The problem of area: Archimedes's quadrature of the parabolamast.queensu.ca/~math120/Fall_Class_Diary_files/lecture-25 slides.pdf1PP 2B is bigger than the area of the triangle 4ABP,

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


The problem of area


b

h Area = b · h

h

b

Area = b · h

b


The problem of area


b

h Area = b · h h

b

Area = b · h

b


The problem of area


b

h Area = b · h h

b

Area = b · h

b


The problem of area


b

h Area = b · h h

b

Area = b · h

b


The problem of area


b

h Area = b · h h

b

Area = b · h

b

h

b

Area = 12b · h


The problem of area


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)


The problem of area


ab

cd

Area =√

(p − a)(p − b)(p − c)(p − d)

p = 12(a + b + c + d)

(Brahmagupta’s formula, 7th century CE)


The problem of area


ab

cd

Area =√

(p − a)(p − b)(p − c)(p − d)

p = 12(a + b + c + d)



The problem of area


ab

cd

Area =√

(p − a)(p − b)(p − c)(p − d)

p = 12(a + b + c + d)



The problem of area


r

Area = πr2

π = circumferencediameter

(due to Archimedes, c. 260 BCE)


The problem of area


rArea = πr2




The problem of area


rArea = πr2




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).


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).





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.




A

B

R

P


farthest from the segment AB.

Then the area of the region R equals 43





A

B

R

P

P0


farthest from the segment AB. Then the area of the region R equals 43



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


Preliminary facts


A

B

P




Preliminary facts


A

B

P

M




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.


Preliminary facts

A

B

P

M

C

D

FACT 3: Every chord CD parallel to AB is bisected by PM

, say at N.


2.



Preliminary facts

A

B

P

M

C

DN



2.



Preliminary facts

A

B

P

M

C

DN



2.



Preliminary facts

A

B

P

M

C

DN



2.



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.



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.



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.



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.



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.


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.



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.



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.



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.


Proof. Bigger and bigger polygons

A

B

P

area(APB)

<area(AP1PP2B)<area(AP3P1P4PP5P2P6B)<. . .<area(R)



A

B

P

P1

P2

area(APB)<area(AP1PP2B)

<area(AP3P1P4PP5P2P6B)<. . .<area(R)



A

B

P

P1

P2

P3

P4

P5

P6

area(APB)<area(AP1PP2B)

<area(AP3P1P4PP5P2P6B)<. . .<area(R)



A

B

P

P1

P2

P3

P4

P5

P6

area(APB)<area(AP1PP2B)<area(AP3P1P4PP5P2P6B)

<. . .<area(R)



A

B

P

P1

P2

P3

P4

P5

P6

area(APB)<area(AP1PP2B)<area(AP3P1P4PP5P2P6B)<. . .

<area(R)



A

B

P

R

area(APB)<area(AP1PP2B)<area(AP3P1P4PP5P2P6B)<. . .<area(R)


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.





Dn = 0.

Proof of the Lemma.


Proof. Exhaustion of RAlthough it looks obvious that this infinite construction leaves out no areaof the region R, we need to prove it...


Dn = 0.

Proof of the Lemma.

A

B

P

M

A’

B’




Dn = 0.

Proof of the Lemma.

A

B

P

M

A’

B’




Dn = 0.

Proof of the Lemma.

A

B

P

M

A’

B’

AA′ ‖ PM ‖ BB ′ and AB ‖ A′B ′ by FACT 1.




Dn = 0.

Proof of the Lemma.

A

B

P

M

A’

B’


Of course area(AA′B ′B) > A.




Dn = 0.

Proof of the Lemma.

A

B

P

M

A’

B’



Moreover area(AA′B ′B) = 2 · area(P0).




Dn = 0.

Proof of the Lemma.

A

B

P

M

A’

B’



Moreover area(AA′B ′B) = 2 · area(P0).

Therefore area(P0) > 12A and D0 <

12A





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.





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 <


12Dn−1 for every






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 <


12Dn−1 for every






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






Dn = 0.



12Dn−1 for every

n ≥ 1

, and this implies that limn→∞Dn = 0.





Dn = 0.



12Dn−1 for every



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.



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.




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.




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.




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.




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.



B

P

M

P2N

M2

R



B

P

M

P2N

M2

R

K

H



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).



B

P

M

P2N

M2

R

K

H


PN/PM = NP22/MB

2= NP2

2/(2MM2)2 =

= NP22/(2NP2)2 = 1/4.



3P2M2.






B

P

M

P2N

M2

R

K

H


PN/PM = NP22/MB

2= NP2

2/(2MM2)2 =

= NP22/(2NP2)2 = 1/4.



3P2M2.






B

P

M

P2N

M2

R

K

H


PN/PM = NP22/MB

2= NP2

2/(2MM2)2 =

= NP22/(2NP2)2 = 1/4.



3P2M2.






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 .



B

P

M

P2N

M2

K

H








B

P

M

P2N

M2

K

H








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).



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).



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)




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)




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)




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)


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.


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.


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.


Documents

The problem of area: Archimedes's quadrature of the parabolamast.queensu.ca/~math120/Fall_Class_Diary_files/lecture-25 slides.pdf1PP 2B is bigger than the area of the triangle 4ABP,