Student #: 65488983

Course: Math 309

__Proposition # 14__

*The surface of any isosceles cone excluding the base is equal to
a circle whose radius is a mean proportional*
*between the side of the cone ( a generator - d) and the radius (
r ) of the circle which is the base of the cone*

Instead of using the proof from the book of Archimedes, I use a different
version which is much shorter.

Let S be the surface of the cone excluding the base.

Let B be the surface of the base of the cone.

Let O be the center of the base

Let T be the apex of the cone

Let A be a point on the base circle

Let OA = r and TA = d

TO is perpendicular to the base circle at O ===> angle
TAO is the angle between the surrounding surface of the cone

and the base circle

===> B = S * cos (TAO)

Since cos(TAO) = OA / TA = r / d. ===>
B = S * (r / d) or S = B * ( d / r) = pi * ( r * r) * ( d / r) =
pi * d * r

or S =
pi * d * r.

So S is equal to the area of a circle whose radius is the square root of d*r

__Proposition # 16__

If an isosceles cone be cut by a plane parallel to the base, the portion
of the surface of the cone between the parallel planes is

equal to a circle whose radius is a mean proportional between (1) the
portion of the side of the cone intercepted by the parallel

planes and (2) the line which is equal to the sum of the radii of the
circles in the parallel planes

Let OAB be a triangle through the axis of a cone, DE its intersection
with the plane cutting off the frustum, and OFC the axis of the

cone. Then the surface of the cone OAB is equal to a circle whose radius
is equal to square root of (OA* AC) ( proposition 14)

Similarly, the surface of the cone ODE is equal to a circle whose radius
is equal to the square root of (OD * DF)

The surface of the frustum is equal to the difference between the two
circles

Now:

OA.AC - OD.DF = (OD + DA) AC - OD. DF = OD. AC + DA. AC - OD.
DF

Since two triangle ODF and OAC are similar

OD. AC = OA.DF and OA / AC = OD / DF

Hence

OA. AC - OD.DF = DA.AC + DA.DF = DA (AC + DF)

And since circles are to one another as the squares of their radii,
it follows that the difference between the circles whose radii

are sqrt (OA.AC), sqrt ( OD.DF) respectively is equal to a circle whose
radius is sqrt (DA ( AC + DF))

Therefore, the surface of the frustum is equal to this circle

__Proposition # 21__

A regular polygon of an even number of sides being inscribed in a circle,
as ABC...A'...C'B'A, so that AA' is a diameter

if two angular points next but one to each other, as B, B', be joined,
and the other lines parallel to BB' and joining pairs of

angular points be drawn, as CC', DD'..., then

(BB' + CC' + .....): AA' = A'B: BA

Let BB', CC', DD' .... meet AA' in F, G, H, ...; and let CB', DC',....
be joined meeting AA' in K, L .... respectively.

Then CB', DC',.... are parallel to one another and to AB.

__Prove CB', DC' parallel to each other and to AB as followings:__*Since arc B'C' = arc C'D' = ( 30 degree) ==> angle KCG = angle
GCL (1)*
*Since AA' is diameter of the circle ==> AA' _|_ CC' at G (2)*
*From (1) and (2) ==> two triangles KCG and GCL are equal*
*
==> angle CKG = angle CLG (3)*

*Two right triangles LGC' and CGL are equal because:*
*
CG = GC'*
*
GL common*
*==> angle CLG = angle C'LG (4)*
*From (3) and (4) ==> angle CKG = angle GLC' ==> CK // LC'
==> CB' // DC' (*)*
*Similarly, we will have AB // CB' // DC'*

Hence, by similar triangles:

BF / FA =
B'F / FK

= CG / GK

= G'G / GL

......................................

= E' I / IA'

and summing the antecedents and consequence respectively, we have:

(BB' + CC' + ....) / AA' = BF / FA

Two right triangles AFB and ABA' is similar because they have the same
angle BAF ( = BAA')

so that BF / FA = A'B / BA

==> (BB' + CC' + ....) / AA' = BF / FA = A'B / BA

__Proposition # 22__

I*f a polygon be inscribed in a segment of a circle LAL' so that all
its sides excluding the base are equal and their number even,*
*as LK.....A.....K'L', A being the middle point of the segment, and
if the lines BB', CC', parallel to the base LL' and joining pairs*
*of angular points be drawn, then*
* (BB' + CC' + ......+ LM) : AM = A'B : BA*
*where M is the middle point of LL' and AA' is the diameter through
M*

Joining CB', DC', KD', LK' as in the last proposition, and supposing
that they meet AM in P,Q,J, R

while BB', CC',DD', KK' meet AM in F, G, O, H we have, by similar triangles,

BF / FA = B'F / FP

= CG / PG

= C'G / GQ

= DO / OQ

= D'O / OJ

.= KH / HJ

= K'H / HR

= LM / RM

and summing the antecedents as consequence, we obtain

( BB' + CC' + .....+ LM)
/ AM = BF / FA = A'B / BA ( using result from proposition # 21)

__Proposition # 23__

Take a great circle ABC... of a sphere, inscribe in it a regular polygon
whose sides are a multiple of four in number.

Let AA', MM' be diameters at right angles and joining opposite angular
points of the polygon. Then if the polygon and

great circle revolve together about the diameter AA', the angular points
of the polygon, except A, A', will describe circles

on the surface of the sphere at right angles to the diameter AA'. Also
the sides of the polygon will describe portions of

conical surfaces, e.g. BC will describe a surface forming part of a
cone whose base is a circle about CC' as diameter

and whose apex is the point in which CB, C'B' produced meet each other
and the diameter AA'.

Comparing the hemisphere MAM' and that half of the figure described
by the revolution of the polygon which is included in the

hemisphere, we see that the surface of the hemisphere and the surface
of the inscribed figure have the same boundaries in one

plane (viz. the circle on MM' as diameter), the former surface entirely
includes the latter, and they are both concave in the same

direction.

In other words, according to Euclidean geometry, the shortest distance
between two points is the line segment goes through these

two points. So, for instance, the arc length connects B and C is longer
than the line segment BC. When arc length BC rotate around

AA', it will form a part the sphere that is limited by two circles
whose diameters are BB' and CC'. Meanwhile, the line segment BC

will form a part of the cone that has the same bases as the partial
sphere above. Since the arc length BC is longer than the line

segment BC, it will create a larger area when they both rotate around
AA'.

*Therefore the surface of the hemisphere is greater than that of the
inscribed figure,; and the same is true of the other*
*halves of the figures. Hence the surface of the sphere is greater
than the surface described by the revolution of the polygon*
*inscribed in the great circle about the diameter of the great circle.*

__Proposition # 24__

If a regular polygon AB...A'...B'A, the number of whose sides is a multiple
of four, be inscribed in a great circle of a sphere, and

if BB' subtending two sides be joined, and all the other lines parallel
to BB' and joining pairs of angular points be drawn, then

the surface of the figures inscribed in the sphere by the revolution
of the polygon about the diameter AA' is equal to a circle the

square of whose radius is equal to the rectangle

BA (BB' + CC' + ...).

The surface of the figure is made up of the surfaces of parts of different
cones

Now the surface of the cone ABB' is equal to a circle whose radius is sqrt(BA. 1/2 BB') (Proposition # 16)

The surface of the frustum BB'C'C is equal to a circle of radius sqrt (BC . 1/2 (BB' + CC')) and so on

It follows, since BA = BC = .. that the whole surface is equal to a circle whose radius is equal to

sqrt( BA (BB' + CC'+ .....+
MM' + ... + YY'))

__Proposition # 25__

*The surface of the figure inscribed in a sphere as in the last propositions,
consisting of portions of conical surfaces, is less*
*than four times the greatest circle in the sphere.*

Let AB...A'...B'A be a regular polygon inscribed in a great circle, the number of its sides being a multiple of four

As before, let BB' be drawn subtending two sides, and CC',...,YY'
parallel to BB'.

Let R be a circle such that the square of its radius is equal to

AB
( BB' + CC'+ ....+ YY').

so that the surface of the figure inscribed in the sphere is equal
to R (Proposition #24)

Now

(BB' + CC'+ .....+ YY')
/ AA' = A'B / AB (Proposition # 21)

whence

AB ( BB' + CC'+ ....+ YY') = AA'. A'B

Hence (radius of R)(radius of R) = AA'.A'B
< AA'.AA'

==> the surface of the inscribed figure ( or circle R) < pi (2. AA'/2) (2. AA'/2) = 4 circle AMA'M'

*Therefore the surface of the inscribed figure, or the circle R, is
less than four times of the circle AMA'M'*

__Proposition # 26__

*The figure inscribed as above in a sphere is equal in volume to a
cone whose base is a circle equal to the surface of the*
*figure inscribed in the sphere and whose height is equal to the
perpendicular drawn from the center of the sphere to one*
*side the polygon*

Suppose, as before, that AB....A'...B'A is the regular polygon inscribed
in a great circle, and let BB', CC', ... be joined.

With apex O construct cones whose bases are the circles on BB', CC',
.... as diameters in planes perpendicular to AA'.

Then OBAB' is a solid rhombus, and its volume is equal to a cone whose
base is equal to the surface of the cone ABB'

and whose height is equal to the perpendicular from O on AB. Let the
length of the perpendicular be p. [Proposition 18]

Volume of OBAB' = Surface of the cone ABB' * p (1)

Again, if CB, C'B' produced meet in T, the portion of the
solid figure which is described by the revolution of the triangle

BOC about AA' is equal to the difference between the rhombi OCTC' and
OBTB', i.e. to a cone whose base is equal to

the surface of the frustum BB'C'C and whose height is p. [Proposition
20]

Volume created by rotating OCB around AA' = surface of frustum BB'C'C * p (2)

Proceeding in this manner we have:

Volume created by rotating OCM around AA' = surface of frustum CC'M'M
* p (3)

Volume created by rotating OMN around AA' = surface of frustum NN'M'M
* p (4)

Volume created by rotating ONY around AA' = surface of frustum NN'YY'
* p (5 )

Volume of the rhombus OYA'Y' = surface of the cone A'YY'
* p (6)

Adding up (1),(2),(3),(4),(5),(6) we have:

*The volume of the solid of revolution is equal to p * (surface of
cone ABB' + surface of CC'MM' +.....+ surface of cone A'YY')*
*or equal to p * the surface of the solid created by
rotating the polygon around AA' or equal to a cone whose base is equal
to*
*the surface of the solid and whose height is p.*

__Proposition # 27__

*The figure inscribed in the sphere as before is less than four times
the cone whose base is equal to a great circle of the sphere*
*and whose height is equal to the radius of the sphere.*

By Prop. 26 the volume of the solid figure is equal to a cone whose
base is equal to the surface of the solid and whose height is p,

the perpendicular from O on any side of the polygon. Let R be such
a cone.

Take also a cone S with base equal to the great circle, and height
equal to the radius, of the sphere.

Now, since the surface of the inscribed solid is less than four times
the great circle (Prop 25), the base of the cone is less than four

times the base of the cone S.

Also the height (p) of R is less than the height of S.

Therefore the volume of R is less than four times that of S; and the
proposition is proved

__Proposition # 28__

*Let a regular polygon, whose sides are a multiple of four in number,
be circumscribed about a great circle of a given sphere, as*
*AB...A'...B'A; and about the polygon describe another circle, which
will therefore have the same center as the great circle of*
*the sphere. Let AA' bisect the polygon and cut the sphere in a,
a'.*
*If the great circle and the circumscribed polygon revolve together
about AA', the great circle will describe the surface of a sphere*
*the angular points of the polygon except A, A' will move around
the surface of a larger sphere, the points of contact of the sides*
*of the polygon with the great circle of the inner sphere will describe
circles on that sphere in planes perpendicular to AA', and*
*the sides of the polygon themselves will describe portions of conical
surfaces. The circumscribed figure will thus be greater than*

Let any side, as BM, touch the inner circle in K, and let K' be the
point of contact of the circle with B'M'. Then the circle described

by the revolution of KK' about AA' is the boundary in one plane of
two surfaces.

(1) the surface formed by the revolution of the circular segment KaK',
and (2) the surface formed by the revolution of the part

KB...A...B'K' of the polygon.

Now the second surface entirely includes the first and they are both
concave in the same direction;

therefore the second surface is greater than the first.

The same is true of the portion of the surface on the opposite side
of the circle on KK' as diameter.

Hence, adding, we see that the surface of the figure circumscribed
to the given sphere is greater than

that of the sphere itself.

__Proposition # 29__

*In a figure circumscribed to a sphere in the manner shown in the
previous proposition the surface is equal to a circle the square*
*on whose radius is equal to AB ( BB' + CC' + .........).*

For the figure circumscribed to the sphere is inscribed in a larger
sphere, and the proof of Prop. 24 applies

__Proposition # 30__

The surface of a figure circumscribed as before about a sphere is greater than four times the great circle of the sphere.

Let AB....A'...B'A be the regular polygon of 4n sides which by its revolution
about AA' describes the figure circumscribing

the sphere of which ama'm' is a great circle. Suppose aa', AA' to be
in one straight line.

Let R be a circle equal to the surface of the circumscribed solid.

Now (BB' + CC' + ...) : AA' = A'B : BA (as in Prop. 21)

so that AB ( BB' + CC' + ....) = AA'.A'B

Hence ( radius of R ) = sqrt ( AA'. A'B ) (Prop. 29)

> A'B

But A'B = 2 OP, where P is the point in which AB touches the circle ama'm'.

Therefore (radius of R) > (diameter of circle ama'm');

whence R, and therefore the surface of the circumscribed solid, is
greater than four times the great circle of the given sphere.

__Proposition # 31__

*The solid of revolution circumscribed as before about a sphere is
equal to a cone whose base is equal to the surface of the solid*
*and whose height is equal to the radius of the sphere.*

The solid is, as before, a solid inscribed in a larger sphere; and
since the perpendicular on any side of the revolving polygon is equal

to the radius of the inner sphere, the proposition is identical with
Prop. 26

COR: The solid circumscribed about the smaller sphere is greater than
four times the cone whose base is a greater circle of the sphere

and whose height is equal to the radius of the sphere

For, since the surface of the solid is greater than four times the greater
circle of the inner sphere ( Prop 30 ), the cone whose base is

equal to the surface of the solid and whose height is the radius of
the sphere is greater than four times the cone of the same height which

has the great circle for base (lemma 1)

Hence, by the proposition, the volume of the solid is greater than
four times the latter cone.