Construction of Regular Icosahedron within Given Sphere

Proof

 * Euclid-XIII-16.png

Let $AB$ be the diameter of the given sphere.

Let $AB$ be cut at $C$ where $AC = 4 \cdot CB$.

Let $ADB$ be a semicircle on the diameter $AB$.

Let $CD$ be drawn from $C$ perpendicular to $AB$.

Let $DB$ be joined.

Let the circle $EFGHK$ be set out whose radius equals $DB$.

From :
 * let the regular pentagon $EFGHK$ be inscribed within the circle $EFGHK$.

Let the arcs $EF, FG, GH, HK, KE$ be bisected at the points $L, M, N, O, P$.

Let $LM, MN, NO, OP, PL, EP$ be joined.

It follows that the pentagon $LMNOP$ is regular.

Also, the straight line $EP$ is the side of a regular decagon.

From the points $E, F, G, H, K$ let the straight lines $EQ, FR, GS, HT, KU$ be constructed perpendicular to the plane containing the circle $EFGHK$.

Let each of $EQ, FR, GS, HT, KU$ be equal to the radius of circle $EFGHK$.

Let $QR, RS, ST, TU, UQ, QL, LR, RM, MS, SN, NT, TO, OU, UP, PQ$ be joined.

We have that $EQ$ and $KU$ are both perpendicular to the same plane.

From :
 * $EQ \parallel KU$

But we also have that $EQ = KU$.

So from :
 * $QU = EK$

and:
 * $QU \parallel EK$

But $EK$ is the side of the regular pentagon $EFGHK$ which is inscribed within the circle $EFGHK$.

Therefore $QU$ is also the side of the regular pentagon $EFGHK$ which is inscribed within the circle $EFGHK$.

For the same reason:
 * each of the straight lines $QR, RS, ST, TU$ are the sides of the regular pentagon $EFGHK$ which is inscribed within the circle $EFGHK$.

Therefore $QRSTU$ is itself a regular pentagon which can be inscribed within a circle equal to the circle $EFGHK$.

We have that $\angle QEP$ is a right angle.

Therefore from :
 * $QE^2 + EP^2 = QP^2$

We have that $QE$ equals to the radius of circle $EFGHK$.

From :
 * $QE$ equals the side of a regular hexagon which could be inscribed within the circle $EFGHK$.

We also have that $EP$ equals the side of a regular decagon which could be inscribed within the circle $EFGHK$.

Therefore from :
 * $QP$ equals the side of the pentagon $EFGHK$.

For the same reason:
 * $PU$ equals the side of the pentagon $EFGHK$.

But also:
 * $QU$ equals the side of the pentagon $EFGHK$.

Therefore $\triangle QPU$ is equilateral.

For the same reason:
 * $\triangle QLR, \triangle RMS, \triangle SNT, \triangle TOU$ are all equilateral triangles.

We have that:
 * $QL$ and $LP$ have been proved to equal the side of the pentagon $EFGHK$.

We also have that $LP$ equal the side of the pentagon $EFGHK$.

Therefore $\triangle QLP$ is equilateral.

For the same reason:
 * $\triangle LRM, \triangle MSN, \triangle NTO, \triangle OUP$ are all equilateral triangles.

Let $V$ be the center of the circle $EFGHK$.

Let $VZ$ be constructed perpendicular to the plane containing the circle $EFGHK$.

Let $VZ$ be produced in the other direction as $VX$.

Let $VW$ be cut off $VZ$ equal to the side of a regular hexagon which could be inscribed within the circle $EFGHK$.

Let each of $VX$ and $WZ$ be made equal to the side of a regular decagon which could be inscribed within the circle $EFGHK$.

Let $Q, QW, UZ, EV, LV, LX, XM$ be joined.

We have that $VW$ and $QE$ are both perpendicular to the same plane.

From :
 * $VW \parallel QE$

But we also have that $EQ = KU$.

So from :
 * $EV = QW$

and:
 * $EV \parallel QW$

But $EV$ equals the side of a regular hexagon which could be inscribed within the circle $EFGHK$.

Therefore $QW$ also equals the side of a regular hexagon which could be inscribed within the circle $EFGHK$.

We have that $\angle QWZ$ is a right angle.

Therefore from :
 * $QW^2 + WZ^2 = QZ^2$

Therefore from :
 * $QZ$ equals the side of the pentagon $EFGHK$.

We have that $WU$ equals the side of a regular hexagon which could be inscribed within the circle $EFGHK$.

But $WZ$ equals the side of a regular decagon which could be inscribed within the circle $EFGHK$.

Also $\angle UWZ$ is a right angle.

So for the same reason as before:
 * $UZ$ equals the side of the pentagon $EFGHK$.

But $QU$ also equals the side of the pentagon $EFGHK$.

Therefore $\triangle QUZ$ 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 apex, is equilateral.

We have that:
 * $VL$ equals the side of a regular hexagon which could be inscribed within the circle $EFGHK$

and:
 * $VX$ equals the side of a regular decagon which could be inscribed within the circle $EFGHK$.

We also have that $\angle LVX$ is a right angle.

So for the same reason as before:
 * $LX$ equals the side of the pentagon $EFGHK$.

Suppose $MV$ were joined.

We have that:
 * $MV$ equals the side of a regular hexagon which could be inscribed within the circle $EFGHK$

and:
 * $MX$ equals the side of a regular decagon which could be inscribed within the circle $EFGHK$.

We also have that $\angle MVX$ is a right angle.

So for the same reason as before:
 * $MX$ equals the side of the pentagon $EFGHK$.

Therefore $\triangle LMX$ is equilateral.

For the same reason:
 * each of the remaining triangles of which the straight lines $MN, NO, OP, PL$ are the bases, and the point $X$ the apex, is equilateral.

Thus we have created an icosahedron which is contained by $20$ equilateral triangles.

Next it is to be demonstrated that it can be inscribed in the given sphere.

We have that:
 * $VW$ equals the side of a regular hexagon which could be inscribed within the circle $EFGHK$.

and:
 * $WZ$ equals the side of a regular decagon which could be inscribed within the circle $EFGHK$.

Therefore from :
 * $VZ$ has been cut in extreme and mean ratio whose greater segment is $VW$.

Therefore:
 * $ZV : VW = VW : WZ$

But:
 * $VW = VE$

and:
 * $WZ = VX$

Therefore:
 * $ZV : VE = EV : VX$

We have that $\angle ZVE$ and $\angle EVX$ are right angles.

We have that $\triangle XEZ$ and $\triangle VEZ$ are similar.

Therefore if $EZ$ is joined, $\angle XEZ$ will be a right angle.

For the same reason:
 * $ZV : VW = VW : WZ$

and:
 * $ZW = XW$

and:
 * $VW = WQ$

Therefore:
 * $XW : WQ = QW : WZ$

And for the same reason again:
 * if $QX$ is joined, $\angle Q$ will be a right angle.

Therefore from :
 * the semicircle described on $XZ$ also passes through $Q$.

Let $XZ$ remain fixed, and let that semicircle be carried around and restored to the same position from which it began.

It will also pass through $Q$ and the remaining vertices of the icosahedron.

Thus the icosahedron has been inscribed in a sphere.

Let $VW$ be bisected at $A'$.

We have that $VZ$ has been cut in extreme and mean ratio at $W$, where the lesser segment is $ZW$.

Therefore from :
 * $\left({ZW + WA'}\right)^2 = 5 \cdot WA'^2$

Therefore:
 * $ZA'^2 = 5 \cdot A'W^2$

Also:
 * $ZX = 2 \cdot ZA$

and:
 * $VW = 2 \cdot A'W$

Therefore:
 * $ZX^2 = 5 \cdot WV^2$

We have that:
 * $AC = 4 \cdot CB$

Therefore:
 * $AB = 5 \cdot BC$

But from:

and:

it follows that:
 * $AB : BC = AB^2 : BD^2$

Therefore:
 * $AB^2 = 5 \cdot BD^2$

But:
 * $ZX = 5 \cdot VW$

and:
 * $DB = VW$

because each of them equals the radius of the circle $EFGHK$.

Therefore:
 * $AB = XZ$

Also, $AB$ is the diameter of the given sphere.

Therefore $XZ$ is also the diameter of the given sphere.

Therefore the icosahedron has been be inscribed in the given sphere.

It remains to be demonstrated that the side of the icosahedron is the irrational straight line called minor.

We have that the diameter $AB$ of the given sphere is rational.

We have:
 * $AB^2 = 5 \cdot EV^2$

where $EV$ is the radius of the circle $EFGHK$.

Therefore $EV$ is also rational.

Therefore the diameter of the circle $EFGHK$ is rational.

But from :
 * the side $EF$ of the regular pentagon $EFGHK$ is minor.

But the side $EF$ of the regular pentagon $EFGHK$ is also the side of the icosahedron.

Hence the result.