Josip Plemelj - A Geometrical Construction From His Schooldays

A Geometrical Construction From His Schooldays

Plemelj had shown his great gift for mathematics early in elementary school. He mastered the whole of the high school syllabus by the beginning of the fourth year and began to tutor students for their graduation examinations. At that time he discovered alone series for sin x and cos x. Actually he found a series for cyclometric function arccos x and after that he just inverted this series and then guessed a principle for coefficients. Yet he did not have a proof for that.

Plemelj had great joy for a difficult constructional tasks from geometry. From his high school days originates an elementary problem — his later construction of regular sevenfold polygon inscribed in a circle otherwise exactly and not approximately with simple solution as an angle trisection which was yet not known in those days and which necessarily leads to the old Indian or Babylonian approximate construction. He started to occupy himself with mathematics in fourth and fifth class of high school. Beside in mathematics he was interested also in natural science and especially astronomy. He studied celestial mechanics already at high school. He liked observing the stars. His eyesight was so sharp he could see the planet Venus even in the daytime.

Let us hear about his early days in school in his own words: "It was the April 1891 in fifth class. The class had two rows of desks with crossing in between and I was sitting on the most side inside chair very rear. I think there were only two desks after me. Professor Borštner did not lecture. He had only given a lection from the book for the next lesson. He called to the blackboard two pupils and there he was discussing the subject and furthermore with this he included the whole class for cooperation. He used to have such a habit gratefully to give geometrical constructional tasks which he dictated from some collection he brought with. Once he gave amongst the other a task: Draught a triangle if one side, its altitude and a difference of two angles along it are given. Classmates had appealed to me before the lesson if the task was a little bit hard. They could not solve this task after several lessons and he had asked them the other. Borštner used to come from before the master's desk and he stopped ahead of me, sat toward me in the desk and hence he examined. After sometime he had said we should solve that task. Perhaps he was suspicious about that we had not yet solved it so he turned to me and asked me if I had tried this construction. I had said to him I could not find any path for the solution. Then he said he would show it in the next lesson. This had plucked up my courage to inspect it again. I had found a solution with subsidiary points, lines and so forth which seems to be inaccessible for human mind if the way which had inevitably led me to my aim is hidden from. Next lesson professor Borštner had sat toward me in the desk again. After customary examination of my classmates he said: "Well, let us work out that construction task." I whispered him I had succeeded till then and he said: "So, let me show how had you done this." He thought I would show him written on paper and he said: "Well, all-right." He had stepped aside and we went before the blackboard. I drew a triangle ΔABC with an ordinary analysis: Given side AB = c, its altitude v_c and the difference 0 < α − β < π.

Let us draw a perpendicular AA' from A to side AB and let AA' = 2 v_c. Let us bind A' with C in the way to be A'C = AC = b and draw A'B = m. Along the side A'C let us gather out along A' an angle α and on the left side of a triangle A'B' = c. On this way the risen triangle is ΔA'B'C ~ ΔABC. By B lies an angle <A'B'C = β and an angle <A'CB' = γ.

A triangle ΔBCB' is isosceles and <BCB' = 2 α, so we have < BB'C = π/2 − α. Now it is the angle <A'B'B = π/2 − α + β and we can construct over the side BA' a circumferential angle of a certain circle. We get a point B' at once because it is A'B' = c. Because a triangle ΔBCB' is isosceles a point C lies on a symmetric of the side BB' where it intersects a parallel with AB in a given altitude v_c. With this the triangle ΔABC is drawn. Professor Borštner was gazing when he saw this curious solution and he held of his head: "Aber um Hergottswillen, das ist doch harsträubend, das ist doch doch menschenunmöglich auf so einen Einfall zu kommen; sagen Sie mir doch, was hat Sie zu dieser Idee geführt!" I said to him I had not guessed this strange solution but I had asked myself about a trigonometric determination of a triangle because I could not find the solution in the other way. Geometric interpretation of this solution had led me up to this pure geometrically understandable construction. We did not spoke anymore about this with professor Borštner, but he had after that shown another easier solution, which I could imitate from my own construction and which I had not perceived because I had precisely traced way.

Trigonometric solution is easy: with altitude v_c from point C to side AB it breaks in two parts v_c cot α and v_c cot β. Then we have:

We can then write:

Because α + β = π − γ, this equation is:

From this equation we have to obtain an angle γ. The easiest way is if we introduce a certain subsidiary angle μ. Namely we raise:

2 v_c = m cos μ c = m sin μ.

Both equations give us for μ a uniform certain acute angle and for m a certain positive length. The equation for γ is then:

We can consider this equation as a theorem of the sine for a certain triangle in which c and m are its sides and their opposite angles are γ − μ and π/2 ± (α − β) respectively ( lightgreen triangle on the picture below ). In this quoted construction this triangle is ΔBA'B', where BA' = m and A'B' = c, the angle <BB'A' = π/2 − α + α and the angle <A'BB' = γ − μ, as we can easily see. In my construction this requested triangle is drafted twice. I saw later on we can interpret above equation with a triangle which has a side AB already. This leads us to very beautiful and short construction. Requested triangle is ΔK'AB.

Straight line CK = a is symmetrically displaced in CK' = a and at the same time is AK = AK' = m. I had spoken as a professor in Černovice with two of my students about this elementary geometrical problem and I said that my high school teacher had dictated this task from a certain collection. They brought me a collection indeed where there was the exact picture from Borštner's construction. Unfortunately I had not written down a title of that book which was with no doubt professor Borštner's collection. Our teacher's library at classical gymnasium in Ljubljana does not have this book at present. But I got from there a Wiegand's book entitled Geometrijske naloge za višje gimnazije (Geometrische Aufgaben für Obergymnasien, Geometric exercises for upper gymnasiums), which does not have this exercise. I found in it a task: construct a triangle if we know one length of an angle symmetrical from one point, perpendicular on this symmetrical from the other point and an angle by the third different solutions. The last has annotation: at the night of the January 1, 1940 after the New Year's Eve 1939".