:700px 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 {{EuclidPropLink|book = IV|prop = 11|title = Inscribing Regular Pentagon in Circle}}: :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 {{EuclidPropLink|book = XI|prop = 6|title = Two Lines Perpendicular to Same Plane are Parallel}}: :$EQ \parallel KU$ But we also have that $EQ = KU$. So from {{EuclidPropLink|book = I|prop = 33|title = Lines Joining Equal and Parallel Straight Lines are Parallel}}: :$QU = EK$ 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 {{EuclidPropLink|book = I|prop = 47|title = Pythagoras's Theorem}}: :$QE^2 + EP^2 = QP^2$ We have that $QE$ equals to the radius of circle $EFGHK$. From {{EuclidPropLink|book = VI|prop = 15|qual=Porism|title = Inscribing Regular Hexagon in Circle}}: :$QE$ 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  {{EuclidPropLink|book = XIII|prop = 10|title = Square on Side of Regular Pentagon inscribed in Circle equals Squares on Sides of Hexagon and Decagon inscribed in same Circle}}: :$QP$ 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 {{EuclidPropLink|book = XI|prop = 6|title = Two Lines Perpendicular to Same Plane are Parallel}}: :$VW \parallel QE$ But we also have that $EQ = KU$. So from {{EuclidPropLink|book = I|prop = 33|title = Lines Joining Equal and Parallel Straight Lines are Parallel}}: :$EV = QW$ 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 {{EuclidPropLink|book = I|prop = 47|title = Pythagoras's Theorem}}: :$QW^2 + WZ^2 = QZ^2$ Therefore from {{EuclidPropLink|book = XIII|prop = 10|title = Square on Side of Regular Pentagon inscribed in Circle equals Squares on Sides of Hexagon and Decagon inscribed in same Circle}}: :$QZ$ 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. {{qed|lemma}} 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 {{EuclidPropLink|book = XIII|prop = 9|title = Sides Appended of Hexagon and Decagon inscribed in same Circle are cut in Extreme and Mean Ratio}}: :$VZ$ 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 {{EuclidPropLink|book = III|prop = 31|title = Relative Sizes of Angles in Segments}}: :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 {{EuclidPropLink|book = XIII|prop = 3|title = Area of Square on Lesser Segment of Straight Line cut in Extreme and Mean Ratio}}: :$\left({ZW + WA'}\right)^2 = 5 \cdot WA'^2$ 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: :{{EuclidPropLink|book = VI|prop = 8|title = Perpendicular in Right-Angled Triangle makes two Similar Triangles}} and: :{{EuclidDefLink|V|9|Duplicate Ratio}} 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. {{qed|lemma}} 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 {{EuclidPropLink|book = XIII|prop = 11|title = Side of Regular Pentagon inscribed in Circle with Rational Diameter is Minor}}: :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. {{qed}} {{Euclid Note|16|XIII|Some sources credit this proof to {{AuthorRef|Theaetetus of Athens}}.}}