THE FOUR-POINT PROBLEM. BY T. J. LOWRY, U. S. C. S., SAN FRANCISCO, CAL. PROBLEM: Four points in the same plane being given in position, to determine the position of any other point (or place of observation) in reference to these given points, having on the four points two angles which have no parts common except their vertices. since DFC=2/DGC, and DF-FC hence in DFC we have DC and all the /s to get FC. But the FCE=/ACD->ACE >FCD... in FCE are given FC and CE and the included angle to find FE and >EFC, then in A EGF all the sides are known, hence >GFE can be found. But >GFC = >EFC + >GFE.:. in GFC we have GF and FC and all the angles to find GC. Then in ▲ GCD are given GC, CD and >CGD to get GD. And now GA and GB follow in a very obvious manner. This problem may obviously be solved with fewer equations by using quadrilaterals, instead of triangles, in the solution. Unless the two circles of position are tangent to each other there will be two points (viz, G and G') which will equally well satisfy the conditions of the problem: hence in this case we judge from an approximate knowledge of our position which of the points G or G' we were at when we observed. The manner of sweeping the circles of position, with Protractor and Dividers is easy and expeditious: since > AEB=2>AGB then > ABE 180°->AEB 2 180°2>AGB = 900° AGB, that is >ABE = the complement of the observed >AGB. The rule for plotting the circles of position is now obviously, to lay off from AB at the points A, and B the compliment of the observed angle AGB, and the point of intersection of the produced sides of these angles will be the center of a circle of position. And now from this point as center with radius AE (or BE) sweep the circle of position. And in like manner lay down the other circle of position through C and D. This problem will often be found servicable to the Hydrographer and Explorer when from either accident or necessity only two angles are measured on four objects. SOLUTIONS OF PROBLEMS IN NO. 6. Solutions of problems in No. 6 have been received as follows: From Geo. L. Dake, 25 & 26; R. M. DeFrance, 25 & 26; Prof. A. B. Evans, 25, 26, 27 & 29; Henry Gunder, 25, 26, 27 & 29; Wm. Hoover, 26; Prof. A. Hall, 29; H. Heaton, 29; D. J. McAdam, 25 & 26; Esther Matthews, 26; Artemas Martin, 27 & 29; A. W. Phillips, 25, 26, 27 & 29; L. Regan, 25, 26 & 29; R. L. Selden, 25; Werner Stille, 25, 2ti, 27 & 29; E. B. Seitz, 25, 26, 27 & 29; Prof. J. Scheffer, 26 & 29; Walter Siverly, 27. 25. "Required the sides of an obtuse angled triangle the area of which is 14.048 acres, the obtuse angle 111°15', and one of the acute angles 11°44'10"." SOLUTION BY HENRY GUNDER, GREENVILLE, OHIO. = Putting A 111°15', B = 11°44′10′′, C 57°50′′, and x, y, z or the sides opposite A, B and C, and a = 14.048 acres = 2247.68 sq. rods. Since the product of two sides and the sine of the included angle equals twice the area we have, sin A sin B By applying logarithms, x 156.705 rods, y = 34.1997 rods, z = 141.034 rods. 26. "Find 0 from the equation 15 sin + 12 cos = 17.97240, (1).' SOLUTION BY WILLIAM HOOVER, SOUTH BEND, IND. The given equation may be written, m sin 0 + n cos 0 = q... . . . (2). In (2) put pcos &: =m and p sin =n and it becomes 27. "Four given equal spheres being placed in close contact with each other, it is required to find the volume of the space inclosed between them and the four triangular planes drawn respectively through each three centers." SOLUTION BY E. B. SEITZ, GREENVILLE, O. = Let r = the radius of each sphere. Then 2 the volume of the tetraedron formed by the four planes, and (3cos1 — )3 — the volume of the four equal spherical sectors cut from the spheres by the planes. Hence, the required volume is V = zr3√/2 — (3cos' 6cos-1)=.20775r3. [For want of room we are obliged to defer publishing the solution of 29 till next month.] PROBLEMS. 34. BY PROF. M. L. COMSTOCK, GALESBURG, ILL.-Given xyz = 18, (1); x2 + y2 + 22 = 33, (2); (x2 — yz)3 + (y2 — xz)3 + (z2 — xy)3 — 3(x2 — yz)(y2 —- xz)(22 — xy) = 6561, (3); to find x, y and z. 35. BY CAPT. O. E. MEIBAELIS, PITTSBURGH, PA.-A man let a stone weighing 40 lbs. to a neighbor-the latter broke it accidentally into four parts--and upon returning the fragments consoled the owner by remarking that now he could weigh all numbers between one and forty. In other words, given a +b+c+d=40, to determine such values for a, b, c and d, as will, by association, produce all numbers from one to forty. 36. BY HENRY A. ROLAND, TROY, N. Y.-A perfectly flexible cord of given length is suspended from two points whose coordinates are x', y' and x", y". How must the weight of the cord vary from point to point in order that it may hang in the arc of a circle. Vol. I. THE ANALYST. September, 1874. No. 9. INVOLUTION AND EVOLUTION OF IMAGINARY QUANTITIES CONSIDERED GEOMETRICALLY. BY PROF. W. D. HENKLE, SALEM, OHIO. In the diagram let the radius of the circle be I and OR its initial position. Conceive the radius to revolve as the hands of a clock. Let OP represent any position of the radius. PROPOSITION I.-The following equations indicate the powers of OP: Demonstration.-Since a = cos and b = sin @ we have PROPOSITION 2.-The following equations indicate the roots of OP: = cos + v. I sin 10. = cos + √ sin 0, = OR revolved through 1⁄4 0. n Revolving OR through ↓ of @ gives but one of the roots of OP, which I shall call the first of the n roots. The following proposition -- I roots. shows how to get any one of the remaining - I and <n + I. Let m>I PROPOSITION 3.-The following equations indicate then I of the n roots of OP after the first: The second of the roots of OP=OP revolved through C, We can make the reference to OR by adding 0. The mth of the roots of OP = OR revolved through Ө |