THE MATHEMATICAL MONTHLY. Vol. II.... APRIL, 1860.... No. VII. PRIZE PROBLEMS FOR STUDENTS. I. PROVE that an arithmetic mean is greater than a geometric. II. Let three bodies with velocities V, V, V", move uniformly in the same direction, in the circumference of a circle. Required the time of their conjunction, supposing them to quit a given point at the same time. III. The diameter of a circle inscribed in the quadrant of a second circle is equal to the side of the regular octagon circumscribed about the second circle. - Communicated by Prof. HENRY H. WHITE. IV. Required the locus of the centres of the circles inscribed within all the right-angled triangles which can be inscribed in a given semicircle. - Communicated by Prof. HENRY H. WHITE. V. From a box containing a very large number of white and black balls, of each an equal number, three balls are taken at random and placed in a bag without being seen. A takes a ball at random from the bag, observes its color, and replaces it four times in succession. The ball was white on each of the four drawings. What are the respective probabilities that the bag contains 1, 2, or 3 white balls? Solutions of these problems must be received by June 1, 1860. REPORT OF THE JUDGES UPON THE SOLUTIONS OF THE PRIZE PROBLEMS IN No. IV., Vol. II. THE first Prize is awarded to W. F. OSBORNE, of the Wesleyan University, Middletown, Ct. The second Prize is awarded to M. K. BOSWORTH, of the Sophomore class, Marietta College, Ohio. The third Prize is awarded to E. S. STEARNS, of Chester Institute, Chester, Morris Co., N. J. By W. F. OSBORNE, Wesleyan University, Middletown, Ct. Find the roots of the equation 3-6x= 4, by trigonometry. (1) By trigonometry, sin 30= 3 sin 0-4 sin3 0; whence Substituting in the given equation x = sin 0, it becomes in which any arbitrary value my be assigned to g. Hence, if we 6 put = 3, or § = √ 8, by comparing (1) and (2) we shall have g Hence, sin ✪ = sin (2,”+23), from which all the different 12 values of sin may be obtained by making n = 0 or 1. Therefore the three values of x are —√ 8 sin 15° = 1−√3, √8 cos 15°=1+√3, PRIZE SOLUTION OF PROBLEM IV. V8 cos 45°-2. By W. F. OSBORNE, Wesleyan University, Middletown, Ct. 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? first trial is evi The probability that a heart will be cut on the dently, and against it; on the second trialX against it; on the third trialX X for, and it; and so on to the nth trial, on which the probability that a heart for, and X XX against But A has the first, fifth, ninth, &c. cuts; hence his probability of success is 1 64 ¥ (1 + (})* + (} )® + &c.) = ‡ • 1 = (@)' = ft· (2)* B has the second, sixth, &c. cuts, and hence his probability of success is are 4 . ‡ (1 + (4)* + (†)® + &c.) = ‡ • £ • 1 − (®) • 1 4 In precisely the same manner C's and D's probabilities of success PRIZE SOLUTION OF PROBLEM V. By ASHER B. EVANS, Madison University, Hamilton, N. Y. The notation of Problem V. in the November number of the Monthly being retained, prove that in the plane (1) cot Acot B+ cot C=cot A cot B cot C'; and in the sphere (2) cot A+ cot B+cot C=cot A cot B cot C in which s = · § (a + b + c), n2 — s (s — a) (s — b) (s — c). s = { (a + b + c), n2 = sin s sin (s — a) sin (s — b) sin (s — c). (4) ... cot A+cot B+cot C= But we have (CHAUVENET'S Trig. Eqs. 307, 319, 332) ; -1+tan2r). |