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
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
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
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
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
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 = 12b · h
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
a
b
c Area =√p(p − a)(p − b)(p − c)
p = 12(a + b + c)
(Heron’s formula, 1st century CE)
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
ab
cd
Area =√
(p − a)(p − b)(p − c)(p − d)
p = 12(a + b + c + d)
(Brahmagupta’s formula, 7th century CE)
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
ab
cd
Area =√
(p − a)(p − b)(p − c)(p − d)
p = 12(a + b + c + d)
(Brahmagupta’s formula, 7th century CE)
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
ab
cd
Area =√
(p − a)(p − b)(p − c)(p − d)
p = 12(a + b + c + d)
(Brahmagupta’s formula, 7th century CE)
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
r
Area = πr2
π = circumferencediameter
(due to Archimedes, c. 260 BCE)
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
rArea = πr2
π = circumferencediameter
(due to Archimedes, c. 260 BCE)
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
rArea = πr2
π = circumferencediameter
(due to Archimedes, c. 260 BCE)
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 4 / 22
The problem of area
What about more general curved regions?
Today we will prove a beautiful formula, due to Archimedes, for the areabetween a parabola and a segment whose endpoints are on the parabola.He proved this in a letter (later titled “Quadrature of the parabola”) to hisfriend Dositheus of Pelusium (who succeeded Conon of Samos –also friendof Archimedes– as director of the mathematical school of Alexandria).
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 areabetween a parabola and a segment whose endpoints are on the parabola.He proved this in a letter (later titled “Quadrature of the parabola”) to hisfriend Dositheus of Pelusium (who succeeded Conon of Samos –also friendof 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
Archimedes’s Theorem
Consider the region R between the parabolic arc_AB and the segment AB.
A
B
R
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 43
times the area of the triangle P0 = 4ABP.
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.
A
B
R
P
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 43
times the area of the triangle P0 = 4ABP.
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.
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 43
times the area of the triangle P0 = 4ABP.
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 9 / 22
Preliminary facts
Here are some facts that we will assume as proven (they were known toArchimedes 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.
FACT 2: The line through P and parallel to . . .the . . . .axis of the parabolameets AB in its middle point M
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 toArchimedes 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.
FACT 2: The line through P and parallel to . . .the . . . .axis of the parabolameets AB in its middle point M
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 toArchimedes 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 parabolameets AB in its middle point M
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 10 / 22
Preliminary facts
A
B
P
M
FACT 3: Every chord CD parallel to AB is bisected by PM, say at N.
FACT 4: PN/PM = ND2/MB
2.
We will assume FACTS 1÷4, without proof.
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 11 / 22
Preliminary facts
A
B
P
M
C
D
FACT 3: Every chord CD parallel to AB is bisected by PM
, say at N.
FACT 4: PN/PM = ND2/MB
2.
We will assume FACTS 1÷4, without proof.
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 11 / 22
Preliminary facts
A
B
P
M
C
DN
FACT 3: Every chord CD parallel to AB is bisected by PM, say at N.
FACT 4: PN/PM = ND2/MB
2.
We will assume FACTS 1÷4, without proof.
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 11 / 22
Preliminary facts
A
B
P
M
C
DN
FACT 3: Every chord CD parallel to AB is bisected by PM, say at N.
FACT 4: PN/PM = ND2/MB
2.
We will assume FACTS 1÷4, without proof.
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 11 / 22
Preliminary facts
A
B
P
M
C
DN
FACT 3: Every chord CD parallel to AB is bisected by PM, say at N.
FACT 4: PN/PM = ND2/MB
2.
We will assume FACTS 1÷4, without proof.
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 11 / 22
Proof. Two new triangles
A
B
P
M
Recall: 4ABP was constructed from the chord AB.Now construct two new triangles in the same way:
4APP1 from AP and4PBP2 from PB.
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 12 / 22
Proof. Two new triangles
A
B
P
M
P1
Recall: 4ABP was constructed from the chord AB.Now construct two new triangles in the same way:
4APP1 from AP and4PBP2 from PB.
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 12 / 22
Proof. Two new triangles
A
B
P
M
P1P1
Recall: 4ABP was constructed from the chord AB.Now construct two new triangles in the same way: 4APP1 from AP
and4PBP2 from PB.
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 12 / 22
Proof. Two new triangles
A
B
P
M
P1P1
P2
Recall: 4ABP was constructed from the chord AB.Now construct two new triangles in the same way: 4APP1 from AP
and4PBP2 from PB.
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 12 / 22
Proof. Two new triangles
A
B
P
M
P1P1
P2P2
Recall: 4ABP was constructed from the chord AB.Now construct two new triangles in the same way: 4APP1 from AP and4PBP2 from PB.
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 12 / 22
Proof. A bigger polygon
A
B
P
P1
P2
The area of the polygon AP1PP2B is bigger than the area of the triangle4ABP, but smaller than the area of the parabolic region (because theparabola 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. A bigger polygon
A
B
P
P1
P2
The area of the polygon AP1PP2B is bigger than the area of the triangle4ABP, but smaller than the area of the parabolic region
(because theparabola 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. A bigger polygon
A
B
P
P1
P2
The area of the polygon AP1PP2B is bigger than the area of the triangle4ABP, but smaller than the area of the parabolic region (because theparabola 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. A bigger polygon
A
B
P
P1
P2
The area of the polygon AP1PP2B is bigger than the area of the triangle4ABP, but smaller than the area of the parabolic region (because theparabola 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
A
B
P
area(APB)
<area(AP1PP2B)<area(AP3P1P4PP5P2P6B)<. . .<area(R)
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 14 / 22
Proof. Bigger and bigger polygons
A
B
P
P1
P2
area(APB)<area(AP1PP2B)
<area(AP3P1P4PP5P2P6B)<. . .<area(R)
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 14 / 22
Proof. Bigger and bigger polygons
A
B
P
P1
P2
P3
P4
P5
P6
area(APB)<area(AP1PP2B)
<area(AP3P1P4PP5P2P6B)<. . .<area(R)
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 14 / 22
Proof. Bigger and bigger polygons
A
B
P
P1
P2
P3
P4
P5
P6
area(APB)<area(AP1PP2B)<area(AP3P1P4PP5P2P6B)
<. . .<area(R)
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 14 / 22
Proof. Bigger and bigger polygons
A
B
P
P1
P2
P3
P4
P5
P6
area(APB)<area(AP1PP2B)<area(AP3P1P4PP5P2P6B)<. . .
<area(R)
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 14 / 22
Proof. Bigger and bigger polygons
A
B
P
R
area(APB)<area(AP1PP2B)<area(AP3P1P4PP5P2P6B)<. . .<area(R)
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 14 / 22
Proof. Exhaustion of R
Although it looks obvious that this infinite construction leaves out no areaof the region R, we need to prove it...
Lemma. Let A = area(R). Let Pn be the polygon constructed at the nthstep of the procedure described above, and let Dn = A− area(Pn). Thenlimn→∞
Dn = 0.
Proof of the Lemma.
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 areaof the region R, we need to prove it...
Lemma. Let A = area(R). Let Pn be the polygon constructed at the nthstep of the procedure described above, and let Dn = A− area(Pn). Thenlimn→∞
Dn = 0.
Proof of the Lemma.
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 15 / 22
Proof. Exhaustion of RAlthough it looks obvious that this infinite construction leaves out no areaof the region R, we need to prove it...
Lemma. Let A = area(R). Let Pn be the polygon constructed at the nthstep of the procedure described above, and let Dn = A− area(Pn). Thenlimn→∞
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 RAlthough it looks obvious that this infinite construction leaves out no areaof the region R, we need to prove it...
Lemma. Let A = area(R). Let Pn be the polygon constructed at the nthstep of the procedure described above, and let Dn = A− area(Pn). Thenlimn→∞
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 RAlthough it looks obvious that this infinite construction leaves out no areaof the region R, we need to prove it...
Lemma. Let A = area(R). Let Pn be the polygon constructed at the nthstep of the procedure described above, and let Dn = A− area(Pn). Thenlimn→∞
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 RAlthough it looks obvious that this infinite construction leaves out no areaof the region R, we need to prove it...
Lemma. Let A = area(R). Let Pn be the polygon constructed at the nthstep of the procedure described above, and let Dn = A− area(Pn). Thenlimn→∞
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 RAlthough it looks obvious that this infinite construction leaves out no areaof the region R, we need to prove it...
Lemma. Let A = area(R). Let Pn be the polygon constructed at the nthstep of the procedure described above, and let Dn = A− area(Pn). Thenlimn→∞
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 RAlthough it looks obvious that this infinite construction leaves out no areaof the region R, we need to prove it...
Lemma. Let A = area(R). Let Pn be the polygon constructed at the nthstep of the procedure described above, and let Dn = A− area(Pn). Thenlimn→∞
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) > 12A and D0 <
12A
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 areaof the region R, we need to prove it...
Lemma. Let A = area(R). Let Pn be the polygon constructed at the nthstep of the procedure described above, and let Dn = A− area(Pn). Thenlimn→∞
Dn = 0.
Proof of the Lemma.Now consider the two triangles 4AP1P and 4PP2B added to P0 to formthe polygon P1.
Apply the above argument to each of the two parabolicregions below AP and PB. We get that the area of each of these trianglesis at least half of the area of the corresponding parabolic region. ThereforeD1 <
12D0. Continuing int this way, we obtain that Dn <
12Dn−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
Proof. Exhaustion of R
Although it looks obvious that this infinite construction leaves out no areaof the region R, we need to prove it...
Lemma. Let A = area(R). Let Pn be the polygon constructed at the nthstep of the procedure described above, and let Dn = A− area(Pn). Thenlimn→∞
Dn = 0.
Proof of the Lemma.Now consider the two triangles 4AP1P and 4PP2B added to P0 to formthe polygon P1. Apply the above argument to each of the two parabolicregions below AP and PB.
We get that the area of each of these trianglesis at least half of the area of the corresponding parabolic region. ThereforeD1 <
12D0. Continuing int this way, we obtain that Dn <
12Dn−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
Proof. Exhaustion of R
Although it looks obvious that this infinite construction leaves out no areaof the region R, we need to prove it...
Lemma. Let A = area(R). Let Pn be the polygon constructed at the nthstep of the procedure described above, and let Dn = A− area(Pn). Thenlimn→∞
Dn = 0.
Proof of the Lemma.Now consider the two triangles 4AP1P and 4PP2B added to P0 to formthe polygon P1. Apply the above argument to each of the two parabolicregions below AP and PB. We get that the area of each of these trianglesis at least half of the area of the corresponding parabolic region.
ThereforeD1 <
12D0. Continuing int this way, we obtain that Dn <
12Dn−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
Proof. Exhaustion of R
Although it looks obvious that this infinite construction leaves out no areaof the region R, we need to prove it...
Lemma. Let A = area(R). Let Pn be the polygon constructed at the nthstep of the procedure described above, and let Dn = A− area(Pn). Thenlimn→∞
Dn = 0.
Proof of the Lemma.Now consider the two triangles 4AP1P and 4PP2B added to P0 to formthe polygon P1. Apply the above argument to each of the two parabolicregions below AP and PB. We get that the area of each of these trianglesis at least half of the area of the corresponding parabolic region. ThereforeD1 <
12D0.
Continuing int this way, we obtain that Dn <12Dn−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
Proof. Exhaustion of R
Although it looks obvious that this infinite construction leaves out no areaof the region R, we need to prove it...
Lemma. Let A = area(R). Let Pn be the polygon constructed at the nthstep of the procedure described above, and let Dn = A− area(Pn). Thenlimn→∞
Dn = 0.
Proof of the Lemma.Now consider the two triangles 4AP1P and 4PP2B added to P0 to formthe polygon P1. Apply the above argument to each of the two parabolicregions below AP and PB. We get that the area of each of these trianglesis at least half of the area of the corresponding parabolic region. ThereforeD1 <
12D0. Continuing int this way, we obtain that Dn <
12Dn−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
Proof. Exhaustion of R
Although it looks obvious that this infinite construction leaves out no areaof the region R, we need to prove it...
Lemma. Let A = area(R). Let Pn be the polygon constructed at the nthstep of the procedure described above, and let Dn = A− area(Pn). Thenlimn→∞
Dn = 0.
Proof of the Lemma.Now consider the two triangles 4AP1P and 4PP2B added to P0 to formthe polygon P1. Apply the above argument to each of the two parabolicregions below AP and PB. We get that the area of each of these trianglesis at least half of the area of the corresponding parabolic region. ThereforeD1 <
12D0. Continuing int this way, we obtain that Dn <
12Dn−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
Proof. Computing the area
We proved that the polygons Pn exhaust the parabolic region R asn→∞.
Now let’s compute area(Pn). Consider NP2 ‖ MB andP2M2 ‖ PM. By FACT 2, M2 is the midpoint of MB. LetR = 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
We proved that the polygons Pn exhaust the parabolic region R asn→∞. Now let’s compute area(Pn).
A
B
P
M
P2N
M2
Consider NP2 ‖ MB and P2M2 ‖ PM.
By FACT 2, M2 is the midpoint ofMB. 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
We proved that the polygons Pn exhaust the parabolic region R asn→∞. Now let’s compute area(Pn).
A
B
P
M
P2N
M2
Consider NP2 ‖ MB and P2M2 ‖ PM. By FACT 2, M2 is the midpoint ofMB.
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
We proved that the polygons Pn exhaust the parabolic region R asn→∞. Now let’s compute area(Pn).
A
B
P
M
P2N
M2
R
Consider NP2 ‖ MB and P2M2 ‖ PM. By FACT 2, M2 is the midpoint ofMB. 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
We proved that the polygons Pn exhaust the parabolic region R asn→∞. Now let’s compute area(Pn).
A
B
P
M
P2N
M2
R
Consider NP2 ‖ MB and P2M2 ‖ PM. By FACT 2, M2 is the midpoint ofMB. 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
We proved that the polygons Pn exhaust the parabolic region R asn→∞. Now let’s compute area(Pn).
A
B
P
M
P2N
M2
RR
Consider NP2 ‖ MB and P2M2 ‖ PM. By FACT 2, M2 is the midpoint ofMB. 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
B
P
M
P2N
M2
R
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 17 / 22
Proof. Computing the area
B
P
M
P2N
M2
R
K
H
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 17 / 22
Proof. Computing the area
B
P
M
P2N
M2
R
K
H
By FACTS 3 and 4, we have
PN/PM = NP22/MB
2= NP2
2/(2MM2)2 =
= NP22/(2NP2)2 = 1/4.
Therefore PM = 4PN and NM = 3PN.
We obtain PM = 43NM = 4
3P2M2.
Since we know already that PM = 2RM2,
we have RM2 = 23P2M2 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
Proof. Computing the area
B
P
M
P2N
M2
R
K
H
By FACTS 3 and 4, we have
PN/PM = NP22/MB
2= NP2
2/(2MM2)2 =
= NP22/(2NP2)2 = 1/4.
Therefore PM = 4PN and NM = 3PN.
We obtain PM = 43NM = 4
3P2M2.
Since we know already that PM = 2RM2,
we have RM2 = 23P2M2 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
Proof. Computing the area
B
P
M
P2N
M2
R
K
H
By FACTS 3 and 4, we have
PN/PM = NP22/MB
2= NP2
2/(2MM2)2 =
= NP22/(2NP2)2 = 1/4.
Therefore PM = 4PN and NM = 3PN.
We obtain PM = 43NM = 4
3P2M2.
Since we know already that PM = 2RM2,
we have RM2 = 23P2M2 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
Proof. Computing the area
B
P
M
P2N
M2
R
K
H
By FACTS 3 and 4, we have
PN/PM = NP22/MB
2= NP2
2/(2MM2)2 =
= NP22/(2NP2)2 = 1/4.
Therefore PM = 4PN and NM = 3PN.
We obtain PM = 43NM = 4
3P2M2.
Since we know already that PM = 2RM2,
we have RM2 = 23P2M2 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
Proof. Computing the area
B
P
M
P2N
M2
K
H
area(PBM2) = 2 area(PBP2).
Moreover, area( PBM ) = 2 area(PBM2).
Therefore area(PBP2) = 14area( PBM ).
This means that the “new” triangle 4BPP2 has
area equal to 14 of that of the “old” triangle 4PBM .
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 18 / 22
Proof. Computing the area
B
P
M
P2N
M2
K
H
area(PBM2) = 2 area(PBP2).
Moreover, area( PBM ) = 2 area(PBM2).
Therefore area(PBP2) = 14area( PBM ).
This means that the “new” triangle 4BPP2 has
area equal to 14 of that of the “old” triangle 4PBM .
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 18 / 22
Proof. Computing the area
B
P
M
P2N
M2
K
H
area(PBM2) = 2 area(PBP2).
Moreover, area( PBM ) = 2 area(PBM2).
Therefore area(PBP2) = 14area( PBM ).
This means that the “new” triangle 4BPP2 has
area equal to 14 of that of the “old” triangle 4PBM .
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 18 / 22
Proof. Computing the area
A
B
P
P1
P2
M
area(PBP2) = 14area( PBM ) and, similarly, area(APP1) = 1
4area( APM ).
Therefore the triangles added to P0 = 4ABP to form P1 = AP1PP2Bhave, combined, area equal to 1
4area(P0).
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) = 14area( PBM ) and, similarly, area(APP1) = 1
4area( APM ).Therefore the triangles added to P0 = 4ABP to form P1 = AP1PP2Bhave, 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),
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. 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. 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. 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 + 14 +
142 +
143 + . . . = 4
3
We claim that
1 +1
4+
1
42+
1
43+ . . . =
4
3.
Proof of the claim.
This picture shows that1
4+
1
42+
1
43+ . . . =
1
3.
Add 1 to get4
3.
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 21 / 22
Proof. 1 + 14 +
142 +
143 + . . . = 4
3
We claim that
1 +1
4+
1
42+
1
43+ . . . =
4
3.
Proof of the claim.
This picture shows that1
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:
RA
B
A
B
P
P0
area(R) =4
3area(P0),
where P0 = 4ABP and P is the point of intersection between theparabola and the line passing through the midpoint of AB parallel to theaxis of the parabola.
Math 120 Archimedes’s quadrature of the parabola November 14, 2016 22 / 22