NOTE ON EUCLID, BOOK VI., PROP; 7.

By J. F. WOLFF, Jun.

MR. WALTON, in No. 35 (March, 1868) of the Quarterly Journal, gives a different demonstration of the 7th Proposition of the 6th Book of Euclid.

There is still a more direct demonstration, which shews the whole bearing of the proposition and reduces it to a case of ambiguous equality.

Let the triangles ABC, DEF (fig. 9) have one angle of the one equal to one angle of the other; namely, LBAC-L DEF, and the sides about two other angles ABC, EDF proportionals, so that AB: BC=ED : DF.

=

In the triangle ABC, produce BC to any point L, and from the vertex A make AG equal to AC, and the angle GAL equal to the angle ABC.

From the point A make AM equal to DE, and from M draw MN parallel to BL.

Then BCAL AGL, and L ABC= L GAL.

Therefore the triangle AGL is similar to ABC.

But the triangle AGL is similar to the triangle AMO, because MN is parallel to BL.

Therefore the triangle ABC is similar to the triangle AMO, and AB: BC=AM: AO, but we have also

AB: BC=ED : DF.

Therefore AM: AO-ED: DF.

But AM=DE; therefore AO=DF.

AO is also equal to AN by construction, and the angle AMO being equal to BAC must also be equal to the angle DEF.

Therefore in the triangle AMO, as well as in the triangle AMN, we have two sides (AM and 40 in the one, and AM and AN in the other) equal to the two sides DE and DF in the triangle DEF and adjacent angle AMN in each of the triangles equal to the adjacent angle DEF of the given triangle.

The question now arises whether the triangle DEF is equal to the triangle AMO, or to the triangle AMN. (We know already that AMO is similar to BAC).

The second condition given in Euclid discards all ambiguity in this case, and it is easily proved that the triangle AMO and not AMN must be equal to the triangle DEF.

If the remaining angle is a right angle, AÑ and AO, or AC and AG coincide; and the equality is obvious.

THEOREM ON THE ACTION OF AN ELECTRIC CURRENT IN A HELIX WOUND ON A CYLINDER OF ANY FORM UPON THE POLE OF A MAGNET.

By PERCIVAL FROST, M.A.

LET N (fig. 10) be the north pole of a magnet acted on by a helix wrapped round a cylinder of any form, PQ an element of the helix, h the distance of the threads, NO perpendicular to the plane of the turn of which PQ is an element, the action of PQ takes place perpendicular to iu.PQ sinQPN the plane QPN and is measured by NP i is the intensity of the current and μ the magnetic fluid in N, and the component parallel to the axis of the cylinder by iu.PQ sinQPN UO if OU be perpendicular to PQ.

NP2

UN

where

Consider the action of the portions of the wires lying in the rectangle PP'Q'Q, whose sides are parallel and perpen

dicular to the axis, the number of elements is

PP'

there

fore, the whole action of these portions is

iu.PQ.PP' sinQPN.UO

h.NP2. UN

With N as centre describe a sphere of radius a and project PP'QQ upon it into pp'q'q. The inclination of the areas is that of OU and NP, which are normals to their planes; therefore

OU
NP

area PP'Q'Q.: area pp'q'q:: NP: a" ;

therefore the action parallel to the axis is

Hence the following curious theorem is true:

The whole action of a cylindrical helix on the pole of a magnet in the direction of the axis of the cylinder varies as the projection upon a sphere of all the parts of the cylinder on which the helix lies, doubled if they overlap.

As an application of the theorem take the cyclinder circular and the pole in the axis.

The projection is the area generated by bc (fig. 11) round the axis

=2πa (coscNa-cosbNa),

if r be the radius of the cylinder, 27 its length, the action on N

if n be the number of turns, or nh=21, and N be supposed to be within the cylinder; which is the formula used by Haedenkamp, Pogg. Ann. Bd. 78, 1849.

ON THE EVOLUTES OF CUBIC CURVES.

By HENRY M. JEFFERY, M.A.

1. THE evolute of a curve of the second class (u=0) can be expressed in the form (u,v,= 16▲1), if (v, = 0) denote the equation of the reciprocal curve (Journal, Vol. x., p. 332-336). It has been explained that this form exhibits the foci of the conic and its reciprocal as common to the evolute. Also, if u =0=μ‚μ‚ ̧ + к (4▲3), v1 = 0 = μ‚μ‚ ̧ +λ (4▲3), so that не, е зле зле в denote the foci, this conclusion readily appears from the equation to the evolute. In like manner, if (u=0) (v=0) denote the equations to a cubic curve of the third class, and its reciprocal curve (continuous for Spherical Geometry, but degenerating into points for Plane Geometry), the equation to its evolute may be written u,v,+ 16A*.w= 0, if w denote an auxiliary curve of the fifth class. It will also be shewn that a simple adaptation of this form will exhibit the evolute of a cubic of the third order.

6

2. To investigate the equation to the evolute of a nonsingular cubic of the third class.

Let the equation be written in the canonical form

af3 + bg3 + ch3 + 6dfgh = 0

...... (1).

The equation to the evolute is the eliminant of this and two others, which denote the point of contact and the perpendicularity of the normal :

(af* +2dgh) p+ (bg3 +2dfh) q + (ch2 + 2dfg) r = 0 ... (2),

if P denote a (ap- bq cos C-cr cos B), and Q, R are similarly interpreted, so that pP+qQ+rR = 4Ã2.

For the convenience of the analysis, write (1) in the form

AP3ƒ3 +BQ3g3 + CR3h3 +6DPQRfgh=0. (4).

.....

Also use the subsidiary quantities L, M, N, H thus defined: L= APp+D (Pp Qq- Rr)

The following ratios of fP, gQ, hR may be obtained from (2) and (3),

ƒP(L −√(H)=gQ{M+ √(H)},

gQ (M−√(H) = hR{N+√(H)},
hR (N − √√(II)=ƒP {L + √√(H)}•

Hence

fP

=

=

M+√(II) L-√(H) L+M'

By substituting these ratios in (4)

· A {(M+ √ (H)}3 − B {L− √√(H)}3 +C (L+M)3

By symmetry,

+6D {M + √(H)} {L− √√(H)} (L+ M) = 0.

− B {N+ √(H)}3 − C {M — √(H)}3 + A (M+N)3

+6D{N+V(H)} {M_v(II)} (M+N)=0

−C {L+ √(H)}3 − A {N− √√ (H)}3 +B (N+L)3

+6D{L+v(H)}{N_v(H)}(N+L)=0.

By adding the three last equations

√(H) {A (N2 — M2) + B (L2 − N2) + C (M2 − L3)}

=A(M+N)(H—MN)+B(N+L)(H−NL)+C (L+M)(HI—LM) -6D(M+N){N+L)(L+M).

After squaring, and rejecting a factor, the equation to the evolute is obtained:

A2 (M + N)3 + B2 (N + L)3 + C2 (L+M)3

− 2BCL (L2 +3H) − 2 A CM (M2 +3H) − 2ABN (N2 + 3H)

− 12D {A (M + N) (H− MN)+B (N+L) (H− NL)

+36D2 (M+ N) (N + L) (L+ M) = 0.

+C(L+M) (H− LM)}

Finally, this may be expressed in the form, which presents both the primitive and reciprocal curves,

(AP3p3 +BQ3q3 +CR3,3 +6DPQRpqr)

× {B* C2+ C*A2+ A'B2-2A3BC-2AB C-2ABC-24 ABCD − 24D2 (BC+CA + AB) – 32D3 (A+B+C) − 48D'} + (Pp+Qq+Rr)2 (8D3 + ABC) {6D (APp+BQq+CRr) +(2AC+2AB− BC) Pp + (2AB+2BC − AC) Qq +(2BC+2A C-AB) Rr} = 0.

After substituting for A, B, C their values

this becomes

(ap3 +bq3 + cr3 +6dpqr)

× {b2c3P® + c2a2 Q® + a2b*R®

-24bcd PQR-24acd PQ'R - 24abd'PQR*

-2 Q3 R3 (a2bc+16ad") -2P R3 (ab3c+ 16bd3) −2P3 Q3(abc2+16cd3)

- 24P2 Q*R* (2d* + abcd)}

+ (4▲2)2 (8ď3 + abc) {6d (ap Q2R2 + bqP2R2 + crP2 Q2)

+ (2ac Q3 +2ab R3 — bcP3) Pp + (2abR3 +2bcP3 – ac Q3) Qq

P

a

+(2bcP3 +2ac Q3 — abR3) Rr} = 0, where denotes (ap-bq cos C-cr cos B) and (in Spherics) is proportional to the corresponding tangential coordinate of the tangent to the polar curve referred to the polar triangle of reference.

It should be observed that the factor (8d3+ abc) is the cube root of the discriminant of the cubic. If the factor be zero, the evolute degenerates into the three given coincident lines and their quadrantal pole.