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Old 11-01-2008   #1
Parpadeos Sexis... So hot

 
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Ven, Ven, Laura.

Hola Laura, hablemos del teorema de Desargues.

En geometría proyectiva, el teorema de Desargues, llamado así en honor a Gérard Desargues afirma la siguiente:
En el plano proyectivo, dos triángulos son perspectivos desde un punto si y sólo si son perspectivos desde una recta.
Considere los triángulos ABC y DEF. El que los triángulos sean perspectivos desde un punto significa que las rectas AD, BE y CF concurren en un mismo punto O De manera similar el que los triángulos sean perspectivos desde una recta significa que los pares de lados AB, DE; BC, EF; AC, DF se cortan sobre una misma recta r.
Al punto O se le llama centro de perspectiva y a la recta r, eje de perspectiva.




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Old 11-01-2008   #2
o.O c u next tuesday O.o





 
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Jajajaj.
Limitaciones de la mecánica cuántica

La teoría de campos cuántica corrige varias deficiencias de la mecánica ordinaria cuántica, la que discutiremos brevemente. La ecuación de Schrödinger, en la forma en que comúnmente se la encuentra, es:

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Old 11-01-2008   #3
Parpadeos Sexis... So hot

 
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Si un cuadrilátero está dado por sus cuatro vértices A, B, C, D, el teorema afirma que:
Donde la línea sobre las Letras indica la longitud de los segmentos entre los vértices correspondientes.
Esta relación puede ser expresada de manera verbal de la siguiente forma:
En un cuadrilátero, si la suma de los productos de de las dos parejas de lados opuestos es igual al producto de sus diagonales, entonces el cuadrilátero puede ser inscrito en un círculo.
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Old 11-01-2008   #4
o.O c u next tuesday O.o





 
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Demostración geométrica




Demostración del teorema de Ptolomeo
  1. Sea ABCD un cuadrilátero cíclico.
  2. Note que en el segmento BC, ángulos inscritos ∠BAC = ∠BDC, y en AB, ∠ADB = ∠ACB.
  3. Ahora, por ángulos comunes △ABK es similar a △DBC, y △ABD ∼ △KBC
  4. Por lo tanto AK/AB = CD/BD, y CK/BC = DA/BD,
    1. Por lo tanto AK·BD = AB·CD, y CK·BD = BC·DA;
    2. Lo que implica AK·BD + CK·BD = AB·CD +BC·DA
    3. Es decir, (AK+CK)·BD = AB·CD + BC·DA;
    4. Pero AK+CK = AC, por lo tanto AC·BD = AB·CD + BC·DA; como se quería demostrar.
Note que la demostración es claramente válida solo para cuadriláteros concíclicos simples, si el cuadrilátero es complejo entonces K se encontrará fuera del segmento AC, y por lo tanto AK-CK=±AC, tal como se esperaba.
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Old 11-01-2008   #5
Parpadeos Sexis... So hot

 
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Ahre ;; *

Teorema de la bisectriz

De Wikipedia, la enciclopedia libre

Saltar a navegación, búsqueda
El teorema de la bisectriz del ángulo interno de un triángulo es un teorema de la geometría elemental que es una consecuencia o corolario del Teorema de Tales.
En un triángulo, la razón entre dos lados es igual a la razón de las partes en las que queda dividido el tercer lado por la bisectriz de ángulo interno opuesto.
O lo que es equivalente:
Dado el triángulo ABC, sea AL la bisectriz del ángulo interno A, entonces se cumple la proporción:
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Whats wrong with the people named Jon :/


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Old 11-01-2008   #6
o.O c u next tuesday O.o





 
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Demostración

Dibujese desde C una línea paralela a la recta AL hasta encontrar la prolongación de lado BA a partir del lado A y encontrandose en el punto D. El triángulo ACD es isósceles porque sus ángulos C y D son congruentes:

porque los dos ángulos son alternos internos respecto a las rectas paralelas AL y DC cortadas por la recta transversal AC

porque son correspondientes a las rectas paralelas AL y DC a las cuales corta la recta BD,
además

porque los ángulos creados por la bisectriz son iguales.
Por la propiedad transitiva de la igualdad tenemos que

Por tanto los segmentos AC y AD son congruentes. Por el Teorema de Tales

y ya que AC y AD son congruentes, también se cumple que
se mantiene la proporción:
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Old 11-02-2008   #7
٩(●̮̮̃•̃)۶ xsakamotox



 
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Old 11-02-2008   #8
o.O c u next tuesday O.o





 
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SI VOS NO SOS CARLOS O LAURA, ANDATE DE ACA. :$
OKEI? OSEA GORR FUERA! :$ .- ¬ k
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Old 11-02-2008   #9
ҳ̸Ҳ̸ҳ̸Ҳ̸ҳҲ̸ҳҲ̸ҳ

 
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consequentemente a função seno é também conhecida como "meia corda". Devido a essa relação, muitas das identidades trigonométricas e teoremas conhecidos hoje também o eram aos matemáticos [[Civilização helenística|helênicos]s, mas na sua forma equivalente de corda.[5]
Apesar de que não há nenhuma trigonometria nos trabalhos de Euclides e de Arquimedes, estritamente falando, existem teoremas apresentados de uma forma geométrica que são equivalentes a fórmulas ou leis trigonométricas específicas.[3] Por exemplo, as proposições 12 e 13 dos Elementos são a lei dos cossenos para ângulos agudos e obtusos, respectivamente. Teoremas a respeito do comprimento das cordas são aplicações da lei dos senos. E o teorema de Arquimedes sobre cordas rompidas é equivalente às fórmulas para o seno de somas e diferenças de ângulos.[3] Para compensar a falta de uma tabela de cordas, os matemáticos da época de Aristarco de Samos às vezes usavam um conhecido teorema de que, na notação moderna, sin α/ sin β < α/β < tan α/ tan β sempre que 0° < β < α < 90°, dentre outros.[6]
A primeira tabela trigonométrica foi aparentemente compilada por Hiparco de Nicéia (180 a.C. - 125 a.C.), que é agora conhecido como o "pai da trigonometria."[7] Hipparchus was the first to tabulate the corresponding values of arc and chord for a series of angles.[7][1]
Uma representação medieval de Claudius Ptolomeu
Apesar de não se saber quando o uso sistemático do círculo de 360° passou a fazer parte da matemática, é sabido que sua introdução se deu um pouco depois de Aristarco de Samos ter escrito Sobre os Tamanhos e Distâncias do Sol e da Lua (ca. 260 a.C.), uma vez que ele mede o ângulo em termos da fração de um quadrante.[6] Parece que o uso sistemático do círculo de 360° é em boa medida devido a Hiparco e sua tabela de cordas. Hiparco pode ter tirado a idéia dessa divisão de Hypsicles, que tinha anteriormente dividido o dia em 360 partes, uma divisão do dia que deve ter sido sugerida pela astronomia babilônica.[8] Na astronomia antiga, o zodíaco havia sido dividido em doze "signos" ou 36 "decanos". Um ciclo sazonal de aproximadamente 360 dias pode ter correspondido aos signos e decanos do zodíaco, dividindo cada signo em trinta partes e cada decano em dez partes


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