The problem of area: Archimedes’s quadrature of the parabola
Francesco Cellarosi
Math 120 - Lecture 25 - November 14, 2016
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 1 / 22 Area = b · h Area = b · h h
b
b
The problem of area
We know how to compute the area of simple geometrical figures
h
b
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 2 / 22 Area = b · h h
b
b
The problem of area
We know how to compute the area of simple geometrical figures
h Area = b · h
b
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 2 / 22 Area = b · h
b
The problem of area
We know how to compute the area of simple geometrical figures
h Area = b · h h
b b
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 2 / 22 b
The problem of area
We know how to compute the area of simple geometrical figures
h Area = b · h Area = b · h h
b 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
h Area = b · h Area = b · h h
b b
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
h Area = b · h Area = b · h h
b b
1 h Area = 2 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
h Area = b · h Area = b · h h
b b
a c Area = pp(p − a)(p − b)(p − c) 1 p = 2 (a + b + c) (Heron’s formula, 1st century CE) b
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 2 / 22 Area = p(p − a)(p − b)(p − c)(p − d)
1 p = 2 (a + b + c + d) (Brahmagupta’s formula, 7th century CE)
The problem of area
We know how to compute the area of simple geometrical figures
b a
c d
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 3 / 22 (Brahmagupta’s formula, 7th century CE)
The problem of area
We know how to compute the area of simple geometrical figures
b a Area = p(p − a)(p − b)(p − c)(p − d)
1 p = 2 (a + b + c + d) c d
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 3 / 22 The problem of area
We know how to compute the area of simple geometrical figures
b a Area = p(p − a)(p − b)(p − c)(p − d)
1 p = 2 (a + b + c + d) c d (Brahmagupta’s formula, 7th century CE)
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 3 / 22 Area = πr 2 π = circumference diameter (due to Archimedes, c. 260 BCE)
The problem of area
We know how to compute the area of simple geometrical figures
r
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 4 / 22 (due to Archimedes, c. 260 BCE)
The problem of area
We know how to compute the area of simple geometrical figures
Area = πr 2 r π = circumference diameter
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 4 / 22 The problem of area
We know how to compute the area of simple geometrical figures
Area = πr 2 r π = circumference diameter (due to Archimedes, c. 260 BCE)
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 4 / 22 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 of Archimedes– as director of the mathematical school of Alexandria).
The problem of area
What about more general curved regions?
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 5 / 22 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 of Archimedes– as director of the mathematical school of Alexandria).
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 5 / 22 Math 120 Archimedes’s quadrature of the parabola November 14, 2016 6 / 22 Math 120 Archimedes’s quadrature of the parabola November 14, 2016 7 / 22 Math 120 Archimedes’s quadrature of the parabola November 14, 2016 8 / 22 _ Theorem (Archimedes). Consider the point P on the arc AB which is the 4 farthest from the segment AB. Then the area of the region R equals 3 times the area of the triangle P0 = 4ABP.
Archimedes’s Theorem
_ Consider the region R between the parabolic arc AB and the segment AB.
B
A
R
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 9 / 22 4 Then the area of the region R equals 3 times the area of the triangle P0 = 4ABP.
Archimedes’s Theorem
_ Consider the region R between the parabolic arc AB and the segment AB.
B
A
R
P
_ Theorem (Archimedes). Consider the point P on the arc AB which is the farthest from the segment AB.
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 9 / 22 Archimedes’s Theorem
_ Consider the region R between the parabolic arc AB and the segment AB.
B
A
PR0
P
_ Theorem (Archimedes). Consider the point P on the arc AB which is the 4 farthest from the segment AB. Then the area of the region R equals 3 times the area of the triangle P0 = 4ABP.
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 9 / 22 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
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).
B
A
P
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 10 / 22 FACT 2: The line through P and parallel to the...... axis of the parabola meets AB in its middle point M
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).
B
A
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).
M B
A
P
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
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 10 / 22 FACT 3: Every chord CD parallel to AB is bisected by PM, say at N. 2 2 FACT 4: PN/PM = ND /MB .
We will assume FACTS 1÷4, without proof.
Preliminary facts
B M
A
P
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 11 / 22 , say at N. 2 2 FACT 4: PN/PM = ND /MB .
We will assume FACTS 1÷4, without proof.
Preliminary facts
B M
A D C
P
FACT 3: Every chord CD parallel to AB is bisected by PM
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 11 / 22 2 2 FACT 4: PN/PM = ND /MB .
We will assume FACTS 1÷4, without proof.
Preliminary facts
B M
A N D C
P
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 We will assume FACTS 1÷4, without proof.
Preliminary facts
B M
A N D C
P
FACT 3: Every chord CD parallel to AB is bisected by PM, say at N. 2 2 FACT 4: PN/PM = ND /MB .
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 11 / 22 Preliminary facts
B M
A N D C
P
FACT 3: Every chord CD parallel to AB is bisected by PM, say at N. 2 2 FACT 4: PN/PM = ND /MB .
We will assume FACTS 1÷4, without proof.
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 11 / 22 4APP1 from AP and 4PBP2 from PB.
Proof. Two new triangles
B M
A
P
Recall: 4ABP was constructed from the chord AB. Now construct two new triangles in the same way:
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 12 / 22 4APP1 from AP and 4PBP2 from PB.
Proof. Two new triangles
B M
A
P1 P
Recall: 4ABP was constructed from the chord AB. Now construct two new triangles in the same way:
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 12 / 22 and 4PBP2 from PB.
Proof. Two new triangles
B M
A
P1 P
Recall: 4ABP was constructed from the chord AB. Now construct two new triangles in the same way: 4APP1 from AP
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 12 / 22 and 4PBP2 from PB.
Proof. Two new triangles
B M
A
P2
P1 P
Recall: 4ABP was constructed from the chord AB. Now construct two new triangles in the same way: 4APP1 from AP
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 12 / 22 Proof. Two new triangles
B M
A
P2
P1 P
Recall: 4ABP was constructed from the chord AB. Now construct two new triangles in the same way: 4APP1 from AP and 4PBP2 from PB.
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 12 / 22 The area of the polygon AP1PP2B is bigger than the area of the triangle 4ABP, 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.
Proof. A bigger polygon
B
A
P2 P1 P
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 13 / 22 (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.
Proof. A bigger polygon
B
A
P2 P1 P
The area of the polygon AP1PP2B is bigger than the area of the triangle 4ABP, but smaller than the area of the parabolic region
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 13 / 22 Now that we have 4 new chords AP1, P1P, PP2, P2B we can repeat the construction above and obtain 4 new triangles.
Proof. A bigger polygon
B
A
P2 P1 P
The area of the polygon AP1PP2B is bigger than the area of the triangle 4ABP, but smaller than the area of the parabolic region (because the parabola is concave up!).
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 13 / 22 Proof. A bigger polygon
B
A
P2 P1 P
The area of the polygon AP1PP2B is bigger than the area of the triangle 4ABP, 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.
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 13 / 22 Proof. Bigger and bigger polygons B A P area(APB) Math 120 Archimedes’s quadrature of the parabola November 14, 2016 14 / 22 Proof. Bigger and bigger polygons B A P2 P1 P area(APB) Math 120 Archimedes’s quadrature of the parabola November 14, 2016 14 / 22 Proof. Bigger and bigger polygons B A P6 P3 P2 P1 P5 P4 P area(APB) Math 120 Archimedes’s quadrature of the parabola November 14, 2016 14 / 22 <. . . Proof. Bigger and bigger polygons B A P6 P3 P2 P1 P5 P4 P area(APB) Math 120 Archimedes’s quadrature of the parabola November 14, 2016 14 / 22 Proof. Bigger and bigger polygons B A P6 P3 P2 P1 P5 P4 P area(APB) Math 120 Archimedes’s quadrature of the parabola November 14, 2016 14 / 22 Proof. Bigger and bigger polygons B A R P area(APB) Math 120 Archimedes’s quadrature of the parabola November 14, 2016 14 / 22 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 Dn = 0. n→∞ Proof of the Lemma. Proof. Exhaustion of R Although it looks obvious that this infinite construction leaves out no area of the region R, we need to prove it... Math 120 Archimedes’s quadrature of the parabola November 14, 2016 15 / 22 Proof of the Lemma. Proof. Exhaustion of R Although it looks obvious that this infinite construction leaves out no area of 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 Dn = 0. n→∞ 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 of 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 Dn = 0. n→∞ Proof of the Lemma. M B A B’ P 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 of 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 Dn = 0. n→∞ Proof of the Lemma. M B A B’ P 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 of 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 Dn = 0. n→∞ Proof of the Lemma. 0 0 0 0 M B AA k PM k BB and AB k A B by FACT 1. A B’ P 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 of 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 Dn = 0. n→∞ Proof of the Lemma. 0 0 0 0 M B AA k PM k BB and AB k A B by FACT 1. A Of course area(AA0B0B) > A. B’ P 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 of 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 Dn = 0. n→∞ Proof of the Lemma. 0 0 0 0 M B AA k PM k BB and AB k A B by FACT 1. A Of course area(AA0B0B) > A. Moreover area(AA0B0B) = 2 · area(P ). B’ 0 P 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 of 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 Dn = 0. n→∞ Proof of the Lemma. 0 0 0 0 M B AA k PM k BB and AB k A B by FACT 1. A Of course area(AA0B0B) > A. Moreover area(AA0B0B) = 2 · area(P ). B’ 0 P 1 1 A’ Therefore area(P0) > 2 A and D0 < 2 A Math 120 Archimedes’s quadrature of the parabola November 14, 2016 15 / 22 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 1 1 D1 < 2 D0. Continuing int this way, we obtain that Dn < 2 Dn−1 for every n ≥ 1, and this implies that limn→∞ Dn = 0. Proof. Exhaustion of R Although it looks obvious that this infinite construction leaves out no area of 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 Dn = 0. n→∞ Proof of the Lemma. Now consider the two triangles 4AP1P and 4PP2B added to P0 to form the polygon P1. Math 120 Archimedes’s quadrature of the parabola November 14, 2016 15 / 22 We get that the area of each of these triangles is at least half of the area of the corresponding parabolic region. Therefore 1 1 D1 < 2 D0. Continuing int this way, we obtain that Dn < 2 Dn−1 for every n ≥ 1, and this implies that limn→∞ Dn = 0. Proof. Exhaustion of R Although it looks obvious that this infinite construction leaves out no area of 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 Dn = 0. n→∞ Proof of the Lemma. Now consider the two triangles 4AP1P and 4PP2B 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 Therefore 1 1 D1 < 2 D0. Continuing int this way, we obtain that Dn < 2 Dn−1 for every n ≥ 1, and this implies that limn→∞ Dn = 0. Proof. Exhaustion of R Although it looks obvious that this infinite construction leaves out no area of 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 Dn = 0. n→∞ Proof of the Lemma. Now consider the two triangles 4AP1P and 4PP2B 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 1 Continuing int this way, we obtain that Dn < 2 Dn−1 for every n ≥ 1, and this implies that limn→∞ Dn = 0. Proof. Exhaustion of R Although it looks obvious that this infinite construction leaves out no area of 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 Dn = 0. n→∞ Proof of the Lemma. Now consider the two triangles 4AP1P and 4PP2B 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 1 D1 < 2 D0. Math 120 Archimedes’s quadrature of the parabola November 14, 2016 15 / 22 , and this implies that limn→∞ Dn = 0. Proof. Exhaustion of R Although it looks obvious that this infinite construction leaves out no area of 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 Dn = 0. n→∞ Proof of the Lemma. Now consider the two triangles 4AP1P and 4PP2B 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 1 1 D1 < 2 D0. Continuing int this way, we obtain that Dn < 2 Dn−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 of 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 Dn = 0. n→∞ Proof of the Lemma. Now consider the two triangles 4AP1P and 4PP2B 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 1 1 D1 < 2 D0. Continuing int this way, we obtain that Dn < 2 Dn−1 for every n ≥ 1, and this implies that limn→∞ Dn = 0. Math 120 Archimedes’s quadrature of the parabola November 14, 2016 15 / 22 P M k PM. By FACT 2, M is the midpoint of MB. Let 2 2 Now let’s compute area2 (Pn). Consider NP2 k MB and R = P2M2 ∩ PB. Note that PM = 2RM2 (similar triangles). Let us focus on the triangles 4PBM2 and 4PBP2, sharing the base PB. Proof. Computing the area We proved that the polygons Pn exhaust the parabolic region R as n → ∞. Math 120 Archimedes’s quadrature of the parabola November 14, 2016 16 / 22 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 4PBM2 and 4PBP2, sharing the base PB. Proof. Computing the area We proved that the polygons Pn exhaust the parabolic region R as n → ∞. Now let’s compute area(Pn). M2 B M A N P2 P Consider NP2 k MB and P2M2 k PM. Math 120 Archimedes’s quadrature of the parabola November 14, 2016 16 / 22 Let R = P2M2 ∩ PB. Note that PM = 2RM2 (similar triangles). Let us focus on the triangles 4PBM2 and 4PBP2, sharing the base PB. Proof. Computing the area We proved that the polygons Pn exhaust the parabolic region R as n → ∞. Now let’s compute area(Pn). M2 B M A N P2 P Consider NP2 k MB and P2M2 k PM. By FACT 2, M2 is the midpoint of MB. Math 120 Archimedes’s quadrature of the parabola November 14, 2016 16 / 22 Note that PM = 2RM2 (similar triangles). Let us focus on the triangles 4PBM2 and 4PBP2, sharing the base PB. Proof. Computing the area We proved that the polygons Pn exhaust the parabolic region R as n → ∞. Now let’s compute area(Pn). M2 B M A R N P2 P Consider NP2 k MB and P2M2 k 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 Let us focus on the triangles 4PBM2 and 4PBP2, sharing the base PB. Proof. Computing the area We proved that the polygons Pn exhaust the parabolic region R as n → ∞. Now let’s compute area(Pn). M2 B M A R N P2 P Consider NP2 k MB and P2M2 k 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). M2 B M A R N P2 P Consider NP2 k MB and P2M2 k 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 4PBM2 and 4PBP2, sharing the base PB. Math 120 Archimedes’s quadrature of the parabola November 14, 2016 16 / 22 Proof. Computing the area M2 B M R N P2 P Math 120 Archimedes’s quadrature of the parabola November 14, 2016 17 / 22 Proof. Computing the area M2 B M K R H N P2 P Math 120 Archimedes’s quadrature of the parabola November 14, 2016 17 / 22 Therefore PM = 4PN and NM = 3PN. 4 4 We obtain PM = 3 NM = 3 P2M2. Since we know already that PM = 2RM2, 2 we have RM2 = 3 P2M2 and RM2 = 2P2R. This implies that the heights M2K and P2H are in 2:1 ratio. Hence area(PBM2) = 2 area(PBP2). Proof. Computing the area M2 B By FACTS 3 and 4, we have M 2 2 2 2 PN/PM = NP2 /MB = NP2 /(2MM2) = K 2 2 = NP2 /(2NP2) = 1/4. R H N P2 P Math 120 Archimedes’s quadrature of the parabola November 14, 2016 17 / 22 4 4 We obtain PM = 3 NM = 3 P2M2. Since we know already that PM = 2RM2, 2 we have RM2 = 3 P2M2 and RM2 = 2P2R. This implies that the heights M2K and P2H are in 2:1 ratio. Hence area(PBM2) = 2 area(PBP2). Proof. Computing the area M2 B By FACTS 3 and 4, we have M 2 2 2 2 PN/PM = NP2 /MB = NP2 /(2MM2) = K 2 2 = NP2 /(2NP2) = 1/4. R H Therefore PM = 4PN and NM = 3PN. N P2 P Math 120 Archimedes’s quadrature of the parabola November 14, 2016 17 / 22 Since we know already that PM = 2RM2, 2 we have RM2 = 3 P2M2 and RM2 = 2P2R. This implies that the heights M2K and P2H are in 2:1 ratio. Hence area(PBM2) = 2 area(PBP2). Proof. Computing the area M2 B By FACTS 3 and 4, we have M 2 2 2 2 PN/PM = NP2 /MB = NP2 /(2MM2) = K 2 2 = NP2 /(2NP2) = 1/4. R H Therefore PM = 4PN and NM = 3PN. 4 4 N P2 We obtain PM = 3 NM = 3 P2M2. P Math 120 Archimedes’s quadrature of the parabola November 14, 2016 17 / 22 Proof. Computing the area M2 B By FACTS 3 and 4, we have M 2 2 2 2 PN/PM = NP2 /MB = NP2 /(2MM2) = K 2 2 = NP2 /(2NP2) = 1/4. R H Therefore PM = 4PN and NM = 3PN. 4 4 N P2 We obtain PM = 3 NM = 3 P2M2. Since we know already that PM = 2RM2, P 2 we have RM2 = 3 P2M2 and RM2 = 2P2R. This implies that the heights M2K and P2H are in 2:1 ratio. Hence area(PBM2) = 2 area(PBP2). Math 120 Archimedes’s quadrature of the parabola November 14, 2016 17 / 22 Moreover, area( PBM)=2 area(PBM2). 1 Therefore area(PBP2) = 4 area( PBM). This means that the “new” triangle 4BPP2 has 1 area equal to 4 of that of the “old” triangle 4PBM . Proof. Computing the area M2 B area(PBM2) = 2 area(PBP2). M K H P2 N P Math 120 Archimedes’s quadrature of the parabola November 14, 2016 18 / 22 1 Therefore area(PBP2) = 4 area( PBM). This means that the “new” triangle 4BPP2 has 1 area equal to 4 of that of the “old” triangle 4PBM . Proof. Computing the area M2 B area(PBM2) = 2 area(PBP2). M Moreover, area( PBM)=2 area(PBM2). K H P2 N P Math 120 Archimedes’s quadrature of the parabola November 14, 2016 18 / 22 Proof. Computing the area M2 B area(PBM2) = 2 area(PBP2). M Moreover, area( PBM)=2 area(PBM2). K 1 Therefore area(PBP2) = 4 area( PBM). H This means that the “new” triangle 4BPP2 has P2 1 N area equal to 4 of that of the “old” triangle 4PBM . P Math 120 Archimedes’s quadrature of the parabola November 14, 2016 18 / 22 Therefore the triangles added to P0 = 4ABP to form P1 = AP1PP2B 1 have, combined, area equal to 4 area(P0). Proof. Computing the area B M A P2 P1 P 1 1 area(PBP2) = 4 area( PBM ) and, similarly, area(APP1) = 4 area( APM). Math 120 Archimedes’s quadrature of the parabola November 14, 2016 19 / 22 Proof. Computing the area B M A P2 P1 P 1 1 area(PBP2) = 4 area( PBM ) and, similarly, area(APP1) = 4 area( APM). Therefore the triangles added to P0 = 4ABP to form P1 = AP1PP2B 1 have, combined, area equal to 4 area(P0). Math 120 Archimedes’s quadrature of the parabola November 14, 2016 19 / 22 1 1 area(P ) = 1 + + area · (P ), 2 4 42 0 and in general 1 1 1 1 area(P ) = 1 + + + + ... + · area(P ). n 4 42 43 4n 0 Therefore 1 1 1 1 A = 1 + + + + ... + + ... · area(P ) 4 42 43 4n 0 Proof. Computing the area Repeating the argument above at each stage, we obtain: 1 area(P ) = 1 + · area(P ), 1 4 0 Math 120 Archimedes’s quadrature of the parabola November 14, 2016 20 / 22 and in general 1 1 1 1 area(P ) = 1 + + + + ... + · area(P ). n 4 42 43 4n 0 Therefore 1 1 1 1 A = 1 + + + + ... + + ... · area(P ) 4 42 43 4n 0 Proof. Computing the area Repeating the argument above at each stage, we obtain: 1 area(P ) = 1 + · area(P ), 1 4 0 1 1 area(P ) = 1 + + area · (P ), 2 4 42 0 Math 120 Archimedes’s quadrature of the parabola November 14, 2016 20 / 22 Therefore 1 1 1 1 A = 1 + + + + ... + + ... · area(P ) 4 42 43 4n 0 Proof. Computing the area Repeating the argument above at each stage, we obtain: 1 area(P ) = 1 + · area(P ), 1 4 0 1 1 area(P ) = 1 + + area · (P ), 2 4 42 0 and in general 1 1 1 1 area(P ) = 1 + + + + ... + · area(P ). n 4 42 43 4n 0 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: 1 area(P ) = 1 + · area(P ), 1 4 0 1 1 area(P ) = 1 + + area · (P ), 2 4 42 0 and in general 1 1 1 1 area(P ) = 1 + + + + ... + · area(P ). n 4 42 43 4n 0 Therefore 1 1 1 1 A = 1 + + + + ... + + ... · area(P ) 4 42 43 4n 0 Math 120 Archimedes’s quadrature of the parabola November 14, 2016 20 / 22 4 Add 1 to get . 3 1 1 1 4 Proof. 1 + 4 + 42 + 43 + ... = 3 We claim that 1 1 1 4 1 + + + + ... = . 4 42 43 3 Proof of the claim. 1 1 1 1 This picture shows that + + + ... = . 4 42 43 3 Math 120 Archimedes’s quadrature of the parabola November 14, 2016 21 / 22 1 1 1 4 Proof. 1 + 4 + 42 + 43 + ... = 3 We claim that 1 1 1 4 1 + + + + ... = . 4 42 43 3 Proof of the claim. 1 1 1 1 4 This picture shows that + + + ... = . Add 1 to get . 4 42 43 3 3 Math 120 Archimedes’s quadrature of the parabola November 14, 2016 21 / 22 Summary We have proved Archimedes’s Theorem: B A R B A P0 P 4 area(R) = area(P ), 3 0 where P0 = 4ABP 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