PROPOSITION 17.
Given two spheres about the same centre,
to inscribe in the greater sphere a polyhedral solid which does not touch the lesser sphere at its surface.
Let two spheres be conceived about the same centre
A; thus it is required to inscribe in the greater sphere a polyhedral solid which does not touch the lesser sphere at its surface.
Let the spheres be cut by any plane through the centre; then the sections will be circles, inasmuch as the sphere was produced by the diameter remaining fixed and the semicircle being carried round it; [
XI. Def. 14] hence, in whatever position we conceive the semicircle to be, the plane carried through it will produce a circle on the circumference of the sphere.
And it is manifest that this circle is the greatest possible, inasmuch as the diameter of the sphere, which is of course the diameter both of the semicircle and of the circle, is greater than all the straight lines drawn across in the circle or the sphere.
Let then
BCDE be the circle in the greater sphere, and
FGH the circle in the lesser sphere; let two diameters in them,
BD,
CE, be drawn at right angles to one another; then, given the two circles
BCDE,
FGH about the same centre, let there be inscribed in the greater circle
BCDE an equilateral polygon with an even number of sides which does not touch the lesser circle
FGH, let
BK,
KL,
LM
ME be its sides in the quadrant
BE. let
KA be joined and carried through to
N, let
AO be set up from the point
A at right angles to the plane of the circle
BCDE, and let it meet the surface of the sphere at
O, and through
AO and each of the straight lines
BD,
KN let planes be carried; they will then make greatest circles on the surface of the sphere, for the reason stated.
Let them make such, and in them let
BOD,
KON be the semicircles on
BD,
KN.
Now, since
OA is at right angles to the plane of the circle
BCDE, therefore all the planes through
OA are also at right angles to the plane of the circle
BCDE; [
XI. 18] hence the semicircles
BOD,
KON are also at right angles to the plane of the circle
BCDE.
And, since the semicircles
BED,
BOD,
KON are equal, for they are on the equal diameters
BD,
KN, therefore the quadrants
BE,
BO,
KO are also equal to one another.
Therefore there are as many straight lines in the quadrants
BO,
KO equal to the straight lines
BK,
KL,
LM,
ME as there are sides of the polygon in the quadrant
BE.
Let them be inscribed, and let them be
BP,
PQ,
QR,
RO and
KS,
ST,
TU,
UO, let
SP,
TQ,
UR be joined, and from
P,
S let perpendiculars be drawn to the plane of the circle
BCDE; [
XI. 11] these will fall on
BD,
KN, the common sections of the planes, inasmuch as the planes of
BOD,
KON are also at right angles to the plane of the circle
BCDE. [cf.
XI. Def. 4]
Let them so fall, and let them be
PV,
SW, and let
WV be joined.
Now since, in the equal semicircles
BOD,
KON, equal straight lines
BP,
KS have been cut off, and the perpendiculars
PV,
SW have been drawn, therefore
PV is equal to
SW, and
BV to
KW. [
III. 27,
I. 26]
But the whole
BA is also equal to the whole
KA; therefore the remainder
VA is also equal to the remainder
WA; therefore, as
BV is to
VA, so is
KW to
WA; therefore
WV is parallel to
KB. [
VI. 2]
And, since each of the straight lines
PV,
SW is at right angles to the plane of the circle
BCDE, therefore
PV is parallel to
SW. [
XI. 6]
But it was also proved equal to it; therefore
WV,
SP are also equal and parallel. [
I. 33]
And, since
WV is parallel to
SP, while
WV is parallel to
KB, therefore
SP is also parallel to
KB. [
XI. 9]
And
BP,
KS join their extremities; therefore the quadrilateral
KBPS is in one plane, inasmuch as, if two straight lines be parallel, and points be taken at random on each of them, the straight line joining the points is in the same plane with the parallels. [
XI. 7]
For the same reason each of the quadrilaterals
SPQT,
TQRU is also in one plane.
But the triangle
URO is also in one plane. [
XI. 2]
If then we conceive straight lines joined from the points
P,
S,
Q,
T,
R,
U to
A, there will be constructed a certain polyhedral solid figure between the circumferences
BO,
KO, consisting of pyramids of which the quadrilaterals
KBPS,
SPQT,
TQRU and the triangle
URO are the bases and the point
A the vertex.
And, if we make the same construction in the case of each of the sides
KL,
LM,
ME as in the case of
BK, and further in the case of the remaining three quadrants, there will be constructed a certain polyhedral figure inscribed in the sphere and contained by pyramids, of which the said quadrilaterals and the triangle
URO, and the others corresponding to them, are the bases and the point
A the vertex.
I say that the said polyhedron will not touch the lesser sphere at the surface on which the circle
FGH is.
Let
AX be drawn from the point
A perpendicular to the plane of the quadrilateral
KBPS, and let it meet the plane at the point
X; [
XI. 11] let
XB,
XK be joined.
Then, since
AX is at right angles to the plane of the quadrilateral
KBPS, therefore it is also at right angles to all the straight lines which meet it and are in the plane of the quadrilateral. [
XI. Def. 3]
Therefore
AX is at right angles to each of the straight lines
BX,
XK.
And, since
AB is equal to
AK, the square on
AB is also equal to the square on
AK.
And the squares on
AX,
XB are equal to the square on
AB, for the angle at
X is right; [
I. 47] and the squares on
AX,
XK are equal to the square on
AK. [
id.]
Therefore the squares on
AX,
XB are equal to the squares on
AX,
XK.
Let the square on
AX be subtracted from each; therefore the remainder, the square on
BX, is equal to the remainder, the square on
XK; therefore
BX is equal to
XK.
Similarly we can prove that the straight lines joined from
X to
P,
S are equal to each of the straight lines
BX,
XK.
Therefore the circle described with centre
X and distance one of the straight lines
XB,
XK will pass through
P,
S also, and
KBPS will be a quadrilateral in a circle.
Now, since
KB is greater than
WV, while
WV is equal to
SP, therefore
KB is greater than
SP.
But
KB is equal to each of the straight lines
KS,
BP; therefore each of the straight lines
KS,
BP is greater than
SP.
And, since
KBPS is a quadrilateral in a circle, and
KB,
BP,
KS are equal, and
PS less, and
BX is the radius of the circle, therefore the square on
KB is greater than double of the square on
BX.
Let
KZ be drawn from
K perpendicular to
BV.
Then, since
BD is less than double of
DZ, and, as
BD is to
DZ, so is the rectangle
DB,
BZ to the rectangle
DZ,
ZB, if a square be described upon
BZ and the parallelogram on
ZD be completed, then the rectangle
DB,
BZ is also less than double of the rectangle
DZ,
ZB.
And, if
KD be joined, the rectangle
DB,
BZ is equal to the square on
BK, and the rectangle
DZ,
ZB equal to the square on
KZ; [
III. 31,
VI. 8 and Por.] therefore the square on
KB is less than double of the square on
KZ.
But the square on
KB is greater than double of the square on
BX; therefore the square on
KZ is greater than the square on
BX.
And, since
BA is equal to
KA, the square on
BA is equal to the square on
AK.
And the squares on
BX,
XA are equal to the square on
BA, and the squares on
KZ,
ZA equal to the square on
KA; [
I. 47] therefore the squares on
BX,
XA are equal to the squares on
KZ,
ZA, and of these the square on
KZ is greater than the square on
BX; therefore the remainder, the square on
ZA, is less than the square on
XA.
Therefore
AX is greater than
AZ; therefore
AX is much greater than
AG.
And
AX is the perpendicular on one base of the polyhedron, and
AG on the surface of the lesser sphere; hence the polyhedron will not touch the lesser sphere on its surface.
Therefore, given two spheres about the same centre, a polyhedral solid has been inscribed in the greater sphere which does not touch the lesser sphere at its surface. Q. E. F.
PORISM.
But if in another sphere also a polyhedral solid be inscribed similar to the solid in the sphere
BCDE, the polyhedral solid in the sphere
BCDE has to the polyhedral solid in the other sphere the ratio triplicate of that which the diameter of the sphere
BCDE has to the diameter of the other sphere.
For, the solids being divided into their pyramids similar in multitude and arrangement, the pyramids will be similar.
But similar pyramids are to one another in the triplicate ratio of their corresponding sides; [
XII. 8, Por.] therefore the pyramid of which the quadrilateral
KBPS is the base, and the point
A the vertex, has to the similarly arranged pyramid in the other sphere the ratio triplicate of that which the corresponding side has to the corresponding side, that is, of that which the radius
AB of the sphere about
A as centre has to the radius of the other sphere.
Similarly also each pyramid of those in the sphere about
A as centre has to each similarly arranged pyramid of those in the other sphere the ratio triplicate of that which
AB has to the radius of the other sphere.
And, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents; [
V. 12] hence the whole polyhedral solid in the sphere about
A as centre has to the whole polyhedral solid in the other sphere the ratio triplicate of that which
AB has to the radius of the other sphere, that is, of that which the diameter
BD has to the diameter of the other sphere. Q. E. D.
