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Appendix A: Margherita Beloch Piazzolla: “Alcune Applicazioni Del Metodo Del Ripiegamento Della Carta Di Sundara Row”

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Appendix A: Margherita Beloch Piazzolla: “Alcune Applicazioni Del Metodo Del Ripiegamento Della Carta Di Sundara Row” Appendix A: Margherita Beloch Piazzolla: “Alcune applicazioni del metodo del ripiegamento della carta di Sundara Row” This appendix presents, for the first time, a translation from Italian into English of Beloch’s four-page-long paper “Alcune applicazioni del metodo del ripiegamento della carta di Sundara Row,”1 published in 1934 in “Atti dell’Acc. di Scienze Mediche, Naturali e Matematiche di Ferrara Serie II, Vol. XI”; the translation is organized according to the numeration of the original paper (pp. 186–189). The footnotes presented in this section are Beloch’s, while remarks and explanations regarding the content of the paper are given in square brackets in a smaller font. For pp. 186–188, the Italian text is also given in footnotes in square brackets.2 It is also advisable to look from time to time at Fig. 5.25 in order to follow the (implicit) steps in Beloch’s argument, as Beloch herself does not supply any drawing. *** p. 186: Some Applications of the Method of Paper Folding of Sundara Row A note by Margherita Beloch Piazzolla, Prof. at the University of Ferrara. In an excerpt from my lessons of a complementary mathematics course taught at the University of Ferrara in the academic year 1933–1934, I proposed the following problem as an application of the paper folding method of Sundara Row,3 a problem 1Beloch (1934a). 2Starting from the end of page 188 until the end of the paper (p. 189), Beloch performs several calculations, and as can be seen, the language she uses is formal; hence, the original text is not given. 3Cf. Sundara Row, Geometric Exercises in Paper Folding (Madras, Addison and Co. 1893; London, Court Company, 1917). © Springer International Publishing AG, part of Springer Nature 2018 377 M. Friedman, A History of Folding in Mathematics, Science Networks. Historical Studies 59, https://doi.org/10.1007/978-3-319-72487-4 378 Appendix A: Margherita Beloch Piazzolla: “Alcune applicazioni del... that allows for a simple solution to the classic problem of duplicating the cube; the solution has not yet been noticed, to my knowledge. Another application is a method given concerning the graphical resolution of the equations of the third degree, which completes the well-known procedure of Lill. The problem in question is the following: Construct a square, two of whose opposite edges [or their extensions] pass through two given points, respectively, and two of whose adjacent vertices are respectively on two given straight lines. Let A, B be the two given points and r, s be the two given straight lines. Denote by X, Y the vertices of the square to be built, lying respectively on the lines r, s. One edge of the square would then be XY. Suppose that the second edge of the square, which exits from X, passes through A, and the second edge, which exits from Y, passes through B.4 p. 187: Consider the parabola having its focus as point A and its tangent at its vertex the line r, where we assume that the (unknown) straight line XY is tangent to the parabola. Similarly, consider the parabola having, as its focus, point B, and as tangent at its vertex the line s, where we assume that the same straight line XY is tangent [to the second parabola]. This straight line can then be built as one of the common tangents to the two parabolas, and therefore the points X, Y can be built, and as a result also the required square. Since two parabolas have three common tangents, the problem admits three solutions, including certain real solutions. Those tangents can be found in the simplest way using the aforementioned paper folding method. For this purpose, it is enough to build the directrix of each of the two parabolas, denoted as d1 and d2 respectively, and remembering the property that the [geometric] locus of the symmetrical points regarding the focus [of the parabola] with respect to various tangents to the parabola is the directrix; on this property, the 4[The original text: Stralcio dalle mie lezioni del corso di Matematiche complementari tenuto all’Università di Ferrara nell’anno accademico 1933–1934, il seguente problema, da me proposto come applicazione del metodo del ripiegamento della carta di Sundara Row e che consente una semplice risoluzione mediante il detto metodo del problema classico della duplicazione del cubo, risoluzione che a quanto io sappia non è stata finora notata. Un’altra applicazione è un metodo da me dato per la risoluzione grafica delle equazioni di 3 grado, che completa il noto procedimento di Lill. Il problema di cui si tratta è il seguente: Costruire un quadrato di cui due lati opposti passino rispettivamente per due punti dati, e i due vertici situati sui rimanenti lati stiano rispettivamente su due rette date. Siano A, B i due punti dati: r, s le due rette date. Indichiamo con X, Y i vertici del quadrato da costruirsi, giacenti rispettivamente sulle rette r, s. Un lato del quadrato sarà allora XY. Supponiamo che il secondo lato del quadrato uscente da X passi per A e il secondo lato uscente da Y passi per B.] Appendix A: Margherita Beloch Piazzolla: “Alcune applicazioni del... 379 noticed construction, through the paper folding method, of a parabola with tangents is founded. That is, taking the (straight) edge of a sheet of paper as a directrix of the parabola, and marking the focus at the given distance, it is enough to hold fast the focus and refold the paper on itself so that the folded edge would pass through the focus: the fold of the paper will give a tangent to the parabola, and this way, one can construct all of the tangents. [As Beloch defines it, a symmetric point is the image of the focus P of the parabola after performing the folding of a piece of paper (on which the parabola and P are drawn) along a tangent to the parabola. The collection of all the symmetric points is the directrix (this was also noted by Row; see Sect. 4.2.2.2).] To find a common tangent to the two parabolas (given the directrixes and the foci), it is enough then to fold the paper in such a way that the two given lines (i.e., the directrices) would pass, after folding, through the two given points (i.e., the foci), respectively (which could be achieved by using a transparent piece of paper). The explained operation may be done with the same ease and accuracy with an ordinary drawing, passing a line through two points; this will allow one to find the (real) solutions to the problem.5 p. 188: The problem concerns, as mentioned above, the graphic solution of equations of the third degree, starting from the data graphic process of Lill,6 and offering a new 5[The original text: Si consideri la parabola avente per fuoco il punto A, e per tangente nel vertice la retta r di cui per proprietà nota la retta (incognita) XY è tangente. Similmente si consideri la parabola avente per fuoco il punto B, e per tangente nel vertice la retta s, di cui per la stessa proprietà la retta XY è tangente. Questa retta si può quindi costruire come una delle tangenti comuni alle due parabole, e determinare quindi i punti X, Y e in conseguenza il quadrato richiesto. Siccome due parabole hanno tre tangenti comuni, il problema ammette tre soluzioni, tra cui certo una reale. Le dette tangenti si possono trovare in modo semplicissimo col suddetto metodo del ripiegamento della carta. Basta all’uopo costruire la direttrice di ognuna delle due parabole, e siano rispettivamente d1 e d2, e ricordare la proprietà che il luogo dei punti simmetrici del fuoco rispetto alle varie tangenti della parabola è la direttrice, proprietà su cui si fonda la nota costruzione d’una parabola per tangenti col metodo del ripiegamento della carta. Prendendo cioè l’orlo (rettilineo) d’un foglio di carta come direttrice della parabola, e segnando il fuoco alla data distanza da questa, basta tener fermo il fuoco e ripiegare la carta su se stessa in modo che l’orlo ripiegato venga a passare per il fuoco: la piega della carta darà una tangente della parabola, e questa quindi si potrà costruire per tangenti. Per trovare una tangente comune alle due parabole basta dunque ripiegare la carta in modo che le due rette date nel ripiegarsi vengano a passare rispettivamente per i due punti dati (ciò che si potrà ottenere usando un foglio di carta trasparente). L’operazione spiegata si potrà eseguire con la stessa facilità e precisione con cui, nel disegno comune, si fa passare una riga per due dati punti, e permetterà di trovare le soluzioni (reali) del problema.] 6Cf. Felix Klein, Elementarmathematik vom höheren Standpunkte aus, II, p. 267, (Berlin, Springer 1925). 380 Appendix A: Margherita Beloch Piazzolla: “Alcune applicazioni del... and simple construction, which certainly follows from that which precedes it, and to which I shall have occasion to return. Another application concerns the classical problem of duplicating the cube, which, as is known, cannot be solved with straightedge and compass. To my knowledge, the construction I give here has not yet been noticed. Take the two lines r, s as orthogonal to each other, and O as their common [intersection] point. Assuming that the point A of the general problem lies on s, and the point B lies on r, and applying the construction mentioned above, one can determine the edge XY of the required square, namely, given the known distances OA ¼ a, OB ¼ b, one can determine the unknown distances OX ¼ x and OY ¼ y (hence solving the problem of the two mean proportionals) with the paper folding method, something that Sundara Row thought to be impossible.7 It is well known that: OA : OX ¼ OX : OY ¼ OY : OB, Namely, a : x ¼ x : y ¼ y : b.
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