A history of Greek mathematics Vol.II from Aristarchus to Diophantus by Heath, Thomas Little, Sir, 1921

ON SEMI-REGULAR' POLYHEDRA 101
similar procedure with the icosahedron and dodecahedron
produces P n and P l2
(see Figs. 3, 4 for the case of the icosahedron).
102 ARCHIMEDES
of the two smaller semicircles adjoin, and NP be drawn at
right angles to AB meeting the external semicircle in P, the
area of the apfi-qXos
(included between the three semicircular
arcs) is equal to the circle on PN as diameter (Prop. 4). In
Prop. 5 it is shown that, if a circle be described in the space
between the arcs AP, AN and the straight line PN touching
Fig. 3. Fig. 4.
The two remaining solids P 10
, P 13
cannot be so simply produced.
They are represented in Figs. 5, 6, which I have
Fig. 5. Fig. 6.
taken from Kepler. P l0
is the snub cube in which each
solid angle is formed by the angles of four equilateral triangles
and one square; P 13
is the snub dodecahedron, each solid
angle of which is formed by the angles of four equilateral
triangles and one regular pentagon.
all three, and if a circle be similarly described in the space
between the arcs PB, NB and the straight line PN touching
all three, the two circles are equal. If one circle be described
in the dpftrjXos touching all three semicircles, Prop. 6 shows
that, if the ratio of AN to NB be given, we can find the
relation between the diameter of the circle inscribed to the
dpftrjXos and the straight line AB ; the proof is for the particular
case AN = §BN, and shows that the diameter of the
inscribed circle = -f^AB.
Prop. 8 is of interest in connexion with the problem of
We are indebted to Arabian tradition for
(y)
The Liber Assumptorum.
Of the theorems contained in this collection many are
so elegant as to
afford a presumption that they may really
be due to Archimedes. In three of them the figure appears
which was called dpftrjXos, a shoemaker's knife, consisting of
three semicircles with a common diameter as shown in the
annexed figure. If N be the point at which the diameters
trisecting any angle. If AB be any chord of a circle with
centre 0, and BC on AB produced be made equal to the radius,
draw CO meeting the circle in D, E
;
then will the arc BD be
one-third of the arc AE (or BF, if EF be the chord through E
parallel to AB). The problem is by this theorem reduced to
a v ever is (cf. vol. i, p. 241).