THE MATHEMATICAL MONTHLY. Vol. II.... AUGUST, 1860.... No. XI. PRIZE PROBLEMS FOR STUDENTS. I. PROVE that an equation of the second degree cannot have more than two roots. II. The sum of n numbers in arithmetical progression is a, and the sum of their squares is b. Required, the first term and the common difference. Communicated by ARTEMAS MARTIN, Franklin, Pa. -- III. Given, three lines, a, b, c, drawn from any point within a regular polygon of n sides to any three of its consecutive corners, A, B, C. Required, the side of the polygon, and also a geometrical solution, granting that a polygon of n sides can be constructed when one side is given. - Communicated by PROF. C. A. YOUNG, Western Reserve College, Hudson, Ohio. IV. Deduce the formulæ of Arithmetical and Geometrical Progressions by the method of differences.- Communicated by PROF. HENRY H. WHITE, Kentucky University, Harrodsburg. V. If three points on the surface of a sphere (radius =1) be joined by three arcs of small circles whose angular radii are g,g', g′′, the area of the included triangle will be 0 + 0 + 0" πpcot q'cot g'±g" cot g′′; ୧ in which q, q', q" are the lengths of these arcs, and 0, 0′, 0′′, the angles at which they intersect. Communicated by ASHER B. EVAns. Solutions of these problems must be received by October 1. 1860. REPORT OF THE JUDGES UPON THE SOLUTIONS OF THE PRIZE PROBLEMS IN No. VIII., Vol. II. THE first Prize is awarded to G. B. HICKS, Cleveland, Ohio. The second prize is awarded to ASHER B. EVANS, Hamilton, N. Y. The third prize is awarded to J. G. WEINBERGER, Millersville, Pa. PRIZE SOLUTION OF PROBLEM I. By J. G. WEINBERGER, Normal School, Millersville, Pa. I. If x is a whole number, prove that - is always divisible by 6. Since 23 x= (x − 1) x (x + 1), it appears that when x is a whole number, the given expression is the product of three consecutive whole numbers. But we know that at least one of three successive whole numbers is divisible by 2, because at least one must be even; and also, that one of them must be divisible by 3, because if the first one is not divisible there must be a remainder of either 1 or 2; and hence it follows that either the second or third number will be divisible. Therefore 3-x is divisible by 2.36. 23. PRIZE SOLUTION OF PROBLEM II. By D. G. BINGHAM, Ellicotteville, N. Y. II. Describe a series of circles whose areas shall be one half, one fourth, one eighth, &c. that of a given circle. Draw from C, the centre of the given circle, two radii, C A and CB, perpendicular to each other, and draw A B, the chord of 90°. From C drop a perpendicular, CK, upon this chord, and with C as a centre, and radius CK, describe a circle; then, as is easily seen, (1) CA2 CK2 = 2 : 1. But since the areas of two circles are to each other as the squares of their radii, it follows from (1) that the area of the circle CK is half the circle CA. In precisely the same way we can find a circle which is half of the circle CK, and therefore one fourth of circle CA. The process may be continued as far as we please. PRIZE SOLUTION OF PROBLEM III. By GEORGE B. HICKS, Cleveland, Ohio. III. If R and r be the radii of the circles circumscribing and inscribing any triangle, and D the distance between their centres, then D2=R2-2 Rr. Required, a geometrical demonstration. Let O be the centre of the circumscribed, and 0 of the inscribed circles. Produce A O indefinitely, and O O to meet the circumscribed circle in H and I. Draw CG, CE, EB. Lastly, from E as centre, with radius E O, describe the circle O NF M. This circle will pass through B and C. For ECO is measured by (GB+BE) and EOC by (AG+EC). But GB+BE AGE C. Hence the triangle EOC is isosceles, and ECO E. Similarly B E O E. Also, since AE bisects BA C, we have The chords HI, AE give HOX0I=(R+ D) (R — D) = A0 × OE. But it is shown in Geometry that (R+ D) (R— D) = R2 — D2 ; It remains to show that A OX OE=2 Rr. And from the secants AF, A C, AO X AF = AC' × A C. But since E is on the bisector of the angle BA C, AC' AB; (2) .. AO X AF ABX A C. = Multiply (1) and (2), and AO XO E = since each side of this ex pression represents double the area of AB C'; as required. АВХАСХ ВС =AO XOE = 2 Rr, PRIZE SOLUTION OF PROBLEM IV. By GEORGE B. HICKS, Cleveland, Ohio. IV. If a circle cut a conic section in four points, A, B, C, D, and a second circle cut the same conic section in A, B, E, F, then will CD and EF be parallel. Let y2-mx-na2=S=0 be the equation of any conic, and x cos ay sina-p=a= 0, x cos b+ y sin b— p=ß=0 (1), the equations of the chords AB, CD. ß =0 Then will S+ ka ẞ= 0 (2) be the equation of some figure through the intersections of (1) with the conic, the species of (2) being determined by the value attributed to the arbitrary constant k; for (2) is satisfied by S=0, a=0; S= 0, ẞ= 0, and therefore passes through the intersections of S, a, and B. Sections, p. 211.) S=0, ß (SALMON'S Conic To determine k so that (2) may represent a circle, write for S, a, and ẞ their values at full length, equate the coefficients of 22 and y2, and equate to zero the cofficient of x y. 1+nk cos (a+b), We get k sin (a+b)=0. Now a and b are the angles made with the axis of x by the perpendiculars from the origin on the lines AB, CD. Hence A B and CD are equally inclined to the axis of the conic (this being the axis of x). So also are A B, EF, since the property just proved is evidently general. And therefore CD, EF make equal angles with the axis; .. they are parallel. Mr. TROWBRIDGE adds to his solution of this problem the following: "If two conics of the same kind cut a third conic in the points A, B, C, D and A, B, E, F respectively, then will CD and EF be parallel, provided the transverse axes of the cutting conics are either parallel or perpendicular to each other." PRIZE SOLUTION OF PROBLEM V. By ASHER B. EVANS, Madison University, Hamilton, N. Y. V. In any conic section, let u denote the length of the perpendicular dropped from any point in the curve upon the directrix, r the distance of this point from the nearest focus, a and b the semi-axes; prove that A conic section is the locus of a point whose distances from a given fixed point, called the focus, and a straight line given in position, called the directrix, are always to one another in a constant ratio. This condition gives ru :: e: 1, or r2 = u2 2, where e is the eccentricity of the conic section. Therefore 2=1—2=6 for . u2 for the hyperbola, and zero for the parabola, |