63.-Given the sides AB, AC and BC, and the angles a and B, (see diagram below), to find the sides AD, BD and CD. SOLUTION BY W. W. BEMAN, UNIVERSITY OF MICH., ANN ARBOR, MICH. Conctruction. Construct the triangle ABC. Upon AB construct a segment containing an angle of 24°, and upon BC a segment containing an angle of 16°. The intersection of these arcs will determine the point D, the distance of which from A, B and C can now be measured. It is evident that there will be two solutions, one when B and D are on the same side of AC, and another when they are on opposite sides. Trigonometrical Computation. Designate the angles and sides as in the figure. Then a++ x + y +B=360°, or A tan (x + y) = tan}(−y) +1 1 =tan (+7). y. c sin ABD sin a Hence we can find x and = 15.17 ms. When B and D are on the same side of AC, x + y = B− (a + B), but the remainder of the work is the same. This method of computation is known as Delambre's. 64.-"Find the maximum value of without the aid of the Calculus". SOLUTION BY ARTEMAS MARTIN, ERIE, PA. The given function will be a maximum when its Napierian logarithm is a maximum. .. u=x log a -x log x, and = (x + h) log a − (x + h) log x − ( x + h) (4) 124 x+h The coefficient of h is log a-log xlog x-1, which put 1, which put = 0 and we have ... the maximum value of the given function is e. [This question may be solved by a simpler process as follows: The given expression in logarithms is a log a-x log xa max.,= log m. log a log m; whence a am, and the original x = logm; x = log m Dividing by a, log a a = a max., or a. m maximum when m is expression becomes m" the base of the system. Therefore me, and the 65. "The corner of a page is turned down, and in every position the area of the triangle is two square inches; find the locus of the angular point." SOLUTION BY WALTER SIVERLY, OIL CITY, PA. Let r and be the polar coordinates of the locus referred to the corner before it is turned down and one edge of the leaf. = 4, or p2 = 16 sin The double area of the cos 0 = 8 sin 20 is the triangle = r(tan 0+cot 0) 66.-"A speaks the truth b times out of a; B, d times out of c; and C, n times out of m. C says that B told him that A said a certain event transpired. Required the probability that the event occurred." [Artemas Martin gets for answer to this question, and for authority he refers to Todhunter's History of the Theory of Probability, p. 462; and Todhunter's Algebra, 4th edition, p. 464. E. S. Farrow gets the same result as Mr. Martin. All the other correspondents give for the required chance bdn ÷ acm. If we assume that all the witnesses did actually testify in relation to the event, then we think there is no doubt that Mr Martin's result is correct: For, let it be assumed that the event did in fact occur; and suppose, in a given statement, that C tells the truth, and that B, in informing C, told a lie. Then B must have told C that A said the event occurred, therefore, because B lied, A must have told B that the event did not occur; but by hypothesis the event did occur, therefore A lied. Hence if C asserts that an event has occurred, and if his statement is based on the fact that B has testified to him in relation to the event, whose testimony in turn is based on A's statement in relation to the event; then it follows that, if any two in making their statements tell a lie and the other one tells the truth, the event must have occurred. We contend, however, that the question, as announced, does not warrant the inference that all three necessarily testified in relation to the event; and that when C says that B said that A said a certain thing, if Clies, he may either lie in relation to the character of the statement which he reports, or he may lie in the affirmation that B made a statement; and there is no reason assigned from which we can infer that he would be more likely to lie in respect to the character of the statement than in relation to the fact of the statement. But if we admit that C lied in attributing a statement to B, when no statement of any kind in relation to the event was made by B; und that B in like manner lied in attributing a statement to A, when no statement was made by A; then it is clear that the chance in favor of the event resulting from the concurrence of C and B in a lie with A in the truth, is only one fourth what is assigned to that contingency above And, for the same reason, the chance resulting from the concurrence of C and A, and B and A respectively in telling a lie, with the other one in telling the truth, is one half what is assigned to that contingency above. To warrant Mr. Martin's result we claim therefore that the statement of the question must be modified. Mr. Martin has appended to his solution of 66, a note correcting a misprint in Todhunter's History of the Theory of Probability, which we give below.-Ed.] NOTE, BY ARTEMAS MARTIN. The denominator of the formula for traditional testimony given in Todhunter's History of the Theory of Probability, p. 462, third paragraph, is wrong—a misprint probably. I will quote a portion of the second and third paragraphs: "Suppose a witness to speak truth m times and falsehood n times out of m + n times; let m' and n' have similar meanings for a second witness. Then if they agree in an assertion the probability of its truth is Using the same notation as before if one witness reports a statement from the report of another the probability of its truth is mm' + nn. for the statement is true if they both tell the truth or if they both tell a falsehood."-The latter formula should be mm' + n n' (m + n)(m' + n')' [We are compelled, for want of room, to defer publishing the solution of problem 67 until the issue of the July No.] PROBLEMS. ... 69. By E. P. NORTON, ALLEN, MICH.-Given 22 + y2 = 793, . (1) † (xy3) + † (x2y) = 30, . . . . (2) to find x and y by quadratics. = 70. BY F. P. MATZ, B. E., ED. MATH. DEP. "NATIONAL EDUCATOR", KUTZTOWN, PA.-Given y2+z2 = 2500,... (1), 22 = | (1), a2 = [130 — (x + t)]2 × 1600, ... (2), xy = t[130 — (x + t)], . . . (3), t = y2 + (z — 40)2. . (4) to find x, y and z. 71. BY WILLIAM HOOVER, BELLEFONTAINE, O.-Given the sides a, b, c of a spherical triangle ABC to find the radii, R, r, of the circumscribed and inscribed circles. 72. By A. W. MASON, CEDAR FALLS, IOWA.-Find the maximum cylinder that can be cut from a given oblate spheroid, whose semi-axes are a and b. 73. BY CHRISTINE LADD, CHELSEA, MASS.-Find the whole number of sets of three integers having a constant sum. 74. BY PROF. W. W. HENDRICKSON, U. S. Naval Acad. ANNAPOLIS, MD. Find the equation of the locus of the middle point of a chord to the hyperbola x2-y2=2a2, the chord being of constant length and equal to seven times the transverse axis. QUERY, BY PROF. A. HALL. Into how many parts can a planes divide space. QUERY, BY E. T. T. GAMBIER, O.—Is 65537 (= 216 + 1) a prime? Errata. On page 8, line 16, for "3a"" which occurs in the numerator of the fraction, read a3; and on page 58, line 4, and lines 4 and 6 from bottom, for “” in the numerator of the left member of the equation, read l. Also, on page 59, line 10, multiply the left member of the equation by sin. [Note. Several correspondents have found fault with Mr. Farrow's solution of 53, but no other solution has been offered. It is evident that the formula used by Mr. Farrow to estimate the effect of the atmospheric resistance, does not apply in calculating the velocity generated while the ball is passing from the breech to the muzzle of the gun, which is the case to be considered; for during that time the only pressure on the rear of the ball is that of the gas generated. The result obtained by Mr. Farrow cannot, therefore, be correct.] VOL. II. JULY, 1875. A NEW THEORY OF THE RULE OF SIGNS. No. 4. BY LEVI W. MEECH A. M., HARTFORD, CONN. For all equations of the third degree, the cubic type is decisive; if it does not indicate imaginary roots, there are none; as appears from comparison with Sturm's theorem. For the fourth and higher degrees, it is decisive only so far as it indicates two imaginary or equal roots; leaving the question of more of them, yet to be decided. Example. f' = - 3 × 2 × 70 ÷ 232=— -.794. 3 x 23÷22—— 17.25, g' Since g' and f' are interchangeable, we enter the foregoing Table with 0.8 nearest value of g', and there find that f' must negatively exceed - 14.42, which it does, being — 17.25, and indicating two imaginary roots. Example. x35x2+2x+12= 0. ƒ' = + 6 ÷ 52 = .24, g' - - -180 22: =- 45. Entering the table with ƒ' = 0.20 and interpolating for .04, we find for f' = 0.24, 91 52.56. As g' is within this limit, the roots are all real, and accord with the Rule of signs. The result is confirmed by computing the criterion (C) with the respective values of f' and g'; and the comparative smallness of (C), being +0.6, indicates two roots to be nearly equal. VII. The Biquadratic Type, or five consecutive terms: x+....[Px" + Qx"-1 + Ra"-2 +Sa"-3 + Ta"--4]+ Differentiating n-4 times and proceeding as with former types, also making m n = m', the final derivative takes the form of .... = : 0. |