its sides and angles?
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3 Phi and Triangles
3.1 Phi and the Equilateral Triangle
Chris and Penny at Regina University's Math Central (Canada) show how we can use any circle to construct on it a hexagon and an equilateral triangle. Joining three pairs of points then reveals a line and its golden section point as follows:
- On any circle (centre O), construct the 6 equally spaced points A, B, C, D, E and F on its circumference without altering your compasses, so they are the same distance apart as the radius of the circle. ABCDEF forms a regular hexagon.
- Choose every other point to make an equilateral triangle ACE.
- On two of the sides of that triangle (AE and AC), mark their mid-points P and Q by joining the centre O to two of the unused points of the hexagon (F and B).
- The line PQ is then extended to meet the circle at point R.
Q is the golden section point of the line PR.
Q is a gold point of PRThe proof of this is left to you because it is a nice exercise either using coordinate geometry and the equation of the circle and the line PQ to find their point of intersection or else using plane geometry to find the lengths PR and QR.
The diagram on the left has many golden sections and yet contains only equilateral triangles. Can you make your own design based on this principle?
Chris and Penny's page shows how to continue using your compasses to make a pentagon with QR as one side.
- Equilateral Triangles and the Golden ratio J F Rigby, Mathematical Gazette vol 72 (1988), pages 27-30.
Earlier we saw that the 36°-72°-72° triangle shown here as ABC occurs in both the pentagram and the decagon.Its sides are in the golden ratio (here P is actually Phi) and therefore we have lots of true golden ratios in the pentagram star on the left.
But in the diagram of the pentagram-in-a-pentagon on the left, we not only have the tall 36-72-72 triangles, there is a flatter on too. What about its sides and angles?
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| First Previous 2 to 2 of 2 Next Last |
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3 Phi and Triangles
3.1 Phi and the Equilateral Triangle
Chris and Penny at Regina University's Math Central (Canada) show how we can use any circle to construct on it a hexagon and an equilateral triangle. Joining three pairs of points then reveals a line and its golden section point as follows:
- On any circle (centre O), construct the 6 equally spaced points A, B, C, D, E and F on its circumference without altering your compasses, so they are the same distance apart as the radius of the circle. ABCDEF forms a regular hexagon.
- Choose every other point to make an equilateral triangle ACE.
- On two of the sides of that triangle (AE and AC), mark their mid-points P and Q by joining the centre O to two of the unused points of the hexagon (F and B).
- The line PQ is then extended to meet the circle at point R.
Q is the golden section point of the line PR.
Q is a gold point of PRThe proof of this is left to you because it is a nice exercise either using coordinate geometry and the equation of the circle and the line PQ to find their point of intersection or else using plane geometry to find the lengths PR and QR.
The diagram on the left has many golden sections and yet contains only equilateral triangles. Can you make your own design based on this principle?
Chris and Penny's page shows how to continue using your compasses to make a pentagon with QR as one side.
- Equilateral Triangles and the Golden ratio J F Rigby, Mathematical Gazette vol 72 (1988), pages 27-30.
Earlier we saw that the 36°-72°-72° triangle shown here as ABC occurs in both the pentagram and the decagon.Its sides are in the golden ratio (here P is actually Phi) and therefore we have lots of true golden ratios in the pentagram star on the left.
But in the diagram of the pentagram-in-a-pentagon on the left, we not only have the tall 36-72-72 triangles, there is a flatter on too. What about its sides and angles?
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Message 57 of 59 on the subject |
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