Proof: By Euclid
(related to Proposition: Prop. 13.16: Construction of Regular Icosahedron within Given Sphere)
- And let the circle $EFGHK$ be set down, and let its radius be equal to $DB$.
- And let the equilateral and equiangular pentagon $EFGHK$ have been inscribed in circle $EFGHK$ [Prop. 4.11].
- And let the circumferences $EF$, $FG$, $GH$, $HK$, and $KE$ have been cut in half at points $L$, $M$, $N$, $O$, and $P$ (respectively).
- And let $LM$, $MN$, $NO$, $OP$, $PL$, and $EP$ have been joined.
- Thus, pentagon $LMNOP$ is also equilateral, and $EP$ (is) the side of the decagon (inscribed in the circle).
- And let the straight lines $EQ$, $FR$, $GS$, $HT$, and $KU$, which are equal to the radius of circle $EFGHK$, have been set up at right angles to the plane of the circle, at points $E$, $F$, $G$, $H$, and $K$ (respectively).
- And let $QR$, $RS$, $ST$, $TU$, $UQ$, $QL$, $LR$, $RM$, $MS$, $SN$, $NT$, $TO$, $OU$, $UP$, and $PQ$ have been joined.
- And since $EQ$ and $KU$ are each at right angles to the same plane, $EQ$ is thus parallel to $KU$ [Prop. 11.6].
- And it is also equal to it.
- And straight lines joining equal and parallel (straight lines) on the same side are (themselves) equal and parallel [Prop. 1.33].
- Thus, $QU$ is equal and parallel to $EK$.
- And $EK$ (is the side) of an equilateral pentagon (inscribed in circle $EFGHK$).
- Thus, $QU$ (is) also the side of an equilateral pentagon inscribed in circle $EFGHK$.
- So, for the same (reasons), $QR$, $RS$, $ST$, and $TU$ are also the sides of an equilateral pentagon inscribed in circle $EFGHK$.
- pentagon $QRSTU$ (is) thus equilateral.
- And side $QE$ is (the side) of a hexagon (inscribed in circle $EFGHK$), and $EP$ (the side) of a decagon, and (angle) $QEP$ is a right angle, thus $QP$ is (the side) of a pentagon (inscribed in the same circle).
- For the square on the side of a pentagon is (equal to the sum of) the (squares) on (the sides of) a hexagon and a decagon inscribed in the same circle [Prop. 13.10].
- So, for the same (reasons), $PU$ is also the side of a pentagon.
- And $QU$ is also (the side) of a pentagon.
- Thus, triangle $QPU$ is equilateral.
- So, for the same (reasons), (triangles) $QLR$, $RMS$, $SNT$, and $TOU$ are each also equilateral.
- And since $QL$ and $QP$ were each shown (to be the sides) of a pentagon, and $LP$ is also (the side) of a pentagon, triangle $QLP$ is thus equilateral.
- So, for the same (reasons), triangles $LRM$, $MSN$, $NTO$, and $OUP$ are each also equilateral.
- Let the center, point $V$, of circle $EFGHK$ have been found [Prop. 3.1].
- And let $VZ$ have been set up, at (point) $V$, at right angles to the plane of the circle.
- And let it have been produced on the other side (of the circle), like $VX$.
- And let $VW$ have been cut off (from $XZ$ so as to be equal to the side) of a hexagon, and each of $VX$ and $WZ$ (so as to be equal to the side) of a decagon.
- And let $QZ$, $QW$, $UZ$, $EV$, $LV$, $LX$, and $XM$ have been joined.
- And since $VW$ and $QE$ are each at right angles to the plane of the circle, $VW$ is thus parallel to $QE$ [Prop. 11.6].
- And they are also equal.
- $EV$ and $QW$ are thus equal and parallel (to one another) [Prop. 1.33].
- And $EV$ (is the side) of a hexagon.
- Thus, $QW$ (is) also (the side) of a hexagon.
- And since $QW$ is (the side) of a hexagon, and $WZ$ (the side) of a decagon, and angle $QWZ$ is a right angle [Def. 11.3] , [Prop. 1.29], $QZ$ is thus (the side) of a pentagon [Prop. 13.10].
- So, for the same (reasons), $UZ$ is also (the side) of a pentagon - inasmuch as, if we join $VK$ and $WU$ then they will be equal and opposite.
- And $VK$, being (equal) to the radius (of the circle), is (the side) of a hexagon [Prop. 4.15 corr.] 0.
- Thus, $WU$ (is) also the side of a hexagon.
- And $WZ$ (is the side) of a decagon, and (angle) $UWZ$ (is) a right angle.
- Thus, $UZ$ (is the side) of a pentagon [Prop. 13.10].
- And $QU$ is also (the side) of a pentagon.
- triangle $QUZ$ is thus equilateral.
- So, for the same (reasons), each of the remaining triangles, whose bases are the straight lines $QR$, $RS$, $ST$, and $TU$, and apexes the point $Z$, are also equilateral.
- Again, since $VL$ (is the side) of a hexagon, and $VX$ (the side) of a decagon, and angle $LVX$ is a right angle, $LX$ is thus (the side) of a pentagon [Prop. 13.10].
- So, for the same (reasons), if we join $MV$, which is (the side) of a hexagon, $MX$ is also inferred (to be the side) of a pentagon.
- And $LM$ is also (the side) of a pentagon.
- Thus, triangle $LMX$ is equilateral.
- So, similarly, it can be shown that each of the remaining triangles, whose bases are the (straight lines) $MN$, $NO$, $OP$, and $PL$, and apexes the point $X$, are also equilateral.
- Thus, an icosahedron contained by twenty equilateral triangles has been constructed.
- So, it is also necessary to enclose it in the given sphere, and to show that the side of the icosahedron is that irrational (straight line) called minor.
- For, since $VW$ is (the side) of a hexagon, and $WZ$ (the side) of a decagon, $VZ$ has thus been cut in extreme and mean ratio at $W$, and $VW$ is its greater piece [Prop. 13.9].
- Thus, as $ZV$ is to $VW$, so $VW$ (is) to $WZ$.
- And $VW$ (is) equal to $VE$, and $WZ$ to $VX$.
- Thus, as $ZV$ is to $VE$, so $EV$ (is) to $VX$.
- And angles $ZVE$ and $EVX$ are right angles.
- Thus, if we join straight line $EZ$ then angle $XEZ$ will be a right angle, on account of the similarity of triangles $XEZ$ and $VEZ$. [Prop. 6.8].
- So, for the same (reasons), since as $ZV$ is to $VW$, so $VW$ (is) to $WZ$, and $ZV$ (is) equal to $XW$, and $VW$ to $WQ$, thus as $XW$ is to $WQ$, so $QW$ (is) to $WZ$.
- And, again, on account of this, if we join $QX$ then the angle at $Q$ will be a right angle [Prop. 6.8].
- Thus, the semicircle drawn on $XZ$ will also pass through $Q$ [Prop. 3.31].
- And if $XZ$ remains fixed, and the semicircle is carried around, and again established at the same (position) from which it began to be moved, then it will also pass through (point) $Q$, and (through) the remaining (angular) points of the icosahedron.
- And the icosahedron will have been enclosed by a sphere.
- So, I say that (it is) also (enclosed) by the given (sphere).
- For let $VW$ have been cut in half at $a$.
- And since the straight line $VZ$ has been cut in extreme and mean ratio at $W$, and $ZW$ is its lesser piece, then the square on $ZW$ added to half of the greater piece, $Wa$, is five times the (square) on half of the greater piece [Prop. 13.3].
- Thus, the (square) on $Za$ is five times the (square) on $aW$.
- And $ZX$ is double $Za$, and $VW$ double $aW$.
- Thus, the (square) on $ZX$ is five times the (square) on $WV$.
- And since $AC$ is four times $CB$, $AB$ is thus five times $BC$.
- And as $AB$ (is) to $BC$, so the (square) on $AB$ (is) to the (square) on $BD$ [Prop. 6.8], [Def. 5.9] .
- Thus, the (square) on $AB$ is five times the (square) on $BD$.
- And the (square) on $ZX$ was also shown (to be) five times the (square) on $VW$.
- And $DB$ is equal to $VW$.
- For each of them is equal to the radius of circle $EFGHK$.
- Thus, $AB$ (is) also equal to $XZ$.
- And $AB$ is the diameter of the given sphere.
- Thus, $XZ$ is equal to the diameter of the given sphere.
- Thus, the icosahedron has been enclosed by the given sphere.
- So, I say that the side of the icosahedron is that 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 circle $EFGHK$, the radius of circle $EFGHK$ is thus also rational.
- Hence, its diameter is also rational.
- And if an equilateral pentagon is inscribed in a circle having a rational diameter then the side of the pentagon is that irrational (straight line) called minor [Prop. 13.11].
- And the side of pentagon $EFGHK$ is (the side) of the icosahedron.
- Thus, the side of the icosahedron is that irrational (straight line) called minor.
∎
Thank you to the contributors under CC BY-SA 4.0! 
- Github:
-
- non-Github:
- @Fitzpatrick
References
Adapted from (subject to copyright, with kind permission)
- Fitzpatrick, Richard: Euclid's "Elements of Geometry"
