Mike Bennett
History of Mathematics
Dr. Patti Hunter
Archimedes: Measurement of a Circle (4.A1)
In the textbook section on Archimedes and Apollonius, Fauven and Grey present a proof
by Archimedes regarding the measurement of a circle by comparing it to a triangle of
equal area.
Archimedes’ Proposition 1 is as follows: The area of any circle is equal to a right-angled
triangle in which one of the sides about the right angle is equal to the radius, and the
other to the circumference, of the circle.
Archimedes’ proof consists of constructing a circle ABCD and a triangle K described by
the proposition, and then proving that the triangle cannot be any greater or less in area
than the circle. The diagram for the proof:
Archimedes starts by assuming the circle ABCD’s area is greater than the triangle
K’s. He then inscribes a square in the circle and bisects the segments of arc AB, BC, CD,
DE subtended by the sides of the square. He proceeds to inscribe another polygon on the
bisected points. This process is repeated over and over until the difference in area
between the circle and the inscribed polygon is smaller than the difference between the
area of the circle and the area of the triangle. The polygon is then greater than the
triangle K.
Archimedes then explains that a line from the center of the polygon to the bisection of
one of its sides is shorter than the radius of the circle, and its circumference is smaller
Mike Bennett
History of Mathematics
Dr. Patti Hunter
than the circumference of the circle. This contradicts the statement that the polygon is
greater than the triangle, since the legs of the triangle are made up of the radius and
circumference of the circle. The triangle K cannot be both smaller and larger than
the polygon, and thus cannot be smaller than the circle.
Having proven that the triangle cannot be smaller than the circle, Archimedes proceeds to
prove that the triangle cannot be larger than the circle, either. This is accomplished by
first assuming the triangle K to be larger than the circle ABCD. Then, a square is
circumscribed around the circle so that lines drawn from the center of the circle will go
through the points A, B, C, and D and bisect the corners of the square, one of which
Archimedes labels T.
Archimedes then connects the sides of the square with a tangent line and labels the points
at which the line meets the square G and F. He goes on to say that because TG > GA >
GH, the triangle formed by FTG is larger than half the area of the difference in area
between the square and the circle. Archimedes uses the fact that continual bisecting of
the arc of a circle will produce a polygon with this characteristic to assert that continuing
this method will ultimately produce a polygon around the circle such that the difference
in area between the polygon and the circle is less than the difference in area between the
triangle K and the circle. The polygon is thus less in area than the triangle K
The length of a line from the center of the circle to a side of the polygon is equal to the
radius of the circle. However, the perimeter of the polygon is larger in length than the
circumference of the circle, and since the circumference of the circle is equal to the
length of the longer leg of the triangle, the polygon must be larger in area than the
triangle K. Again, the triangle cannot be both larger and smaller than the polygon,
so the triangle cannot be larger than the circle.
Archimedes’ proof by contradiction is now complete. Having shown that the triangle
with legs equal to the radius and circumference of a given circle is not greater or less in
area than that circle, it is shown that the two must be equal in area. Archimedes uses the
contradiction method to prove other theorems, such as the idea that a segment of a
parabola is equal in area to four thirds the triangle with the same base and height.