The same answered by the Rev. Thomas P. Irving, Newbern, North-Carolina.

Sirs, Mr. Walsh, I think, must own,
One horse is useless lumber;

Unless he'll make three odds combin'd
Produce an even number.

OTHERWISE.

I think Mr. Walsh Alexander, must own
That Bucephalus sleeps out o' doors;
Unless he has skill to rectangle three threes,
That shall do what is done by four fours.

III. QUEST. 21. Answered by Ebenezer R. White, Danbury, Connecticut.

r

After correcting a typographical error, which, in this question, made n=7, instead of n=17; let the first of the given equations, x=√am+b", be transformed to

(+1)x: from which, the first of

the general rules required is self-evident. Hence the value of x will be found as follows:

log. x=log. 127321=5•1048999

Hence x=127321 nearly; and by a similar process we obtain the second general rule; from which is found, y=126657 nearly. In the same manner, the value of a binomial surd may often be easily found, when the common methods of calculation would render the operation exceedingly troublesome.

IV. QUEST. 22. Answered by Diarius Yankee, Bunker's Hill.

The learned do not expect the commencement of a sempiternal spring: for, in nature, no cause, cacapable of producing such an effect, has hitherto been discovered.* It may therefore be said, that the learned all agree, that, should an eternal spring commence, at some future period, the distance of that period from the present century is now unknown. Hence the first six lines of this question may be considered as an ingenious enigma, signifying that the length of the fish's head is not given. Let, now, 2y=length of the body, and xthe length of the head; then by the question y+x=the length of the tail, and 3y+2x=the length of the fish. Also by the question 2 y=y+2x+0; whence y=2x; and consequently if x=1, y=2, y+x=3, 2y=4, and 3y+ 2x=8; or the length of the head, tail, body and fish are as 1, 3, 4 and 8. Therefore the question admits of an indefinite number of answers.

* See Prop. 34, Emerson's Centripetal Forces.

V. QUEST. 23. Answered by G. Baron, New-York.

LEMMA.

If in a straight line A B, drawn on the plane E AD, a point P be taken between A and B, and nearer to B than to A; the locus of all the points P, p, on the same plane, is a circle, when the straight line p P always bisects the rectilineal angle Ap B: and the centre of this circle is in the straight line A B produced towards D..

P

A

E

D

DEMONSTRATION.

Let the point P, in moving from P to p, describe the line Prp. Then, since by hypothesis the straight line pP bisects the rectilineal angle Ap B; AP: PB:: App B, by prop. 25, book 2, Emerson's Geom. 2d edit. Again, by hypothesis, AP is greater than PB; and therefore the increment of A p is always greater than the cotemporary increment of p B: of consequence A p-p B continually increases. But when A p-p B becomes equal to A B, or when Ap

AB+PB; the triangle A p B vanishes, the point p coincides with D, and the straight lines A p and pB respectively coincide with, and become equal to AD and B D. Hence AP: PB:: AppB::AD: DB;

and therefore, by the prop. above quoted, the straight line p D bisects Ep B, the exterior angle of the triangle A p B. Lastly, since p P bisects the angle Ap B, Ep D+A p P P p D-a right angle; and consequently the curve Prp D is a semicircle, whose centre O is in A B, produced towards D. 2. E. D.

COROLLARY.

When AP P B, the triangle A p B will be isoceles; and the locus of the point P will be a straight line. But if AP be less than P B, the locus of P is a circle, whose centre is in B A, produced beyond A.

N. B. The demonstration of this lemma might have been inferred from prob. 13, const. geom. probs. Simpson's Geom. 4th edit. and the reverse might have been drawn from prop. 23, book 4, Emerson's Geom. 2d edit.

SOLUTION.

By the data of the question, and the principles of perspective, a straight line, drawn from any point in the circular fence to the middlemost tree, must bisect the angle formed by straight lines drawn from the same point to each of the other two trees. But, since one of the trees grows in the fence, the third tree must be planted in a straight line with the first and second; otherwise, a straight line, drawn from some point in the fence, through the middlemost tree, might also pass through one of the other trees, without passing through the third; which is contrary to the conditions of the question. Now, by the above lemma, if B, P, and A, represent the places of the three trees, in a straight line A B, Prp Dp r will represent the circular fence, whose centre O is in

A B, produced beyond B: for, by the question, the third tree A was to be planted outside the fence. Also by the lemma DB: AD:: PB: PA; and by a theorem in proportion D B-P B=2 BO : A З PA=2OP :: PB: PA, or OB: OP:: PB: PA. Again, by the question and a well known theorem, 1×660X66

the radius O P=√

3-14159, &c.

117.75219 feet: and

since P B is given=90 feet, O B=OP-PB=27.75219 feet. Hence, in the four proportionals, the three first terms O B, OP, and P B, are given to find the fourth PA=381.8692 feet: and consequently the place of the third tree, A, is known.

COROLLARY.

OA is a third proportional to OB and OP: for OB:OP:: PB:PA::O B÷P B=OP:0P+ PA=O A.

VI. QUEST. 24. Answered by the Proposer, the Rev. J. Blackburn, Cambridge, England.

Because equilateral triangles are equiangular, and the angles, at the circumference of a circle, standing on the same arc, are equal one to another, the angles BDA=BCA-ABC-ADC: and hence AD bisects the an- F gle BDC. In C D produced,

B

A

E

C

take D F=D B, and join B F. Then the angle DBF =DFB by 5, 1, Euc.=ADC by 32, 1, Euc. for it has been shown A D, that bisects the angle BD C. But the angle B D F BAC by 22, 3, Euc. ABC =ADC=DBF=DF B, by what has already been