Cones or Cylinders are Equal iff Bases are Reciprocally Proportional to Heights
Theorem
In the words of Euclid:
- In equal cones and cylinders the bases are reciprocally proportional to the heights; and those cones and cylinders in which the bases are reciprocally proportional to the heights are equal.
(The Elements: Book $\text{XII}$: Proposition $15$)
Proof
Let there be cones and cylinders which are similar.
Let the circles $\map c {ABCD}$ and $\map c {EFGH}$ be their bases.
Let $KL$ and $MN$ be the axes of the cones and cylinders.
Let $L$ and $N$ be the apices of the cones.
Thus:
- let $KL$ be the height of the cone $\map c {ABCDL}$ and the cylinder $AO$
- let $MN$ be the height of the cone $\map c {EFGHM}$ and the cylinder $EP$.
It is to be shown that:
- $\map c {ABCD} : \map c {EFGH} = MN : KL$
That is, the bases of the cones and cylinders are reciprocally proportional to their heights.
Either $LK = MN$ or $LK \ne MN$.
First suppose $LK = MN$.
We have that $AO = EP$.
- $\map c {ABCD} = \map c {EFGH}$
Thus:
- $\map c {ABCD} : \map c {EFGH} = MN : KL$
Without loss of generality, suppose $LK \ne MN$.
Suppose $MN > LK$.
Let $QN$ be cut off from $MN$ equal to $KL$.
Let the cylinder $EP$ be cut by the plane $TUS$ through $Q$ parallel to the planes holding the circles $\map c {EFGH}$ and $\map c {RP}$.
Let the cylinder $ES$ be described with the circle $\map c {EFGH}$ as its base and with height $NQ$.
We have that the cylinder $AO$ equals the cylinder $EP$.
Therefore from Proposition $7$ of Book $\text{V} $: Ratios of Equal Magnitudes:
- $AO : ES = EP : ES$
- $AO : ES = \map c {ABCD} : \map c {EFGH}$
- $EP : ES = MN : QN$
Therefore from Proposition $11$ of Book $\text{V} $: Equality of Ratios is Transitive:
- $\map c {ABCD} : \map c {EFGH} = MN : QN$
But:
- $QN = KL$
Therefore:
- $\map c {ABCD} : \map c {EFGH} = MN : KL$
Therefore in the cylinders $AO$ and $EP$, the bases are reciprocally proportional to their heights.
$\Box$
Let the bases of the cylinders $AO$ and $EP$ be reciprocally proportional to their heights:
- $\map c {ABCD} : \map c {EFGH} = MN : KL$
It is to be proved that cylinders $AO$ and $EP$ are equal.
Let $QN$ be cut off from $MN$ equal to $KL$.
Let the cylinder $EP$ be cut by the plane $TUS$ through $Q$ parallel to the planes holding the circles $\map c {EFGH}$ and $\map c {RP}$.
Let the cylinder $ES$ be described with the circle $\map c {EFGH}$ as its base and with height $NQ$.
We have that $KL = QN$.
Thus:
- $\map c {ABCD} : \map c {EFGH} = MN : QN$
- $\map c {ABCD} : \map c {EFGH} = AO : ES$
- $MN : QN = EP : ES$
Therefore from Proposition $11$ of Book $\text{V} $: Equality of Ratios is Transitive:
- $AO : ES = EP : ES$
Therefore from Proposition $9$ of Book $\text{V} $: Magnitudes with Same Ratios are Equal:
- $AO = EP$
$\blacksquare$
Historical Note
This proof is Proposition $15$ of Book $\text{XII}$ of Euclid's The Elements.
Sources
- 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 3 (2nd ed.) ... (previous) ... (next): Book $\text{XII}$. Propositions