To “square” a figure, for the Greeks, was to find a square, or another rectilinear figure, with the same area. Squaring the circle was the impossible problem par excellence. Archimedes managed to square a curved figure: the segment of a parabola, the region between a parabola and a chord.

Every segment bounded by a straight line and a section of a right-angled cone is four thirds of the triangle that has the same base and the same height as the segment.

— Archimedes, Quadrature of the Parabola, letter to Dositheus

“Section of a right-angled cone” was the name Archimedes gave to the parabola. The book is dedicated to Dositheus, after the death of Conon, to whom Archimedes would have shown the result first.

Triangles inside triangles

Draw the chord and, on the arc, the point where the tangent is parallel to the chord. The three points form a triangle, of area $T$, which fills most of the segment. Two small regions are left over, between the sides of the triangle and the curve. In each of them, do the same thing: a new triangle with its third vertex on the parabola. Four smaller regions are left over, and so on.

Archimedes’s discovery is that each pair of new triangles has, together, one quarter of the area of the triangle that generated them. So the first stage gives $T$, the second $\frac{T}{4}$, the third $\frac{T}{16}$, and the total area is

$$T + \frac{T}{4} + \frac{T}{16} + \frac{T}{64} + \cdots$$

Adding infinitely many terms

Archimedes did not speak of an “infinite sum”, and carefully avoided the infinite. Instead, he proved a fact about finite sums: for any number of terms,

$$1 + \frac{1}{4} + \frac{1}{16} + \cdots + \frac{1}{4^n} + \frac{1}{3}\cdot\frac{1}{4^n} = \frac{4}{3}.$$

In other words, what is missing to reach $\frac{4}{3}$ is always one third of the last term, which can be made as small as one wishes. Then, with a double argument by contradiction, he showed that the area of the segment can be neither greater nor less than $\frac{4}{3}T$. This is the method of exhaustion, inherited from Eudoxus, and it is the direct ancestor of the concept of limit.

Checking with calculus

With today’s tools, the calculation is immediate. For the parabola $y = 1 - x^2$ and the chord on the $x$-axis, between $-1$ and $1$:

$$\int_{-1}^{1} (1 - x^2)\,dx = \left[x - \frac{x^3}{3}\right]_{-1}^{1} = \frac{4}{3}.$$

The inscribed triangle has base 2 and height 1, that is, area $T = 1$. The ratio is exactly $\frac{4}{3}$, as Archimedes proved nineteen centuries before Newton and Leibniz.

Exercise: Archimedes's sum

Check Archimedes’s identity for $n = 2$: $1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{3}\cdot\frac{1}{16} = \frac{4}{3}$.

With denominator 48: $\frac{48}{48} + \frac{12}{48} + \frac{3}{48} + \frac{1}{48} = \frac{64}{48} = \frac{4}{3}$. For any $n$, the proof is by induction: adding the term $\frac{1}{4^{n+1}}$ and replacing $\frac{1}{3}\cdot\frac{1}{4^n}$ by $\frac{1}{3}\cdot\frac{1}{4^{n+1}}$ does not change the sum, because $\frac{1}{4^{n+1}} + \frac{1}{3\cdot 4^{n+1}} = \frac{4}{3\cdot 4^{n+1}} = \frac{1}{3\cdot 4^n}$.

Exercise: the first two smaller triangles

On the parabola $y = 1 - x^2$, the first triangle has vertices $(-1, 0)$, $(1, 0)$ and $(0, 1)$. The second, on the right, has vertices $(0, 1)$, $(1, 0)$ and the point of the parabola with $x = \frac{1}{2}$. Compute its area and check that the two triangles of the second stage add up to $\frac{1}{4}$.

The third vertex is $\left(\frac{1}{2}, \frac{3}{4}\right)$. By the area formula with coordinates, $\frac{1}{2}\left|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\right|$, with $(0,1)$, $(1,0)$, $\left(\frac{1}{2}, \frac{3}{4}\right)$: $\frac{1}{2}\left|0 + 1\cdot\left(\frac{3}{4} - 1\right) + \frac{1}{2}(1 - 0)\right| = \frac{1}{2}\cdot\frac{1}{4} = \frac{1}{8}$. By symmetry, the one on the left also has $\frac{1}{8}$, and together they add up to $\frac{1}{4}$ of the first.

Sources: Archimedes, Quadrature of the Parabola, propositions 14–24 · Thomas L. Heath, The Works of Archimedes (1897) · E. J. Dijksterhuis, Archimedes (1956), ch. X · Carl B. Boyer, The History of the Calculus and Its Conceptual Development (1949).

All topics on Archimedes · ← The measurement of the circle and the number π · The sphere and the cylinder →

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