# Volumes of Similar Cones and Cylinders are in Triplicate Ratio of Diameters of Bases

## Theorem

In the words of Euclid:

Similar cones and cylinders are to one another in the triplicate ratio of the diameters in their bases.

## Proof

Let there be cones and cylinders which are similar.

Let the circles $c \left({ABCD}\right)$ and $c \left({EFGH}\right)$ be their bases.

Let $BD$ and $FH$ be the diameters of the bases $c \left({ABCD}\right)$ and $c \left({EFGH}\right)$.

Let $KL$ and $MN$ be the axes of the cones and cylinders.

Let $L$ and $N$ be the apices of the cones.

It is to be demonstrated that:

$c \left({ABCDL}\right) : c \left({EFGHN}\right) = BD^3 : FH^3$

where $c \left({ABCDL}\right)$ and $c \left({EFGHN}\right)$ are the cones on $c \left({ABCD}\right)$ and $c \left({EFGH}\right)$ respectively.

Suppose it is not the case that:

$c \left({ABCDL}\right) : c \left({EFGHN}\right) = BD^3 : FH^3$

Then:

$c \left({ABCDL}\right) : O = BD^3 : FH^3$

where $O$ is either greater than $c \left({EFGHN}\right)$ or less than $c \left({EFGHN}\right)$.

First suppose that $O < c \left({EFGHN}\right)$.

Let the square $\Box EFGH$ be inscribed in $c \left({EFGH}\right)$.

$\Box EFGH$ is greater than half $c \left({EFGH}\right)$.

Let a pyramid be set up on the base $\Box EFGH$ of equal height as the cone.

From:

Proposition $6$ of Book $\text{XII}$: Sizes of Pyramids of Same Height with Polygonal Bases are as Bases

and:

Square Inscribed in Circle is greater than Half Area of Circle

it follows that:

the pyramid so set up is greater than half the cone.

Let the arcs $EF, FG, GH, HE$ be bisected at $P, Q, R, S$.

Let $EP, PF, FQ, QG, GR, RH, HS, SE$ be joined.

Using the reasoning of Proposition $2$ of Book $\text{XII}$: Areas of Circles are as Squares on Diameters:

each of $\triangle EPF, \triangle FQG, \triangle GRH, \triangle HSE$ is greater than half the segment of the circle $EFGH$ around it.

On each of $\triangle EPF, \triangle FQG, \triangle GRH, \triangle HSE$, let pyramids be set up with the same apex as the cone.

Again from the reasoning in Proposition $2$ of Book $\text{XII}$: Areas of Circles are as Squares on Diameters:

each of these pyramids is greater than half the segment of the cone around it.

The operation:

bisecting the arcs remaining, joining the points of bisection with straight lines and setting up pyramids on the resulting triangles with the same apex as the cone

can be repeated indefinitely.

eventually we will leave some segments of the cone which will be less than the amount by which $c \left({EFGHN}\right)$ is greater than $O$.

Let such segments be left, and let them be $EP, PF, FQ, QG, GR, RH, HS, SE$.

Therefore the remainder, the pyramid $EPFQGRHSN$ whose base is the polygon $EPFQGRHS$ and whose apex is $N$, is greater than $O$.

Let the polygon $ATBUCVDW$ similar to $EPFQGRHS$ be inscribed in $c \left({ABCD}\right)$.

Let a pyramid $ATBUCVDWL$ be set up on the base $ATBUCVDW$ with the same apex as the cone $c \left({ABCDL}\right)$.

Let $\triangle LBT$ be one of the triangular faces of $ATBUCVDWL$.

Let $\triangle NFP$ be one of the triangular faces of $EPFQGRHSN$.

Let $KT$ and $MP$ be joined.

We have that $c \left({ABCDL}\right)$ is similar to $c \left({EFGHN}\right)$.

$BD : FH = KL : MN$

But:

$BD : FH = BK : FM$

Therefore:

$BK : FM = KL : MN$
$BK : KL = FM : MN$
$\triangle BKL$ is similar to $\triangle FMN$.

Again, we have:

$BK : KT = FM : MP$
$\triangle BKT$ is similar to $\triangle FMP$.

We have that:

$BK : FM = KL : MN$

while:

$BK = KT$

and

$FM : PM$

We have that:

$\angle TKL = \angle PMN$

as both are right angles.

$\triangle LKT$ is similar to $\triangle NMP$.

We have that:

$\triangle LKB$ is similar to $\triangle NMF$

and so:

$LB : BK = NF : FM$

Also:

$\triangle BKT$ is similar to $\triangle FMP$.

and so:

$KB : BT = MF : FP$
$LB : BT = NF : FP$

We also have that:

$\triangle LTK$ is similar to $\triangle NPM$.

So:

$LT : TK = NP : PM$

Also:

$\triangle TKB$ is similar to $\triangle PMF$.

So:

$KT : TB = MP : PF$
$LT : TB = NP : PF$

But it has already been proved that:

$TB : BL = PF : FN$
$TL : LB = PN : NF$

Therefore in $\triangle LTB$ and $\triangle NPF$ the sides are proportional.

$\triangle LTB$ and $\triangle NPF$ are equiangular.

Therefore from Book $\text{VI}$ Definition $1$: Similar Rectilineal Figures:

$\triangle LTB$ and $\triangle NPF$ are similar.

Therefore from Book $\text{XI}$ Definition $9$: Similar Solid Figures:

the pyramid $BKTL$ is similar to the pyramid $FMPN$.
$BKTL : FMPN = BK^3 : FM^3$

In a similar manner, lines can be drawn:

from $A, W, D, V, C, U$ to $K$

and:

from $E, S, H, R, G, Q$ to $M$

and pyramids can be drawn with each of the triangles as bases which have the same apices as the cones.

Hence it can be proved that the ratio of each of the similarly arranged pyramids to each of the similarly arranged pyramids is a ratio triplicate of that which $BD$ has to $FH$.

$BKTL : FMPN = ATBUCVDWL : EPFQGRHSN$

Hence:

$ATBUCVDWL : EPFQGRHSN = BD^3 : FH^3$

But by hypothesis:

$c \left({ABCDL}\right) : O = BD^3 : FH^3$

Therefore:

$c \left({ABCDL}\right) : O = ATBUCVDWL : EPFQGRHSN$
$c \left({ABCDL}\right) : ATBUCVDWL = O : EPFQGRHSN$

But:

$c \left({ABCDL}\right) > ATBUCVDWL$

Therefore:

$O > EPFQGRHSN$

But it has already been shown that:

$O < EPFQGRHSN$

Therefore it cannot be the case that:

$c \left({ABCDL}\right) : O = BD^3 : FH^3$

where $O < c \left({EFGHN}\right)$.

In a similar way it can be proved that it cannot be the case that:

$c \left({EFGHN}\right) : O = BD^3 : FH^3$

where $O < c \left({ABCDL}\right)$.

Next it is to be shown that it is not the case that:

$c \left({ABCDL}\right) : O = BD^3 : FH^3$

where $O > c \left({EFGHN}\right)$.

Suppose it were possible that:

$c \left({ABCDL}\right) : O = BD^3 : FH^3$

where $O > c \left({EFGHN}\right)$.

Therefore:

$O : c \left({ABCDL}\right) = FH^3 : BD^3$

But we have that:

$O : c \left({ABCDL}\right) = c \left({EFGHN}\right) : X$

where $X$ is some solid figure such that $X < c \left({ABCDL}\right)$.

Therefore:

$c \left({EFGHN}\right) : X = FH^3 : BD^3$

where $X < c \left({ABCDL}\right)$.

But this has been demonstrated to be impossible.

Therefore it is not the case that:

$c \left({ABCDL}\right) : O = BD^3 : FH^3$

where $O > c \left({EFGHN}\right)$.

But it has also been shown that:

$c \left({ABCDL}\right) : O = BD^3 : FH^3$

where $O < c \left({EFGHN}\right)$.

Therefore:

$c \left({ABCDL}\right) : c \left({EFGHN}\right) = BD^3 : FH^3$

$\Box$

the cylinder on $c \left({ABCD}\right)$ is three times the cone on $c \left({ABCD}\right)$

and:

the cylinder on $c \left({EFGH}\right)$ is three times the cone on $c \left({EFGH}\right)$.

Therefore the ratio of the cylinder on $c \left({ABCD}\right)$ to the cylinder on $c \left({EFGH}\right)$ equals the ratio triplicate of $BD$ to $FH$.

$\blacksquare$

## Historical Note

This proof is Proposition $12$ of Book $\text{XII}$ of Euclid's The Elements.