from Ex. 2, art. 656, Marrat's Mechanics, that ax): (r2+2x+3=2) x (4-4). From these equa · p2 + r x + x2· tions, I find (by the rule of double position) = 1.081 ft. nearly; and hence the solidity of the part E Q R to be cut off is 9.97 feet, as required.. Answers were also given by Messrs. Rylando, P B 18. Qu. (73) Answered by Mr. J. Butterworth, Haggate, Let AP and BP be two tangents intersecting in P, and draw the radii O A, O B, and join OP; then by the question A PBP is given, and AO and OB are given; therefore, A P + A ¤2 + BP2+ OB2 = 20 P2 is given; hence OP is given; and O is given, therefore the locus of P will be a circle concentric with the given circles. This is only a particular case of question 11. In the same manner it was answered by Messrs. Baines, Brooke, Eyres, Gawthorp, Hine, Maffett, Nesbit, Putsey, Whitley, Winward, and Vayheeg. 19. Qu. (74) Answered by Messrs. Gawthorp, Nesbit, Rylando, and Whitley. Retaring the notation made use of in art. 786, Marrat's Mechanics, we have by trig c: A —√(1 − x2), hence √(1 − x2) с (1 - x2) :: 1 : Am x n n the friction on the inclined plane; which, by the ques tion, is to be the least possible; therefore its fluxion m n and √ (c2 m2 + n2)3 n = 3) √(9+c2) or, (art. 773, ibid. when m 1 and This result is different from that given at art. 19, p. 191, Young's Analysis, but that is, in all probability, a press error. 20. Qu. (75) Answered by Rylando. Composition. Demonstration. From any point P in PD draw the tangents PBG, PCA meeting GEC in the points G, Č; it is to be proved that BG is equal CA. Draw the other lines as per figure, then if OP meet A B in I, and OD meet EF in F, we shall have by the wellknown properties of the figure OFx OD=EO' and OIX OP A O2 = 0 E2; therefore OE × OD= OLX OP; or OI:OE:: OD: OP; thus it appears that the point F is in the right line A B; therefore AFC BFG, but the angles GBO, GFO are right angles, and therefore the points G, B, F, O, are in a circle, wherefore BFG BOG. In the same way exactly it may be shown that the angle AFC = 4 AOC, therefore ▲ BOG — ▲ A OC, and because the right angled triangles O BG, OAC have the BOGAO C and O BAO, they are, therefore, equal in all respect, and consequently BG is equal to A C. 2. E. D. Again, by Mr. Butterworth." equal to C B, and draw the tangents A G, AK; also let the tangent A G meet the tangent BH in D'; draw the radii OG, OH, and join O D'; also, join GK, and HI, and it is well-known that they will intersect each other, and the lines DE, and OC in the same point P. Then, because AC and CB are equal, the arcs HG and KI will be equal, therefore the angle HPG=HOG, consequently the circumference of a circle will pass through the points G, P, H, D'; and the angle GPD EPIH PD, therefore, the angle HPD HOD'; consequently the points D, D' coincide in D. Again, the tangents DH, DG, EK, and E I, are evidently all equal, therefore HD EI. 2. E. D. Again, by Mr. Whitley. From O, the centre of the given circle BG AF, demit on PE the straight line given in position, the perp. OE: draw the tangents EF, EG; then through the points of contact G,F, draw the indefinite P B E right line CD, which will be the line required. Demonstration. From any point P in PE, draw the tangents PA, PB, meeting C D in the points D, C; then will A D be equal to B C. For drawing the lines as in the figure, let A B meet O P in I, and CD meet OE in S; then the triangles OS F, OFE will be similar, as will also the triangles OIB, OBP; therefore OS: OF:: OF:: OE, and OI: OB OF::OB OF:OP; hence by equality OI:OS:: OE: OP; thus it evidently appears that the point S is in the straight line AB. Again: since the angles OSD, O AD are right angles, a circle might pass through the points O, S, A, D; and for the same reason a circle might pass through the points O, S, C, B; therefore the angle ODA is equal to the angle OSB, that is, equal to the angle OCB in the same segment; hence, since in the triangles OAD, OBC, the angle O D A is equal to the angle O C B, the angle O A D equal to OB C, each being a right angle, and the side OA equal to the side O B... the triangles are identical; and therefore the segment A D is equal to BC. 2. E. D. It was also answered by Mr. Gawthorp. Mr. Jesse Winward is requested to send for any book, the price of which does not exceed Half à Guinea. NEW MATHEMATICAL QUESTIONS TO BE ANSWERED IN NO. IX. 1. Qu. (97) By Mr. Rt. Maar. Extract the .6' root of .0005. 2. Qu. (98) By the same. 6 places of decimals. 3. Qu. (99) By Mr. J. Baines, jun. At what time of the day, on the 9th of Dec. 1812, in lat. 53°, 42′ N. will the sun's alt. above the horizon be equal to his azimuth from the south? 4. Qu. (100) By the same, The square of any prime number of the form 3n+ 1, is of the form 3 a + b2; where b is a prime, and a composite number: required proof. 5. Qu. (101) By Mr. W. Pulsey, Pickering. It is proposed to divide the beam of a steelyard, or to find the points of division where the weights of 1, 2, 3, 4, &c. s. on one side, will just balance a constant weight of 95 bs. at the distance of two inches on the other side of the centre of suspension; the weight of the beam being 10 lbs. and its whole length 36 inches. (Hutton's Course.) 6. Qu. (102) By Mr. Nesbit, Farnley. There is a field in form of a trapezoid, the parallel ends of which, perp. to one side (the base) are 9 and 15 chains, and the base 40 chains; divide it equally among three persons, by fences, parallel to the perpendicular sides. 7. Qu. (103) By Mr. Rt. Maar. Every thing being as in question 8 (63), to determine when and where the bodies will both be in a right line parallel to the horizon. 8. Qu. (104) By Mr. Gawthorp. Determine a point in the side of a cone, the perp. alt. and base of which are respectively 3 and 2 feet, to which if a string be fastened, and the cone suspended by it, the axis of the cone may make an angle of 25° 15' with a plane parallel to the horizon. 9. Qu. (105) By Mr. Nesbit. If the greatest altitude of the piston of a common sucking pump above the surface of the water be 18, and the least alt. 15 feet; how high will the water rise in the pump by 3 strokes of the piston, admitting the height of a column of water, equivalent to the pressure of the atmosphere, to be 32 feet? 10. Qu. (106) By Mr. J. Winward, R. L. Militia, Plymouth. Determine the area of the curve, whose equation is y = (a2 + x2)ž + (a2 + a x + x2)ž. 11. Qu. (107) By Mr. J. Baines, jun. Horbury Bridge. Reduce a given triangle to a trapezium, having the same area and perimeter. 12. Qu. (108) By Mr. J. Whitley, Masbro'. Determine a triangle whose three sides shall be integers, in arithmetical progression, such, that the area and the diameters of the inscribed and circumscribed circles, shall be all integers. 13. Qu. (109) By the same. Let ABC be a plane triangle, and let A F, BE, be tangents to the circumscribing circle at A, B; draw CEF perp. to A B, cutting A F, BE, in F, E; the square of AC is to the square of BC, as CF to CE; required a demonstration. 14. Qu. (110) By the same. Find p and q in general terms, so that p2 + q2 +2 and p2 q2 1 may be squares. 15. Qu. (111) By the same. Find expressions for the principal diameters of the maximum ellipse inscribed in a given sector of a circle; |