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
√
√
2p6 P6
24 3
√ = 48 − 24 3
P12 =
=
p6 + P 6
3+2 3
and
q
q
p
√
√
p12 = p6 P12 = 6(48 − 24 3) = 12 2 − 3.
p
√
√
The arithmetic mean of these two values
is
24−12
3+6
2
−
3, which
p
√
√
implies a value for π of 12 − 6 3 + 3 2 − 3 ≈ 3.16.
√
Using the areas, we use elementary geometry to compute a6 = 3 3/2 and
√
A6 = 2 3. Then using Archimedes’ formulas we have
s
√
3 3 √
a12 =
2 3=3
2
and
A12
√
√
2(2 3)(3)
= 24 − 12 3.
= √
2 3+3
√
The arithmetic mean of these two values is 13 12 − 6 3, which implies
√
a value for π of 13 12 − 6 3 ≈ 3.1077.
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
p
x2 + y 2
p
= r ⇒ x2 + y 2 = r 2 [(x − p)2 + (y − q)2 ]
2
2
(x − p) + (y − q)
⇒
⇒
x2 + y 2 = r 2 x2 − 2r 2 px + r 2 p2 + r 2 y 2 − 2r 2 qy + r 2 q 2
(1 − r 2 )x2 + 2r 2 px − r 2 p2 + (1 − r 2 )y 2 + 2r 2 qy − r 2 q 2 = 0
p2 r 2
2qr 2
q2r2
2pr 2
2
x
−
+
y
+
y
−
=0
1 − r2
1 − r2
1 − r2
1 − r2
2 2
pr 2
qr 2
p2 r 4
q2r4
p2 r 2
q2r2
⇒ x+
+
y
+
=
+
+
+
1 − r2
1 − r2
(1 − r 2 )2 (1 − r 2 )2 1 − r 2 1 − r 2
2 2
qr 2
r 2 (p2 + q 2 )
pr 2
⇒ x+
+ y+
=
.
1 − r2
1 − r2
(1 − r 2 )2
⇒
x2 +
This is the equation for a circle. (Note that the fact that the distances
are unequal ensures that r =
6 1, so that we can divide by 1 − r 2 .)
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
⇒
So the two numbers are 7 and 3.
x2 − 10x + 21 = (x − 7)(x − 3) = 0.
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
of a cistern in one day. Thus one cistern is filled in
2 3 4
12
12
of one day.
25