THE MATHEMATICAL MONTHLY. Vol. II.... JANUARY, 1860.... No. IV. PRIZE PROBLEMS FOR STUDENTS. I. A and B can do a piece of work in m days; B and C in n days; in what time can A and C do the same, it being supposed that A can do p times as much as B in a given time? III. Find the roots of the equation -6x=4, by trigonometry. IV. Four persons, A, B, C, D, in order, beginning with A, cut a pack of cards, replacing them after each cut, on condition that the first who cuts a heart shall win. What are their respective probabilities of success? V. The notation of Problem V. in the November number of the Monthly being retained, prove that in the plane cot A+ cot B+ cot and in the sphere C=cot A cot B cot C; cot 4+cot B+cot C=cot A cot B cot Cos (p+8) cos (p −8) & C cos2r cosp The solutions of these problems must be received by March 1, REPORT OF THE JUDGES UPON THE SOLUTIONS OF THE PRIZE PROBLEMS IN No. I., Vol. II. The first Prize is awarded to J. D. VAN BUREN, Rensselaer Polytechnic Institute, Troy, N. Y. The second Prize is awarded to O. B. WHEELER, Sophomore class, University of Michigan, at Ann Arbor. The third Prize is awarded to GEORGE H. TOWER, Classical High School, Petersham, Mass. PRIZE SOLUTION OF PROBLEM II. By PERRIN B. PAGE, Nunda, N. Y. In any plane triangle, prove that the sines of the angles are inversely as the perpendiculars let fall from them upon the opposite sides. Let A B C be the triangle, and denote by a, b, c, the perpendiculars dropped respectively from the angles A, B, C. By trigonometry By geometry, twice the area of A ABC is (2.) BCX a ACX b= ABX c. = Multiplying equations (1) and (2) together, member by member, ... sin A : sin Bb: a BC: AC, which not only proves the proposition, but also that the sines of the angles are proportional to their opposite sides. This is a slight modification of W. C. HENCK'S Solution. PRIZE SOLUTION OF PROBLEM III. By ISAAC H. TURRELL, Mt. Carmel, Indiana. Having given the diagonals of a quadrilateral inscribed in a given circle to determine its sides geometrically, when the diagonals intersect each other at right angles. Communicated by Professor D. W. HOYT. Let AB and CD be the given diagonals. Inscribe AB in the given circle. Next, draw parallel to AB and at a distance from the centre equal to one half the other diagonal CD, the line CC. From either extremity of this line, as C, draw a line perpendicular to AB and produce it till it meets the circumference in D. The line CD is therefore inscribed in the circle, perpendicular to AB, and A CBD is the required quadrilateral. CHARLES B. BOUTELLE'S solution is essentially the same as the above. Adding and subtracting (1) and (3), (1) and (2), and also multiplying (3) and (2) together, we obtain (9.) y z (x2 + w2) + x w (z2 + y2) = a b ď2. Combining the products of (5) and (6), and (7) and (8), with (9) and (4) respectively, we get , x2 — y2 = ± b√ď2 — d2, y2—w2 =±a√b2 — ď3‚ x2 + w2 = d2 = y2+2. Therefore, 2x2 = d2±b √ d2 — a2 ± a √ b2 — d2, Whence we obtain the values of the required quantities by dividing by 2 and extracting the square root. Similar solutions were given by WILLIAM W. JOHNSON and O. B. WHEELER. O. B. WHEELER remarks that the equations of Problem IV. are readily obtained from the figure of Problem III. From geometry we have, "The rectangle of the two diagonals is equivalent to the sum of the rectangles of the opposite sides," which is (1). We have (2) and (3) by the proposition, "In any triangle the rectangle contained by two sides is equivalent to the rectangle contained by the diameter of the circumscribed circle and the perpendicular let fall on the third side;" and (4) from the sum of the equations The connection between Problems III. and IV. was also noticed by W. C. CLEVELAND, who gave the following geometrical demonstration of (4). Let the chord CC be drawn parallel to AB; then A CBC =x, angle C' CD 90°; .. C'D is a diameter; ··· x2 + w2 = d2. Similarly y2+2 = d2; whence equation (4). PRIZE SOLUTION OF PROBLEM V. By J. D. VAN BUREN, Rensselaer Polytechnic Institute, Troy, N. Y. If, in a plane or spherical triangle, A, B, C denote the angles, and a, b, c the opposite sides respectively; if r, e denote the radii of the circumscribed and inscribed circles, and the distance between the centres of these circles; then in the plane triangle and in the spherical triangle sin A+ sin B+ sin C 4 cos A cos B cos C C, cos & cos r cos p Spherical Triangle. We have, from spherical trigonometry, |