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Archimedes And The Parabola

Among the many physical and mathematical results proved by Archimedes (287–212 BCE), one of his more interesting was the determination of the area of the parabolic segment, that portion of a parabola cut off by a chord. Each chord determines a unique parabolic segment. Each chord also determines a unique inscribed triangle, the triangle with side and third vertex P whose tangent line is parallel to (see figure 1). Archimedes proved that for any chord, the area of the parabolic segment is precisely 4/3 the area of the inscribed triangle.

So why is this result interesting?

First, it's a result proved by one of the greatest mathematicians of all time, Archimedes of Syracuse, and it's always interesting to study the methods of the masters. And though he found the area under a curve without the use of the integral calculus, he employed some distinctly calculus-like techniques (which makes you wonder what Archimedes would have done had he been around in the 17th century.or later). Finally, it's fascinating that the area under the parabola, a curved figure, compared to the area of the inscribed triangle, yields a simple 4 to 3 ratio.

It does not detract from the genius of Archimedesfs proof if we reframe parts of it using modern mathematical notation; his underlying logic is still there. Hence, we begin with a modern definition of a parabola: it is the curve defined by the quadratic y = a0x2 + b0x + c0, where a0 ‚ 0. This equation can be rewritten in the vertex form y = a0 (x . H)2 + k, which by a suitable rigid transformation can be transformed into y = b . Ax2, a downward opening parabola whose axis Of symmetry is the y-axis. This will make subsequent calculations easier.

To determine the inscribed triangle MPR, where the tangent line at vertex P is parallel to the chord , we make use of the Mean Value Theorem (MVT).

The MVT as applied to the parabola f(x) = b – ax2 says that on the open interval (x1, x2) there is a point x3 such that

If we plug in expressions for the quadratic f and its derivative f we obtain

Solving yields

the midpoint. This means that the tangent line at point P is parallel to the chord if and only if the “plumb-line” , which is parallel to the axis of symmetry (the y-axis), bisects chord at W. (See figure 2.) This is the first proposition used in Archimedes’s proof.

Proposition 1: From point P on a parabola, drop the straight line parallel to the axis of symmetry meeting the chord Parallel to the tangent at P if and only if W bisects the chord .

Suppose next that chords and are parallel, with midpoints V and W respectively. We will see that there is a quadratic relationship between the ratio of the lengths of the “plumb-line” segments and and the corresponding ratio for the “half-chords” and .

This follows by direct calculation. Let M and R be the points (x1, b – ax1 ) and (x2, b – ax2 ); let N and Q be the points (x3, b – ax3 ) and (x4, b – ax4 ). The coordinates of P, V, and W are easily computed, the latter two using midpoint formulas.By direct calculation

Since

it follows that

Thus the second proposition used by Archimedes:

Proposition 2: From a point P on a parabola draw a straightLine parallel to the axis of symmetry meeting two chords and that are parallel to the tangent line P, at points V and W respectively. Then

With these two propositions in hand, Archimedes had the tools to erect the scaffolding needed to prove the main result.

As shown in figure 4, the inscribed triangle with sides and determines two more inscribed triangles and . Archimedes proved (see Proposition 3 below) that.Continuing on, for there are two more inscribed triangles whose areas are . The same is true for . So with seven inscribed triangles, the area of the parabolic segment can be approximated by

This process can be extended indefinitely to obtain

Summing the geometric series will yield Archimedes’s result:

To show we consider only the right half of ; that is, we’ll show .

Proposition 3: .

Proof (see figure 5): From points N and Q, drop parallel plumb-lines that by Proposition 1 bisect and at Points X and Y, respectively. Thus, is similar to , and so the line is parallel to .

It also follows that . Thus, V bisects , and so by Proposition 1 is parallel to the tangent line at P and hence parallel to . Since

implies

by Proposition 2.

To continue, three triangles (see figure 6) are now used: , which is contained in , and , which is half of .

Since

it follows that

Moreover, since is similar to ,

Now put it all together, starting with :

Corollary: By a similar argument, . Therefore, .

Extending the inscribed triangle process indefinitely gives better and better approximations to the area, so that in the limit we have the result:

However, Archimedes did not sum the series. Instead, he showed that the area of the parabolic segment could neither be greater than nor less than , leaving only the remaining possibility, that the area of the parabolic segment must equal .

His method depended on showing that inscribed triangles could be used to get arbitrarily close to the area of the parabolic segment. In figure 7, it is seen that the area of the inscribed triangle is half the circumscribed parallelogram.hus, is greater than half the parabolic segment.

Since each added inscribed triangle accounts for over half the remaining area for that parabolic segment, it is possible to use 2n . 1 inscribed triangles to construct a 2n + 1 sided inscribed polygon such that the difference between the area of the polygon and the parabolic segment is less than any To see how this can be accomplished, note first that the area of the inscribed 2n + 1 sided polygon, denoted by Tn, is given by the finite sum

Next, observe that the difference between Tn and is For if we add them,

Since

the finite sum telescopes to 4/3.

Now it’s time to finish off Archimedes’s argument!

Let A be the area of the parabolic segment. Let

be the area of the 2n + 1 sided polygon formed from 2n – 1 triangles inscribed under the parabola. Let

where can be made arbitrarily small by making n sufficiently large.

1. Assume , so the area of the parabolic segment is larger. Then or Construct a 2n + 1 sided inscribed polygon such that the difference between area of the parabolic segment and the polygon is less than ; that is, or Then Contradiction!

2. Assume , so the area of the parabolic segment is smaller. Then or Choose a value n such that Then

Contradiction!

Therefore, since both and are false, must be true; Archimedes’s result follows.

Further Reading

Archimedes’s full proof can be found in The Works of Archimedes, by Thomas L. Heath (Dover Press, 1953). Over the course of 24 propositions, Archimedes actually gives two proofs for the area of the parabola, a mechanical one (Proposition 17) and a mathematical one (Proposition 24).Proofs of his first three propositions, which include the result obtained from the MVT and the plumb-line and half-chord squared ratio result are not given; Archimedes simply states “and these propositions are proved in the elements of conics,” referring to treatises on conics by Euclid and Aristaeus.

Exercise 1: Using the integral calculus, verify Archimedes’s result by showing that the area between the curve y = b – ax2 and the x-axis is 4/3 times the area of the inscribed triangle.

Exercise 2: Verify Archimedes’s result using the area between the parabola y = 4 – x2 and the line y = 2 – x. Use the MVT to find the vertex of the inscribed triangle and determine its altitude by constructing a line perpendicular to the base of the triangle.

About the author: Brian J. Shelburne is an associate professor of mathematics and computer science at Wittenberg University in Springfield, Ohio.

email: bshelburne@wittenberg.edu

So why is this result interesting?

First, it's a result proved by one of the greatest mathematicians of all time, Archimedes of Syracuse, and it's always interesting to study the methods of the masters. And though he found the area under a curve without the use of the integral calculus, he employed some distinctly calculus-like techniques (which makes you wonder what Archimedes would have done had he been around in the 17th century.or later). Finally, it's fascinating that the area under the parabola, a curved figure, compared to the area of the inscribed triangle, yields a simple 4 to 3 ratio.

It does not detract from the genius of Archimedesfs proof if we reframe parts of it using modern mathematical notation; his underlying logic is still there. Hence, we begin with a modern definition of a parabola: it is the curve defined by the quadratic y = a0x2 + b0x + c0, where a0 ‚ 0. This equation can be rewritten in the vertex form y = a0 (x . H)2 + k, which by a suitable rigid transformation can be transformed into y = b . Ax2, a downward opening parabola whose axis Of symmetry is the y-axis. This will make subsequent calculations easier.

To determine the inscribed triangle MPR, where the tangent line at vertex P is parallel to the chord , we make use of the Mean Value Theorem (MVT).

The MVT as applied to the parabola f(x) = b – ax2 says that on the open interval (x1, x2) there is a point x3 such that

If we plug in expressions for the quadratic f and its derivative f we obtain

Solving yields

the midpoint. This means that the tangent line at point P is parallel to the chord if and only if the “plumb-line” , which is parallel to the axis of symmetry (the y-axis), bisects chord at W. (See figure 2.) This is the first proposition used in Archimedes’s proof.

Proposition 1: From point P on a parabola, drop the straight line parallel to the axis of symmetry meeting the chord Parallel to the tangent at P if and only if W bisects the chord .

Suppose next that chords and are parallel, with midpoints V and W respectively. We will see that there is a quadratic relationship between the ratio of the lengths of the “plumb-line” segments and and the corresponding ratio for the “half-chords” and .

This follows by direct calculation. Let M and R be the points (x1, b – ax1 ) and (x2, b – ax2 ); let N and Q be the points (x3, b – ax3 ) and (x4, b – ax4 ). The coordinates of P, V, and W are easily computed, the latter two using midpoint formulas.By direct calculation

Since

it follows that

Thus the second proposition used by Archimedes:

Proposition 2: From a point P on a parabola draw a straightLine parallel to the axis of symmetry meeting two chords and that are parallel to the tangent line P, at points V and W respectively. Then

With these two propositions in hand, Archimedes had the tools to erect the scaffolding needed to prove the main result.

As shown in figure 4, the inscribed triangle with sides and determines two more inscribed triangles and . Archimedes proved (see Proposition 3 below) that.Continuing on, for there are two more inscribed triangles whose areas are . The same is true for . So with seven inscribed triangles, the area of the parabolic segment can be approximated by

This process can be extended indefinitely to obtain

Summing the geometric series will yield Archimedes’s result:

To show we consider only the right half of ; that is, we’ll show .

Proposition 3: .

Proof (see figure 5): From points N and Q, drop parallel plumb-lines that by Proposition 1 bisect and at Points X and Y, respectively. Thus, is similar to , and so the line is parallel to .

It also follows that . Thus, V bisects , and so by Proposition 1 is parallel to the tangent line at P and hence parallel to . Since

implies

by Proposition 2.

To continue, three triangles (see figure 6) are now used: , which is contained in , and , which is half of .

Since

it follows that

Moreover, since is similar to ,

Now put it all together, starting with :

Corollary: By a similar argument, . Therefore, .

Extending the inscribed triangle process indefinitely gives better and better approximations to the area, so that in the limit we have the result:

However, Archimedes did not sum the series. Instead, he showed that the area of the parabolic segment could neither be greater than nor less than , leaving only the remaining possibility, that the area of the parabolic segment must equal .

His method depended on showing that inscribed triangles could be used to get arbitrarily close to the area of the parabolic segment. In figure 7, it is seen that the area of the inscribed triangle is half the circumscribed parallelogram.hus, is greater than half the parabolic segment.

Since each added inscribed triangle accounts for over half the remaining area for that parabolic segment, it is possible to use 2n . 1 inscribed triangles to construct a 2n + 1 sided inscribed polygon such that the difference between the area of the polygon and the parabolic segment is less than any To see how this can be accomplished, note first that the area of the inscribed 2n + 1 sided polygon, denoted by Tn, is given by the finite sum

Next, observe that the difference between Tn and is For if we add them,

Since

the finite sum telescopes to 4/3.

Now it’s time to finish off Archimedes’s argument!

Let A be the area of the parabolic segment. Let

be the area of the 2n + 1 sided polygon formed from 2n – 1 triangles inscribed under the parabola. Let

where can be made arbitrarily small by making n sufficiently large.

1. Assume , so the area of the parabolic segment is larger. Then or Construct a 2n + 1 sided inscribed polygon such that the difference between area of the parabolic segment and the polygon is less than ; that is, or Then Contradiction!

2. Assume , so the area of the parabolic segment is smaller. Then or Choose a value n such that Then

Contradiction!

Therefore, since both and are false, must be true; Archimedes’s result follows.

Further Reading

Archimedes’s full proof can be found in The Works of Archimedes, by Thomas L. Heath (Dover Press, 1953). Over the course of 24 propositions, Archimedes actually gives two proofs for the area of the parabola, a mechanical one (Proposition 17) and a mathematical one (Proposition 24).Proofs of his first three propositions, which include the result obtained from the MVT and the plumb-line and half-chord squared ratio result are not given; Archimedes simply states “and these propositions are proved in the elements of conics,” referring to treatises on conics by Euclid and Aristaeus.

Exercise 1: Using the integral calculus, verify Archimedes’s result by showing that the area between the curve y = b – ax2 and the x-axis is 4/3 times the area of the inscribed triangle.

Exercise 2: Verify Archimedes’s result using the area between the parabola y = 4 – x2 and the line y = 2 – x. Use the MVT to find the vertex of the inscribed triangle and determine its altitude by constructing a line perpendicular to the base of the triangle.

About the author: Brian J. Shelburne is an associate professor of mathematics and computer science at Wittenberg University in Springfield, Ohio.

email: bshelburne@wittenberg.edu