:800px 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 {{EuclidDefLink|XI|14|Sphere}}: :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 {{EuclidPropLink|book = XII|prop = 16|title = Construction of Equilateral Polygon with Even Number of Sides in Outer of Concentric Circles}}: :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 {{EuclidPropLink|book = XI|prop = 18|title = Plane through Straight Line Perpendicular to other Plane is Perpendicular to that Plane}}: :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 {{EuclidPropLink|book = XI|prop = 11|title = Construction of Straight Line Perpendicular to Plane from point not on Plane}}: :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 {{EuclidDefLink|XI|4|Plane at Right Angles to Plane}}: :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: :{{EuclidPropLink|book = III|prop = 27|title = Angles on Equal Arcs are Equal}} and: :{{EuclidPropLink|book = I|prop = 26|title = Triangle Side-Angle-Angle Equality}} it follows that: :$PV = SW$ and: :$BV = KW$ But: :$BA = KA$ Therefore: :$VA = WA$ and so: :$BV : VA = KW : WA$ Therefore from {{EuclidPropLink|book = VI|prop = 2|title = Parallel Line in Triangle Cuts Sides Proportionally}}: :$WV \parallel KB$ We have that each of $PV$ and $SW$ is perpendicular to $BCDE$. Therefore from {{EuclidPropLink|book = XI|prop = 6|title = Two Lines Perpendicular to Same Plane are Parallel}}: :$PV \parallel SW$ But we have that: :$PV = SW$ So from {{EuclidPropLink|book = 1|prop = 33|title = Lines Joining Equal and Parallel Straight Lines are Parallel}}: :$WV = SP$ and: :$WV \parallel SP$ From {{EuclidPropLink|book = XI|prop = 9|title = Lines Parallel to Same Line not in Same Plane are Parallel to each other}}: :$SP \parallel KB$ From {{EuclidPropLink|book = XI|prop = 7|title = Line joining Points on Parallel Lines is in Same Plane}}: :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 {{EuclidPropLink|book = XI|prop = 2|title = Two Intersecting Straight Lines are 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. {{qed|lemma}} It remains to be demonstrated that this polyhedron will not touch the inner sphere. Using {{EuclidPropLink|book = XI|prop = 11|title = Construction of Straight Line Perpendicular to Plane from point not on Plane}}: :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 {{EuclidDefLink|XI|3|Line at Right Angles to Plane}}: :$AX$ 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 {{EuclidPropLink|book = I|prop = 47|title = Pythagoras's Theorem}}: :$AX^2 + XB^2 = AB^2$ 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: :{{EuclidPropLink|book = III|prop = 31|title = Relative Sizes of Angles in Segments}} :{{EuclidPropLink|book = VI|prop = 8|title = Perpendicular in Right-Angled Triangle makes two Similar Triangles}} and: :{{EuclidPropLink|book = VI|prop = 8|qual=Porism|title = 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: :$BA^2 = KA^2$ From {{EuclidPropLink|book = I|prop = 47|title = Pythagoras's Theorem}}: :$BX^2 + XA^2 = BA^2$ :$KZ^2 + ZA^2 = KA^2$ 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. {{qed}} {{Euclid Note|17|XII}}