The (almost really) Complete Works of Lewis Carroll

Prop. I. Theorem

The area of a Circle is less than four times, and greater than twice, the Square on its radius.

Let $ABCD$ be a Circle whose centre is O: and let $AC$, $BD$ be two diameters at right angles to each other: and through A, B, C, D, let tangents be drawn to the Circle: and join $AB$, $BC$, $CD$, $DA$.

Then it is evident that $OE$ is the Square on $OA$, and that $EFGH$ is 4 times this Square and $ABCD$ twice it.

Hence the area of the Circle, being $<EFGH$ and $>ABCD$, is $<4$ times, and > twice, the Square on its radius. Q.E.D.

We will now prove this by another method, similar to what we shall use in the next Proposition.

Take a Line $OA$, and from A draw $AB$ at right angles to $OA$ and equal to it. With centre O, at distance $OA$, describe the Circle $ACQ$: join $OB$, cutting this Circle at C: and from C draw $CD$ at right angles to $OA$.

$$\because \{\begin{array}{c}tanAOB=\frac{AB}{OA}=\frac{OA}{OA}=1,\\ \text{and that it also}=\frac{CD}{OD},\end{array}$$

$$\therefore \frac{CD}{OD}=1\text{; i.\hspace{0.17em}e. }CD=OD\text{.}$$

also, $\because tanAOB=1$,

$$\therefore \text{angle }AOB={tan}^{-1}=\text{45°}\text{;}$$

∴ the Sector $AOC=\frac{1}{8}$ of the Circle; i. e. the area of the Circle = 8 times that of the Sector.

Again, ∵ the Triangle $OAB$ has its altitude $AB$ = its base $OA$,

∴ it = $\frac{1}{4}$ the Square on $OA$;

similarly, the Triangle $ODC$ = $\frac{1}{2}$ of the Square on $OD$.

Again, ∵ the Square on $OC$ = the sum of the Squares on $OD$, $DC$, [Euc. I. 47

∴ it = twice the Square on $OD$;

∴ the Square on $OD$ = $\frac{1}{2}$ of the Square on $OC$;

= $\frac{1}{2}$ of the Square on $OA$;

∴ the Triangle $ODC$ = $\frac{1}{2}\text{ of }\left(\frac{1}{2}\text{ of the Square on }OA\right)$;

i. e. it = $\frac{1}{4}$ of the Square on $OA$.

Now the area of the Sector $OAC$ is evidently < the Triangle $OAB$, and > the Triangle $ODC$;

i. e. it is $<\frac{1}{2}$, and $>\frac{1}{4}$, of the Square on $OA$;

∴ the area of the Circle, being 8 times that of the Sector, is $<8\times \frac{1}{2}$, and $>8\times \frac{1}{4}$, of the Square on $OA$;

i. e. it is $<4$ times, and > twice, the Square on its radius. Q.E.D.

Prop. 2. Theorem

The area of a Circle is less than $3\frac{1}{3}$, and greater than $2\frac{2}{3}$, of the Square on its radius.

Take a Line $OA$, and from A draw $AB$ at right angles to $OA$ and equal to $\frac{1}{2}$ of it. With centre O, at distance $OA$, describe the Circle $ACQ$: join $OB$, cutting this Circle at C: and from C draw $CD$ at right angles to $OA$. Also from C draw $CE$ at right angles to $OC$ and equal to $\frac{1}{3}$ of it.

Hence $CE$ falls wholly outside the Circle. [Euc. III. 16

Join $OE$, cutting the Circle at F. Also from $OC$ cut off $OG=OD$, and from G draw $GH$ at right angles to $OC$.

$$\text{Now, }\because \{\begin{array}{c}tanAOB=\frac{AB}{OA}=\frac{\frac{1}{2}\text{ of }OA}{OA}=\frac{1}{2},\\ \text{and that it also}=\frac{CD}{OD},\end{array}$$

$$\therefore \frac{CD}{OD}=\frac{1}{2}\text{; i.\hspace{0.17em}e. }CD=\frac{1}{2}\text{ of }OD\text{;}$$

similarly, $tanCOE=\frac{1}{3}$, and $HG=\frac{1}{3}\text{ of }OE$, $=\frac{1}{3}\text{ of }OD$;

hence also, $HG$ is not greater than $CD$.

Also, ∵ in the right-angled Triangles $OGH$, $ODC$, $OG=OD$, and $HG$ is not greater than $CD$,

∴ $OH$ is not greater than $OC$;

∴ H is not outside the Circle;

∴ the Triangle $OGH$ is within the Circle.

Now $tanAOE=tan\left(AOB+COE\right)$,

$$=\frac{tanAOB+tanCOE}{1-tanAOB.tanCOE}\text{,}$$

$$=\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}\cdot \frac{1}{3}}=\frac{\frac{5}{6}}{\frac{5}{6}}=1\text{;}$$

∴ angle $AOE={tan}^{-1}1=\text{45°}$;

∴ the Sector $AOF=\frac{1}{8}$ of the Circle; i. e. the area of the Circle = 8 times that of the Sector.

Again, ∵ the Triangle $OAB$ has its altitude $AB=\frac{1}{2}$ of its base $OA$,

$$\therefore \text{ it }=\frac{1}{2}\text{ of }\left(\frac{1}{2}\text{ of the Square on }OA\right)\text{;}$$

similarly, the Triangle $OCE=\frac{1}{2}\text{ of }\left(\frac{1}{3}\text{ of the Square on }OC\right)$

$$=\frac{1}{2}\text{ of }\left(\frac{1}{3}\text{ of the Square on }OA\right)\text{;}$$

∴ the sum of the Triangles $OAB$, $OCE$, $=\frac{1}{2}\text{ of }\left\{\left(\frac{1}{2}+\frac{1}{3}\right)\text{ of the Square on }OA\right\}$,

$$=\frac{5}{12}\text{ of the Square on }OA\text{.}$$

Similarly, the sum of the Triangles $ODC$, $OGH$, $=\frac{5}{12}\text{ of the Square on }OD$.

Again, ∵ the Square on $OC$ = the sum of the Squares on $OD$, $DC$, [Euc. I. 47

∴ it = the Square on $OD$ with the Square on $\frac{1}{2}$ of $OD$;

i. e. it $=\left(1+\frac{1}{4}\right)$ of the Square on $OD$;

$$=\frac{5}{4}\text{ of the Square on }OD\text{;}$$

∴ the Square on $OD$ = $\frac{4}{5}$ of the Square on $OC$;

= $\frac{4}{5}$ of the Square on $OA$;

∴ the Sum of the Triangles $ODC$, $OGH$, $=\frac{5}{12}\text{ of the Square on }OA\times \frac{4}{5}$,

$$=\frac{1}{3}\text{ of the Square on }OA\text{.}$$

Now the area of the Sector $AOF$ is evidently < the sum of the Triangles $OAB$, $OCE$, and > the sum of the Triangles $ODC$, $OGH$; i. e. it is $<\frac{5}{12}$, and $>\frac{1}{3}$, of the Square on $OA$.

∴ the area of the Circle, being 8 times that of the Sector, is $<8\times \frac{5}{12}$, and $>8\times \frac{1}{3}$, of the Square on $OA$.

That is, the area of a Circle is $<3\frac{1}{3}$ and $>2\frac{2}{3}$, of the Square on its radius. Q.E.D.

We have thus found certain Limits, between which the area of Circle must lie: the Superior Limit being $3\frac{1}{3}$ of the Square on the radius, and the Inferior Limit being $2\frac{2}{3}$ of the same Square. Also we are now able to find fresh pairs of Limits, closer and closer together, ad libitum: for, by mere inspection of the preceeding Theorem, we can obtain a rule for finding such pairs, whenever we can resolve the angle 45° into the sum of a series of angles whose tangents are of the form $\frac{1}{a}$, $\frac{1}{b}$, &c., where a is not greater than any other of the denominators.

The Limits are $$4\times \left(\frac{1}{a}+\frac{1}{b}+\text{\&c.}\right),\text{ and }4\times \left(\frac{1}{a}+\frac{1}{b}+\text{\&c.}\right)\times \left(1-\frac{1}{a^2+1}\right)$$ of the Square on the radius. Hence, if we know that $$4\times \left(\frac{1}{a}+\frac{1}{b}+\text{\&c.}\right)<X,\text{ and }>x\text{,}$$ we may take X times the Square on the radius as a Superior Limit and $\left(x-\frac{X}{a^2+1}\right)$ of the same Square as an Inferior Limit.

As an instance of this rule, let us resolve 45° into the sum of 3 angles.

We know already that $$\text{45°}={tan}^{-1}\frac{1}{2}+{tan}^{-1}\frac{1}{3}\text{;}$$

$$\begin{array}{rl}\text{now }{tan}^{-1}\frac{1}{2}-{tan}^{-1}\frac{1}{3}& ={tan}^{-1}\frac{\frac{1}{2}-\frac{1}{3}}{1+\frac{1}{2}\cdot \frac{1}{3}}\\ & ={tan}^{-1}\frac{1}{7}\text{;}\end{array}$$

$$\therefore {tan}^{-1}\frac{1}{2}={tan}^{-1}\frac{1}{3}+{tan}^{-1}\frac{1}{7}\text{;}$$

$$\therefore \text{45°}={tan}^{-1}\frac{1}{3}+{tan}^{-1}\frac{1}{3}+{tan}^{-1}\frac{1}{7}\text{.}$$

Hence, if $4\times \left(\frac{1}{3}+\frac{1}{3}+\frac{1}{7}\right)<X$ and $>x$, we may take, as Limits, X times, and $\left(x-\frac{X}{3^2+1}\right)$ times, the Square on the radius.