Construction of Polyhedron in Outer of Concentric Spheres

Proof

 * Euclid-XII-17.png

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.

By :
 * 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.

From :
 * 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 $BCDE$.

Therefore from :
 * 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.

Using :
 * 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 $BCDE$.

So from :
 * 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:

and:

it follows that:
 * $PV = SW$

and:
 * $BV = KW$

But:
 * $BA = KA$

Therefore:
 * $VA = WA$

and so:
 * $BV : VA = KW : WA$

Therefore from :
 * $WV \parallel KB$

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

Therefore from :
 * $PV \parallel SW$

But we have that:
 * $PV = SW$

So from :
 * $WV = SP$

and:
 * $WV \parallel SP$

From :
 * $SP \parallel KB$

From :
 * 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.

But from :
 * $\triangle URO$ 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.

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

Using :
 * 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 $KBPS$.

So from :
 * $AX$ 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 $AB = AK$.

Thus from :
 * $AX^2 + XB^2 = AB^2$

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:

and:

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:
 * $BA^2 = KA^2$

From :
 * $BX^2 + XA^2 = BA^2$
 * $KZ^2 + ZA^2 = KA^2$

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.