these are distinct, for the centers c7 coalesce and unite with the incenter of the configuration. The points co will then lie on a circle, and the points cs will be affected somewhat. Fig. 1. It is worth investigation to see how these comparatively few points c5 behave in the case where the quadrilateral is considered as a special case of eight directed lines; and it is interesting to note in Figure 1 their resemblance to the configuration discussed above; namely, the rectangular net formed by the incenters of an inscribed quadrilateral. The figure of the circles, C5, is also of interest in this case, (Figure 2). As their centers are constrained to the rectangular net it is to be expected that they are in families of four; in addition they all pass through a point. III. The configuration suggested by the figures is readily proved. It is convenient to consider first a quadrilateral with two sides counted both ways; by repeating this figure the whole configuration may be obtained, but all that is essential is contained in Figure 3. Here there are six directed lines. By the general theorem there is a centercircle for all six, expressed in the manner of Loud (l.c., p. 325), as where the constants are complex and t an orthogonal number. The six circles Cs are included in the double infinity of circles which are known to pass through a point. Figure 2 is thus confirmed; all the circles C's pass through a point, which turns out to be the Clifford point of the quadrilateral. On examination the six points c, in Figure 3 are seen to fall into two sets; four on a rectangle, and two separate. Consider now a triangle with two sides counted both ways; by drawing the centers c, it is found that the incenter cs coincides with the circumcenter of the triangle. Here there are two such triangles, formed by omitting the single lines 3 and 4 in turn. It follows that the two separate points noticed above are the circumcenters of those triangles, and in passing to the case of Figure 1, the four points c5 which do not lie on the rectangular net are found to be the circumcenters of the four triangles of the original quadrilateral. They each count for 3 of the cs, making up with the other 16 the total of 28. It is well known that the circumcenters of a quadrilateral lie on a circle;1 in this particular case the node or Clifford point is also on this circle. 1 C5 1 Steiner, Gesammelte Werke, Band 1, p. 323. That the other four centers in Figure 3 are rectangular, may be proved from the fact that all the circles of which they are centers pass through the node. Through the node there are a pair of natural rectangular axes,1 indicated in the figure, to which the rectangle is parallel. These axes govern the grouping of the circles in Figure 2. be incenters of the four triangles whose centers have been noted as on a circle. To prove this, draw any of the circles C6, and note that the four points c5 corresponding to each of the triangles of the inscribed quadrilateral are mutually orthocentric.2 The rather pretty configuration resulting from the consideration of the quadrilateral as a special case in the theory of directed lines suggests the investigation of other cases. The centers c4 of three lines counted both ways, for example, reduce to the nine point arrangement on a circle of Feuerbach. Similar configurations will hold for five or more lines, but the complication will increase. In the general treatment the analytical method of Professor Morley's memoir (l.c.), (which is followed in the paper by Dr. Loud (l.c.)), involves much less fatigue in the transference of thought than does the geometrical argument as here given. 1 These are called the incentric lines by Clawson, l.c., p. 246. 2 For a study of the properties of an orthocentric group of points see the article by Professor Altshiller-Court in the last number of this MONTHLY, On the Orthocentric Quadrilateral, pp. 199-202. -EDITOR. QUESTIONS AND DISCUSSIONS. EDITED BY W. A. HURWITZ, Cornell University, Ithaca, N. Y. NEW QUESTIONS. The following interesting questions were formulated by Professor S. Kakeya at the request of the editors, and transmitted through Professor Birkhoff. Professor Kakeya has treated some problems of similar nature in various papers in the Science Reports of the Tôhoku Imperial University. They involve considerations of decided difficulty in the analytic formulation of geometric relations. 39. There are certain problems in geometry which are simple in statement but can be reduced only to very complicated problems in transcendental analysis. Following are several examples of the type of problem in question. 1. What is the smallest plane area within which a given figure can be turned through a complete revolution? It is not implied that the figure should revolve about a fixed point, but merely that in the course of its motion the figure should have every possible orientation in the plane. The problem may be modified by considering only convex areas. An interesting special case is that in which the given figure is a segment of a straight line. In this case it has been conjectured by Professors Osgood and Kubota that the smallest area may be bounded by a three-cusped hypocycloid; if we consider only convex areas, perhaps the result will be an equilateral triangle. I have no indication of a proof. 2. For every closed convex curve of area P there is an n-sided circumscribed polygon of least area Q and an n-sided inscribed polygon of greatest area R. For a fixed value of the integer n and for all closed convex curves, what is the upper limit of Q/P and what is the lower limit of R/P? I have succeeded only in proving that for the case n = 3, the upper limit of Q/P is 2. 3. Let the area of a given simple closed curve A be a. Remove from A the greatest possible area a similar to another given simple closed curve B. From the remaining figure remove the greatest possible area a2 similar to B. Continue this process indefinitely. Is it or is it not true that I have proved the statement to be true in the special case that A is convex and B is a circle. 4. Let a given closed convex curve K have the property that a given triangle whose angles are incommensurable with can be revolved completely within K (see part 1 of this question), always remaining inscribed in K. What may the curve K be? Can any other curve except a circle satisfy the conditions? DISCUSSIONS. For a special type of quintic equation whose coefficients involve two parameters, Mr. C. B. Haldeman, in the first discussion following, obtains an algebraic solution by means of elementary operations and the extraction of roots; he also derives expressions for the roots in trigonometric form, and gives a graphical construction by the use of a certain cubic curve and a circle. Professor Trevor, who has previously given instances of the use of mathematics in thermodynamics (1919, 444-447; 1920, 55-57), presents an interesting set of results in connection with the study of a state of thermodynamic equilibrium in a homogeneous fluid. Several transformations of independent variables in a fundamental relation yield corresponding interpretations in the physical problem. Professor Ransom gives a simple formula which may be used as a check in connection with the ambiguous case in the solution of plane triangles. Of course it applies only to problems which yield two solutions. Apparently there is a similar formula for the analogous case of spherical triangles: tan (c' + c') = and another for the dual case of ambiguity. tan b cos A, Professor Dick discusses a relationship between the dimensions of the "King's Chamber" in the Great Pyramid, the regular pentagon, and the regular icosahedron and dodecahedron. The implication of a connection between this relationship, the occurrence of the numbers of Fibonacci in plant life, and certain mystic statements of the ancient mathematicians will perhaps seem to most readers to be somewhat far-fetched. It should also be stated that the relation between a regular pentagon and the inscription of a square in a semicircle is clearly implied by the usual construction. I. RESOLUTION OF A CERTAIN QUINTIC EQUATION AND A GEOMETRICAL CONSTRUCTION FOR ITS ROOTS. By C. B. HALDEMAN, Ross, Butler County, Ohio. will reduce the equation y5+ 5ay3 + 5a2y + 2b = 0 to 10 + 2bx5 — a3 = 0; from which we find x, and consequently If b2 + a5 be negative it appears the roots of the given equation are trigonometrical functions of the coefficients; for the transformation represents a real trident of Newton1 when b2 + a3 is negative, and is the equation of a circle, which will be real when a is negative. from (1) by means of (2), the result may be placed under the form (y3 + 5ay3 + 5a2y + 2b) (a2y + 2b) = 0. (1) (2) Eliminating x The first of these factors is identical with the quintic expression given above and has five real roots when b2 + a is negative. From this it appears the five real roots may be represented by the ordinates of the intersections of a trident and a circle. The five intersections, whose ordinates are the five real roots of this equation, are the vertices of a regular pentagon inscribed in a circle whose radius is 2 √— a, as may be seen by reference to the above values of y. II. CERTAIN MATHEMATICAL FEATURES OF THERMODYNAMICS. By J. E. TREVOR, Cornell University. Let e, v, s denote the energy, volume, and entropy of unit mass of a body of homogeneous fluid in a state of thermodynamic equilibrium under the pressure p at the temperature 0. The energy e is then a continuous function of v, s, and equilibrium subsists when and only when. The state of equilibrium is stable, with respect to small displacements, when 1 One of the four canonical forms (no. 108) to which Newton has reduced the general equation of cubic curves in his Enumeratio linearum tertii ordinis, first printed in Newton's Optics, London, 1704.-EDITOR. |