Construction of Polyhedron in Outer of Concentric Spheres

== Theorem ==
=== Porism ===
== 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.
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.
== Sources ==
* : Book $\text{XII}$. Propositions
Category:Spheres
Category:Polyhedra