Para os interessados pela genealogia destes piauienses que viveram nos séculos XVIII e XIX convém ver:
quarta-feira, 27 de setembro de 2017
terça-feira, 26 de setembro de 2017
O PT NÃO PRECISA ERRAR TANTO
Pensei que fosse notícia falsa no facebook. Mas agora vejo na TRIBUNA DA INTERNET, a foto abaixo:
Fonte: http://veja.abril.com.br/politica/defesa-de-lula-entrega-recibos-de-aluguel-com-datas-inexistentes/
Errar é humano, todos nós erramos. Mas prá que errar tanto, meu Deus? E agora, o próprio Palocci está se sentindo mal com este vexame todo, pedindo para sair do PT. O partido ainda tem uma boa estrutura organizacional, não poderia mais estar atrelado a esta situação ridícula que está arrasando e arrastando o partido para o buraco. Se alguns anéis foram perdidos, para que perder todos os dedos? Não faz sentido.
sábado, 23 de setembro de 2017
Terremoto na Coreia do Norte
Virou moda. Mais um terremoto a zero metros de profundidade na Coreia do Norte. Está na cara que é teste de bomba. Qual tipo de bomba, é que é o mistério. Como pensam que os EUA vai atacar, estão freneticamente construindo todos os tipos imagináveis de bombas atômicas. Realmente estão decididos a enfrentar uma guerra atômica em larga escala.
quinta-feira, 21 de setembro de 2017
Limit points of golden spirals of a golden rectangle are vertices of inner golden rectangle
By Lohans de Oliveira Miranda (UFPI) and Lossian Barbosa Bacelar Miranda (IFPI)
We know that from a golden rectangle we can construct four golden spirals that converge to four distinct limit points, which form a rectangle. What we did not know was that this rectangle, inner to the initial, is golden rectangle.
We know that from a golden rectangle we can construct four golden spirals that converge to four distinct limit points, which form a rectangle. What we did not know was that this rectangle, inner to the initial, is golden rectangle.
PROPOSITION. The four points limits of the four golden spirals of golden rectangle are vertices of a golden rectangle. In addition, the area of this is a fifth of the area of the initial golden rectangle.
Proof. Figure 1 below represents a golden rectangle, where φ = (1+√5)/2, is the golden number. Let P be the point of intersection of the segments AB and CD. It is known that P is the limit point of the golden spiral OQR. Another three limit points can be obtained from three other spirals, which can be obtained from OQR by 180° rotation. The rectangle that has these four limit points as vertices has the largest side equal to (2+ 3φ)(1+3φ)⁻¹a-[a+φ⁻¹a-(2+3φ)(1+3φ)⁻¹a] and or smaller side equal to a-2φ(1 + 3φ)⁻¹a. Therefore, the quotient between the largest and the smallest side is equal to φ.
The quotient between the smaller side of the initial rectangle and the smaller side of the inner rectangle is equal to a/[a-2φ(1+3φ)⁻¹a]=2φ-1=√5. This indicates a reduction factor of 1/√5. Therefore, the quotient between the area of the initial golden rectangle and the area of the inner golden rectangle is equal to (2φ-1)⁻¹=5.
Immediate consequences: On the construction of each of the four golden spirals there is the construction of a sequence of golden rectangles. From the result we have just seen, within each of the golden rectangles of this sequence there is a sequence of golden rectangles in decreasing chain by inclusions. And all these golden rectangles are buildable by ruler and compass. What is the measure and the dimension of the set formed by these limit points of spirals? How would they be distributed throughout the plan? And if we join to this set the vertices of the other golden rectangles that can be constructed from the elaboration of the spirals, how should the above questions be answered? Moreover, the four spirals can extend, since their limit points are also the origins of new spirals, so that the junctions of these spirals are curves that converge towards the center of the initial golden rectangle.
Note: We have just seen a beautiful and didactic earlier explanation of Jeffrey R. CHASNOV (https://www.youtube.com/watch?v=jMFv0MRmEuo).
The quotient between the smaller side of the initial rectangle and the smaller side of the inner rectangle is equal to a/[a-2φ(1+3φ)⁻¹a]=2φ-1=√5. This indicates a reduction factor of 1/√5. Therefore, the quotient between the area of the initial golden rectangle and the area of the inner golden rectangle is equal to (2φ-1)⁻¹=5.
Immediate consequences: On the construction of each of the four golden spirals there is the construction of a sequence of golden rectangles. From the result we have just seen, within each of the golden rectangles of this sequence there is a sequence of golden rectangles in decreasing chain by inclusions. And all these golden rectangles are buildable by ruler and compass. What is the measure and the dimension of the set formed by these limit points of spirals? How would they be distributed throughout the plan? And if we join to this set the vertices of the other golden rectangles that can be constructed from the elaboration of the spirals, how should the above questions be answered? Moreover, the four spirals can extend, since their limit points are also the origins of new spirals, so that the junctions of these spirals are curves that converge towards the center of the initial golden rectangle.
Note: We have just seen a beautiful and didactic earlier explanation of Jeffrey R. CHASNOV (https://www.youtube.com/watch?v=jMFv0MRmEuo).
quarta-feira, 20 de setembro de 2017
Temer vai perder de novo no STF
“Não cabe [ao STF] proferir juízo antecipado a respeito de eventuais teses defensivas”, disse Fachin há pouco (http://www.msn.com/pt-br/noticias/brasil/stf-julga-futuro-de-den%c3%bancia-contra-temer/ar-AAsgLq7?li=AAggV10&ocid=wispr). Foi um voto técnico irretocável que deve ser seguido pelos demais colegas, pois este é o entendimento majoritário do supremo. A defesa de Temer fica se "queimando", perdendo desnecessariamente no STF, agindo de modo parecido com Lula junto ao juízo comandado por Moro. Creio que a melhor estratégia seja deixar o barco andar e decidir politicamente, que é o forte do governo, conforme ficou provado na primeira denúncia. Usar os argumentos políticos, da necessidade de levar o governo aos trancos e barrancos até 2018 e fazer a eleição normalmente. Apoiar um bom candidato e rezar pela vitória dele. Ficar colecionando derrotas e chateando o STF com pedidos todo dia não é bom caminho.
NOTA. Gilmar Mendes está se "estribuchando", mas ele já está previsível demais. Vai fazer companhia ao Temer, vai perder também, de novo.
NOTA. Gilmar Mendes está se "estribuchando", mas ele já está previsível demais. Vai fazer companhia ao Temer, vai perder também, de novo.
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