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But SP = yo + SN? = (2ar — x?) + (a +Va6%)'

a
which reduces to
a- 6%

2 abo. (a IV a bo) x' + 2ao

SP =

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-6° + 2a Va? -62 .. (2)
which being identical with (1) when x = x', we have

SP = NP, and ..

SP = NP. Q. E. D. The geometrical proof, although easier, is not grounded on such obvious and general principles as this.

112. Euler, who invented this theorem, conceived it to be universally applicable; but we shall presently see that it is true only in a very particular class of polyedrons.

Case 1. Let the polyedron be without perforation, and consist of cne surface only.

Let F, S, E, be the number of faces, solid angles and edges of the polyedron, respectively.

Then supposing any one face projected upon a given plane in such a manner that the interior angles of the projection may be each less than two right angles, and all the others upon the same plane, so that their projections may be wholly within the former, (which is evidently possible), the projection thus formed of the polyedron will consist of F plane polygons, united by the S points of concourse of the E different sides, one of them circumscribing the rest. Now, supposing n, n,, N., ..... n; to represent the number of edges in the respective faces or polygons; then, since it can evidently make no difference in the sum of the inner angles of the interior polygons, whether or no any of them be greater than two right angles, (for they equally fill space about the points of concourse), we have 2n, R – 4R

4R

= 4 (S – n) R + 2n R – 4R
&c.
2nF R – 4R
Or a + m + 2 (F – 1) = 2S – N – 2.

2n, R

}

But, since each of the sides, except those of the circumscribing polygon, is common to two of the polygons, we have

n, + N2 + = 2 (E – n) .. 2E

2F = 28 :: S + F = E + 2...... (1) which shews the Theorem to be true in this case.

Case 2. Let the polyedron be supposed to be hollow, i. e., to have two surfaces consisting of plain faces, one interior with respect to the other.

Then, supposing s, f, e; s', f", é', to be the number of solid angles, faces and edges of these surfaces, we get, by case 1,

s + f = e + 22

s' + f =é + 25
and s + s' = S, f + g = F, and e + é=E

:. in this case

S + F= E + 4; and, generally, if there be h interior boundaries, we get

S + F = E + 2 (h + 1)..... (2).

Case 3. Let the polyedron be annular, or have a perforation.

Then, supposing a plane to cut it into two polyedrous of the first class, making (m) new edges, and :: (m) new solid angles, and two new faces in each, we have (using the above notation)

s + f = e + 22

$' + fré' + 2S
is + s + f + f = e+ e + 4
Or S + m + F + 4 = E + m + 4

::S + F=E;
and, generally, if there be p such perforations, we have
S + F = E + 2 - 2p.

(3).

Case 4. Suppose the polyedron formed by the union of two others of the first class, in making the planes of two of their unequal faces coincide, so that the face of the one may be wholly within that of the other.

Let S, F, E, be the number of solid angles, faces, and edges of

the compound polyedron: s, f,e; s', f, é', those of the components, then by case (1)

s + f = e + 2? ::8 + s + f + f = et é' + 4

s' + f' =é + 2S
But s + s' = S, f + f = F + 1, e té = E

:. S + F = E + 3; and similarly, if the polyedron be formed by the union of n others, upon any face of one of them, we have

S+F= E + 2 + n. Hence, generally, if ni, ng, Na, .... n be the number of such formations upon

each of m faces, we have S + F =E + 2 + m + x + m + Nm Hence, then, Euler's Theorem may be generalized, by stating it, If a Polyedron have (h) interior surfaces, (p) perforations, and Phy, Mong Ng, ... Nem augments upon m of its faces severally, then

S + F = E + 2 + 2 (h - p) + n+ no + п•

113. In the axis AB of the cycloid AP (Fig. 59,) of which C is the centre of its generating o, let CN=AM, then if MP be 1 AB, PN be joined, &c., as in the fig. required to shew that the sector ANP = A MQB.

Let AC = r, AM = x, PM = y, then the equation to the cycloid is y=AQ + QM = vers.-" x + N 2rx x2 x 2r

s v2rx — 22 x dx N a =SQM Xd. AM=area AMQ; i.e., Am P=AMQ. Hence sector ANP = AMP PNM Рq x AC CM X MQ

MN X PM = PM X AM

+

2 PM X (2AM – MN).- PQ X AC CM X MQ

-- Szdy = f*

X dx =

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2

2

2

2

MQ X AC CM X MQ

+ 2 = A BMQ. Q.E.D.

MB X QM

2

TANGENTS AND ASYMPTOTES.

114. Let x = AE,

then

y r. sin. a :. Subtangent =

ydx

= tan. X.

dy And the area = Sydx = r sdx . sin. x = c - r cos. x .

Let x = 0. Then c -- go?, and the area = 2 r cos. x; let x = AC, then the whole area = 72.

115. Let AP (Fig. 60,) be any curve whatever, and suppose ordinates PP I AN, to Le erected on AP as a line of abscissæ, tracing out the curve AP ; then T being the intersection of the tangents at P, P, and p'p the next position of the ordinate, if P'm' be drawn parallel to TPp we have, by similar A

PT : PP :: Pm : mp
or PT : ý :: d. AP : dy
and calling AP = s, we get
PT =
y'ds

(1) dy which will always give the value of PT, when y' is a function of s, and the nature of AP is given.

In the problem, y = as"

And AP is a circle,

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:: PT is known, and joining P'T we get the tangent PT. In the case of the cycloid AP is the generating circle, and n = 1, and ;. PT = PP'.

Hence / PPT = supplement of TPP = TPN = 2 TPA by property of the circle. See Prob. 26, p. 22.

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116. Let the base AB of the A ABC = a, and the angle A = nB, then AC being the radius vector p, and ZA = 0 the angle traced by it, we have p: a :: sin. B : sin. C = sin. (A + B)

:: sin. nd : sin. (n + 1) 0
a sin, ne
sin.(n+)

(1). the polar equation to the locus of C, which will be a straight line, lyperbola, &c., according as n = 1, 2, &c.

Now Subtangent (See Appendix to new edit. of Simpson, or Lacroix). pode

a sin. 2 no de n cos.nl. sin. n+1.6-(n+1) sin. n 0 cos. n +10 a sin.? no

the subtangent at any n sin. A sin. n 8. cos. n + 10 point (p, 0). P = 00, then C = - (n + 1) 0 = 0, :: 0

n+1' :. the subtangent to the asymptote is

a sin.

n+1

ST =

Let

n

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n + 1. sin.

n+1 which is also given in position by its inclination to AB, which is

-1 2

n+1 n+1

n

.

2

2

Hence drawing a I at its extremity, we shall have the asymptote required.

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