ARTICLE XXII. SOLUTIONS of the QUESTIONS proposed in ARTICLE XXI. I. QUEST. 62. Solved by A. Rabbit, at Uncle Sam's, Downbelow. Because C's part was lost by his death, it is plain 43 the share of A: the share of that : B. Hence 7:4 : £100000 : £571425-A's share; and 7:5:: £100000 £428574-B's share. In answering this question, Shepherd first divided the given sum between A, B, and the ghost of C, in the ratios of, and; and afterwards divided the share of the ghost between A and B, in the ratio of to. This is the curious process, by which Shepherd created the fractional tails of his answers. II. QUEST. 63. Solved by Enoch Lewis, Westown, Pennsylvania. Let the required number of Royalists, Federalists, Clintonians, and Lewisites, be respectively represented by x, xy, xy2 and xy3; then by the question and the nature of differentials, we have (y3 + y2+y+1)xx = 30, and (33y+3y-1)xx = 2. From these two equations we have 33.23-1 : y3+y2+y+1:1:15; or 77323y2+22y=8; which resolved, gives y=2. Hence is easily found x=2, xy=4, xy-8, and xy3=16: consequently our 30 political sages consist of 2 Royalists, 4 Federalists, 8 Clintonians, and 16 Lewisites. P III. QUEST. 64. Solved by the proposer, Henry Smith, New-York. It is plain that the hour and minute hands of a watch form one continued straight line at the hour of six. And, since the lengths of the two hands may be considered as equal; if the length of the arc described by the hour hand in one hour, be put 1; the length of that described by the minute hand, in the same time, will be = 12; and the difference of these two arcs will be 11. Now, when this difference becomes 12, the hands will again form a continued straight line: therefore by proportion 11: 12: 1 hour: 1 hour 5 min. Further, since the angular velocities of the two hands are always invariable, the required times are evidently in arithmetic progression; and their common difference is 1 hour, 5fr min. and 4 54, the hour and minute hands of a watch are diametrically opposite to each other, or form one continued straight line. IV. QUEST. 65. Solved by Enoch Lewis. a b It is evident that + is an algebraic divisor of m m a+b; assume m=2, 3, 4, &c. and the expressions are still algebraical; whence, as the assumed value of m may be infinitely varied, the number of algebraic divisors of a+b, is infinite. The same, solved by John Capp, Harrisburg, Pennsylvania. It is easily proved, that when n denotes any odd number whatever, a+b is a divisor of a"+". Now, In In I *1 because a a and b=6" it follows that a"+b" is a divisor of a+b. Hence since n is any odd number, and an infinity of odd numbers exists, a+b has an infinite number of algebraic divisors. V. QUEST. 66. Solved by Jacob Winooski, among the Green Mountains, Vermont. This question is only a particular case of the following general PROBLEM. In a given arc of a circle, to find a point from which a straight line drawn to the middle of the chord of that arc, shall be a mean proportional between two straight lines drawn from the required point to the extremities of the same chord. Let ABC be the given arc of F a circle ABD; and AC its chord. DEMONSTRATION. H E Draw the chords GC, GA and GB; and through G and O draw a straight line to intersect the circumference of the circle in H. Make the arc GI=BG; and draw the chords IG and IC. By construction, and the cor. to prop. 1. book 3. Sim. Euc. BD pas ses through the centre of the circle; and hence by prop. 30. book 3. Euc. the arc AFB-BGC. Again, BD passing through the centre of the circle, cuts AC and its parallel FG at right angles; therefore by prop. 3. book 3. Euc. FE=EG, and by prop. 30. book 3. Euc. the arc FB-BG, by construction=GI. But the are AFB=BGC, conséquently the arc AF= GC. In the triangles BGE and EGO, right angled at E, the side BE-EO by construction, and EG is common; therefore by prop. 4. book 1. Euc. BG= GO, and the angle BGF=FGH; whence by prop. 26. book 3. Euc. the arc FH-FB, by what has been shown,=GI.. But it has been shown that the are AF GC, consequently the arc AH-CI; and by prop. 27. book. 3. Euc. the angle AGH-IGC. Further by prop. 21. book 3. Euc. the angle GAC= CIG. From what has been already demonstrated, it follows that the triangles AGO and GCI are similar, and hence AG: GO : : GI: GC. But by construction, the arc GI=GB, and therefore by prop. 29. book 3. Euc. the chord GI=GB, which has been proved GO; consequently AG : GO : : GO : GC, as was to be demonstrated. VI. QUEST. 67. Solved by James McGinness, Harris burg, Pennsylvania. Let ABCD be the trapezium, touching the cir cle EFGH in the points E, F, G and H. From C draw a straight line cutting the circumference of the circle in the points I and K. Then by prop. 36. book 3. Euc. CI.CK =CG, and also=CF' A H E D B F consequently CF=CG. In a similar manner it may be demonstrated that FB-BE, AH-AE, and DH -DG. Whence CF+BF+AH+HD=CB+AD= CG+BE+AE+DG=AB+DC, as was to be de monstrated. Corollary. BC1AB=DC~ AD, and BCDC= AB AD. VII. QUEST. 68. Solved by John Capp. Let 614 the given base, and x=the altitude, or perpendicular height of the triangle. well known that square; that = bx Then it is the side of the inscribed 6+x bx=the area of the triangle; and that the area of the inscribed square. 2bx+x2=0: consequently x=b=14; and by plain trigonometry, the two required sides of the triangle are easily found to be 17.5184 and 14-4235. VIII. QUEST. 69. Solved by William Lenhart, Balti more. Let ABC be a plane rectilineal triangle, in which are given the area=a=840, D the point in the middle of the base AC, the length of the straight line BD-6=39, and the vertical angle ABC-61° 56'. E Draw CE perpendicular to AB, and put t=15332029 the natural cotangent of ABC, the radius being=1; then by trigonometry CE: BE:: 1:t, or BA.CE: BA.BĖ: 1: t. But by mensuration BA.CE=2a; and by Quest. 46. Math. Corr. BA.BE=BD-DC' 6- DC3. Whence by substitution 2a : b2-DC2 :: 1: t; and by reduction DC√6 · 2at= |