---
book: 13
number: 16
id: "XIII.16"
kind: "construction"
uses: ["[[book-11/proposition-6]]", "[[book-1/proposition-33]]", "[[book-13/proposition-10]]", "[[book-4/proposition-15]]", "[[book-13/proposition-9]]", "[[book-6/proposition-8]]", "[[book-3/proposition-31]]", "[[book-13/proposition-3]]", "[[book-13/proposition-11]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:13.prop.16"
license: "CC-BY-SA-4.0"
---

# XIII.16

*To construct an icosahedron and comprehend it in a sphere*, *like the aforesaid figures; and to prove that the side of the icosahedron is the irrational straight line called minor.*

## Proof

Let the diameter *AB* of the given sphere be set out, and let it be cut at *C* so that *AC* is quadruple of *CB*, let the semicircle *ADB* be described on *AB*, let the straight line *CD* be drawn from *C* at right angles to *AB*, and let *DB* be joined; let the circle *EFGHK* be set out and let its radius be equal to *DB*, let the equilateral and equiangular pentagon *EFGHK* be inscribed in the circle *EFGHK*, let the circumferences *EF*, *FG*, *GH*, *HK*, *KE* be bisected at the points *L*, *M*, *N*, *O*, *P*, and let *LM*, *MN*, *NO*, *OP*, *PL*, *EP* be joined.

Therefore the pentagon *LMNOP* is also equilateral, and the straight line *EP* belongs to a decagon.

Now from the points *E*, *F*, *G*, *H*, *K* let the straight lines *EQ*, *FR*, *GS*, *HT*, *KU* be set up at right angles to the plane of the circle, and let them be equal to the radius of the circle *EFGHK*, let *QR*, *RS*, *ST*, *TU*, *UQ*, *QL*, *LR*, *RM*, *MS*, *SN*, *NT*, *TO*, *OU*, *UP*, *PQ* be joined.

Now, since each of the straight lines *EQ*, *KU* is at right angles to the same plane, therefore *EQ* is parallel to *KU*. [[book-11/proposition-6|XI. 6]]

But it is also equal to it; and the straight lines joining those extremities of equal and parallel straight lines which are in the same direction are equal and parallel. [[book-1/proposition-33|I. 33]]

Therefore *QU* is equal and parallel to *EK*.

But *EK* belongs to an equilateral pentagon; therefore *QU* also belongs to the equilateral pentagon inscribed in the circle *EFGHK*.

For the same reason each of the straight lines *QR*, *RS*, *ST*, *TU* also belongs to the equilateral pentagon inscribed in the circle *EFGHK*; therefore the pentagon *QRSTU* is equilateral.

And, since *QE* belongs to a hexagon, and *EP* to a decagon, and the angle *QEP* is right, therefore *QP* belongs to a pentagon; for the square on the side of the pentagon is equal to the square on the side of the hexagon and the square on the side of the decagon inscribed in the same circle. [[book-13/proposition-10|XIII. 10]]

For the same reason *PU* is also a side of a pentagon.

But *QU* also belongs to a pentagon; therefore the triangle *QPU* is equilateral.

For the same reason each of the triangles *QLR*, *RMS*, *SNT*, *TOU* is also equilateral.

And, since each of the straight lines *QL*, *QP* was proved to belong to a pentagon, and *LP* also belongs to a pentagon, therefore the triangle *QLP* is equilateral.

For the same reason each of the triangles *LRM*, *MSN*, *NTO*, *OUP* is also equilateral.

Let the centre of the circle *EFGHK* the point *V*, be taken; from *V* let *VZ* be set up at right angles to the plane of the circle, let it be produced in the other direction, as *VX*, let there be cut off *VW*, the side of a hexagon, and each of the straight lines *VX*, *WZ*, being sides of a decagon, and let *QZ*, *QW*, *UZ*, *EV*, *LV*, *LX*, *XM* be joined.

Now, since each of the straight lines *VW*, *QE* is at right angles to the plane of the circle, therefore *VW* is parallel to *QE*. [[book-11/proposition-6|XI. 6]]

But they are also equal; therefore *EV*, *QW* are also equal and parallel. [[book-1/proposition-33|I. 33]]

But *EV* belongs to a hexagon; therefore *QW* also belongs to a hexagon.

And, since *QW* belongs to a hexagon, and *WZ* to a decagon, and the angle *QWZ* is right, therefore *QZ* belongs to a pentagon. [[book-13/proposition-10|XIII. 10]]

For the same reason *UZ* also belongs to a pentagon, inasmuch as, if we join *VK*, *WU*, they will be equal and opposite, and *VK*, being a radius, belongs to a hexagon; [[book-4/proposition-15|IV. 15, Por.]] therefore *WU* also belongs to a hexagon.

But *WZ* belongs to a decagon, and the angle *UWZ* is right; therefore *UZ* belongs to a pentagon. [[book-13/proposition-10|XIII. 10]]

But *QU* also belongs to a pentagon; therefore the triangle *QUZ* is equilateral.

For the same reason each of the remaining triangles of which the straight lines *QR*, *RS*, *ST*, *TU* are the bases, and the point *Z* the vertex, is also equilateral.

Again, since *VL* belongs to a hexagon, and *VX* to a decagon, and the angle *LVX* is right, therefore *LX* belongs to a pentagon. [[book-13/proposition-10|XIII. 10]]

For the same reason, if we join *MV*, which belongs to a hexagon, *MX* is also inferred to belong to a pentagon.

But *LM* also belongs to a pentagon; therefore the triangle *LMX* is equilateral.

Similarly it can be proved that each of the remaining triangles of which *MN*, *NO*, *OP*, *PL* are the bases, and the point *X* the vertex, is also equilateral.

Therefore an icosahedron has been constructed which is contained by twenty equilateral triangles.

It is next required to comprehend it in the given sphere, and to prove that the side of the icosahedron is the irrational straight line called minor.

For, since *VW* belongs to a hexagon, and *WZ* to a decagon, therefore *VZ* has been cut in extreme and mean ratio at *W*, and *VW* is its greater segment; [[book-13/proposition-9|XIII. 9]] therefore, as *ZV* is to *VW*, so is *VW* to *WZ*.

But *VW* is equal to *VE*, and *WZ* to *VX*; therefore, as *ZV* is to *VE*, so is *EV* to *VX*.

And the angles *ZVE*, *EVX* are right; therefore, if we join the straight line *EZ*, the angle *XEZ* will be right because of the similarity of the triangles *XEZ*, *VEZ*.

For the same reason, since, as *ZV* is to *VW*, so is *VW* to *WZ*, and *ZV* is equal to *XW*, and *VW* to *WQ*, therefore, as *XW* is to *WQ*, so is *QW* to *WZ*.

And for this reason again, if we join *QX*, the angle at *Q* will be right; [[book-6/proposition-8|VI. 8]] therefore the semicircle described on *XZ* will also pass through *Q*. [[book-3/proposition-31|III. 31]]

And if, *XZ* remaining fixed, the semicircle be carried round and restored to the same position from which it began to be moved, it will also pass through *Q* and the remaining angular points of the icosahedron, and the icosahedron will have been comprehended in a sphere.

I say next that it is also comprehended in the given sphere.

For let *VW* be bisected at *A'*.

Then, since the straight line *VZ* has been cut in extreme and mean ratio at *W*, and *ZW* is its lesser segment, therefore the square on *ZW* added to the half of the greater segment, that is *WA'*, is five times the square on the half of the greater segment; [[book-13/proposition-3|XIII. 3]] therefore the square on *ZA'* is five times the square on .

And *ZX* is double of *ZA'*, and *VW* double of ; therefore the square on *ZX* is five times the square on *WV*.

And, since *AC* is quadruple of *CB*, therefore *AB* is five times *BC*.

But, as *AB* is to *BC*, so is the square on *AB* to the square on *BD*; [[book-6/proposition-8|VI. 8]], [[book-5/definitions#Definition 9|V. Def. 9]] therefore the square on *AB* is five times the square on *BD*.

But the square on *ZX* was also proved to be five times the square on *VW*.

And *DB* is equal to *VW*, for each of them is equal to the radius of the circle *EFGHK*; therefore *AB* is also equal to *XZ*.

And *AB* is the diameter of the given sphere; therefore *XZ* is also equal to the diameter of the given sphere.

Therefore the icosahedron has been comprehended in the given sphere

I say next that the side of the icosahedron is the irrational straight line called minor.

For, since the diameter of the sphere is rational, and the square on it is five times the square on the radius of the circle *EFGHK*, therefore the radius of the circle *EFGHK* is also rational; hence its diameter is also rational.

But, if an equilateral pentagon be inscribed in a circle which has its diameter rational, the side of the pentagon is the irrational straight line called minor. [[book-13/proposition-11|XIII. 11]]

And the side of the pentagon *EFGHK* is the side of the icosahedron.

Therefore the side of the icosahedron is the irrational straight line called minor.

Porism. From this it is manifest that the square on the diameter of the sphere is five times the square on the radius of the circle from which the icosahedron has been described, and that the diameter of the sphere is composed of the side of the hexagon and two of the sides of the decagon inscribed in the same circle. Q. E. D.