MATHEMATICS AND ORIGAMI

Jesús de la Peña Hernández 7- Therefore, trapezium ABCD has an area half of the rhomboid ADGH. 8- So: area ABCD = 2 1 DG × IJ = ...

Mathematics and Origami 7.6 MAXIMA AND MINIMUMS The value of x that will make z’ = 0 is the minimum for AB we are trying to find ...

Jesús de la Peña Hernández Let ́s set the hypothesis that z’ = 0 when its last factor ́s value is 0. As the previous fac- tor is ...

Mathematics and Origami 7.7 RESOLUTION OF A QUADRATIC EQUATION Let the equation x^2 +mx+n= 0 (1) As it ́s well known, any equati ...

Jesús de la Peña Hernández 1 1 1 x ac Ax=x −b+ ; substituting in (2): (^)       = − + 1 1 1 2 1 (^22) x ac a x x x b a^ ...

Mathematics and Origami 7.8 SQUARE ROOT OF A NUMBER This is a particular case of Point 7.7: it ́s a matter of solving a quadrati ...

Jesús de la Peña Hernández 7.10 PARABOLA ASSOCIATED TO THE FOLDING OF A QUADRATIC EQUATION Point 1.2.4 explained how in folding ...

Mathematics and Origami 2 120 1 y= 30 + x → 2 120 1 30 3 130 y+ = + x Which for y = 0 gives 30120 40 3 130  =±     x=  − W ...

Jesús de la Peña Hernández t d xF c 2 = − 2 a xB = 2 d yF= 2 at yB=b− On the other hand, B F F B y y x x t − − = , i.e.: bt at d ...

Mathematics and Origami 7.13 PARABOLAS ASSOCIATED TO THE FOLDING OF A COMPLETE EQUATION OF 3rd DEGREE Folding operation in fig 1 ...

Jesús de la Peña Hernández In this figure we may observe: Points D, C have been obtained by calculus (Point 7.11). From focuses ...

Mathematics and Origami 7.14 FUNDAMENT OF ORTHOGONAL BILLIARDS GAME (H. HUZITA) This game is an ingenious recreation of conventi ...

Jesús de la Peña Hernández Play to two tablesides In this case we shall fix our attention only in orthogonal billiards (Fig. 5) ...

Mathematics and Origami Process is the inverse if we want to get the square root of b: Start with points C (-1, 0) and B (b,0). ...

Jesús de la Peña Hernández 7.14.3 THE ORTHOGONAL SPIRAL OF POWERS (H.H.) Observing the process pursued along the two previous Po ...

Mathematics and Origami Start in I (IO = 1) a series of coefficient vectors according to these criteria: IO = 1: coefficient of ...

Jesús de la Peña Hernández 3 1 4 ( 1 ) 1 x x − − = − ; 3 1 4 ( 2 ) 2 x x − − = − In general: x^2 + 4x + 3 = 0 7.14.5 RESOLUTION ...

Mathematics and Origami Let ́s figure out the three real roots of equation t^3 +t^2 − 2 t− 1 = 0 that was already solved with a ...

Jesús de la Peña Hernández and rotate it 180º around O, Fig. 2 of Point 7.13 is obtained. Both figures give the same results for ...

Mathematics and Origami 1 2 (^21) − = ED t ; 1 2 (^12) − = t ED t ED 2 − 1 = t t 2 1 2 2 2 − − = ; (t^2 − 1 )t=t−(t^2 − 1 ) (t^ ...