Proposition 34 – A Few Comments
In proposition 34, Archimedes needs to find sides of an inscribed and
circumscribed polygon, say β and δ, with β > δ, such that β3/δ3 < the ratio of
two volumes, V1 and V2. We can certainly find β and γ such that β/γ < V1/V2
as that was proven in proposition 3. So Archimedes begins with β and γ
chosen using proposition 3 and then proceeds to construct δ. Although
Archimedes omits the proof that β3/δ3 < β/γ < V1/V2, we have Eutocius’ proof
that with δ as defined by Archimedes, β3/δ3 < β/γ . In this note we try to view
Eutocius’ proof from the perspective of arithmetic and geometric
sequences. As far as possible we will stick to the notation of Heath’s
version of Eutocius’ proof.
Let ε=( β + γ )/2. And let δ =(ε + β)/2. Now let η = β – δ . From this start it is
apparent that δ = β – η and ε = β – 2η and γ = β – 4η. Thus β, δ, ε, and γ are all
terms in a descending arithmetic sequence.
Eutocius (or Heath) also define x and y as follows. First x is chosen so
that δ is the mean proportional between x and β and then y is chosen so
that x is the mean proportional between y and δ. If we let r= δ/β, we can
show that β, δ, x, and y form a descending geometric sequence.
These tables summarize:
β
δ
β
β–η
β
β
δ
βr
ε
β – 2η
γ
β –3 η
η= β – δ
x
βr2
y
βr3
r= δ/β
Before we try to show that β3/δ3 < β/γ, there is one further fact to recall.
Let a and b be any two positive numbers. Then √ab ≤ (a+b)/2.
We now proceed to prove that β3/δ3 < β/γ.
Proof:
√βx = δ = (β + ε)/2. But √βx < (β + x)/2. Thus ε < x.
And δ–x > x – y (since r<1). Thus δ–ε > x – y. And then ε– γ > x – y.
Now since ε < x, we have that γ < y and then γ/β < y/β or β/γ > β/y.
But now since y = βr3 we have that β/γ > 1/ r3 or β/γ >β3/δ3
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