In a problem proposed for solution Cesàro gave yet another form of expression3 for the ratio of the areas: "Three lines drawn through the vertices of a triangle determine on the opposite sides six segment such that the difference between the product of three non-consecutive segments and the product of the three others is In this expression a', b', c' denote the sides of the triangle formed by the three lines; l, m, n the segments of these lines, contained between the vertices and the sides of the first triangle." 4 The methods of solution of our problem by Steiner and Cesàro are those of elementary geometry. As Cesaro's solution is in a source inaccessible to many MONTHLY readers it is herewith reproduced: Let us seek first the ratios of the segments determined on each of the lines AD, BE, CF by the two others. The triangles ADC, ADB are cut respectively by the transversals BE, CF, giving 2 See also Steiner's Gesammelte Werke, Volume 1, 1881, p. 166; slips in the result as printed in Crelle's Journal are here corrected. 3 Nouvelle correspondance mathématique, Volume 6, March, 1880, pp. 143–144, Question 545. 4 Neuberg seemed to be the first to show (Nouvelle correspondance mathématique, Oct., 1880, Volume 6, p. 472) that where AF/FB = C1/C2, BD/DC = a1/a2, CE/EA = b1/b2, and A' is the area of the triangle DEF. This result was also stated by Gentry in Nouvelles annales de mathématiques, November, 1880, Vol. 39 (2), p. 528. 340 (Calculus) [1913, 196; 1919, 213]. Proposed by C. N. SCHMALL, New York, N. Y. A pencil of parallel rays of light is incident upon a lens whose faces have the radii r1, r2, respectively. Show that the distance of the principal focus from the center of the first face of the lens will be a maximum or a minimum when SOLUTION BY H. S. UHLER, Yale University. It is necessary to remark at the very beginning that the given result appears incorrect as it does not involve the thickness of the lens, and the statement of the problem is ambiguous since it does not specify whether the first or the second principal focus is meant. By straightforward analysis, involving Snell's law, but too long to deserve publication in this place, I found x = t + r2 μr 22 (u − 1)[μ(rı + r2) (μ - 1)t]' (1) where, to fix the ideas, a double convex lens was contemplated, x = distance from the center of the first (or incidence) face to the second principal focus (the focus beyond the emergence face), t = axial thickness of lens, μ = index of refraction of the material (glass) of the lens with respect to that of the surrounding medium (air), and r1 and r2 denote the arithmetical values of the radii of curvature of the first and second faces of the lens, respectively. Formula (1) may be checked by referring to James P. C. Southall's The Principles and Methods of Geometrical Optics, either edition, page 275, equation (170); and by changing his d, n, and r2 to t, u, and r2, respectively. Now what quantity is to be the independent variable? Let it be assumed that t is to vary while μ, 71, and T2 remain constant. Then the necessary condition for a maximum or a minimum is which leads directly to dx dt = t 1 μr 22 Lu(r1 + r2) (μ = 0, To determine which sign has physical meaning we may proceed as follows: x be negative, otherwise the principal focus will fall inside the material of the lens. in (1) the expression for (μ — 1) t given by (2) we find (2) t must not Substituting according as the upper or lower sign in (2) be taken. Since, under all ordinary conditions, μ > 1 the upper sign alone is acceptable. Since this result is positive the thickness given by (2) corresponds to a minimum focal length, as might be anticipated from purely physical considerations. It is thus clear that a reasonable interpretation of the statement of the problem leads to an analytical condition which is quite different from the given answer. The thickness of the lens cannot be considered negligible (as is often done in very elementary discussions of problems in geometrical optics) because the numerator of (2) cannot vanish when the upper sign is taken. In other words, t cannot be avoided in the final condition. In conclusion, attention may be called to the fact that each of dx/ər1 = = == 0, O leads to a result which bears no relation to the given answer and hence t is the only sensible independent variable. 349 (Calculus) [1913, 312]. Proposed by C. N. SCHMALL, New York City. If y = a cos (log x) + b sin (log x), eliminate the constants a and b and obtain the equation SOLUTION BY T. E. MERGENDAHL, Tufts College. Taking the derivative of y, as given in the problem, and multiplying the result by x, we have Applying the same process to both sides of this equation, we find Also solved by W. W. BEMAN, T. M. BLAKSLEE, P. J. DA CUNHA, H. T. DAVIS, H. H. DOWNING, G. H. GRAVES, P. HANSEN, A. M. HARDING, W. L. MISER, A. PELLETIER, ELIJAH SWIFT, and H. S. UHLER. 385 (Calculus) [1915, 161; 1919, 73]. Proposed by H. B. PHILLIPS, Massachusetts Institute of Technology. If f(x) is continuous between a and x, show that T. H. GRONWALL, New York City, offers the following criticism and completion of the solution already published. First, the passage to the limit for n = ∞, which forms the last step (p. 74), requires justification (this is however easily done by using Taylor's theorem with remainder term). Second, the problem as proposed assumes only that f(x) is continuous, so that the use of the derivatives of f(x) is not permissible, since they may not exist. The following proof requires f(x) to be bounded in the interval from a to x and to be continuous only at the point a, but not in the whole interval. It is readily shown (integration by parts and complete induction) that writing & = a + t(x − a) and ƒ(§) = ❤(t), the formula to be proved becomes 1 A proof of this relation may be found in Goursat-Hedrick's Mathematical Analysis, Vol. 2, pt. II, p. 36.-EDITORS. (under the assumption that (t) is continuous at t = 0), or what is the same Let > 0 be as small as we please; (t) being continuous at t = O, it is then possible to choose a ♪, 0 < 8 < 1, and so small that |❤(t) — ❤(0)| < e/2 for 0≤t≤8. Let M be greater than (t) − y(0)| for 0 ≤t≤1. Then, decomposing the integral into two parts and using the first mean value theorem, Now N may be chosen so large that M(1 — 8)" < e/2 for n > N, so that finally 192 (Number Theory) [1913, 196; 1919, 214]. Proposed by the late ARTEMAS MARTIN. Find rational values of v, w, and x that will satisfy simultaneously the conditions (m2 + n2) (v2 + w2 + x2)2 — 4m2n2 v2 + m2n2 (m2 + n2) = 0, (m2 + n2) (v2 + w2 + x2)2 — 4m2n2w2 + m2n2(m2 + n2) = 0, (m2 + n2) (v2 + w2 + x2)2 — 4m2n2x2 + m2n2(m2 + n2) = 0, m and n being known rational quantities. DISCUSSION BY H. S. UHLER, Yale University. The following analysis shows that the problem is impossible. Any set of values of v, w, and x that fulfil the given conditions must also satisfy the sum of the three equations, that is, must satisfy 3(m2 + n2) (v2 + w2 + x2)2 — 4m2n2 (v2 + w2 + x2) + 3m2n2 (m2 + n2) = 0. In general, the right hand member is irrational, and the sum of the squares of any number of rational numbers cannot have an irrational value; therefore, the problem is impossible. = We may proceed farther and find explicit expressions for v, w, and x. By inspection, or by forming the differences between the given conditions, taken in pairs, we see that v2 w2 = x2. Consequently, Proposed by E. L. REES, University of Kentucky. 2734 [1918, 444]. Given two circles tangent to each other externally. From the extremity of a diameter through the point of tangency draw a secant such that the segment between the circles shall be equal to a given segment. F E B SOLUTION BY THE PROPOSER. = == d', x, AB d, AC Then we have, y2 = Let PB be the required position of the secant, assuming the problem having been solved. Draw PQ perpendicular to AC and let BP = y, AQ and RP = a, the given segment. (d + x)2 + x(d' - x). Also (y — a)/d (d + x)/y. Eliminating x, we get y22a'y d2 = 0, where a' a(2d + d')/2(d+d'). Hence, y = a' ± √a12 + ď2. = = The analysis suggests the following construction: With B as a center, BC as a radius, describe the arc cutting the horizontal diameter produced of the lower circle in D. Draw BD cutting the common tangent in E. Take EF = a. Draw FG perpendicular to AC. Then AG will equal a'. Make AH = AG. Draw BH and produce it to J, making HJ AH. With B as a center and BJ as radius, describe an arc cutting the lower circle in P. Then BP is the required secant. Also solved by P. J. DA CUNHA. 2741 [1919, 35]. Proposed by H. L. OLSON, New Hampshire College. = Prove or disprove the following statement: If the three sides and the area of a triangle are integers, at least one of the three altitudes is an integer. I. SOLUTION BY FRANK IRWIN, University of California. The statement is not true since the triangle whose sides are 5, 29, 30 has area 72, and altitudes 144/5, 144/29, and 24/5. II. REMARKS BY J. L. RILEY, Stephenville, Texas. On page 12 of Carmichael's Diophantine Analysis we find the following theorem: A necessary and sufficient condition that rational numbers x, y, z shall represent the sides of a rational triangle is that they shall be proportional to numbers of the form n(m2 + h2), m(n2 + h2), (m + n)(mn − h2), where m, n, h are positive rational numbers and mn > h2. 2= In deriving this result it is pointed out that if x = n(m2 + h2), y = m(n2 + h2), and (m + n)(mn — h2), the area is hmn(m + n)(mn — h2) and the altitude upon z is 2hmn. If m, n, and h are integers (mn > h2) we have a series of triangles for which the statement of Mr. Olson is true. 2742 [1919, 36]. Proposed by C. N. SCHMALL, New York City. In Gregory's Examples in the Differential and Integral Calculus, 1841, Chap. VII, p. 124, ex. 22, I find the following celebrated problem: "To find a point within a triangle from which if lines be drawn to the angular points their sum may be the least possible." The author remarks that "the direct solution of this problem is long and complicated, etc." Required a simple, brief solution. I. SOLUTION BY F. V. MORLEY, Johns Hopkins University. Consider the three points, a, ß, y (complex variable) as on a unit or base circle. Then th sum of the distances from the point x is |