II. If a circle be described touching the base of a triangle and the sides produced, and a second circle be inscribed in the triangle; prove that the points where the circles touch the base are equidistant from its extremities, and that the distance between the points where they touch either of the sides is equal to the base. III. Inscribe the maximum rectangle between the conchoid and its directrix. Communicated by Prof. DANIEL KIRKWOOD. IV. Given a cask containing a gallons of wine. Through a cock at the bottom of the cask wine flows out at the rate of b gallons per minute, and through a hole at the top water flows in at the same rate. Supposing the water, as fast as it flows in, to mingle perfectly with the wine, how long before the quantities of wine and water in the cask will be equal? and how much wine will be left in the cask at the end of t minutes? - Communicated by Prof. C. A. YOUNG. V. Two circles being given in a plane, find geometrically the locus of the points from which chords of similar arcs in the two circles will be seen under the same angle, the chords being perpendicular to the lines of vision drawn through the centres of the given circles. - Communicated by Prof. WM. CHAUVENET. The solutions of these Problems must be received by the 1st of May, 1860. REPORT OF THE JUDGES UPON THE SOLUTIONS OF THE PRIZE PROBLEMS IN No. III., Vol. II. THE first Prize is awarded to JOHN Q. HOLLISTON, Sophomore Class, Hamilton College, Clinton, N. Y. The second Prize is awarded to F. E. TOWER, Senior Class, Amherst College, Amherst, Mass. The third Prize is awarded to FRANK N. DEVEREUX, Boston, Mass. PRIZE SOLUTION OF PROBLEM I. By FRANK N. DEVEREUX, Boston, Mass. If two circles touch each other, any straight line passing through the point of contact cuts off similar parts of their circumferences. The line joining the centres C and C will pass through the point of contact B. Let A and A' be the points in which the line passing through the point of contact meets the circumferences. Join A and C, A' and C'. The triangles A B C and A'B C' thus formed are isosceles and similar, as is easily seen. The angles at the centres, A CB and A' C' B, are therefore equal, and are measured by similar parts of the circumferences. Hence the proposition is true. This proposition applies whether the circles are tangent externally or internally, a fact not noticed by any of the competitors. PRIZE SOLUTION OF PROBLEM II. Find the four roots of the recurring equation x1 — § x2 + 2x2 — § x + 1 = 0. Dividing the given equation by 22 it becomes. ... x2+1=0; or x=±√−1. ±√−1. ··· x2 All the competitors gave one or the other of the above solutions. 2 cos n 0 = u” + ; and then find the sum of the series, cos + cos 20+ cos 30 un Denoting by 2, the sums relatively to n, we have since u +u2 + &c., and ++ &c., are geometrical series with ratios u and REMARK. น И u2 By dividing the terms of the first fraction, and multi plying those of the second, by ut, we get This form of the result was not given by any of the competitors. PRIZE SOLUTION OF PROBLEM IV. JOHN Q. HOLLISTON, Hamilton College, Clinton, N. Y. Having given the Right Ascensions and Declinations of two stars, to find the formula for the distance between them. Also, find what the distance becomes, when for one star A. R. is 8h 12m 38.17, and Dec. 17° 23′ 49′′.8 north, and for the other A. R. is 13h 28m 19.92, and Dec. 21° 12′ 37′′.2 south. Let and be the declinations of the two stars; then 90° and 90° will be their co-declinations; and since the right ascensions are measured on the equator, their difference will measure the angle at the pole made by the meridians passing through the two stars. Let H denote this angle, and the distance sought. We have then a spherical triangle in which the sides 90° and 90° - include the angle H, and 4, the third side, may be found from the formula cos (1) = cos (90° — 8) cos (90°—♪)+ sin (90° —♪) sin (90° —♪) cos H, = sin 8 sin &+ cos & cos & cos H, (2) sin 8 cos & (sin+cot 8 cos H). = Assume Acos d' Formulas (3) and (4) need only tables of logarithmic sines, cosines, &c.; and we find 86° 24′ 12′′.2. PRIZE SOLUTION OF PROBLEM V. By ASHER B. EVANS, Madison University, Hamilton, N. Y. In a frustum of any pyramid or cone, the area of a section, parallel to the two bases and equidistant from them, is the arithmetical mean of the arithmetical and geometrical means of the areas of the two bases. The three sections are evidently similar figures; hence their areas will be as the squares of any homologous lines. Let x, xy, and x+y, represent any homologous lines in the upper base, middle section, and lower base, respectively. Then their areas may be represented by m(x − y)2, m x2, m (x + y)2, respectively. The arithmetical mean of the two bases is m (x − y)2 + m (x + y)2 their geometrical mean is 2 = m (x2 + y2); √ m2 (x − y)2 (x + y)2 = m (x2 — y2); and the arithmetical mean of m (x2 + y2) and m (x2 — y2) is m x2, as was to be shown. SIMON NEWCOMB. W. P. G. BARTLETT. TRUMAN HENRY SAFFORD. |