Cylinder, Cone and Sphere Problems: Very Difficult Problems with Solutions

Solution:
Let [tex]R\gt r[/tex] be the radii of the bases, [tex]h[/tex] the height, [tex]l[/tex] the generatrix and [tex]\rho[/tex] the radius of the sphere. The axial section is an isosceles trapezoid [tex]ABCD[/tex] ([tex]AB=2R[/tex], [tex]CD=2r[/tex], [tex]AD=BC=l[/tex]) with a great circle of the sphere inscribed in it; by symmetry this circle touches [tex]AB[/tex] at its midpoint [tex]O_1[/tex] and [tex]CD[/tex] at its midpoint [tex]O_2[/tex].
The tangent segments from [tex]A[/tex] are equal to [tex]AO_1=R[/tex], and those from [tex]D[/tex] are equal to [tex]DO_2=r[/tex], so [tex]l=AD=R+r[/tex]. Let [tex]DH[/tex] be the height of the trapezoid; then [tex]AH=R-r[/tex] and
[tex]h^2=l^2-(R-r)^2=(R+r)^2-(R-r)^2=4Rr[/tex]
The sphere touches both bases, so [tex]h=2\rho[/tex] and [tex]\rho^2=Rr[/tex]. The volumes of the ball and of the truncated cone are
[tex]V_1=\frac{4}{3}\pi\rho^3, \ V_2=\frac{1}{3}\pi h(R^2+Rr+r^2)=\frac{2}{3}\pi\rho(R^2+Rr+r^2)[/tex]
The condition [tex]V_1=\frac{1}{2}V_2[/tex] gives
[tex]\frac{4}{3}\pi\rho^3=\frac{1}{3}\pi\rho(R^2+Rr+r^2)\Rightarrow 4\rho^2=R^2+Rr+r^2[/tex]
[tex]4Rr=R^2+Rr+r^2\Rightarrow R^2+r^2=3Rr[/tex]
Since [tex]DH[/tex] is parallel to the axis, it is perpendicular to the plane of the larger base. Hence [tex]AH[/tex] is the orthogonal projection of [tex]AD[/tex] on this plane and [tex]\beta=\angle DAH[/tex]. From the right triangle [tex]AHD[/tex]:
[tex]\tan \beta=\frac{DH}{AH}=\frac{h}{R-r}\Rightarrow \tan ^2\beta=\frac{4Rr}{R^2-2Rr+r^2}=\frac{4Rr}{3Rr-2Rr}=4[/tex]
Since [tex]\beta[/tex] is acute, [tex]\tan \beta=2[/tex].

Solution:
Let [tex]h=\frac{a\sqrt{3}}{2}[/tex] be the height of the triangle and [tex]m[/tex] the axis. The line parallel to [tex]AC[/tex] at distance [tex]h[/tex] on the side of [tex]B[/tex] passes through [tex]B[/tex], so [tex]m[/tex] lies on the other side of [tex]AC[/tex]: the points [tex]A[/tex] and [tex]C[/tex] are at distance [tex]h[/tex] from [tex]m[/tex], and [tex]B[/tex] is at distance [tex]2h[/tex].
Let [tex]A'[/tex], [tex]B'[/tex], [tex]C'[/tex] be the feet of the perpendiculars from [tex]A[/tex], [tex]B[/tex], [tex]C[/tex] to [tex]m[/tex]; then [tex]A'C'=a[/tex] and [tex]B'[/tex] is the midpoint of [tex]A'C'[/tex], because the foot of the height from [tex]B[/tex] is the midpoint of [tex]AC[/tex]. Rotating the trapezoids [tex]A'ABB'[/tex] and [tex]B'BCC'[/tex] gives two congruent truncated cones with radii [tex]h[/tex] and [tex]2h[/tex], height [tex]\frac{a}{2}[/tex] and generatrix [tex]a[/tex]; rotating the rectangle [tex]A'ACC'[/tex] gives a cylinder with radius [tex]h[/tex] and height [tex]a[/tex].
The triangle is obtained from the two trapezoids by removing the rectangle, so the solid consists of the two truncated cones with a cylindrical hole. The volume [tex]V_1[/tex] of one truncated cone and the volume [tex]V_2[/tex] of the cylinder are (using [tex]h^2=\frac{3a^2}{4}[/tex]):
[tex]V_1=\frac{1}{3}\pi\times \frac{a}{2}\left(h^2+h\times 2h+(2h)^2\right)=\frac{7\pi ah^2}{6}=\frac{7\pi a^3}{8}[/tex]
[tex]V_2=\pi h^2a=\frac{3\pi a^3}{4}[/tex]
[tex]V=2V_1-V_2=\frac{7\pi a^3}{4}-\frac{3\pi a^3}{4}=\pi a^3[/tex]
The surface consists of the lateral surfaces of the two truncated cones (described by [tex]AB[/tex] and [tex]BC[/tex]) and the lateral surface of the cylinder (described by [tex]AC[/tex]):
[tex]S=2\times \pi(h+2h)a+2\pi ha=8\pi ah=8\pi a\times \frac{a\sqrt{3}}{2}=4\sqrt{3}\pi a^2[/tex]
Hence [tex]V=\pi a^3[/tex] and [tex]S=4\sqrt{3}\pi a^2[/tex].

Solution:
The angle does not depend on the size, so let the radius of the spheres be [tex]1[/tex]. Let [tex]O_1[/tex], [tex]O_2[/tex], [tex]O_3[/tex] be the centers of the spheres touching the base, [tex]O_4[/tex] the center of the fourth sphere and [tex]S[/tex] the apex of the cone. The centers of two touching spheres are at distance [tex]2[/tex], so [tex]O_1O_2O_3O_4[/tex] is a regular tetrahedron with edge [tex]2[/tex].
The points [tex]O_1[/tex], [tex]O_2[/tex], [tex]O_3[/tex] are at distance [tex]1[/tex] from the base, so the plane [tex]O_1O_2O_3[/tex] is parallel to the base and perpendicular to the axis; let the axis meet it at [tex]G[/tex]. Since [tex]O_4[/tex] lies on the axis, [tex]O_4G[/tex] is perpendicular to this plane, and the equal oblique segments [tex]O_4O_1=O_4O_2=O_4O_3[/tex] have equal projections, i.e. [tex]G[/tex] is the center of the equilateral triangle [tex]O_1O_2O_3[/tex]. From the right triangle [tex]O_4GO_1[/tex]:
[tex]GO_1=\frac{O_1O_2}{\sqrt{3}}=\frac{2}{\sqrt{3}}, \ O_4G=\sqrt{O_4O_1^2-GO_1^2}=\sqrt{4-\frac{4}{3}}=\frac{2\sqrt{6}}{3}[/tex]
Consider the axial section through [tex]O_1[/tex] and let [tex]SA[/tex] be the generatrix in it on the side of [tex]O_1[/tex]. By the rotational symmetry, the point of the lateral surface nearest to [tex]O_1[/tex] lies in this plane, so the distance from [tex]O_1[/tex] to the line [tex]SA[/tex] equals the radius [tex]1[/tex]. The same holds for [tex]O_4[/tex], which also lies in this plane.
Both centers lie inside the cone, on the same side of [tex]SA[/tex] and at the same distance [tex]1[/tex] from it, hence [tex]O_4O_1\parallel SA[/tex]. Therefore the half-angle at the apex [tex]\theta=\angle ASO[/tex] ([tex]O[/tex] is the center of the base) equals [tex]\angle GO_4O_1[/tex]:
[tex]\tan \theta=\frac{GO_1}{O_4G}=\frac{2}{\sqrt{3}}\times \frac{3}{2\sqrt{6}}=\frac{\sqrt{2}}{2}[/tex]
The angle at the apex of the axial section is [tex]\varphi=2\theta[/tex], so
[tex]\cos\varphi=\frac{1-\tan ^2\theta}{1+\tan ^2\theta}=\frac{1-\frac{1}{2}}{1+\frac{1}{2}}=\frac{1}{3}[/tex]
(Such a cone exists: [tex]\sin \theta=\frac{1}{\sqrt{3}}[/tex], so the apex is at distance [tex]SO_4=\frac{1}{\sin \theta}=\sqrt{3}\gt 1[/tex] from [tex]O_4[/tex], above the fourth sphere.)

Solution:
Let [tex]R[/tex] be the radius of the ball and [tex]O[/tex] its center. The cylinder of maximal volume is inscribed in the ball, i.e. both its base circles lie on the sphere (otherwise it could be enlarged inside the ball).
The center of each base circle is the foot of the perpendicular from [tex]O[/tex] to its plane, so [tex]O[/tex] lies on the axis of the cylinder. The two base circles have equal radii [tex]r[/tex], so their planes are equidistant from [tex]O[/tex] and [tex]O[/tex] is the midpoint of the axis. If [tex]h[/tex] is the height, the axial section is a rectangle with sides [tex]2r[/tex] and [tex]h[/tex] inscribed in a great circle, so
[tex]r^2+\frac{h^2}{4}=R^2,\qquad 0\lt h\lt 2R[/tex]
The volume of the cylinder as a function of [tex]h[/tex] and its derivative are
[tex]V(h)=\pi r^2h=\pi\left(R^2h-\frac{h^3}{4}\right),\qquad V'(h)=\pi\left(R^2-\frac{3h^2}{4}\right)[/tex]
[tex]V'(h)\gt 0[/tex] for [tex]h\lt\frac{2R}{\sqrt{3}}[/tex] and [tex]V'(h)\lt 0[/tex] for [tex]h\gt\frac{2R}{\sqrt{3}}[/tex], so the maximum is at [tex]h=\frac{2R}{\sqrt{3}}[/tex], where [tex]r^2=R^2-\frac{R^2}{3}=\frac{2R^2}{3}[/tex]:
[tex]V_{\max}=\pi\times \frac{2R^2}{3}\times \frac{2R}{\sqrt{3}}=\frac{1}{\sqrt{3}}\times \frac{4}{3}\pi R^3[/tex]
The volume of the ball is [tex]\frac{4}{3}\pi R^3=\sqrt{3}[/tex], hence [tex]V_{\max}=\frac{\sqrt{3}}{\sqrt{3}}=1[/tex] cubic inch.

Solution:
Each lateral face touches the lateral surface of the cylinder along a generatrix, so it is parallel to the axis. Hence the lateral edges are perpendicular to the bases and the parallelepiped is a right prism [tex]ABCDA_1B_1C_1D_1[/tex] with height [tex]5[/tex], whose base [tex]ABCD[/tex] is a parallelogram circumscribed about the base circle of the cylinder (center [tex]O[/tex], radius [tex]1[/tex]).
In a circumscribed quadrilateral the sums of opposite sides are equal: [tex]AB+CD=BC+AD[/tex]; since [tex]AB=CD[/tex] and [tex]BC=AD[/tex], we get [tex]AB=BC[/tex], so [tex]ABCD[/tex] is a rhombus. The center of the rhombus (the intersection of its diagonals) is the center [tex]O[/tex] of the inscribed circle, and the height of the rhombus equals the diameter [tex]2[/tex].
The volume of the cylinder is [tex]\pi\times 1^2\times 5=5\pi[/tex], so for the volume [tex]V[/tex] of the parallelepiped and the area of its base:
[tex]V=5\pi\times \frac{5}{\pi}=25,\qquad S_{ABCD}=\frac{25}{5}=5=2\times AB\Rightarrow AB=\frac{5}{2}[/tex]
Let [tex]AC=d_1\ge BD=d_2[/tex]. The diagonals of a rhombus are perpendicular and bisect each other, so
[tex]\frac{d_1d_2}{2}=5,\qquad \left(\frac{d_1}{2}\right)^2+\left(\frac{d_2}{2}\right)^2=\left(\frac{5}{2}\right)^2\Rightarrow d_1d_2=10,\quad d_1^2+d_2^2=25[/tex]
[tex](d_1+d_2)^2=45,\quad (d_1-d_2)^2=5\Rightarrow d_1=2\sqrt{5},\quad d_2=\sqrt{5}[/tex]
The longest diagonal of the parallelepiped is [tex]AC_1[/tex]; its orthogonal projection on the base is the longer diagonal [tex]AC[/tex] (because [tex]CC_1\perp(ABC)[/tex]). From the right triangle [tex]ACC_1[/tex]:
[tex]AC_1=\sqrt{AC^2+CC_1^2}=\sqrt{20+25}=3\sqrt{5}[/tex]
All points of [tex]AC_1[/tex] lie between the planes of the bases, so a point of [tex]AC_1[/tex] is inside the cylinder exactly when its orthogonal projection on the base is inside the base circle. The projection [tex]AC[/tex] passes through [tex]O[/tex], and its points inside the circle form a diameter [tex]MN[/tex] of length [tex]2[/tex]. Let [tex]M'[/tex] and [tex]N'[/tex] be the points of [tex]AC_1[/tex] that project onto [tex]M[/tex] and [tex]N[/tex]; the required part is [tex]M'N'[/tex].
Since [tex]MM'\parallel NN'\parallel CC_1[/tex], by the intercept theorem
[tex]\frac{M'N'}{AC_1}=\frac{MN}{AC}\Rightarrow M'N'=\frac{2\times 3\sqrt{5}}{2\sqrt{5}}=3[/tex]
The required length is [tex]3[/tex] in.

Solution:
Let [tex]O[/tex] be the center of the square [tex]ABCD[/tex] and [tex]r[/tex] the radius of the cylinder; its height is [tex]2r[/tex]. We have [tex]OA=\frac{a\sqrt{2}}{2}[/tex], and from the right triangle [tex]AOM[/tex]
[tex]MO=\sqrt{a^2-\frac{a^2}{2}}=\frac{a\sqrt{2}}{2}=OA\Rightarrow \angle MAO=45^{\circ}[/tex]
The cylinder lies on the base plane, touching it along [tex]AC[/tex]; its bases are perpendicular to [tex]AC[/tex] and meet [tex]AC[/tex] at points [tex]K[/tex] and [tex]L[/tex], [tex]KL=2r[/tex], with [tex]K[/tex] nearer to [tex]A[/tex]. By the symmetry with respect to the plane [tex]ACM[/tex], the base through [tex]K[/tex] touches the faces [tex]ABM[/tex] and [tex]ADM[/tex], the base through [tex]L[/tex] touches [tex]CBM[/tex] and [tex]CDM[/tex], and [tex]AK=CL=x[/tex].
The plane of the base through [tex]K[/tex] is perpendicular to [tex]AC[/tex] (hence parallel to [tex]BD[/tex]) and cuts [tex]AB[/tex], [tex]AD[/tex], [tex]AM[/tex] at points [tex]P[/tex], [tex]Q[/tex], [tex]N[/tex]. Since [tex]\angle KAB=\angle KAD=45^{\circ}[/tex], we get [tex]KP=KQ=x[/tex] and [tex]PQ=2x[/tex]. The plane [tex]ACM[/tex] is perpendicular to the base, so [tex]NK\perp PQ[/tex], and from the right triangle [tex]AKN[/tex] with [tex]\angle NAK=45^{\circ}[/tex] we get [tex]NK=x[/tex].
So [tex]PNQ[/tex] is an isosceles right triangle with hypotenuse [tex]PQ=2x[/tex] and legs [tex]NP=NQ=x\sqrt{2}[/tex]. The base circle of the cylinder lies in its plane and touches [tex]PQ[/tex] (at [tex]K[/tex]), [tex]NP[/tex] and [tex]NQ[/tex], i.e. it is the incircle of [tex]\triangle PNQ[/tex]:
[tex]r=\frac{NP+NQ-PQ}{2}=x(\sqrt{2}-1)\Rightarrow x=r(\sqrt{2}+1)[/tex]
From [tex]AC=AK+KL+LC[/tex]:
[tex]a\sqrt{2}=2r(\sqrt{2}+1)+2r=2r(\sqrt{2}+2)\Rightarrow r=\frac{a}{2(\sqrt{2}+1)}=\frac{a(\sqrt{2}-1)}{2}[/tex]
Then [tex]x=\frac{a}{2}\lt OA[/tex], so [tex]K[/tex] lies between [tex]A[/tex] and [tex]O[/tex]; the sections of the pyramid perpendicular to [tex]AC[/tex] grow from [tex]A[/tex] to [tex]O[/tex] and shrink symmetrically towards [tex]C[/tex], so the whole cylinder is inside the pyramid. Its volume is
[tex]V=\pi r^2\times 2r=2\pi r^3=\frac{\pi a^3(\sqrt{2}-1)^3}{4}=\frac{\pi a^3(5\sqrt{2}-7)}{4}[/tex]

Solution:
Let [tex]H[/tex] be the center of the square [tex]ABCD[/tex] ([tex]SH[/tex] is the height of the pyramid) and let [tex]P[/tex], [tex]Q[/tex] be the midpoints of [tex]SB[/tex], [tex]SD[/tex]. The ball touches [tex]SB[/tex] at [tex]P[/tex], so [tex]OP\perp SB[/tex], i.e. [tex]O[/tex] lies in the plane through [tex]P[/tex] perpendicular to [tex]SB[/tex], the perpendicular bisector plane of [tex]SB[/tex]. Since [tex]AS=AB=a[/tex] and [tex]CS=CB=a[/tex], the points [tex]A[/tex] and [tex]C[/tex] also lie in this plane.
In the same way [tex]O[/tex], [tex]A[/tex], [tex]C[/tex] lie in the perpendicular bisector plane of [tex]SD[/tex]. The two planes are different and both contain the line [tex]AC[/tex], so they meet in [tex]AC[/tex] and [tex]O\in AC[/tex]. Moreover [tex]AC\perp BD[/tex] and [tex]AC\perp SH[/tex], so [tex]AC\perp(SBD)[/tex] and [tex]OH\perp HP[/tex].
In the right triangle [tex]SHB[/tex] the median to the hypotenuse is [tex]HP=\frac{SB}{2}=\frac{a}{2}[/tex]. Let [tex]OH=x[/tex]; from the right triangle [tex]OHP[/tex], [tex]OP^2=x^2+\frac{a^2}{4}[/tex], and [tex]AH=\frac{a\sqrt{2}}{2}[/tex]. The radii [tex]OA[/tex] and [tex]OP[/tex] are equal. If [tex]O[/tex] were on the ray [tex]HC[/tex], then
[tex]OA^2=\left(\frac{a\sqrt{2}}{2}+x\right)^2=\frac{a^2}{2}+\sqrt{2}ax+x^2\gt x^2+\frac{a^2}{4}=OP^2[/tex]
which is impossible. So [tex]O[/tex] lies on the ray [tex]HA[/tex] and
[tex]\left(\frac{a\sqrt{2}}{2}-x\right)^2=x^2+\frac{a^2}{4}\Rightarrow \frac{a^2}{2}-\sqrt{2}ax=\frac{a^2}{4}\Rightarrow x=\frac{a\sqrt{2}}{8}[/tex]
Thus [tex]O[/tex] lies between [tex]H[/tex] and [tex]A[/tex], and
[tex]CO=CH+HO=\frac{a\sqrt{2}}{2}+\frac{a\sqrt{2}}{8}=\frac{5a\sqrt{2}}{8}=\frac{5}{8}CA[/tex]
Since [tex]C[/tex] lies in the plane [tex](SCD)[/tex], the distance from [tex]O[/tex] to [tex](SCD)[/tex] is [tex]\frac{5}{8}[/tex] of the distance from [tex]A[/tex] to [tex](SCD)[/tex]. The pyramids [tex]OSCD[/tex] and [tex]ASCD[/tex] have the common base [tex]SCD[/tex], so [tex]V_{OSCD}=\frac{5}{8}V_{ASCD}[/tex]. From the right triangle [tex]SHA[/tex], [tex]SH=\sqrt{a^2-\frac{a^2}{2}}=\frac{a\sqrt{2}}{2}[/tex], hence
[tex]V_{ASCD}=V_{SACD}=\frac{1}{3}\times \frac{a^2}{2}\times \frac{a\sqrt{2}}{2}=\frac{\sqrt{2}a^3}{12}[/tex]
[tex]V_{OSCD}=\frac{5}{8}\times \frac{\sqrt{2}a^3}{12}=\frac{5\sqrt{2}a^3}{96}[/tex]

Solution:
Let [tex]O_1[/tex] and [tex]O_2[/tex] be the centers of the spheres with radii [tex]10[/tex] and [tex]3\sqrt{5}[/tex], and let their common circle have center [tex]Q[/tex] and radius [tex]6[/tex] and lie in the plane [tex]\alpha[/tex]. The foot of the perpendicular from the center of a sphere to a cutting plane is the center of the section circle, so [tex]O_1Q\perp\alpha[/tex] and [tex]O_2Q\perp\alpha[/tex], i.e. [tex]O_1[/tex], [tex]O_2[/tex] and [tex]Q[/tex] lie on one line perpendicular to [tex]\alpha[/tex].
Take a point [tex]P[/tex] on the circle: [tex]QP=6[/tex], [tex]O_1P=10[/tex], [tex]O_2P=3\sqrt{5}[/tex]. From the right triangles [tex]O_1QP[/tex] and [tex]O_2QP[/tex]
[tex]O_1Q=\sqrt{10^2-6^2}=8, \quad O_2Q=\sqrt{45-36}=3[/tex]
If the centers are on different sides of [tex]\alpha[/tex], then [tex]O_1O_2=O_1Q+QO_2=8+3=11[/tex]; if they are on the same side, then [tex]O_1O_2=O_1Q-O_2Q=8-3=5[/tex].
Both cases are possible: if [tex]O_2[/tex] is a point of the line [tex]O_1Q[/tex] at distance [tex]3[/tex] from [tex]Q[/tex] (on either side of [tex]\alpha[/tex]), every point of the circle is at distance [tex]\sqrt{3^2+6^2}=3\sqrt{5}[/tex] from [tex]O_2[/tex], so the circle lies on both spheres, and two different spheres meet in at most one circle.
Therefore the distance between the centers is [tex]5[/tex] or [tex]11[/tex].

Solution:
Let [tex]M[/tex] be the apex, [tex]O[/tex] the center of the base and [tex]AB[/tex] a diameter of the base; the axial section is the isosceles triangle [tex]ABM[/tex] with [tex]MA=MB=l[/tex] and [tex]AB=2r[/tex]. By symmetry the center [tex]I[/tex] of the ball lies on the axis [tex]MO[/tex], so the plane of the axial section cuts the ball in a great circle, which touches [tex]AB[/tex], [tex]MA[/tex] and [tex]MB[/tex]. Hence the radius [tex]\rho[/tex] of the ball is the inradius of triangle [tex]ABM[/tex].
From the right triangle [tex]AOM[/tex] the height is [tex]h=\sqrt{l^2-r^2}[/tex]. The area of triangle [tex]ABM[/tex] is [tex]\frac{1}{2}\times 2r\times h=rh[/tex] and its semiperimeter is [tex]l+r[/tex], so
[tex]\rho=\frac{rh}{l+r}=\frac{r\sqrt{(l-r)(l+r)}}{l+r}=r\sqrt{\frac{l-r}{l+r}}[/tex]
The volume of the ball is
[tex]V=\frac{4}{3}\pi\rho^3=\frac{4}{3}\pi r^3\times \frac{l-r}{l+r}\sqrt{\frac{l-r}{l+r}}=\frac{4\pi r^3(l-r)}{3(l+r)}\sqrt{\frac{l-r}{l+r}}[/tex]

Solution:
Let [tex]P[/tex] be the orthogonal projection of [tex]D[/tex] on the plane [tex]ABC[/tex] and [tex]h=DP[/tex]. The segments [tex]PA[/tex], [tex]PB[/tex], [tex]PC[/tex] are the projections of [tex]DA[/tex], [tex]DB[/tex], [tex]DC[/tex], so [tex]\angle DAP=\angle DBP=60^{\circ}[/tex], [tex]\angle DCP=45^{\circ}[/tex], and from the right triangles [tex]DPA[/tex] and [tex]DPB[/tex]
[tex]PA=PB=h\cot 60^{\circ}=\frac{h}{\sqrt{3}}[/tex]
Let [tex]K[/tex] be the foot of the perpendicular from [tex]P[/tex] to the line [tex]CA[/tex]. By the theorem of three perpendiculars [tex]DK\perp CA[/tex], so [tex]\angle DKP=60^{\circ}[/tex] is the angle between the face [tex]DCA[/tex] and the base, and [tex]PK=h\cot 60^{\circ}=PA[/tex]. If [tex]K\ne A[/tex], the leg [tex]PK[/tex] of the right triangle [tex]PKA[/tex] would be shorter than the hypotenuse [tex]PA[/tex]; hence [tex]K=A[/tex], i.e. [tex]PA\perp CA[/tex] and [tex]DA\perp CA[/tex]. In the same way [tex]PB\perp CB[/tex] and [tex]DB\perp CB[/tex].
So [tex]\angle DAC=\angle DBC=90^{\circ}[/tex], and [tex]A[/tex] and [tex]B[/tex] lie on the sphere with diameter [tex]DC[/tex]. This sphere passes through all four vertices, so it is the circumscribed sphere and [tex]DC=2\times 3=6[/tex]. From the right triangle [tex]DPC[/tex] with [tex]\angle DCP=45^{\circ}[/tex]
[tex]h=DP=PC=\frac{6}{\sqrt{2}}=3\sqrt{2}, \quad PA=PB=\frac{3\sqrt{2}}{\sqrt{3}}=\sqrt{6}[/tex]
and from the right triangles [tex]PAC[/tex] and [tex]PBC[/tex]
[tex]AC=BC=\sqrt{PC^2-PA^2}=\sqrt{18-6}=2\sqrt{3}[/tex]
Since [tex]PA=PB[/tex] and [tex]CA=CB[/tex], the line [tex]PC[/tex] is the perpendicular bisector of [tex]AB[/tex]; let [tex]E=AB\cap PC[/tex]. The segment [tex]AE[/tex] is the altitude to the hypotenuse of the right triangle [tex]PAC[/tex], so
[tex]AE=\frac{PA\times AC}{PC}=\frac{\sqrt{6}\times 2\sqrt{3}}{3\sqrt{2}}=2, \quad CE=\frac{AC^2}{PC}=\frac{12}{3\sqrt{2}}=2\sqrt{2}[/tex]
Hence [tex]AB=2AE=4[/tex] and
[tex]S_{ABC}=\frac{1}{2}\times AB\times CE=\frac{1}{2}\times 4\times 2\sqrt{2}=4\sqrt{2}, \quad V=\frac{1}{3}S_{ABC}\times DP=\frac{1}{3}\times 4\sqrt{2}\times 3\sqrt{2}=8[/tex]

Solution:
Let the truncated pyramid have lower base [tex]A_1A_2\ldots A_n[/tex], upper base [tex]B_1B_2\ldots B_n[/tex] and lateral edges [tex]A_iB_i[/tex] (indices are taken modulo [tex]n[/tex], so [tex]A_{n+1}=A_1[/tex]); it is cut from the pyramid [tex]A_1A_2\ldots A_nS[/tex] by a plane parallel to the base. Let [tex]Q[/tex] be the center of the sphere. All tangent segments from a point [tex]X[/tex] to the sphere are equal (each is a leg of a right triangle with hypotenuse [tex]XQ[/tex] whose other leg is the radius).
The plane of the lower base cuts the sphere in a circle [tex]k_1[/tex] with center [tex]O_1[/tex] (the foot of the perpendicular from [tex]Q[/tex]) and radius [tex]r[/tex]. Each side [tex]A_iA_{i+1}[/tex] lies in this plane and has exactly one common point [tex]M_i[/tex] with the sphere, so it touches [tex]k_1[/tex] at [tex]M_i[/tex]. Hence [tex]k_1[/tex] is the inscribed circle of the base [tex]A_1A_2\ldots A_n[/tex]. Similarly, the upper base has an inscribed circle [tex]k_2[/tex] whose center [tex]O_2[/tex] is the foot of the perpendicular from [tex]Q[/tex]; since the bases are parallel, [tex]O_1[/tex], [tex]O_2[/tex] and [tex]Q[/tex] lie on one line perpendicular to the bases.
The homothety with center [tex]S[/tex] that maps the plane of the lower base onto the plane of the upper base maps [tex]A_i[/tex] to [tex]B_i[/tex], so it maps [tex]k_1[/tex] onto a circle inscribed in [tex]B_1B_2\ldots B_n[/tex]. A convex polygon has at most one inscribed circle (its center is the intersection of the bisectors of two adjacent angles), so the image of [tex]k_1[/tex] is [tex]k_2[/tex] and the image of [tex]O_1[/tex] is [tex]O_2[/tex]. Hence [tex]S[/tex] lies on the line [tex]O_1O_2[/tex], i.e. [tex]SO_1\perp(A_1A_2\ldots A_n)[/tex].
Let [tex]a_i[/tex] be the length of the tangent segments from [tex]A_i[/tex] and [tex]t\gt 0[/tex] the length of the tangent segments from [tex]S[/tex]. The edge [tex]A_iB_i[/tex] touches the sphere at a point [tex]T_i[/tex] of the segment [tex]SA_i[/tex], so [tex]SA_i=ST_i+T_iA_i=t+a_i[/tex]. Moreover [tex]A_iM_i=a_i[/tex], [tex]O_1M_i=r[/tex] and [tex]O_1M_i\perp A_iA_{i+1}[/tex]. From the right triangles [tex]O_1M_iA_i[/tex] and [tex]SO_1A_i[/tex]:
[tex]O_1A_i^2=r^2+a_i^2,\qquad SA_i^2=SO_1^2+O_1A_i^2[/tex]
[tex]\Rightarrow (t+a_i)^2=SO_1^2+r^2+a_i^2\Rightarrow 2ta_i=SO_1^2+r^2-t^2[/tex]
The right-hand side does not depend on [tex]i[/tex], so [tex]a_1=a_2=\ldots=a_n=a[/tex]. Then all sides are equal, [tex]A_iA_{i+1}=A_iM_i+M_iA_{i+1}=2a[/tex], and all vertices are at the same distance [tex]O_1A_i=\sqrt{r^2+a^2}[/tex] from [tex]O_1[/tex].
So the base is inscribed in a circle with center [tex]O_1[/tex] and has equal sides; equal chords subtend equal arcs, so all its angles are equal as well and the base is a regular polygon with center [tex]O_1[/tex]. Since [tex]S[/tex] projects onto [tex]O_1[/tex], the pyramid [tex]A_1A_2\ldots A_nS[/tex] is regular, and therefore the truncated pyramid cut from it by a plane parallel to the base is regular.

Solution:
Let [tex]K[/tex] and [tex]L[/tex] be the midpoints of [tex]AB[/tex] and [tex]CD[/tex]. In the isosceles triangle [tex]ABM[/tex] the median [tex]MK[/tex] is also an altitude, [tex]MK\perp AB[/tex]; since [tex](ABM)\perp(ABCD)[/tex], it follows that [tex]MK\perp(ABCD)[/tex], so [tex]MK=h[/tex] is the height of the pyramid.
We will use this fact: if a plane [tex]\beta[/tex] is perpendicular to a plane [tex]\gamma[/tex] and [tex]O'[/tex] is the orthogonal projection of a point [tex]O[/tex] on [tex]\gamma[/tex], then the distance from [tex]O[/tex] to [tex]\beta[/tex] equals the distance from [tex]O'[/tex] to the line [tex]\beta\cap\gamma[/tex] (the line [tex]OO'[/tex] is parallel to [tex]\beta[/tex], and the perpendicular in [tex]\gamma[/tex] from [tex]O'[/tex] to [tex]\beta\cap\gamma[/tex] is perpendicular to [tex]\beta[/tex]).
Suppose a sphere with center [tex]O[/tex] and radius [tex]\rho[/tex] is inscribed in the pyramid; then [tex]O[/tex] is inside the pyramid and at distance [tex]\rho[/tex] from all five faces.
First, [tex]AB\perp KL[/tex] and [tex]AB\perp MK[/tex], so [tex]AB\perp(MKL)[/tex], and also [tex]CD\perp(MKL)[/tex] because [tex]CD\parallel AB[/tex]. Hence the planes [tex](ABCD)[/tex], [tex](ABM)[/tex], [tex](CDM)[/tex] are perpendicular to [tex](MKL)[/tex] and meet it in the lines [tex]KL[/tex], [tex]KM[/tex], [tex]LM[/tex]. The projection of the pyramid on [tex](MKL)[/tex] is the triangle [tex]MKL[/tex] ([tex]A[/tex] and [tex]B[/tex] go to [tex]K[/tex], [tex]C[/tex] and [tex]D[/tex] go to [tex]L[/tex]), so the projection [tex]O'[/tex] of [tex]O[/tex] is inside this triangle at distance [tex]\rho[/tex] from its three sides. Thus [tex]O'[/tex] is the incenter of [tex]\triangle MKL[/tex] and [tex]\rho=r_1[/tex], the inradius of [tex]\triangle MKL[/tex].
Second, [tex]BC\perp AB[/tex] and [tex]BC\perp MK[/tex], so [tex]BC\perp(ABM)[/tex], and also [tex]AD\perp(ABM)[/tex]. Hence the planes [tex](ABCD)[/tex], [tex](BCM)[/tex], [tex](ADM)[/tex] are perpendicular to [tex](ABM)[/tex] and meet it in [tex]AB[/tex], [tex]BM[/tex], [tex]AM[/tex]. The projection of the pyramid on [tex](ABM)[/tex] is the triangle [tex]ABM[/tex], and in the same way [tex]\rho=r_2[/tex], the inradius of [tex]\triangle ABM[/tex]. So we would need [tex]r_1=r_2[/tex].
Both triangles have area [tex]\frac{ah}{2}[/tex]: [tex]\triangle MKL[/tex] is right-angled at [tex]K[/tex] with legs [tex]h[/tex] and [tex]KL=a[/tex], and [tex]\triangle ABM[/tex] has base [tex]a[/tex], height [tex]h[/tex] and legs [tex]AM=BM=\sqrt{h^2+\frac{a^2}{4}}[/tex]. By the formula [tex]r=\frac{2S}{P}[/tex] ([tex]S[/tex] the area, [tex]P[/tex] the perimeter)
[tex]r_1=\frac{ah}{a+h+\sqrt{a^2+h^2}}, \quad r_2=\frac{ah}{a+\sqrt{a^2+4h^2}}[/tex]
But
[tex]\left(h+\sqrt{a^2+h^2}\right)^2=a^2+2h^2+2h\sqrt{a^2+h^2}\gt a^2+2h^2+2h\times h=a^2+4h^2[/tex]
so [tex]h+\sqrt{a^2+h^2}\gt\sqrt{a^2+4h^2}[/tex] and [tex]r_1\lt r_2[/tex]. This contradiction shows that no sphere can be inscribed in the pyramid.