A history of Greek mathematics - Wilbourhall.org

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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 be 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
?bfB> I PEQ >^ B;
and he proceeds to prove that AB 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
T^R >IFCG > aio^-
The first relation is obvious, for, since GO > E0,
Z PEQ > Z FGG.
Next, the perimeter of any polygon inscribed in the circle
AOB is less than - 4
T
4 - GO (i.e. -~- times the diameter)
Therefore AB < ^•-\ 4 - CO or T^g CO,
and, a fortiori, AB < tJq CO.
Now, the triangles GAM, COF being equal in all
respects,
AM= OF, so that AB = 20F= (diameter of sun) > CH+ OK,
since the diameter of the sun is greater than that of the earth
;
therefore
for
CH + OK < ^CO, and HK > ^CO.
And CO > CF, while HK < EQ, so that EQ > T %%CF.
We can now compare the angles OCF, OEQ
;
Z OCF
TOEQ
> tan OCFtan
0EQ_
EQ
> CF
> T
9
o
9
o, a fortiori.
Doubling the angles, we have
IFGG >^%.IPEQ
> 2omo^ since I PEQ > ^i?,
^ 203
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