**Archimedes'
Estimate of π Activity**

The formula _{} for circle area is
only as good as the approximation of π used in it. In Proposition 3 of his
treatise, *On the Measurement of the
Circle,* the Greek mathematician Archimedes (287-212 BCE) gave better
estimates for π than were known previously.

**Proposition 3. ***The
ratio of the circumference of any circle to its diameter is less than 3 1/7 but
greater than 3 10/71*
(Dunham, 97; Katz, 109).

What Archimedes is saying is that 3
10/71 < p < 3 1/7, or, rounding off to three
places after the decimal point, 3.141 < p < 3.143. Archimedes' approximation of p was the first in history that was
correct to two places after the decimal point!

Archimedes made the approximations 3
10/71 < p < 3 1/7 by estimating the
circumference of a circle using the perimeters of inscribed and circumscribed
polygons, such as those shown in Figure 1. If *C* is the circumference of the circle, *r* is its radius, and *P*_{insc} and *P*_{circ}
are the perimeters of the inscribed and circumscribed polygons, respectively,
then

*P*_{insc} < *C*
< *P*_{circ}, or

*P*_{insc} < 2p*r* < *P*_{circ},
so that

*P*_{insc} /2*r*
< p < *P*_{circ}
/2r.

If we take the
radius of the circle to be 1 (*r* = 1),
then

*P*_{insc} /2 < p < *P*_{circ}
/2.

Archimedes started with inscribed
and circumscribed regular hexagons. Since each of the six sides of a regular
hexagon inscribed in a circle of radius 1 has length 1, then *P*_{insc}
= 6 in this case (see Problem 1a). Likewise, since each side of a regular
hexagon circumscribed about a circle of radius 1 has length _{}, then *P*_{circ}
= _{} (see Problem 1b).
Hence, *P*_{insc}
/2 < π < *P*_{circ}
/2 yields _{}, or _{}.

Archimedes then doubled the number
of sides of each polygon to 12, obtaining an inscribed regular dodecagon of
perimeter *P*_{insc}
= _{} (see Problem 1c), and
a circumscribed regular dodecagon of perimeter *P*_{circ} = _{}. This yields _{}, or 3.1058 < p < 3.2154. Archimedes proceeded to double the number of
sides three more times until he had inscribed a regular 96-gon inside his
circle and circumscribed another regular 96-gon about his circle, obtaining 3
10/71 < p < 3 1/7 (Dunham, 97-98; Johnson and Mowry, 404-405, 410).

Archimedes' overestimate of p, p » 3 1/7 =
22/7, was used for hundreds of years afterwards by Greek mathematicians and by
mathematicians of the next great mathematical civilization, the Arab Empire.
Heron of Alexandria, a Greek mathematician who lived during the first century,
wrote a very influential practical geometry book called the *Metrica**.* Heron's formula for circle area in the *Metrica* was _{} which assumes p = 22/7 (Katz, 160, 162). This formula
for circle area appeared in mathematical works for hundreds of years, including
the work of one of the first great Arab mathematicians, Muhammad ibn Musa
al-Khwarizmi, in the early ninth century (Katz, 268).

Besides simply using Archimedes'
estimate of p, later mathematicians also improved upon
it by carrying out his technique with larger and larger numbers of inscribed
and circumscribed polygons. In about 250 CE, the Chinese mathematician Liu Hui,
who probably was not aware of Archimedes' work, used an inscribed polygon with
192 sides to obtain a slightly better approximation of p than Archimedes had made. Liu Hui or
possibly a later Chinese mathematician then used a polygon with 3072 sides to
obtain the estimate p » 3.1416. The Chinese mathematician Zu Chongzhi (or Tsu Ch'ung Chi, 429-500 CE) may
have used a polygon with 24,576 sides (or possibly a more efficient method) to
get 3.1415926 < p <
3.1415927 (Straffin, 173). French mathematician
Francois Viete (1540-1603) used Archimedes' technique
to estimate p correctly to 9 places. Viete began with a hexagon and doubled the number of sides
of successive polygons 16 times to obtain a regular polygon with 393,216 sides
(Dunham, 107). Just a few years later, in about 1600, Dutch mathematician Ludolph van Ceulen began with a
square and doubled the number of sides of successive polygons 60 times to
obtain a regular polygon with 2^{62} sides and an estimate of p correct to 35 places (Dunham, 107-108).

Seventeenth century Japanese
mathematicians may have estimated circle area, and hence p, using the method illustrated in Figure
2 (Beckmann, 125-127). Since the radius of the circle
shown is 1, its area is p and the
area of one quarter of the circle is p/4. By estimating the area of the quarter circle as the sum
of the areas of the rectangular strips, one estimates p/4 and hence p. Using sums of areas of rectangles to
estimate areas of regions having curved, rather than straight, boundaries is
one of the fundamental ideas of integral calculus.

Beginning
in the fifteenth century with the Indian mathematician Madhava,
mathematicians have used infinite series (sum) formulas from calculus to obtain
improved estimates of π. Today researchers are using computers to estimate
p to billions of digits of accuracy.

In
the following problems, you will use the method of Archimedes to estimate p.

**1.** We use a
regular hexagon inscribed in a circle of radius 1 to obtain the estimate

p » 3. See Figure 3. Since the hexagon is regular, its perimeter, *P*_{insc}, is given by *P*_{insc}
= 6*s,* where *s* is the length of any one of its sides.

**1a.** How large is
each of the six angles at the center of the circle?

**1b.** How large are
the other two angles (the base angles) in the isosceles triangle with sides
labeled 1, 1, and *s*?

**1c.** Find *s* and *P*_{insc}.

**1d.** Use the
approximation *C* » *P*_{insc} to estimate p.

**2.** Now we use a
regular hexagon circumscribed about a circle of radius 1 to obtain the estimate
π » _{}. See Figure 4a. Since the hexagon is regular, its perimeter,
*P*_{circ}, is given by *P*_{circ} = 6*s,* where *s* is the length
of any one of its sides. As you discovered in Problem 1, the six triangles
formed by the sides of the hexagon are equilateral triangles. This time,
however, each side has length *s*
rather than 1. If we divide one of the equilateral triangles in half using a
radius, as shown in Figure 4a, we obtain the right triangle with sides of
lengths 1, (1/2)*s,* and *s* shown in Figure 4b.

**2a.** Use the
Pythagorean Theorem to find *s.*

**2b.** Compute *P*_{circ}, then use the
approximation *C* » *P*_{circ}
to estimate p.

**3.** In this
problem, we use a regular dodecagon inscribed in a circle of radius 1 to obtain
the estimate π » _{} » 3.1058. See Figure 5a. Since the
dodecagon is regular, its perimeter, *P*_{insc}, is given by *P*_{insc} = 12*s,* where *s* is the length
of any one of its sides. We begin with the regular hexagon in Figure 3 and form
a regular dodecagon by creating two new sides for each existing side, as shown
in Figure 5a. Note that the new radius shown in Figure 5a bisects one side of
the regular hexagon.

**3a.** Refer
to Figure 5b. Find *y,* then apply the
Pythagorean Theorem to the right triangle with sides of lengths *x, y,* and 1 to find *x.*

**3b.** Find
*z,* then apply the Pythagorean Theorem
to the right triangle with sides of lengths *y,
z,* and *s* to find *s.*

**3c.** Compute *P*_{insc},
then use the approximation *C* » *P*_{insc} to estimate p.

**4.** In
this problem, we obtain estimates of π using circumscribed and inscribed
squares.

**4a.** Find
the side length *s* and perimeter, *P*_{circ}, of a square
circumscribed about a circle of radius 1, then use the approximation *C* » *P*_{circ}
to estimate p. See Figure 6a.

**4b.** Find
the side length *s* and perimeter, *P*_{insc},
of a square inscribed in a circle of radius 1, then use the approximation *C* » *P*_{insc} to estimate p. See Figure 6b.

**4c.** Find
the average of your two estimates from parts (a) and (b). Is the resulting
estimate too large or too small?

**5.** In this
problem, we use a regular octagon inscribed in a circle of radius 1 to obtain
the estimate π » _{} » 3.0615. See Figure 7a. Since the octagon
is regular, its perimeter, *P*_{insc}, is given by *P*_{insc} = 8*s,* where *s* is the length
of any one of its sides. We begin with the square in Figure 6 and form a regular
octagon by creating two new sides for each existing side, as shown in Figure
7a. Note that the new radius shown in Figure 7a bisects one side of the square.

**5a.** Refer to Figure
7b. Find *x* and *y.*

**5b.** Find
*z,* then apply the Pythagorean Theorem
to the right triangle with sides of lengths *y,
z,* and *s* to find *s.*

**5c.** Compute *P*_{insc},
then use the approximation *C* » *P*_{insc} to estimate p.

**6.** Show that the
formula _{} for circle area is
correct, if you replace

p by 22/7.

Instructor
Notes

**Objective: **Students will investigate how Archimedes
used perimeters of regular polygons to estimate p.

**How to Use: **Share with students, or have students
read, information about the life and work of Archimedes.

For
Problems 1-7, students will need to use the theorem that the measures of the
angles in any triangle sum to 180 degrees, along with the Pythagorean Theorem.
You might remind them of the statements of these theorems, as well as of the
definition of regular polygon. Note that Problems 1-3 form a complete set, as
do Problems 4 and 5.

Before or after completing Problems
3 and 5, you could have students show that if a regular *n*-gon inscribed in a circle of radius 1
has side length *S*_{n}, then
the side length *S*_{2n} of a
regular 2*n*-gon inscribed in the same
circle is given by

_{}.

By repeated
use of this formula, students may replicate Archimedes’ underestimate of p using the perimeter of an inscribed
96-gon, and may improve upon it by using inscribed polygons with even larger
numbers of sides (Dunham Problems, 18; other sources for Problems 1-7: Harrigan, 1802-1804; Johnson and Mowry,
404-405, 410).

**Related Activity:** Hold a Pi Day celebration on March 14,
preferably at

**Solutions:**

**1a.** 360/6 = 60
degrees.

**1b.** Since the
triangle is isosceles, the two unknown angles are equal, let's say of measure

_{}. Since 60 + 2_{} = 180, then _{} = 60 degrees.

**1c.** By part (b), the
triangle is equilateral, so *s* = 1.
Then *P*_{insc}
= 6*s* = 6.

**1d.** *C* » *P*_{insc}_{ }implies 2p » 6, which implies p » 3.

**2a.** That *s*^{2} = 1^{2} + ((1/2)*s*)^{2} implies *s* = _{}.

**2b.** *P*_{circ} = 6*s* = _{}.

*C*
» *P*_{circ}
implies 2p » _{}, which implies p » _{}.

**3a.** Since the radius
labeled *x* + *z* = 1 bisects a line segment of length 1, *y* = 1/2.

Then
*x*^{2} + *y*^{2} = 1^{2} becomes *x*^{2} + (1/2)^{2} = 1, yielding *x* = _{}.

**3b.** Since _{}, then *s*^{2}
= *y*^{2} + *z*^{2} becomes _{},

yielding
*s* = _{}.

**3c.** *P*_{insc}
= 12*s* = _{}, so that *C* » *P*_{insc}_{ }implies 2p » _{},

which implies p » _{} »
3.1058. _{}

_{ }

**4a.** We have *s* = 2, so that *P*_{circ} = 4 x 2 = 8 and p » 4.

**4b.** By the
Pythagorean Theorem, *s*^{2} =
1^{2} + 1^{2}, so that *s*
= _{}.

Then
*P*_{insc}
= _{} and p » _{} » 2.8284.

**4c.** Averaging the
two estimates gives 2 + _{} » 3.4142, an overestimate of p.

**5a.** Since
the radius labeled *x* + *z* = 1 bisects a line segment of length _{}, *y* = _{}. Use the Pythagorean Theorem, or note that the triangle with
sides of lengths *x, y,* and 1 is
isosceles with legs of lengths *x* and *y,* to obtain *x* = _{}.

**5b.** Since _{}, then *s*^{2}
= *y*^{2} + *z*^{2} becomes _{},

yielding
*s* = _{}.

**5c.** *P*_{insc}
= 8*s* = _{}, so that *C* » *P*_{insc}_{ }becomes 2p » _{},

which
implies p » _{} or 3.0615.

**6.** If we replace *r* by *D/*2
in the formula _{} we obtain _{}

If
we replace p by 22/7 in the formula _{} we obtain _{}

**References: **Activity from *Lengths, Areas, and Volumes, *by J. Beery, C. Dolezal, A. Sauk, and
L. Shuey, in *Historical Modules for the
Teaching and Learning of Secondary Mathematics,* Mathematical Association of

Beckmann, Petr, *A History of Pi,*

Dunham, William, *Journey Through Genius: The Great Theorems of Mathematics,* John
Wiley and Sons,

Dunham,
William, *Problems for Great Theorems* (unpublished
problem sets to accompany *Journey Through
Genius: The Great Theorems of Mathematics*),

Harrigan, Mary, *Activities for Mathematics 109,* *The
Mathematical Experience* (unpublished classroom activities),

Johnson, David, and Thomas Mowry, *Mathematics: A Practical Odyssey,* PWS
Publishing,

Katz, Victor J., *A History of
Mathematics: An Introduction,* Addison-Wesley,

Straffin, Philip D., “Liu Hui and the First
Golden Age of Chinese Mathematics,” *Mathematics
Magazine,* 71 (1998), 163-181.