Theorems of Euclidean Geometry through Calculus Martin Buysse∗ Facult´ed'architecture, d'ing´enieriearchitecturale, d'urbanisme { LOCI, UCLouvain We re-derive Thales, Pythagoras, Apollonius, Stewart, Heron, al Kashi, de Gua, Terquem, Ptolemy, Brahmagupta and Euler's theorems as well as the inscribed angle theorem, the law of sines, the circumradius, inradius and some angle bisector formulae, by assuming the existence of an unknown relation between the geometric quantities at stake, observing how the relation behaves under small deviations of those quantities, and naturally establishing differential equations that we integrate out. Applying the general solution to some specific situation gives a particular solution corresponding to the expected theorem. We also establish an equivalence between a polynomial equation and a set of partial differential equations. We finally comment on a differential equation which arises after a small scale transformation and should concern all relations between metric quantities. I. THALES OF MILETUS II. PYTHAGORAS OF SAMOS Imagine that Newton was born before Thales. When If he was born before Thales, Newton was born before considering a triangle with two sides of lengths x and Pythagoras too, so that we do not have to make any y, he could have fantasized about moving the third further unlikely hypothesis. Imagine that driven by his side parallel to itself and thought: "Well, I am not success in suspecting the existence of a Greek theorem, an ancient Greek geometer but I am rather good in he moved to consider a right triangle of legs of lengths x calculus and I feel there might be some connection and y and of hypotenuse of length z. between the way x and y vary in such circumstances." He might have been tempted He would have materialized to speculate about the link, his suspicion in a function if any, between x, y and z in every right triangle. And y = y(x) (1) again, as calculus master, he could have postulated that connecting x and y what- ever the position of the third side, as long as it is moved z = z(x; y) (6) parallel to itself. In partic- a relation that must be true for any x, y and z in a ular, after a slight displace- right triangle. In particular, after a slight increase in the ment resulting in small devi- length of x, leading to a small deviation δx, while δy = 0, ations δx and δy, he would he would have found, at first order, that have had at first order δz = @xz δx (7) δy = y0(x)δx (2) But the lengths δx and δy of the small added segments must themselves obey equa- tion (1), that is arXiv:2007.00002v1 [math.GM] 29 Jun 2020 δy = y(δx) (3) Here, δx is the length of the hypotenuse of a right trian- To see it, translate those segments to the (x; y) vertex. gle with one leg of length δz. At first order, this small Developing the right-hand-side member of equation (3) triangle is similar to the initial one. Using the Thales at first order and noticing that y(0) = 0, we have theorem that he has just found out, δy = y0(0)δx (4) x which, compared to eq. (2), implies that y0(x) is con- δz = δx (8) stant. Integrating y0 = k, k being a positive constant z since y(x) is an increasing and smooth function, gives Substituting this result to δz in eq. (7) would have led the Thales theorem [1, Book VI, Prop. II] him to the partial differential equation y(x) = kx (5) z@xz = x (9) whose general solution is ∗ [email protected] z2(x; y) = x2 + k(y) (10) 2 where k(y) is an arbitrary function of y. But the function The δ`-length arc is also z has to be symmetric in x and y (i.e. he could have made the first order hypotenuse of the same reasoning with a non-zero δy while δx = 0) and another small right triangle z(x; 0) = x. Hence with one leg of length δd, 2 2 2 which is similar to the tri- z (x; y) = x + y (11) angle with a d-length hy- potenuse and whose corre- that is, the Pythagorean theorem, which, like Thales', sponding leg is a segment was probably discovered long before { and published a starting from the foot of the few centuries later by Euclid [1, Book I, Prop. XLVII]. median and parallel to { and As will be shown in the last section, the same result can thus half of the length of { be obtained by considering a small rotation of the hy- the h -length height, so that potenuse around one of its extremities followed by an y infintesimal scale transformation. This proof is known h and is published in a slightly different form in [2, 3]. δd = y δ` (15) 2d Inserting those deviations in eq. (13) leads to the partial III. APOLLONIUS OF PERGA differential equation x d@ d = (16) x 2 Thales and Pythagoras the- orems are not the only whose general solution is ones that are named be- x2 fore famous Greek geome- d2(x; y; z) = + k(y; z) (17) 2 ters. Newton could have gone a step further { east- where k(y; z) is a function of y and z. But d(x; y; z) has wards, a few centuries later to be symmetric in x and y (i.e. we can make the same { and assumed that in any reasoning with a non-zero δy while δx = 0). Hence triangle of sidelengths x, y x2 + y2 and z, the length d of the d2(x; y; z) = + c(z) (18) median relative to the z- 2 length side is a smooth func- with c(z) a function of z. Furthermore, if x = 0 (or tion of x, y and z, i.e. y = 0), y = z (or x = z) and d = z=2. This yields c(z) = −z2=4, which can alternatively be found by invoking the d = d(x; y; z) (12) Pythagorean theorem for y = x. The particular solution reads After an infinitesimal rota- x2 + y2 − 2(z=2)2 tion of the y-length side d2(x; y; z) = (19) around the (y; z) vertex re- 2 sulting in a small deviation that is, Apollonius's theorem, to be found in a slightly δx, with δy = 0 and δz = 0, more elaborate form in [4]. at first order: δd = @xd δx (13) IV. MATTHEW STEWART In order to get an expression for δx and δd and then Suppose Newton was born before Stewart, an 18th- a differential equation leading to the would-be theorem, century Scottish mathematician (and reverend). Well, consider the infinitesimal arc travelled by the moved ver- he was. Perhaps he was not interested, or did not tex, of length δ`. have the time, otherwise he could have used this At first order, it can be seen tool to generalize Apollonius's theorem to any cevian. as the hypotenuse of a small right triangle with one leg of length δx, which is sim- ilar to a larger right triangle whose corresponding leg is the hy-length height relative to the y-length side, and the hypotenuse is the x-length side, so that h δx = y δ` (14) x 3 In a triangle of sidelengths x, y and z, assume that the V. HERON OF ALEXANDRIA length d of a cevian dividing the side of length z in two segments of lengths m and n, is a smooth function of x, Intoxicated by his findings, Newton could have switched y, m and n, that is to a more elaborate, though older, challenge { as proba- bly did an Ancient Greek Roman Egyptian mathemati- d = d(x; y; m; n) (20) cian... What if, for any triangle, the area A could be a smooth function of the sides lengths x, y and z? He After an infinitesimal rotation of the y-length side around would have assumed the (y; z) vertex resulting in a small deviation δx, with δy = 0, δm = 0 and δn = 0, at first order: A = A(x; y; z) (27) After an infinitesimal rotation of the y-length side around δd = @xd δx (21) the (y; z) vertex resulting in a small deviation δx, while δy = 0 and δz = 0, at first order: δA = @xA δx (28) First note that the hz-length height relative to the z- length side divides the ini- tial triangle in two right triangles of horizontal legs of lengths t and z − t re- spectively. One can ex- press hz as a result of the Pythagorean theorem in Using the same similarities as for the Apollonius's theo- both right triangles. rem, with the unique difference that the foot of the cevian 2 2 2 2 is not necessarily the middle of the (m + n)-length side Equating those expressions yields x − t = y − (z − t) but falls at a distance n from its right vertex, we find and hence x2 − y2 + z2 y2 − x2 + z2 h nh t = z − t = (29) δx = y δ` δd = y δ` (22) 2z 2z x (m + n)d hy being the length of the height relative to y, we have the partial differential equation n d@ d = x (23) x m + n whose general solution is n d2(x; y; m; n) = x2 + k(y; m; n) (24) m + n where k(y; m; n) is a function of y, m and n. But d(x; y; m; n) must be symmetric in (x; m) and (y; n) (i.e.