# Construction of Polyhedron in Outer of Concentric Spheres

## Theorem

In the words of Euclid:

Given two spheres about the same centre, to inscribe in the greater sphere a polyhedral solid which does not touch the lesser sphere.

### Porism

In the words of Euclid:

But if in another sphere also a polyhedral solid be inscribed similar to the solid in the sphere $BCDE$, the polyhedral solid in the sphere $BCDE$ has to the polyhedral solid in the other sphere the ratio triplicate of that which the diameter of the sphere $BCDE$ has to the diameter of the other sphere.

## Proof

Let two spheres be described about the same center $A$.

It is required that a polyhedron is inscribed within the greater (outer) sphere which does not touch the smaller (inner) sphere.

Let the spheres be cut by any plane through the center.

a sphere is produced by rotating a semicircle around its semicircle.

Thus the intersections of the spheres with this plane will be circles.

Let $BCDE$ be the circle in the greater sphere.

Let $FGH$ be the circle in the lesser sphere.

Let two diameters $BD$ and $CE$ be drawn perpendicular to one another.

let a polygon with an even number of sides be inscribed within $BCDE$ which does not touch $FGH$.

Let $BK, KL, LM, ME$ be its sides in the quadrant $BE$.

Let $KA$ be joined and carried through to $N$.

Let $AO$ be set up from $A$ perpendicular to the plane containing $BCDE$.

Let $AO$ meet the outer sphere at the point $O$.

Let planes be described:

containing $AO$ and $BD$
containing $AO$ and $KN$

Consider the circles which are the intersections of the outer sphere with these planes.

Let $BOD$ and $KON$ be the semicircles on $BD$ and $KN$ formed from these circles.

We have that $AO$ is perpendicular to the plane containing $BCDE$.

all the planes through $OA$ are also perpendicular to the plane containing $BCDE$.

Hence $BOD$ and $KON$ are perpendicular to the plane containing $BCDE$.

We have that $BED$, $BOD$ and $KON$ are all on equal diameters $BD$ and $KN$.

Therefore:

$BED = BOD = KON$

Therefore the quadrants $BE, BO, KO$ are also equal.

Therefore as many straight lines can be drawn in the quadrants $BO$ and $KO$ equal to $BK, KL, LM, ME$ in $BE$ as there are sides of the polygon that has been inscribed within $BCDE$.

Let them be inscribed.

Let them be:

$BP, PQ, QR, RO$

and:

$KS, ST, TU, UO$

Let $SP, TQ, UR$ be joined.

Let perpendiculars be drawn from $P$ and $S$ to the plane containing $BCDE$.

We have that the planes containing $BOD$ and $KON$ are perpendicular to the plane containing $BCDE$.

the perpendiculars drawn from $P$ and $S$ will fall on the common sections $BD$ and $KN$.

Let them fall at $V$ and $W$, thereby making the perpendiculars $PV$ and $SW$.

Let $VW$ be joined.

We have that $BP$ and $KS$ are equal chords that have been drawn in the semicircles $BOD$ and $KON$.

We also have the perpendiculars $PV$ and $SW$.

So from:

Proposition $27$ of Book $\text{III}$: Angles on Equal Arcs are Equal

and:

Proposition $26$ of Book $\text{I}$: Triangle Side-Angle-Angle Equality

it follows that:

$PV = SW$

and:

$BV = KW$

But:

$BA = KA$

Therefore:

$VA = WA$

and so:

$BV : VA = KW : WA$
$WV \parallel KB$

We have that each of $PV$ and $SW$ is perpendicular to $BCDE$.

$PV \parallel SW$

But we have that:

$PV = SW$
$WV = SP$

and:

$WV \parallel SP$
$SP \parallel KB$
the quadrilateral $KBPS$ lies all in one plane.

For the same reason:

the quadrilateral $SPQT$ lies all in one plane
the quadrilateral $TQRU$ lies all in one plane.
$\triangle URO$ lies all in one plane.

Let straight lines be imagined from $P, Q, R, S, T, U$ to $A$.

Then a polyhedron will be formed between the arcs $BO$ and $KO$ which consists of pyramids whose bases are $KBPS, SPQT, TQRU$ and $\triangle URO$, and whose apices are the point $A$.

The same construction can be made in the case of each of the lines $KL, LM, ME$ as for $BK$.

Then the same construction again can be made for each of the other three quadrants.

Thus a polyhedron will be constructed which is inscribed within the outer sphere consisting of pyramids such that:

the described quadrilaterals $KBPS, SPQT, TQRU$ and $\triangle URO$ and those corresponding to them are the bases

and

$A$ are their apices.

$\Box$

It remains to be demonstrated that this polyhedron will not touch the inner sphere.

Let $AX$ be drawn from $A$ perpendicular to the plane holding the quadrilateral $KBPS$.

Let $X$ be the point at which $AX$ meets $KBPS$.

Let $XB$ and $XK$ be joined.

We have that $AX$ is perpendicular to $KBPS$.

$AX$ is perpendicular to all other straight lines which meet it and are in the plane holding the quadrilateral $KBPS$.

Therefore $AX \perp BX$ and $AX \perp XK$.

We have that $AB = AK$.

$AX^2 + XB^2 = AB^2$

and:

$AX^2 + XK^2 = AK^2$

Therefore:

$AX^2 + XB^2 = AX^2 + XK^2$

and so:

$BX^2 = XK^2$

so:

$XB = XK$

Similarly it can be proved that:

$XP = XK = XB = XS$

Therefore the circle described with center $X$ and whose radius is $XB$ and $XK$ will also pass through $P$ and $S$.

Thus $KBPS$ is a cyclic quadrilateral.

We have that:

$KB > WV$

while:

$WV = SP$

Therefore:

$KB > SP$

But:

$KB = KS = BP$

Therefore:

$KS > SP$

and:

$BP > SP$

So we have that:

$KBPS$ is a cyclic quadrilateral
$KB = BP = KS$
$PS < KB$ etc.
$BX$ is the radius of the circle around $KBPS$.

Therefore:

$KB^2 > 2 \cdot BX^2$

Let $KZ$ be drawn from $K$ perpendicular to $BV$.

We have that:

$BD < 2 \cdot DZ$

Also:

$BD : DZ = DB \cdot BZ : DZ \cdot ZB$

Let a square be described on $BZ$.

Let the parallelogram on $ZD$ be completed.

Then:

$DB \cdot BZ < 2 \cdot DZ \cdot ZB$

Let $KD$ be joined.

From:

Proposition $31$ of Book $\text{III}$: Relative Sizes of Angles in Segments
Proposition $8$ of Book $\text{VI}$: Perpendicular in Right-Angled Triangle makes two Similar Triangles

and:

Porism to Proposition $8$ of Book $\text{VI}$: Perpendicular in Right-Angled Triangle makes two Similar Triangles

it follows that:

$DB \cdot BZ = BK^2$

and:

$DZ \cdot BZ = KZ^2$

Therefore:

$KB^2 < 2 \cdot KZ^2$

But:

$KB^2 > 2 \cdot BX^2$

Therefore:

$KZ^2 > BX^2$

We have that:

$BA = KA$

so:

$BA^2 = KA^2$
$BX^2 + XA^2 = BA^2$
$KZ^2 + ZA^2 = KA^2$

Therefore:

$BX^2 + XA^2 = KZ^2 + ZA^2$

Of these:

$KZ^2 > BX^2$

Therefore:

$ZA^2 < XA^2$

Therefore:

$AX > AZ$

and so

$AX \gg AG$

We have that:

$AX$ is the perpendicular to one face of the polyhedron.

while:

$AG$ is the radius of the inner sphere.

Hence the polyhedron will not touch the inner sphere.

$\blacksquare$

## Historical Note

This proof is Proposition $17$ of Book $\text{XII}$ of Euclid's The Elements.