History of Mathematics Homework 3 Due Wednesday, October 6 1. Beginning, As Did Archimedes, with a Regular Hexagon Inscribed In

History of Mathematics Homework 3 Due Wednesday, October 6

1. Beginning, as did Archimedes, with a regular hexagon inscribed in a circle, use an Archimedean recursion algorithm to ﬁnd either p12 and P12 or a12 and A12. What value of π would be implied by the arithmetic mean of your answers?

Using the perimeters, we use elementary geometry to compute that p6 = 6 and P6 =4√3. Then using Archimedes’ formulas we have

2p6P6 24√3 P12 = = = 48 24√3 p6 + P6 3+2√3 − and

p12 = p6P12 = 6(48 24√3) = 12 2 √3. − − q q The arithmetic meanp of these two values is 24 12√3+6 2 √3, which − − implies a value for π of 12 6√3+3 2 √3 3.16. − − ≈ p Using the areas, we use elementary geometryp to compute a6 =3√3/2 and A6 =2√3. Then using Archimedes’ formulas we have

3√3 a12 = 2√3=3 s 2 and 2(2√3)(3) A12 = = 24 12√3. 2√3+3 − 1 The arithmetic mean of these two values is 13 2 6√3, which implies 1 − a value for π of 13 6√3 3.1077. 2 − ≈

2. Prove the theorem concerning the “circle of Apollonius;” that is, show that the locus of points whose distance from two ﬁxed points are unequal but are in ﬁxed ratio, is a circle.

Suppose the points are (0, 0) and (p, q), and let r be the fixed ratio. Then for (x, y) to be on the curve, we require that

x2 + y2 = r x2 + y2 = r2[(x p)2 +(y q)2] (x p)2 +(y q)2 ⇒ − − −p − p x2 + y2 = r2x2 2r2px + r2p2 + r2y2 2r2qy + r2q2 ⇒ − − (1 r2)x2 +2r2px r2p2 + (1 r2)y2 +2r2qy r2q2 =0 ⇒ − − − − 2pr2 p2r2 2qr2 q2r2 x2 + x + y2 + y =0 ⇒ 1 r2 − 1 r2 1 r2 − 1 r2 2 − −2 − − pr2 qr2 p2r4 q2r4 p2r2 q2r2 x + + y + = + + + ⇒ 1 r2 1 r2 (1 r2)2 (1 r2)2 1 r2 1 r2 − − − − − − 2 2 pr2 qr2 r2(p2 + q2) x + + y + = . ⇒ 1 r2 1 r2 (1 r2)2 − − − This is the equation for a circle. (Note that the fact that the distances are unequal ensures that r =1, so that we can divide by 1 r2.) 6 −

3. Solve the “Problem of Apollonius” for the case of two points and a line. In other words, describe a construction that will accomplish the following: Given two points and a line not containing either of the points, to draw a circle passing through the two points and tangent to the line.

The set of points equidistant from a point and a line is a parabola. Thus using the line and each point gives us two parabolas. These will meet in a single point, which will be equidistant from all three. This is the center of the circle. The radius is then the segment joining this center to either of the other two points.

4. How can one account for the fact that the period of the rise of Greek trigonometry was a time of decline in Greek geometry?

5. Solve the problem of Diophantus in which it is required to ﬁnd two numbers such that their sum is 10 and the sum of their cubes is 370.

Assume the numbers are x and 10 x. Then the sum of their cubes is − x3 + (10 x)3 = 1000 300x + 30x2. − − Setting this equal to 370 yields the quadratic equation

30x2 300x +630= 0 x2 10x +21=(x 7)(x 3)=0. − ⇒ − − − So the two numbers are 7 and 3. 6. Solve the following problem from the Greek Anthology of Simplicius (fl. 520): if one pipe can fill a cistern in one day, a second in two days, a third in three days, and a fourth in four days, how long will it take all four running together to fill it?

Note that the first pipe one cistern a day, the second fills 1/2 a cistern per day, the third fills 1/3 a cistern per day, and the fourth fills 1/4 a cistern per day. Hence the four working together can fill 1+ 1 1 1 25 2 + 3 + 4 = 12 of a cistern in one day. Thus one cistern is filled in 12 25 of one day.