1

BD. Then by prop. 25 and 22, book 2, Emerson's Geometry, AB+BC-AC+2.AB.BE=4.DC2+ 2.AB.BE. But by prop. 28, book 2, Emerson's. Geometry,AB2+BC2=2.DC2+2.BD2; consequent ly 4.DC2+2.AB.BE=2.DC2+2.BD2, or DC+ AB.BE=BD': whence it is evident that AB.BE. BD DC, as was to be proved.

Corollary. From the above demonstration it is plain that BD is greater or less than DC, according as the angle B is acute or obtuse.

VIII. QUEST. 47. Solved by the Proposer, James North,
Philadelphia.

Put c=100 gallons, the content of the vessel, s=10 seconds of time, and x=the quantity of wine remaining in the vessel at the end of any variable time t

then from the nature of the question s:t::1
: : 1 :i=the

8

quantity of the mixture discharged in the infinitely
small particle of time i; and c:x:::*:
:: whence

8

whose fluent corrected gives hyp. log.

t

nt

x=hyp. log. c. Now, assume t=3600 the number of seconds in one hour, and put n=43429448; and we have com. log.x=com.log.c =4365399; consequently x=2.7.3237 gallons, the quantity of wine required.

CS

Corollary. The wine would never be entirely exhausted in the vessel; for in the above general equa tion it is easy to perceive that when x=o, t is infinite,

IX. QUEST. 48. Solved by Ben. Cheetham, Billiard-Hall

Let ABCD represent the two plantations separated by the crooked fence EFG, and let BC and AD be straight lines. By a well known problem a straight line GI may be drawn, which will separate these enclosures, without varying their respective areas. Then, if BC be parallel to AD, a straight line passing through the middle of Gl, and meeting BC and AD at right angles, will evidently be the required fence.. But when BC and AD are not parallel, they may be produced to meet in some point P. In this case find PI-a mean proportional between P/ and PG; make PK-PI, and join KI, which will be the fence required. For in the triangles P/G and PIK, the angle. P is common, and by construction PPG PI2= PI.PK, or Pl: PI::PK:PG; therefore, by prop. 15, book 6, Euclid, the triangles P/G and PIK have equal areas; and consequently KI separates the twoplantations, without changing their respective areas. Also, since PI PK, or the angle PIK-PKI; by the lemma to prob. 68, Appendix to Simpson's Algebra, the straight line KI is a minimum, as was required.

Corollary. In every case of this problem the angle BKI is equal to the angle AIK.

I. QUEST. 49. Solved by Article 12, Dr. Hutton's Miscellanea Mathematica.

Let ABCD be the frustum of a DEC right cone, EF its axis, and GH a variable straight line intersecting

H

I

the axis EF, and parallel to DC the
diameter of the base of the frus-
tum. Put DC=a, AB=b, AB-
DC=d, EF-h, •785398 &c.=c, and EI=x.

Then

dx+ah
h

=GH; hence cx

362+2ab+a2

Xh=the distance of the

b2+ab+a2

minating d) centre of gravity from the smaller end of the frustum. And the same theorem will serve for the frustum of any right pyramid, by putting a and b for any like dimensions, in the two ends of the frustum.

Corollary. When a=b, the frustum becomes a cylinder, and the distance of the centre of gravity from either end=h.

Corollary 2. When a=0, the frustum becomes a cone, and the distance of the centre of gravity from the vertex=3h.

XI. QUEST. 50. Solved by John D. Craig, Philadelphia.

The first equation gives x+y=a➡%,

The second gives

And the third gives

x+y×z=b―xy,

C

xy=z'

From the first of these equations, multiplied by ˆz, and compared with the second, we have a-zxz—b -xyb, by the third: hence z3-az2+bz=c.

C

Lastly, since in each of the original equations the three unknown quantities are equally concerned; therefore the three roots of this last cubic equation will be the values of x, y, and z.

XII. PRIZE QUEST. 51. Solved by John Smithis, Philadelphia.

Let the annexed figure be the stereographic projection of the convex northern hemisphere of the earth, on the plane of the equator ABCD; and let P represent the north pole, PA, PB, PC, and PD meridians meeting the equator in the points A, B, C, and D. Imagine any meridian

PA to revolve uniformly about the pole P towards D, while a point moves from P along PA, so as to describe the curve Plocus, such that. Plof, any length of the curve, may be always equal to AEDF, the cor

responding difference of longitude; then, from the question it is plain that the ship must describe a curve of this nature on the surface of the sea Infinitely near a coincidence with the meridian PF draw another meridian PG, intersecting the curve Plofc in the point g, and through the point ƒ draw the infinitely small arc fe parallel to the equatorial arc FG. Put r=the radius of the earth, Plof AEDF= z, Pf the complement of the latitude of f=L, sine of L-y, and the sine of the course the sine of the angle Ffg-sine of the angle fge=x, to the same radius r; then FG=fg=z, ge=L, cosine of L= r2 y2, and the sine of the angle efg=cosine of the course= √x2. In the infinitely small triangle fge, right angled at e, r:x::;:*2=fe; and by similar spherical

r

sectors y:r:: XZ : *%=FG=%: hence x=y; and since the sines x and y have the common radius r, the course is constantly equal to the complement of the latitude. Again, in the right angled triangle feg, r: √r2x2 ;.%√r2—y2=ge=L; therefore

ri

::z:

r

Now let L be considered as equal to an

arc of latitude reckoned from the equator; then by example 11, prop. 13, sect. 1, of the third edition of Emerson's Fluxions, the fluent z=the meridional parts of L, which needs no correction; for when z=0, Land the meridional parts of L are each 0. Hence, when a ship is steered according to the hypothesis of this question, the difference of longitude, or the distance of the vessel from the nearest pole, measured on the curve she describes, is always equal to the meridional parts of a latitude, whose complement is No. 6.

M