A History of Mathematics From Mesopotamia to Modernity

76 A History ofMathematics

through Z lines NXO parallel to AG and PRS parallel to BD. Draw perpendicularTY fromto

NXO and perpendicular KFQ from K to PRS.

Now since the sun traverses circleKLM with uniform motion, it will traverse arcKin94^12

days and arc KL in 92^12 days. In 94^12 days its mean motion is aproximately 93;9◦and in 92^12 days

91;11◦. [It covers 360 degrees in a year of 365^14 days, so slightly less than 1◦per day; this is where

these figures come from.] Therefore arcKL=184;20◦and by subtraction of the semi-circle NPO,

arc N+arc LO=4;20◦

So arcNY=2 arcN=4;20◦andY=Crd arcNY=4;32p, where the diameter of the

eccentre is 120p. [Remember that our tables deal with a circle of radius 60, diameter 120.] And

EX=T=^12 Y=2; 16p.

Now since arcNPK=93;9◦and arcN=2;10◦and quadrant NP= 90 ◦, by subtraction,

arc PK=0;59◦, and arc KPQ=2.arc PK=1;58◦. So KFQ=Crd arc KPQ=2;4p, and ZX=KF=

1

2 KFQ=1;2

p. And we have shown that EX=2;16pin the same units.

Now since EZ^2 =EX^2 +ZX^2 ,EZ=2; 29^12

p

where the radius of the eccentre is 60p. Therefore the

radius of the eccentre is approximately 24 times the distance between the centres of the eccentre

and the ecliptic.

[This completes the first half of the calculation, showing how far the Earth is from the centre of

the eccentric circle. It remains to find the direction of the line EZ so as to situate Z exactly; as you

can see, this follows from the ratio of EX to EZ; we would use the tangent, but Ptolemy has to use

the chord function again. The answer is that angle ZEX is 24;30◦.]

Solutions to exercises

1. If(x,y)is on the two curves, thenx^3 =x.x^2 =x.y= 2 a^3 ;sox=a.^3

√

2. The description of
   what Menaechmus did does not read quite like this—for a plausible version, see Knorr. If you
   replace 2 bymin the equation of the hyperbola, then you solve the problem of increasing the
   volume bym(and so the side by^3

√

m), similarly.

2. Suppose C and D are constructed. Then (rules about ratios, think of them as fractions), A : B=
   (A:C).(C:D).(D:B). The three ratios in brackets are equal, so this is(A:C)^3 .If B:A=m, then
   (cube on B):(cube on A) ism^3. So (cube on C):(cube on A) ism.


3. Straightforward; the radius is the height of the equilateral triangle whose side is the side of the
   hexagon. So by Pythagoras’s theorem, the side is 2/

√

3 times the radius; and the perimeter of

the hexagon is 6.r.( 2 /

√

3 )= 4 r.

√

3.

4. This depends on Euclid VI.3, which you may not know. This says that in triangle ABC, if AD
   bisects angle A and meets BC at D, then AD : AB=BD : AB. (Look it up, or try to work out why it
   is true.) In our case, this gives (looking at the bisected angle at the centre of the circle in Fig. 4)
   A′:B=A−A′: C. Manipulating ratios (componendo, see Appendix A), A′:B=A:B+C
   as required.


5. (a) (See Fig. 11) In the picture, CDOE is a square, so all of its sides equalr. Hence, AE=b−r,
   BD=a−r. But by the property of tangents, this means that AF=b−rand BF=a−r.
   Hence, AB=AF+FB=a−r+b−r, and the result follows. (b) From the factorization,
   1
   2 (a+b+c)=35,

1

2 (a+b−c)=6; soa+b=41. This is why we square 41, and get

(a+b)^2 = 1681. But also the areasrequals 35× 6 = 210, and this is (by a different