84 
ARCHIMEDES 
the sun when it has just cleared the horizon. Draw from E 
two tangents EP, EQ to the circle with centre 0, and from 
C let CF, CG he drawn touching the same circle. With centre 
G and radius GO describe a circle: this will represent the path 
of the centre of the sun round the earth. Let this circle meet 
the tangents from G in A, B, and join AB meeting GO in M. 
Archimedes’s observation has shown that 
T ^R > Z PEQ >^o E; 
and he proceeds to prove that A B is less than the side of a 
regular polygon of 656 sides inscribed in the circle AOB, 
but greater than the side of an inscribed regular polygon of 
1,000 sides, in other words, that 
TI4-R >lFGG > 2^0 R- 
The first relation is obvious, for, since CO > EO, 
Z PEQ > Z FGG. 
Next, the perimeter of any polygon inscribed in the circle 
AOB is less than -\ 4 - GO (i.e. - 2 ? 2 - times the diameter); 
Therefore AB < 00 or xxis 00, 
and, a fortiori, AB < T |-g GO. 
Now, the triangles GAM, GOF being equal in all respects, 
AM = OF, so that AB —2 OF — (diameter of sun) > GH + OK, 
since the diameter of the sun is greater than that of the earth; 
therefore CH+OK < CO, and 11 K > t 9 5 9 q GO. 
And GO > CF, while 11K < EQ, so that EQ > CF. 
We can now compare the angles OCF, OEQ; 
tan OGF' 
_ tan 0EQ_ 
*EQ 
GF 
> to 9 oi a fortiori. 
Doubling the angles, we have 
ZFGG> T %%.APEQ 
> 20150^ sin c e ¿-PEQ >-2*6-5E, 
' > -263 E- 
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