Set out the diameter AB of the given sphere, and cut it at C so that AC is quadruple CB, describe the semicircle ADB on AB, draw the straight line CD from C at right angles to AB, and join DB.
Set out the circle EFGHK, and let its radius be equal to DB. Inscribe the equilateral and equiangular pentagon EFGHK in the circle EFGHK, bisect the circumferences EF, FG, GH, HK, and KE at the points L, M, N, O, and P, and join LM, MN, NO, OP, PL, and EP.
Therefore the pentagon LMNOP is also equilateral, and the straight line EP belongs to a decagon.
Now from the points E, F, G, H, and K set up the straight lines EQ, FR, GS, HT, and KU at right angles to the plane of the circle, and make them equal to the radius of the circle EFGHK. Join QR, RS, ST, TU, UQ, QL, LR, RM, MS, SN, NT, TO, OU, UP, and PQ.
Now, since each of the straight lines EQ and KU is at right angles to the same plane, therefore EQ is parallel to KU.
But it is also equal to it, and the straight lines joining those ends of equal and parallel straight lines which are in the same direction are equal and parallel. 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, and 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 equals the sum of the square on the side of the hexagon and the square on the side of the decagon inscribed in the same circle.
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, and TOU is also equilateral.
And, since each of the straight lines QL and 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, and OUP is also equilateral.
Take the center V of the circle EFGHK, set VZ up from V at right angles to the plane of the circle, and produce it in the other direction VX. Cut off VW, the side of a hexagon, and each of the straight lines VX and WZ, sides of a decagon. Join QZ, QW, UZ, EV, LV, LX, and XM.
Now, since each of the straight lines VW and QE is at right angles to the plane of the circle, therefore VW is parallel to QE. But they are also equal, therefore EV and QW are also equal and parallel.
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.
For the same reason UZ also belongs to a pentagon, for if we join VK and WU, then they will be equal and opposite, and VK, being a radius, belongs to a hexagon, 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.
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, and 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.
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, and 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.
Since VW belongs to a hexagon, and WZ to a decagon, therefore VZ is cut in extreme and mean ratio at W, and VW is its greater segment. Therefore as ZV is to VW as VW is to WZ.
But VW equals VE, and WZ equals VX, therefore ZV is to VE as EV is to VX.
And the angles ZVE and EVX are right, therefore, if we join the straight line EZ, then the angle XEZ will be right since the triangles XEZ and VEZ are similar.
For the same reason, since ZV is to VW as VW is to WZ, and ZV equals XW, and VW equals WQ, therefore XW is to WQ as QW is to WZ.
And for this reason again, if we join QX, then the angle at Q will be right, therefore the semicircle described on XZ will also pass through Q.
And if, XZ remaining fixed, the semicircle is carried round and restored to the same position from which it began to be moved, then it will 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.
Bisect VW at A'.
Then, since the straight line VZ is 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. Therefore the square on ZA' is five times the square on A'W.
And ZX is double ZA', and VW is double A'W, therefore the square on ZX is five times the square on WV. And, since AC is quadruple CB, therefore AB is five times BC.
But AB is to BC as the square on AB is to the square on BD, 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 equals VW, for each of them equals the radius of the circle EFGHK, therefore AB also equals XZ. And AB is the diameter of the given sphere, therefore XZ also equals 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.
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 is inscribed in a circle which has its diameter rational, then the side of the pentagon is the irrational straight line called minor.
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.
Unlike most of the Euclid’s illustrations, the diagram he used for this proposition is highly schematic; it is not intended to be an accurate projection of the icosahedron. Of course, it could be that the diagram changed over the centuries of copying, but his diagram has the advantage of spreading out the vertices to be readable. The figure shown in the proof above is a standard orthogonal projection of the icosahedron. The same icosahedron is shown directly below without all the auxiliary lines.
Pacioli’s constructionLuca Pacioli (1445&ndah;1517) discussed the construction of the regular polygons in his De divina proportione, entitled for his term for the extreme and mean ratio. In the preceding diagram, the 12 vertices of icosahedron appear in three groups of four vertices, each group being the vertices of three mutually perpendicular rectangles, LMTU, OXRZ, and NPSQ.Pacioli recognized that the sides of each of those rectangles are in extreme and mean ratio, and used them as the construction of the icosahedron.
