---
book: 12
number: 17
id: "XII.17"
kind: "construction"
uses: ["[[book-11/proposition-18]]", "[[book-11/proposition-11]]", "[[book-3/proposition-27]]", "[[book-1/proposition-26]]", "[[book-6/proposition-2]]", "[[book-11/proposition-6]]", "[[book-1/proposition-33]]", "[[book-11/proposition-9]]", "[[book-11/proposition-7]]", "[[book-11/proposition-2]]", "[[book-1/proposition-47]]", "[[book-3/proposition-31]]", "[[book-6/proposition-8]]", "[[book-12/proposition-8]]", "[[book-5/proposition-12]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:12.prop.17"
license: "CC-BY-SA-4.0"
---

# XII.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.*

## Proof

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; [[book-11/definitions#Definition 14|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*; [[book-11/proposition-18|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*; [[book-11/proposition-11|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. [[book-11/definitions#Definition 4|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*. [[book-3/proposition-27|III. 27]], [[book-1/proposition-26|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*. [[book-6/proposition-2|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*. [[book-11/proposition-6|XI. 6]]

But it was also proved equal to it; therefore *WV*, *SP* are also equal and parallel. [[book-1/proposition-33|I. 33]]

And, since *WV* is parallel to *SP*, while *WV* is parallel to *KB*, therefore *SP* is also parallel to *KB*. [[book-11/proposition-9|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. [[book-11/proposition-7|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. [[book-11/proposition-2|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*; [[book-11/proposition-11|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. [[book-11/definitions#Definition 3|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; [[book-1/proposition-47|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*; [[book-3/proposition-31|III. 31]], [[book-6/proposition-8|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*; [[book-1/proposition-47|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; [[book-12/proposition-8|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; [[book-5/proposition-12|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.