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The problem of area: Archimedes’s quadrature of the parabola

Francesco Cellarosi Math 120 - Lecture 25 - November 14, 2016

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The problem of area

We know how to compute the area of simple geometrical figures b h

The problem of area

We know how to compute the area of simple geometrical figures b h Area = b · h

The problem of area

We know how to compute the area of simple geometrical figures b h Area = b · h h b

The problem of area

We know how to compute the area of simple geometrical figures b h Area = b · h h b Area = b · h

The problem of area

We know how to compute the area of simple geometrical figures b h Area = b · h h b Area = b · h b

Math 120 Archimedes’s quadrature of the parabola November 14, 2016 2 / 22

The problem of area

We know how to compute the area of simple geometrical figures b h Area = b · h h b Area = b · h b h b Area = 1

2b · h

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The problem of area

We know how to compute the area of simple geometrical figures b h Area = b · h h b Area = b · h b a b c Area =

• p(p − a)(p − b)(p − c)

p = 1

2(a + b + c)

(Heron’s formula, 1st century CE)

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The problem of area

We know how to compute the area of simple geometrical figures a b c d

The problem of area

We know how to compute the area of simple geometrical figures a b c d Area =

• (p − a)(p − b)(p − c)(p − d)

p = 1

2(a + b + c + d)

The problem of area

We know how to compute the area of simple geometrical figures a b c d Area =

• (p − a)(p − b)(p − c)(p − d)

p = 1

2(a + b + c + d)

(Brahmagupta’s formula, 7th century CE)

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The problem of area

We know how to compute the area of simple geometrical figures r

The problem of area

We know how to compute the area of simple geometrical figures r Area = πr2 π = circumference diameter

The problem of area

We know how to compute the area of simple geometrical figures r Area = πr2 π = circumference diameter (due to Archimedes, c. 260 BCE)

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The problem of area

What about more general curved regions?

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The problem of area

What about more general curved regions? Today we will prove a beautiful formula, due to Archimedes, for the area between a parabola and a segment whose endpoints are on the parabola. He proved this in a letter (later titled “Quadrature of the parabola”) to his friend Dositheus of Pelusium (who succeeded Conon of Samos –also friend

• f Archimedes– as director of the mathematical school of Alexandria).

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Archimedes’s Theorem

Consider the region R between the parabolic arc

⌢

AB and the segment AB. A B R

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Archimedes’s Theorem

Consider the region R between the parabolic arc

⌢

AB and the segment AB. A B R P Theorem (Archimedes). Consider the point P on the arc

⌢

AB which is the farthest from the segment AB.

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Archimedes’s Theorem

Consider the region R between the parabolic arc

⌢

AB and the segment AB. A B R P P0 Theorem (Archimedes). Consider the point P on the arc

⌢

AB which is the farthest from the segment AB. Then the area of the region R equals 4

3

times the area of the triangle P0 = △ABP.

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Preliminary facts

Here are some facts that we will assume as proven (they were known to Archimedes since the had already been proved by Euclid and Aristarchus). A B P

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Preliminary facts

Here are some facts that we will assume as proven (they were known to Archimedes since the had already been proved by Euclid and Aristarchus). A B P FACT 1: The tangent line to the parabola at P is parallel to AB.

Math 120 Archimedes’s quadrature of the parabola November 14, 2016 10 / 22

Preliminary facts

Here are some facts that we will assume as proven (they were known to Archimedes since the had already been proved by Euclid and Aristarchus). A B P M FACT 1: The tangent line to the parabola at P is parallel to AB. FACT 2: The line through P and parallel to ... the .... axis of the parabola meets AB in its middle point M

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Preliminary facts

A B P M

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Preliminary facts

A B P M C D FACT 3: Every chord CD parallel to AB is bisected by PM

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Preliminary facts

A B P M C D N FACT 3: Every chord CD parallel to AB is bisected by PM, say at N.

Math 120 Archimedes’s quadrature of the parabola November 14, 2016 11 / 22

Preliminary facts

A B P M C D N FACT 3: Every chord CD parallel to AB is bisected by PM, say at N. FACT 4: PN/PM = ND

2/MB 2.

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Preliminary facts

A B P M C D N FACT 3: Every chord CD parallel to AB is bisected by PM, say at N. FACT 4: PN/PM = ND

2/MB 2.

We will assume FACTS 1÷4, without proof.

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• Proof. Two new triangles

A B P M Recall: △ABP was constructed from the chord AB. Now construct two new triangles in the same way:

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• Proof. Two new triangles

A B P M P1 Recall: △ABP was constructed from the chord AB. Now construct two new triangles in the same way:

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• Proof. Two new triangles

A B P M P1 P1 Recall: △ABP was constructed from the chord AB. Now construct two new triangles in the same way: △APP1 from AP

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• Proof. Two new triangles

A B P M P1 P1 P2 Recall: △ABP was constructed from the chord AB. Now construct two new triangles in the same way: △APP1 from AP

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• Proof. Two new triangles

A B P M P1 P1 P2 P2 Recall: △ABP was constructed from the chord AB. Now construct two new triangles in the same way: △APP1 from AP and △PBP2 from PB.

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• Proof. A bigger polygon

A B P P1 P2

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• Proof. A bigger polygon

A B P P1 P2 The area of the polygon AP1PP2B is bigger than the area of the triangle △ABP, but smaller than the area of the parabolic region

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• Proof. A bigger polygon

A B P P1 P2 The area of the polygon AP1PP2B is bigger than the area of the triangle △ABP, but smaller than the area of the parabolic region (because the parabola is concave up!).

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• Proof. A bigger polygon

A B P P1 P2 The area of the polygon AP1PP2B is bigger than the area of the triangle △ABP, but smaller than the area of the parabolic region (because the parabola is concave up!). Now that we have 4 new chords AP1, P1P, PP2, P2B we can repeat the construction above and obtain 4 new triangles.

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• Proof. Bigger and bigger polygons

A B P area(APB)

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• Proof. Bigger and bigger polygons

A B P P1 P2 area(APB)<area(AP1PP2B)

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• Proof. Bigger and bigger polygons

A B P P1 P2 P3 P4 P5 P6 area(APB)<area(AP1PP2B)

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• Proof. Bigger and bigger polygons

A B P P1 P2 P3 P4 P5 P6 area(APB)<area(AP1PP2B)<area(AP3P1P4PP5P2P6B)

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• Proof. Bigger and bigger polygons

A B P P1 P2 P3 P4 P5 P6 area(APB)<area(AP1PP2B)<area(AP3P1P4PP5P2P6B)<. . .

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• Proof. Bigger and bigger polygons

A B P R area(APB)<area(AP1PP2B)<area(AP3P1P4PP5P2P6B)<. . .<area(R)

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• Proof. Exhaustion of R

Although it looks obvious that this infinite construction leaves out no area

• f the region R, we need to prove it...

Math 120 Archimedes’s quadrature of the parabola November 14, 2016 15 / 22

• Proof. Exhaustion of R

Although it looks obvious that this infinite construction leaves out no area

• f the region R, we need to prove it...
• Lemma. Let A = area(R). Let Pn be the polygon constructed at the nth

step of the procedure described above, and let Dn = A − area(Pn). Then lim

n→∞ Dn = 0.

Math 120 Archimedes’s quadrature of the parabola November 14, 2016 15 / 22

• Proof. Exhaustion of R

Although it looks obvious that this infinite construction leaves out no area

• f the region R, we need to prove it...
• Lemma. Let A = area(R). Let Pn be the polygon constructed at the nth

step of the procedure described above, and let Dn = A − area(Pn). Then lim

n→∞ Dn = 0.

Proof of the Lemma. A B P M A’ B’

Math 120 Archimedes’s quadrature of the parabola November 14, 2016 15 / 22

• Proof. Exhaustion of R

Although it looks obvious that this infinite construction leaves out no area

• f the region R, we need to prove it...
• Lemma. Let A = area(R). Let Pn be the polygon constructed at the nth

step of the procedure described above, and let Dn = A − area(Pn). Then lim

n→∞ Dn = 0.

Proof of the Lemma. A B P M A’ B’

Math 120 Archimedes’s quadrature of the parabola November 14, 2016 15 / 22

• Proof. Exhaustion of R

Although it looks obvious that this infinite construction leaves out no area

• f the region R, we need to prove it...
• Lemma. Let A = area(R). Let Pn be the polygon constructed at the nth

step of the procedure described above, and let Dn = A − area(Pn). Then lim

n→∞ Dn = 0.

Proof of the Lemma. A B P M A’ B’ AA′ PM BB′ and AB A′B′ by FACT 1.

Math 120 Archimedes’s quadrature of the parabola November 14, 2016 15 / 22

• Proof. Exhaustion of R

Although it looks obvious that this infinite construction leaves out no area

• f the region R, we need to prove it...
• Lemma. Let A = area(R). Let Pn be the polygon constructed at the nth

step of the procedure described above, and let Dn = A − area(Pn). Then lim

n→∞ Dn = 0.

Proof of the Lemma. A B P M A’ B’ AA′ PM BB′ and AB A′B′ by FACT 1. Of course area(AA′B′B) > A.

Math 120 Archimedes’s quadrature of the parabola November 14, 2016 15 / 22

• Proof. Exhaustion of R

Although it looks obvious that this infinite construction leaves out no area

• f the region R, we need to prove it...
• Lemma. Let A = area(R). Let Pn be the polygon constructed at the nth

step of the procedure described above, and let Dn = A − area(Pn). Then lim

n→∞ Dn = 0.

Proof of the Lemma. A B P M A’ B’ AA′ PM BB′ and AB A′B′ by FACT 1. Of course area(AA′B′B) > A. Moreover area(AA′B′B) = 2 · area(P0).

Math 120 Archimedes’s quadrature of the parabola November 14, 2016 15 / 22

• Proof. Exhaustion of R

Although it looks obvious that this infinite construction leaves out no area

• f the region R, we need to prove it...
• Lemma. Let A = area(R). Let Pn be the polygon constructed at the nth

step of the procedure described above, and let Dn = A − area(Pn). Then lim

n→∞ Dn = 0.

Proof of the Lemma. A B P M A’ B’ AA′ PM BB′ and AB A′B′ by FACT 1. Of course area(AA′B′B) > A. Moreover area(AA′B′B) = 2 · area(P0). Therefore area(P0) > 1

2A and D0 < 1 2A

Math 120 Archimedes’s quadrature of the parabola November 14, 2016 15 / 22

• Proof. Exhaustion of R

Although it looks obvious that this infinite construction leaves out no area

• f the region R, we need to prove it...
• Lemma. Let A = area(R). Let Pn be the polygon constructed at the nth

step of the procedure described above, and let Dn = A − area(Pn). Then lim

n→∞ Dn = 0.

Proof of the Lemma. Now consider the two triangles △AP1P and △PP2B added to P0 to form the polygon P1.

Math 120 Archimedes’s quadrature of the parabola November 14, 2016 15 / 22

• Proof. Exhaustion of R

Although it looks obvious that this infinite construction leaves out no area

• f the region R, we need to prove it...
• Lemma. Let A = area(R). Let Pn be the polygon constructed at the nth

step of the procedure described above, and let Dn = A − area(Pn). Then lim

n→∞ Dn = 0.

Proof of the Lemma. Now consider the two triangles △AP1P and △PP2B added to P0 to form the polygon P1. Apply the above argument to each of the two parabolic regions below AP and PB.

Math 120 Archimedes’s quadrature of the parabola November 14, 2016 15 / 22

• Proof. Exhaustion of R

Although it looks obvious that this infinite construction leaves out no area

• f the region R, we need to prove it...
• Lemma. Let A = area(R). Let Pn be the polygon constructed at the nth

step of the procedure described above, and let Dn = A − area(Pn). Then lim

n→∞ Dn = 0.

Proof of the Lemma. Now consider the two triangles △AP1P and △PP2B added to P0 to form the polygon P1. Apply the above argument to each of the two parabolic regions below AP and PB. We get that the area of each of these triangles is at least half of the area of the corresponding parabolic region.

Math 120 Archimedes’s quadrature of the parabola November 14, 2016 15 / 22

• Proof. Exhaustion of R

Although it looks obvious that this infinite construction leaves out no area

• f the region R, we need to prove it...
• Lemma. Let A = area(R). Let Pn be the polygon constructed at the nth

step of the procedure described above, and let Dn = A − area(Pn). Then lim

n→∞ Dn = 0.

Proof of the Lemma. Now consider the two triangles △AP1P and △PP2B added to P0 to form the polygon P1. Apply the above argument to each of the two parabolic regions below AP and PB. We get that the area of each of these triangles is at least half of the area of the corresponding parabolic region. Therefore D1 < 1

2D0.

Math 120 Archimedes’s quadrature of the parabola November 14, 2016 15 / 22

• Proof. Exhaustion of R

Although it looks obvious that this infinite construction leaves out no area

• f the region R, we need to prove it...
• Lemma. Let A = area(R). Let Pn be the polygon constructed at the nth

step of the procedure described above, and let Dn = A − area(Pn). Then lim

n→∞ Dn = 0.

Proof of the Lemma. Now consider the two triangles △AP1P and △PP2B added to P0 to form the polygon P1. Apply the above argument to each of the two parabolic regions below AP and PB. We get that the area of each of these triangles is at least half of the area of the corresponding parabolic region. Therefore D1 < 1

• 2D0. Continuing int this way, we obtain that Dn < 1

2Dn−1 for every

n ≥ 1

Math 120 Archimedes’s quadrature of the parabola November 14, 2016 15 / 22

• Proof. Exhaustion of R

Although it looks obvious that this infinite construction leaves out no area

• f the region R, we need to prove it...
• Lemma. Let A = area(R). Let Pn be the polygon constructed at the nth

step of the procedure described above, and let Dn = A − area(Pn). Then lim

n→∞ Dn = 0.

Proof of the Lemma. Now consider the two triangles △AP1P and △PP2B added to P0 to form the polygon P1. Apply the above argument to each of the two parabolic regions below AP and PB. We get that the area of each of these triangles is at least half of the area of the corresponding parabolic region. Therefore D1 < 1

• 2D0. Continuing int this way, we obtain that Dn < 1

2Dn−1 for every

n ≥ 1, and this implies that limn→∞ Dn = 0.

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• Proof. Computing the area

We proved that the polygons Pn exhaust the parabolic region R as n → ∞.

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• Proof. Computing the area

We proved that the polygons Pn exhaust the parabolic region R as n → ∞. Now let’s compute area(Pn). A B P M P2 N M2 Consider NP2 MB and P2M2 PM.

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• Proof. Computing the area

We proved that the polygons Pn exhaust the parabolic region R as n → ∞. Now let’s compute area(Pn). A B P M P2 N M2 Consider NP2 MB and P2M2 PM. By FACT 2, M2 is the midpoint of MB.

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• Proof. Computing the area

We proved that the polygons Pn exhaust the parabolic region R as n → ∞. Now let’s compute area(Pn). A B P M P2 N M2 R Consider NP2 MB and P2M2 PM. By FACT 2, M2 is the midpoint of

• MB. Let R = P2M2 ∩ PB.

Math 120 Archimedes’s quadrature of the parabola November 14, 2016 16 / 22

• Proof. Computing the area

We proved that the polygons Pn exhaust the parabolic region R as n → ∞. Now let’s compute area(Pn). A B P M P2 N M2 R Consider NP2 MB and P2M2 PM. By FACT 2, M2 is the midpoint of

• MB. Let R = P2M2 ∩ PB. Note that PM = 2RM2 (similar triangles).

Math 120 Archimedes’s quadrature of the parabola November 14, 2016 16 / 22

• Proof. Computing the area

We proved that the polygons Pn exhaust the parabolic region R as n → ∞. Now let’s compute area(Pn). A B P M P2 N M2 R Consider NP2 MB and P2M2 PM. By FACT 2, M2 is the midpoint of

• MB. Let R = P2M2 ∩ PB. Note that PM = 2RM2 (similar triangles).

Let us focus on the triangles △PBM2 and △PBP2, sharing the base PB.

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• Proof. Computing the area

B P M P2 N M2 R

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• Proof. Computing the area

B P M P2 N M2 R K H

Math 120 Archimedes’s quadrature of the parabola November 14, 2016 17 / 22

• Proof. Computing the area

B P M P2 N M2 R K H By FACTS 3 and 4, we have PN/PM = NP2

2/MB 2 = NP2 2/(2MM2)2 =

= NP2

2/(2NP2)2 = 1/4.

• Proof. Computing the area

B P M P2 N M2 R K H By FACTS 3 and 4, we have PN/PM = NP2

2/MB 2 = NP2 2/(2MM2)2 =

= NP2

2/(2NP2)2 = 1/4.

Therefore PM = 4PN and NM = 3PN.

• Proof. Computing the area

B P M P2 N M2 R K H By FACTS 3 and 4, we have PN/PM = NP2

2/MB 2 = NP2 2/(2MM2)2 =

= NP2

2/(2NP2)2 = 1/4.

Therefore PM = 4PN and NM = 3PN. We obtain PM = 4

3NM = 4 3P2M2.

• Proof. Computing the area

B P M P2 N M2 R K H By FACTS 3 and 4, we have PN/PM = NP2

2/MB 2 = NP2 2/(2MM2)2 =

= NP2

2/(2NP2)2 = 1/4.

Therefore PM = 4PN and NM = 3PN. We obtain PM = 4

3NM = 4 3P2M2.

Since we know already that PM = 2RM2, we have RM2 = 2

3P2M2 and RM2 = 2P2R. This implies that the heights

M2K and P2H are in 2:1 ratio. Hence area(PBM2) = 2 area(PBP2).

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• Proof. Computing the area

B P M P2 N M2 K H area(PBM2) = 2 area(PBP2).

• Proof. Computing the area

B P M P2 N M2 K H area(PBM2) = 2 area(PBP2). Moreover, area( PBM ) = 2 area(PBM2).

• Proof. Computing the area

B P M P2 N M2 K H area(PBM2) = 2 area(PBP2). Moreover, area( PBM ) = 2 area(PBM2). Therefore area(PBP2) = 1

4area( PBM ).

This means that the “new” triangle △BPP2 has area equal to 1

4 of that of the “old” triangle △PBM .

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• Proof. Computing the area

A B P P1 P2 M area(PBP2) = 1

4area( PBM ) and, similarly, area(APP1) = 1 4area( APM ).

Math 120 Archimedes’s quadrature of the parabola November 14, 2016 19 / 22

• Proof. Computing the area

A B P P1 P2 M area(PBP2) = 1

4area( PBM ) and, similarly, area(APP1) = 1 4area( APM ).

Therefore the triangles added to P0 = △ABP to form P1 = AP1PP2B have, combined, area equal to 1

4area(P0).

Math 120 Archimedes’s quadrature of the parabola November 14, 2016 19 / 22

• Proof. Computing the area

Repeating the argument above at each stage, we obtain: area(P1) =

• 1 + 1

4

• · area(P0),

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• Proof. Computing the area

Repeating the argument above at each stage, we obtain: area(P1) =

• 1 + 1

4

• · area(P0),

area(P2) =

• 1 + 1

4 + 1 42

• area · (P0),

Math 120 Archimedes’s quadrature of the parabola November 14, 2016 20 / 22

• Proof. Computing the area

Repeating the argument above at each stage, we obtain: area(P1) =

• 1 + 1

4

• · area(P0),

area(P2) =

• 1 + 1

4 + 1 42

• area · (P0),

and in general area(Pn) =

• 1 + 1

4 + 1 42 + 1 43 + . . . + 1 4n

• · area(P0).

Math 120 Archimedes’s quadrature of the parabola November 14, 2016 20 / 22

• Proof. Computing the area

Repeating the argument above at each stage, we obtain: area(P1) =

• 1 + 1

4

• · area(P0),

area(P2) =

• 1 + 1

4 + 1 42

• area · (P0),

and in general area(Pn) =

• 1 + 1

4 + 1 42 + 1 43 + . . . + 1 4n

• · area(P0).

Therefore A =

• 1 + 1

4 + 1 42 + 1 43 + . . . + 1 4n + . . .

• · area(P0)

Math 120 Archimedes’s quadrature of the parabola November 14, 2016 20 / 22

• Proof. 1 + 1

4 + 1 42 + 1 43 + . . . = 4 3

We claim that 1 + 1 4 + 1 42 + 1 43 + . . . = 4 3. Proof of the claim. This picture shows that 1 4 + 1 42 + 1 43 + . . . = 1 3.

Math 120 Archimedes’s quadrature of the parabola November 14, 2016 21 / 22

• Proof. 1 + 1

4 + 1 42 + 1 43 + . . . = 4 3

We claim that 1 + 1 4 + 1 42 + 1 43 + . . . = 4 3. Proof of the claim. This picture shows that 1 4 + 1 42 + 1 43 + . . . = 1

• 3. Add 1 to get 4

3.

Math 120 Archimedes’s quadrature of the parabola November 14, 2016 21 / 22

Summary

We have proved Archimedes’s Theorem: R A B A B P P0 area(R) = 4 3area(P0), where P0 = △ABP and P is the point of intersection between the parabola and the line passing through the midpoint of AB parallel to the axis of the parabola.

Math 120 Archimedes’s quadrature of the parabola November 14, 2016 22 / 22