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Font size minimum qu' il d^passe ensuite pour retoumer k la valeur ISO''.

Pour construire g^om^triquement le point M r^pondant k une
valeur donn^ de Y angle M, il suffit de determiner la position du
point B. L' angle QBR ^tant connu, et les points Q et R ^tant fixes,
on d^crira sur RQ un segment de cercle capable' de 1' angle donn^.
Les points d' intersection de ce segment avec la circonference donnee
donnent les points B cherch^s. On voit que si les points sont Fun int^-
rieur et 1* autre ext^rieur, il y aura toujours au-dessus du diamfetre QR
un point B r^pondant k la question. Si les points sont tous deux
int^rieurs ou tous deux ext^rieurs, il pourra y avoir deux positions,
une seule, ou aucune suivant les cas. L' ^tude des conditions de
possibilite de cette construction g^m^trique du point B conduirait
aux r^sultats etablis plus haut par une autre voie.

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III. Les triangles semblables PMQ et RBQ donnent

BR^QR^QB .

PM QP QM'

on, puisque BK Â« PA,

PA QR

PM^QP*

Le dernier rapport est constant, puisque les points Q et R sont fixes.

Le point P ^tant fixe, il r^sulte de cette ^galit^ que le point M est sur

une circonf^rence homoth^tique k la circonf^rence donn^ par rapport

au point P, qui est centre d' homoth^tie directe.

r, . QB QR

On a aussi ^ - =Â» 5â€” .
QM QP

Cette ^galit^ montre que le point Q est le centre d' homoth^tie inverse

des deux circonf^rences.

Oes r^ultats se rapportent k la figure 45 ; ils seraient renvers^ pour

le cas de la figure 46. Dans tous les cas P et Q sont les centres

d' homoth^tie directe et inverse de la circonf 4rence donn^ O, et de la

circonf^rence lieu du point M, que nous appellerons circonf^rence O'.

Oonsid^rons par exemple la figure 51. Les points M et A' sont anti-

homologues. La tangente en A' au cercle et la tangente en M ^ la

circonf^rence O' font done avec MA' des angles ^gaux, et se coupent

en I sur Y axe radical des deux cercles. Pour la mdme raison, les

points M et B' ^tant anti-homologues, la tangente en B' au cercle O

rencontre MI sur V axe radical, c'est-4-dire au point I, et les angles

1MB' et IB'M sont ^ux. Le point I est done situ^ sur les perpen-

diculaires ^lev^s au milieu de MA' et au milieu de MB' ; c' est done

le centre du cercle circonscrit au triangle MA'B'. II r^sulte des

constructions pr^c^entes que ce point est toujours sur V axe radical

des circonf^rences et 0'. Done le lieu demand^ est cet axe radical.

. The BquUateral and the Bquiajigulor Polygon.

By R. E. Allardicb, M.A.

The Equilateral Polygon.

Since an tzrgon is determined by 2n >- 3 conditions, and n-l con-
ditions are involved in its being equilateral, there are still in the case
of an equilateral n-gon w - 2 conditions to be determined. These
n - 2 conditions cannot all be given in terms of the angles, since an

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infinite number of n-gons may alwayB be described similar to, but not
necessarily congruent with, any given equilateral n-gon. Hence only
n - 3 of the angles of an equilateral n-gon may be assigned arbitrarily,
and there must .therefore be 3 independent relations connecting the
angles of any equilateral r^-gon. These three conditions may be
obtained by projecting the perimeter of the n-gon on any three lines.

Let a be the common length of the sides; A, B, G L, the

exterior angles of the n-gon.

Projection on one of the sides adjacent to the angle A gives the
relation

acosA + acos(A + B) + aco8(A + B + C) + =0;

cosA + cos(A + B) + co8(A + B + C) + =0. (1),

Similarly projection on the other sides gives

cosB + cos(B + C) + co8(B + + D) + = 0, (2),

cosC + co8(0 + D) + co3(0 + D + E) + =0, &c. ; (3),

and projection on lines perpendicular to these gives

sinA + sin(A + B) + sin(A + B + C) + =0, (4),

sinB + sin(B + C) + sin(B + + D) + =0. (5).

There thus arise 2n equations ; but, as has been seen, only three
of these are independent. This may be proved analytically in the
following manner.

Assume equations (1), (2), and (4) above, and assume also

8inB + sin(B + C) + sin(B + C + D) + =/?. (6).

Let cos A + tsinA Â« e*^ = a, cosB + isinB = e*^ = j8, &c. ;
then cos( A + B) + ttin( A + B) = e*<^+Â»> = e* VÂ« = ap.
Hence equations (1), (2), (4), and (6) are equivalent to
a + a/? + a/?7+ +{<^ky '^) = 0,

l3+Py+ky^+ (fiy^ Â«)=i>Â»

a-(a'jSy A)=-apt;

but a Â» cosA + tsinA is not equal to 0,

i-(a/5r ^)= -Ph

co8(A + B+......L) + tsin(A + B+ L)=l+pi,

cos(A + B+ L)=:l, sin(A + B + L)=i?,

A + B + L = 2n7r,^=:0, andajS A=l.

Hence /3 + /3y + /3y5+ {pyB Aa) = 0.

Hence also, since /? is not equal to 0,

l+7 + y6+ {yS Aa) = 0,

that is, (a/Jy A) + 7 + y5+ {yS Aa) = 0,

7 + 7Â«+ (yS Aa/3) = 0;

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and in the same way the truth of all the other relations may be
established.

These equations may also be written in the form

l+p + l3y+ (fiy A) = 0,&c.

In order now to prove that the equations do in general involve
three independent conditions, it will be sufficient to show that in the
case where n is 3, they are sufficient to determine the actual values
of A, B and 0.

The equations may be assumed in the form

cosA + cos( A + B) + cos(A + B + C) = 0,
cosB + cos(B + C) + cos(B + C + A) = 0,

A + B + C = 2nir;
which lead at once to the equations

1 + cos A + cosC = 0,

I + cosB + cos A = ;

whence also I + cosC + cosB = 0.

These equations give cosA = cosB = cosO = â€” J ; the only solution
of which lying between and trisA = B = C = 2ir/3.

From the equations in the case of the quadrilateral may be deduced
the equations cosA â€” cosO and cosB = cosD. The only equilateral
quadrilateral in the ordinary sense is of course the rhombus ; but the
above equations include the limiting cases of two straight lines which
meet at a point taken twice over, and a single straight line taken four
times over.

It should be noticed that in general it is not allowable to take the
angles in an arbitrary order. This is exemplified in the case of the
. quadrilateral ; but it may be shown that in general two angles cannot
change places.

Suppose that an equilateral polygon may be formed with certain

angles taken in the order ABCDEF and also with the same

angles taken in the order EBCD AF. .....

Then a + a/J + a^y + a/JyS + a^y& + =0,

and â‚¬ + â‚¬^ + â‚¬^y + â‚¬^yS + â‚¬^ySa+ =0

(a-ÂŁ)(l+/J + j8y + ^yS) = 0.
Hence the angles cannot change places unless they are equal, or the
intervening angles satisfy the relation l+/3 + /3y + /SyS = 0. This
condition involves that the sides between the two vertices considered,
themselves form a closed polygon ; for it implies that the sum of the
projections of these sides on each of two straight lines at right angles

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to one another is zero. This equation which is equivalent to two
conditions, since it involves the imaginary unit, only contains (n-l)
of the angles in the general case, and is therefore not in itself sufficient
to determine that the n angles may be the angles of an equilateral
polygon.

It may be noted further that three conditions of the form

l+co8A + cos(A + B) + ^0

are not sufficient ; for they may be satisfied by angles whose sum is

not a multiple of 27r. For let n equal lines PiPj, PgPs PnPn+iÂ» bÂ©

drawn making angles A, B, K with one another (the first

w - 1 of the angles) ; then the first condition will be satisfied if F^^i
lies in the perpendicular to PjPj through the point Pj. Let now
another straight line Pn+iPÂ«+2 ^ drawn making with Pâ€žPâ€ž+i an
angle L; then the second condition will be satisfied if the point Pâ€ž^j
lies in the perpendicular to PsPs through the point P, ; and so on for
a third, fourth, etc. condition, the next line making with that last
considered an angle A, the next again an angle B, and so on. Thus
in the case where n is 3, the conditions are satisfied by the angles of
a square, that is, by the values A = B = C = 7r/2 ; in the case where
n is 4, the four conditions of the form considered are satisfied by the
angles of a regular hexagon, that is, by the values, A = B = C = D = 7r/3 ;
and, in the general case, the n conditions are satisfied by the angles
of a regular polygon of 2(n-l) sides, that is, by the values

A = B = = =7r/(n-l).

The three conditions may be given in the form

l+cosA + cos(A + B) + =0

l+cosB + cos(B + C) + =0

A + B + C+ = 27r.

Assume 1 + sinA + 8in(A + B) + =p

l+8inB + sin(B + C)+ =g.

Then l+a + a/J+ =^pi

l+j8 + /37+ ^^qi

^h = 1.

pi-qia = 0;

jtn - g't(cosA + tsinA) = ;

.'. if sin A is not equal to 0, ^ =: and p = 0.
If the equations are assumed in the form

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l-l-co8A + co8(A + B)+ =0

l+smA + 8m(A + B)+ =0

A + B + C+ Â»2ir,

or in the equivalent form

l + a + a/3+ =0

o^y Â«1.

the truth of all the other equations is at once obvious.

From the above it is seen that when the conditions are given in
any of the forms discussed here, either two of the three conditions
must be derived from the projection of the sides of the polygon on
two different straight lines, or one of them must be that the sum of
the exterior angles is 2m7r, where m is some integer.

If n angles be found which may be made the angles of an equi-
lateral polygon, then these angles may be combined in various ways
so as to form the angles of other equilateral polygons. Suppose, for
example, that n is 5 ; then the conditions connecting the angles may
be written

l + a + aj8 + aj8y + 0^878 =

which may be written in the form

1 + (Â«)8) + (a)8)(y8) + (aj8)(y8)(*a) + (a/3)(yS)(Â«aX^y) =
(Â«i8)(y8)(Â«Â«)08y)(&) = l;
and these equations show that an equilateral polygon may be made of
which the exterior angles are, A + B, C + D, E + A, B + C, D + K
In the same way, if n is not divisible by 3, the angles may be added
together in sets of three to form the angles of a new equilateral
polygon. And it may easily be'seen that the angles may be combined
in various ways. For let a radius vector rotate from the position
OX through an angle A into the position OA, then through an angle
B into the position OB, and so on. Now if the radius vector be
conceived to start from OX and to rotate into the position of any of
the lines just considered, OP say, then form OP into another position,
OQ say, and so on, the different angles through which it rotates may
be made the angles of an equilateral polygon, it being supposed that
the radius vector always rotates in the same direction and never stops
in the same position twice. This may be proved in the same way as
the particular cases given above ; but it is seen even more easily as a
consequence of the laws of the composition of vectors.

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Thb Equiangular Poltoon.

In the fact that an n-gon is equiangular (n â€” 1) conditions are
involved ; and since the other (n - 2) conditions may all be given in
terms of the sides, it follows that there must be two relations con-
necting the sides of an equiangular polygon.

The exterior angle of a regular n-gon is 2irln or 2mv/n = Ay
say, where A may be assumed not to be a multiple of w ; and hence
a is not real and cannot vanish, where

a = 6^* =s oosA + tsinA.

Let the sides be denoted by a, 6, c k; and let the perimeter be

projected on the side a and on a line perpendicular to a.

Then a + 6cosA+ccos2A+ +A;cos(w- 1)A=0;

6sinA + csin2A+ -|-A;sin(w- 1)A = 0.

a + 6a + ca*+ +^aÂ»-* = 0; (1)

a"=l.
Multiply equation (1) by a*^^ ; then

6 + ca + c?a^+ aa"-* = 0;

and in the same way all the other similar equations may be deduced.
It may easily be shown that the two equations may be assumed
in the form

a + ftcosA + ccos2A + = ;

6 + ccosA+c?cos2A+ =0.

The two conditions may be expressed in terms oi the sides alone
as follows : â€”

If the sides are represented by a^ a,, a, aÂ«, the conditions are

ai + a|a + a,a*+......+aâ€ža"-^=:0

Oi + a,a + a4a'+ +aia**-*=Â»0

Â»Â« + Â«ia + Oja*+ +Â«Â«.

= 0.

Hence

Â«1 Â«8

an

Â«8 ^4

But a is not real, while Oi, c^ a^ are all real; and therefore

both numerator and denominator in the value of a must vanish. It
is obvious that all the other first minors of the circulant deter-
minant (oiO, a^) must also vanish. The conditions may be

expressed in the form â€”

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= 0.

Conversely, if the condition

ai + a^ + o^a' + Â«Â»Â«*"* =

be given, where a"= 1, it may be shown that an equiangular polygon

can be formed with the sides o^, a^ a^. For if a polygon be

formed with these sides and with (n-l) exterior angles, each equal
to 27r/n, the above condition shows that the polygon will be a closed
one, and the condition a"= 1 that the last exterior angle will also be
2?r/n.

In the reasoning of last paragraph a, instead of being coB29r/n
+ isin2ir/w, may have any one of the values cos2A;ir/n + tsin2fo/n,
where k has any integral value from to (n - 1). It would thus
appear that there are n distinct species of equiangular n-gons, distin-
guished by the magnitude of the exterior angle ; but some of these
are degenerate cases and others are not distinct.

Thus if a = cosO + tsinO

then (ii + a2+ +a^^0;

and since none of the sides are negative each must be zero.

Again if n is even and

a^cosir + tsinir,
the condition becomes

ai-Â«3+ +Â«*-!- aÂ« = 0;

and the polygon is a flat one, such as ABODEFQHA (fig. 52).
Further the two values

a = cos2nr/w + tsin2rfr/w,
a = cos2(n - r)v/n + isin2(n - r)ir/n,
give the same polygon, the angles regarded as the exterior angles in
the two cases being the conjugates of one another.

All the other polygons are distinct ; and hence it follows that if
n is odd there are J(w- 1), and if n is even i(w - 2), equiangular
n-gons of different species ; all of which, of course, with the excep.
tion of one, are crossed polygons.

A particular case of the above is that in which the polygons are
regular. In this case, however, if n is not prime, some of the poly-
gons cpnsist of those with a smaller number of sides taken several
times over. Thus one of the regular octagons consists of a square

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taken twice over. These may be called degenerate cases ; but there
are always certain regular w-gons which are not degenerate cases ;
one such being the ordinary regular non-crossed n-gon. The number
of non-degenerate regular polygons is easily seen to be half the

number of special roots of the equation af* - 1 = 0. Now if jt?, 5^, r

are primes, and if n=p^g^r^ , the number of special roots of

this equation is

nil-l/p)il-l/q)

Hence the number of regular w-gons is

Hi-i/;>)(i-i/g)

where jt?, q, r are the prime factors of n, including n if n be

prime.

The number of regular w-gons may also be seen very easily by
consideration of a circle divided into n equal parts. Each point may
be joined to the next, or to the next but one, or to the next but
two, and so on ; and a regular 7^-gon will be formed in each case
provided every point of division is included.

If an equiangular n-gon can be formed with the sides, o^,

Os a^ taken in a defbiite order, then an equiangular polygon of

any other species may be formed by taking the same sides in a
different order, namely in the order, 1st, (r + 1)*^, (2r 4- 1)***. . . provided
neither of the polygons be of a species which degenerates when it
becomes regular. For if an equiangular polygon of the kind con-
sidered can be formed with the sides o^, a^ aâ€ž, taken in that

order, then

ai + a^a+ +aÂ«a"-* = 0,

where a is a special root of the equation cc" - 1 = 0. If )8 is any
other special root of this equation, then

^ = 0", )SÂ« = a% &c.,

Oi + ar+ii8 + Â«2r+ii8'+ =0;

which shows that an equiangular polygon may be formed with the
same sides taken in the order, Oi, a^^, o^^i . . . Figs. 53, 54, 55,
represent the three species of equiangular nonagons which can be
formed by taking the same lines in different orders. The first is the
ordinary non-crossed nonagon the exterior angle of which is 27r/9 ;
the second is formed from the first by taking every second side, and
has iv/9 for its exterior angle ; and the third is formed from the
first by taking every fourth side, and has 87r/9 for its exterior angle.
It is in general impossible to form an equiangular nonagon of the

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remaining (third) species with the same sides, as this species
degenerates when it becomes regular, becoming in that case an
equilateral triangle taken thrice over. The same thing is indicated
by the ÂŁact that it is impossible to get all the sides of a nonagon by
going round it and taking every third side.

The propositions of last paragraph may also be proved by means
of the laws of composition of vectors. For if vectors be drawn
through any point parallel to the sides of an equiangular polygon^
these vectors will make equal angles with one another, and if they

be compounded in the order, 1st, (r+1)***, (2r+l)*** they will

again form an equiangular polygon, provided all the lines are
included, when they are compounded in this way, that is, provided n
be not a multiple of r.

It may now be shown that conversely an equiangular polygon

may be formed with the sides Oj, a, a^, if the matrix of the cir-

culant 0(0103 oâ€ž) vanishes.

It is sufficient to show that if this matrix vanishes, then

Oi + 02a+ +o^a''~"^ = 0, where a"=l.

Now if ;the above matrix vanishes, then the circulant 0(oiOa o,)

or this divided by (o^ + o^ + +0^) also vanishes.

But 0(010^ oâ€ž) = n(oi + 02a^+ +oâ€žaP"*),

where a^ is a root of aJ" - 1 = 0, and r has every value from 1 to w.

Hence ai-^ro^a-^ +oâ€ža*^^ = 0,

where a is some root of aJ" - 1 = ; and therefore an equilateral
triangle may be formed with the sides taken in the above order, the
exterior angle being one of the values of ^hrjn.

It is obvious that the sides may not be taken in any arbitrary
order and that in general only one of the angles 2kirjn may be taken.
As a matter of fact, if the above-mentioned matrix vanishes, at

least two of the factors (oj + Oaa^-l- +oâ€žaJ~^) must vanish;

namely those in which the quantities a^ are conjugate imaginaries.
A possible case when n is even, included in the above, is that of the
flat polygon mentioned before. This case arises when a is equal to
w and the factor which vanishes is then {ai-a2'\-az- )â€˘

The condition that o^, Og o^ be the sides of an equiangular

polygon may be represented as follows : â€”

Mat.0(oia2 aÂ«) = 0.

Now, if the sides taken in the order o^, a,.^i, o^r+i > form an equi-
angular polygon, which they will do if r is not a factor of n, then
Mat.C(aiO,+iO^+, .) = 0.

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Hence if one of these matrices vanishes so mast the other. Tn fact
it may easily be proved by means of the identity

0(aiaa a^) = n(ai + Oja + Oja' )

that the two circulants are equal ; that is the circulant 0(aias a^)

is not altered if the letters be written in the order, 1st, (r+1)^,
(2r +1)*** &c., if r be not a factor of n.

It follows from the above that if

0(aia, aâ€ž) = 0, and Oi-k-a^-k- +aÂ« is not equal to 0,

then Mat.O(aias aÂ«) = 0.

For in this case, for some value of a

ai + flBaa+ +aâ€ža'*-^ = 0;

and therefore,''as has been seen before, Mat 0(0103 â‚¬tn) â€” 0*

It is assumed that if n is even Oi-Oz-k-a^- is not equal to 0;

and that o^, o, o^ are all real.

Again, if three consecutive sides be increased by a, - 2a?cosA, x,
the polygon still remains equiangular, A being the exterior angle of
the polygon. Hence if the first matrix of a circulant vanish and if
three consecutive letters p, q^ r, be changed to p + x, p- 2a:cos A,
r + Xj where x is arbitrary and A is a certain one of the angles 2A;7r/n,
the circulant will still vanish.

This may also be proved analytically ; for let the factor of

0(0103 oj which vanishes be

Oi + Oaa+ oâ€ža"-^

where a = co^^pTrjn + isni2pirjn.

Then o^+ia*" + o^ja'^"^ + o^.+sa'^

becomes ar+i***" + ^r+g**^^ + ^r+S*^**^'

+ a'"(fl5 - 2iC(icos2pn-/n + sea') ;
and X - 2a;acos2/Mr/w + xa^

= x{l- 2(cos2/?7r/n + ism2pirln)cos2pTr/n + cosipir/n + isinipir/n}

= a;{(l - 2cos^2jE?7r/7i + cos4cpir/n) + {{miipir/n - sin4j97r/w)}

= 0.

Oonversely this transformation will indicate which of the factors

of 0(oiO3 oâ€ž) vanishes when the circulant itself vanishes.

Particular cases. In the simpler cases the general conditions given
above reduce to the following : â€”
Triangle 01 = 02 = 03.

Quadrilateral 01 = ^3') 03 = 04.

Pentagon. {^a-{b + c + d + e)+ j5{b-c-d + e)}-=^0;

{{b - ef + (<; - cf) -\- Jmb- ef -{d- cY) = 0.

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By means of these equations it may easily be seen that an equi-
angular pentagon can only have its sides commensurable if it be
regular.

Hexagon. a, + Oa = ^4 + Os ;

Oj + Â«8 = ^ + Â®Â«Â«
Octagon, (aa-Â«4-Â«Â« + Â«8)+ N/2(ai - ^s) = ;

(aa + a4-08-Â«8)+ ^2(03 - o^) = 0.
In an equiangular octagon with commensurable sides, opposite
sides are equal.
Decagon, ^al - Og) + (oa ~ Oj + 04 - a^- o^ + Og - Oj + 0,0)

+ ^5(0,4- a8-04-a5-ay-a8 + 09 + a,o) = 0;
(01 + 05-07 -Â«io)' + K + a4-Â«8-Â«Â»)'

+ V5{(08 + a4-08-Oj)'-(o, + 05-Oy-Oio)'}-0.

In an equiangular decagon with commensurable sides,
Oj - OeÂ«=07 - aa = Oj - 08==0j - 04 = 05 - Oio ;
where o, and a^ o, and o^, <fec., are opposite sides.
Dodecagon. 2(oi - o^) + (o, - Oj - Oj + o^) + ^3(o, - Oj - Og + 0^) =- ;

2(04 - Oio) + (O, + Og - Os - Ou) + J3{<h + O5 - Oj - Ou) = 0.

In an equiangular dodecagon with commensurable sides,
o, -03 = 010-04 =3 Og-Oi2;Oi-Oy = Oj-o, = Oa-Ou;
where o^ and 03, 04 and o^q, <fec., are opposite sides.

Third Meeting, Ja/nua/ry lith, 1887.

W. J. Macdonald, Esq., M.A., Vice-President, in the Ohair.

On Certain Inverse Boolette Problems.

By Pbofessob Chrystal.

The problem of designing cams or centrodes to produce any
given motion in one plane is one of some practical importance ; and
it seems worth while to illustrate by examples some simple methods
by which the solution can in certain cases be arrived at. These
methods are founded, for the most part, on the use of the so-called
Pedal Equation (or j9-r-equation), which has great advantages in the
present investigation, inasmuch as it depends on the form but not
on the position of the curve which it represents.

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Few of the results arrived at are absolutely new. Most of them
have been found directly by Clerk Maxwell in a paper published in
the Transactions of the Rdyal Society of Edinburgh, Vol. XVI.,
1849.

If we suppose a plane 11 to slide upon a fixed plane 11', it is
obvious that the motion of IT is determined if the space loci of two
points of IT, say P and Q, be given. There will therefore be an in-
finite number of ways of causing 11 to move so that any point P in it
may have a given locus. In other words, the problem to generate a
given plane curve as a roulette is indeterminate.

I. If, however, there be given a fixed curve 0', then it is a deter-
minate problem to find what curve in 11 must roll on 0' in order
that the point P in 11 may trace a given curve R.

II. Again, if there be given a curve in II, then it is a deter-
minate problem to find a curve C such that, if roll on C, then P
shall trace a given curve R.

Owing to the fact that the point P is fixed in 11, and thus afibrds
an origin of reference for the curve 0, or body centrode, as it is called,
the first of these problems is in general easier than the second. We
can at all events in general find, without much difficulty, an equation
connecting the radius vector (r) from P and the perpendicidar (p)
from P on the tangent to 0.

In fact, if K (Fig. 56) be the point of contact of the body and
space centrodes and 0', then, if P be the corresponding position of
the point which traces out the given curve B, we know that PK is
normal to R. Moreover, the tangent to 0' at K is the tangent to C
at K. Hence, if PM be perpendicular to this tangent, we have (P
being fixed with reference to 0) PK = r, PM =p.

Now the two curves 0' and JEl are given in the plane 11' ; and from
their properties we can deduce a relation between PM, and PK, P
being a variable point on R, and PK normal to R.

If this relation be/(PM, PK) = 0, then the jo-r-equation to the
curve 0, with respect to P as origin, will be

AP.r)^0 (1).

Since l/p*= l/f' + (dr/r'd6)', (1) gives us the differential equation

The obtaming of the relation (1) is a comparatively simple matter
in many cases ; but as a rule the integration of the equation (2) pre-
sents great difficulty.

Digitized by