value of the function when x = 1, that is, the sum of the coefficients of the function. As the arms are turned from 0° to 90°, keeping the slides properly adjusted and all the cords taut, the distance of this slide, R, from its initial position will be continuously the value of the function as a varies continuously from 1 to 0. ON THE ORTHOCENTRIC QUADRILATERAL.1 By NATHAN ALTSHILLER-COURT, University of Oklahoma. Introduction. (a) The altitudes AD, BE, CF of a triangle ABC meets in a point H, the orthocenter of ABC. The triangle DEF formed by the feet D, E, F, of the altitudes is frequently called the orthic triangle of ABC. To Carnot2 is due the credit for having called attention to the almost obvious fact that each of the four points H, A, B, C, is the orthocenter of the triangle formed by the other three. The points A, B, C, H are referred to as an orthocentric group of points, or an orthocentric quadrilateral, and the four triangles determined by these four points as an orthocentric group of triangles. (b) In 1821 Brianchon and Poncelet showed that the circumcircle (N) of the orthic triangle DEF of ABC passes through the mid-points A', B', C', of the sides BC, CA, AB, of ABC, and also through the mid-points P, Q, R, of the segments AH, BH, CH respectively.3 That the circle through the first six points mentioned passes also through the last three becomes obvious if we observe that DEF is the orthic triangle not only of ABC, but of each of the four triangles of the orthocentric group ABCH. (c) In 1822 Feuerbach proved that the circle (N) is tangent to the four circles which touch the sides of the triangle ABC. It was not until 1861 that Sir William R. Hamilton pointed out that (N) is also tangent to the circles touching the sides of the triangles BCH, CHA, HAB.5 Now since the orthic triangle DEF is common to the four triangles of the orthocentric group ABCH, the circumcircle (N) of DEF is the nine-point circle of each of these four triangles, and therefore Hamilton's extension of Feuerbach's theorem becomes self-evident. Read before the American Mathematical Society, St. Louis, December 31, 1919. Readers of this article will be interested in comparing it with the first part of the author's earlier article "On the I-centre of a triangle" (1918, 241-246)-EDITOR. 2 Carnot, De la corrélation des figures de Géométrie, 1801, p. 102. 3 For the proof, see, for instance, J. Casey, A Sequel to Euclid, second edition, 1882, p. 58, or C. V. Durell, Plane Geometry for Advanced Students, vol. 1, pp. 30-31. 4 For a proof see Casey, l.c., pp. 58-61, or Durell, l.c., pp. 46-47 and pp. 149–150. 5 In making this statement Professor Altshiller-Court was possibly misled by Casey's reference to the result as "Sir William Hamilton's Theorem" (Quarterly Journal of Mathematics, 1861, p. 249) and by the fact that Sir William proposed the result as a problem in Nouvelles annales de mathématiques, 1861, vol. 20, p. 216. The result was not, however, given originally by Sir William, but by T. T. Wilkinson, as prize-problem 1883 in Lady's and Gentleman's Diary, London, 1854, p. 72 (Solutions, Diary, 1855, pp. 67-69).-EDITOR. These examples suggest that in certain connections it may be fruitful to consider the circle (N) as belonging not to the triangle ABC, but to the orthocentric group ABCH. The following considerations are based on this remark. 1. The center N of the nine-point circle (N) of the triangle ABC was shown by Benjamin Bevan, in 1804,' to lie midway between the orthocenter H and the circumcenter O of the triangle ABC. In other words, the circumcenter O of ABC is the symmetric of H with respect to N. But, as has been pointed out above, (N) is also the nine-point circle of the triangle BCH, whose orthocenter is the point A, hence the circumcenter 01 of BCH is the symmetric of A with respect to N. Similarly for the circumcenters 02, 03, of the triangles CHA, HAB. Consequently: The four circumcenters of an orthocentric group of triangles form an orthocentric group which is the symmetric of the given group with respect to the nine-point center. 2. From the symmetry of the two groups of points ABCH and 0102030 follow immediately all the known properties of the circumcenters 01, 02, 03, 0. For instance: (a) The triangles 010203 and ABC are congruent and furthermore, their sides are respectively parallel. It may also be observed that these properties hold for the pairs of triangles 02030 and BCH; 03001 and CHA; 00102 and HAB. 1 For the proof compare Casey, or Durell, l.c. 2 Durell, l.c., p. 36, exercise 89. (b) The point H is the circumcenter of the triangle 010203.1 Similarly the points A, B, C, are the respective circumcenters of the triangles 02030, 03001, 00102. (c) The lines BO3 and CO2 are parallel1 and similarly for other pairs of analogous lines. 3. A wealth of other propositions, so far unobserved or unannounced, may be derived from the two symmetrical figures. We shall call attention to the following. In the symmetrical transformation considered the center of symmetry N is a double point, and the nine-point circle (N) is transformed into itself, hence: An orthocentric group of triangles and the orthocentric group of their circumcenters have the same nine-point circle. 4. The orthic triangle D'E'F' of the orthocentric group 0010203 is the symmetric of the orthic triangle DEF of the orthocentric group ABCH, the pairs of points D, D'; E, E'; F, F'; being diametrically opposite on the circle (N). Thus we find a geometric interpretation of three new points of the nine-point circle of the triangle ABC. 5. The nine-point circle (N) of the orthocentric group 0010203 is tangent to the sixteen circles which touch the sides of the four triangles of this group, according to Hamilton's extension of Feuerbach's theorem (Introduction). These sixteen circles are the symmetric, with respect to N, of the analogous sixteen circles of the orthocentric group ABCH. Thus we find sixteen new circles which are tangent to the nine-point circle of the triangle ABC.2 6. The orthocentric group 0010.03 has been derived by symmetry from the given orthocentric group ABCH. But the process may be reversed, and the group HABC may be derived from the group 0010203 considered as given. Consequently: The vertices of a given orthocentric group of triangles may be considered as the four circumcenters of a second orthocentric group of triangles, the two orthocentric groups having the same nine-point circle and being symmetrical with respect to its center. 7. The point of intersection G of the medians of ABC, often referred to as the centroid of ABC, lies, according to a theorem of Euler,3 'on the line joining the orthocenter H to the circumcenter 0 of ABC, and we have GO/GH = 1/2. Since the nine-point center N is the midpoint of OH, we have NG|NH 1/3, the points G and H being on opposite sides of N. In other words, the point G 1 Durell, l.c., p. 36, exercise 88. = 2 The results of paragraphs 1, 2, 3, 4 and 5 were given by T. T. Wilkinson in Mathematical Questions with their Solutions from the Educational Times, London, Vol. 1, 1864, pp. 6-7; see also Mathematical Questions, etc., Vol. 6, 1866, pp. 25-26. T. T. Wilkinson seems to have been the first to discover an infinite series of circles tangent to the nine-point circle of a triangle (Lady's and Gentleman's Diary, London, 1857, p. 86; 1858, 87): "If the radical centers of the inscribed and escribed circles of any triangle be taken, and circles be inscribed and escribed to the triangles formed by joining these radical centers, and the radical centers of the latter system of circles be again taken and circles inscribed and escribed to the triangles thus formed, and so on ad infinitum, the infinite number of circles thus found, as well as the original system of inscribed and escribed circles, always touch the circle drawn through the middle points of the [sides of the] first triangle."-EDITOR. 3 Durell, l.c., p. 41. corresponds to H in a similitude of ratio - 1/3, the center of similitude being N.1 But N is also the nine-point center of the triangle BCH, whose orthocenter is A, hence the centroid G1 of BCH corresponds to A in a similitude of ratio - 1/3 with N as center of similitude. Similarly for the centroids G2, G3, of the triangles CHA, HAB. Consequently: The four centroids of an orthocentric group of triangles form an orthocentric group, the two groups being similar and similarly placed. 8. Since the centroids G, G1, G2, G3, form an orthocentric group, all the properties of such a group immediately follow, as, for instance, that G is the orthocenter of the triangle G1G2Gз, etc. Again the similitude of the two groups GG1G2G3 and HABC puts into evidence a great many properties, as for instance, that G1G2 is parallel to AB and is equal to 1/3 of its length; that the point of intersection of GG1 and G2G3, which will be represented by (GG1, G2G3), is collinear with N and D = (HA, BC); etc. The reader may find it interesting to formulate a number of these propositions. 9. In the similitude (7) by which the group GG1G2G3 is derived from the group HABC, the center of similitude N is a double point. Hence: An orthocentric group of triangles and the orthocentric group of their centroids have the same nine-point center. 10. The orthocentric group GG1G2Gз has been derived from the given orthocentric group HABC by a similitude of center N and ratio - 1/3. But the process may be reversed, and the orthocentric group HABC may be derived from the orthocentric group GG1G2G3, considered as given, by a similitude of ratio - 3, the center remaining the same. Consequently: The four points of an orthocentric group may be considered as the centroids of another orthocentric group of triangles, the two groups having the same nine-point center, this point being a center of similitude of the two groups, the ratio of similitude being — 3. 11. Since from (1) the two groups HABC and 0010203 are symmetrical about the center N, therefore it follows from (10) that the two groups GG1G2G3 and 0010203 admit N as a center of similitude, the ratio of similitude being + 3. Hence: The centroids and the circumcenters of an orthocentric group of triangles form two orthocentric groups of points having the same nine-point center, this point being a center of similitude of these two groups, the ratio of similitude being + 3. 1720 C. Maclaurin's Geometria organica sive descriptio linearum currarum universalis, published at London-G. Poleni's De mathesis in rebus physicis utilitate praelectio habita・・・, published at Patavia-Second edition of L'Hospital's Traité analytiques des sections coniques, published at Paris-Alexandre Savérien, author of Dictionnaire universel de mathématiques et de physique (2 vols., Paris, 1753), born July 16. 1 Euler, Novi comment. acad. sc. Petrop., vol. 11 (1765), 1767, p. 114.-EDITOR. A GRAPHIC SOLUTION OF THE CUBIC EQUATION.1 By J. P. BALLANTINE, University of Maine. The purpose of this paper is to suggest a graphic method of finding the real roots of a numerical cubic equation, a method which will involve comparatively little work when there are a large number of problems. (1) Let us consider the general cubic reduced to the standard form u3 + u2 + xu = y. This can easily be effected by dividing by the coefficient of u3, and then by dividing the roots of the equation by the coefficient of u2. It may happen in a particular case that this latter coefficient is zero, in which case the form obtained will differ from the form given above in that the u2 term is missing; but a similar discussion is easily worked out for that case. All the cubics of the standard form are easily set in a one to one correspondence with the points of the plane. Let us say that the x and the y of the cubic are the coördinates of the corresponding point. The necessary and sufficient condition that v be a root of the standard cubic is that: The same is a necessary and sufficient condition that the point (x, y) lie on a certain straight line, of which the above is said to be the equation. The points therefore corresponding to a set of isoradical cubics2 lie on a straight line, and the common root of the cubics is the slope of the line. Let us next consider the locus of points corresponding to cubics with two equal roots. The necessary and sufficient condition that v be a double root of = 1 While the essential features of this solution are not new they are probably sufficiently unfamiliar to readers of the MONTHLY to render Mr. Ballantine's presentation of his rediscovery of them a matter of general interest. As applied to the equation u3 + xu + y O there is a recent discussion in J. Lipka, Graphical and Mechanical Computation, 1918, pp. 35-36. A similar discussion is found in C. Runge, Graphical Methods, New York, 1912, pp. 59-60 (also Praxis der Gleichungen, Leipzig, 1900, p. 159), and L. E. Dickson incorrectly attributes the method to Runge (Dickson, Elementary Theory of Equations, New York, 1914, p. 17). This method originated with Léon Lalanne who made the discovery in 1843 and regarded the cubic as a special case of the trinomial equation um + xun + y = 0 ("Sur les tables graphiques et sur la géométrie anamorphique . . .,” Annales des ponts et chaussées, 1846, 1er semestre, pp. 27-29; Comptes rendus de l'académie d. Sc., Paris, tome 81, 1875, pp. 1187-1188; and "Méthodes graphiques pour l'expression des lois à trois variables," Notices réunies par le ministere des travaux publics à l'occasion de l'Exposition universelle de Paris, 1878). Lalanne found that his form of the cubic equation led to the semi-cubical parabola 4x3 + 27y2 = 0 as an envelope. The subject is also discussed in J. de la Gournerie, Traité de Géométrie descriptive, seconde édition, troisième partie, Texte, Paris, 1885, pp. 211-212, Atlas, 3, fig. 455 (filling a quarto page), plate 46; and in M. d'Ocagne, Traité de Nomographie, Paris, 1899, pp. 43-44.-EDITOR. Two cubics are said to be isoradical if they have a root in common. A straight line is said to be isoradical if its points correspond to isoradical cubics. |