Prisms and Pyramids Problems: Very Difficult Problems with Solutions

Solution:
Let [tex]P[/tex] be the midpoint of [tex]BC[/tex] and [tex]O[/tex] the center of the base; since the pyramid is regular, [tex]O[/tex] is the foot of its height [tex]MO[/tex] and lies on the median [tex]AP[/tex]. As [tex]MB=MC[/tex] and [tex]AB=AC[/tex], we have [tex]BC\perp MP[/tex] and [tex]BC\perp AP[/tex], so [tex]BC\perp(AMP)[/tex] and in particular [tex]BC\perp AM[/tex].
Hence the plane through [tex]BC[/tex] perpendicular to [tex]AM[/tex] passes through [tex]P[/tex], and [tex]D[/tex] is the foot of the perpendicular from [tex]P[/tex] to [tex]AM[/tex] in the plane [tex](AMP)[/tex]; also [tex]BC\perp PD[/tex]. Since [tex]AD\perp(BCD)[/tex], the segment [tex]AD[/tex] is the height of the pyramid [tex]ABCD[/tex] to the base [tex]BCD[/tex]:
[tex]V=\frac{1}{3}\times \frac{BC\times PD}{2}\times AD=\frac{1}{3}PD\times AD[/tex]
Put [tex]c=\cos\alpha[/tex]. From the right triangle [tex]MPC[/tex] with [tex]PC=1[/tex] we get [tex]AM=MC=\frac{1}{c}[/tex]. Let [tex]\beta=\angle MAP[/tex]; from the right triangle [tex]AOM[/tex] with [tex]AP=\sqrt{3}[/tex] and [tex]AO=\frac{2}{3}AP=\frac{2\sqrt{3}}{3}[/tex]:
[tex]\cos\beta=\frac{AO}{AM}=\frac{2\sqrt{3}}{3}c[/tex]
From the right triangle [tex]ADP[/tex] (right angle at [tex]D[/tex]):
[tex]AD=AP\cos\beta=2c,\quad PD=\sqrt{AP^2-AD^2}=\sqrt{3-4c^2}[/tex]
[tex]V=\frac{2}{3}c\sqrt{3-4c^2}=\frac{2}{3}\sqrt{3c^2-4c^4}[/tex]
The point [tex]D[/tex] lies on the edge [tex]AM[/tex] exactly when [tex]AD\le AM[/tex], i.e. [tex]2c\le\frac{1}{c}[/tex], [tex]c^2\le\frac{1}{2}[/tex]. For [tex]t=c^2\in\left(0, \frac{1}{2}\right][/tex] the quadratic function [tex]f(t)=3t-4t^2[/tex] attains its maximum at the vertex [tex]t=\frac{3}{8}[/tex], which lies in this interval. Therefore
[tex]\cos^2\alpha=\frac{3}{8}\Rightarrow\cos\alpha=\frac{\sqrt{6}}{4},\quad V_{\max}=\frac{2}{3}\sqrt{\frac{9}{8}-\frac{9}{16}}=\frac{1}{2}[/tex]

Solution:
Let [tex]DL[/tex] be the height of triangle [tex]BCD[/tex] from [tex]D[/tex]. The planes [tex](BCD)[/tex] and [tex](ABC)[/tex] are perpendicular and [tex]DL[/tex] is perpendicular to their common line [tex]BC[/tex], so [tex]DL\perp(ABC)[/tex]: [tex]DL[/tex] is the height of the tetrahedron and [tex]L[/tex] lies on the line [tex]BC[/tex].
Drop [tex]DM\perp AB[/tex] and [tex]DN\perp AC[/tex]. The right triangles [tex]AMD[/tex] and [tex]AND[/tex] have the common hypotenuse [tex]AD[/tex] and an angle of [tex]60^{\circ}[/tex] at [tex]A[/tex], so [tex]AM=AN=\frac{AD}{2}[/tex]. By the theorem of the three perpendiculars, [tex]LM\perp AB[/tex] and [tex]LN\perp AC[/tex]. The right triangles [tex]AML[/tex] and [tex]ANL[/tex] have the common hypotenuse [tex]AL[/tex] and [tex]AM=AN[/tex], hence [tex]\angle LAM=\angle LAN[/tex], i.e. [tex]AL[/tex] is the bisector of [tex]\angle BAC[/tex]. By the angle bisector theorem
[tex]\frac{BL}{LC}=\frac{AB}{AC}=\frac{2}{3}\Rightarrow BL=\frac{2}{5}BC=\frac{8}{5}[/tex]
By the law of cosines in triangle [tex]ABC[/tex]
[tex]\cos\angle B=\frac{AB^2+BC^2-AC^2}{2\times AB\times BC}=\frac{4+16-9}{16}=\frac{11}{16},\quad\sin \angle B=\sqrt{1-\frac{121}{256}}=\frac{3\sqrt{15}}{16}[/tex]
From the right triangle [tex]BML[/tex], and then from the right triangle [tex]AMD[/tex]:
[tex]BM=BL\cos\angle B=\frac{11}{10},\quad LM=BL\sin \angle B=\frac{3\sqrt{15}}{10}[/tex]
[tex]AM=AB-BM=\frac{9}{10},\quad DM=AM\tan 60^{\circ}=\frac{9\sqrt{3}}{10}[/tex]
From the right triangle [tex]DLM[/tex] we find the height of the tetrahedron:
[tex]DL=\sqrt{DM^2-LM^2}=\sqrt{\frac{243}{100}-\frac{135}{100}}=\frac{\sqrt{108}}{10}=\frac{3\sqrt{3}}{5}[/tex]
The area of the base and the volume are
[tex]S_{ABC}=\frac{1}{2}AB\times BC\sin \angle B=\frac{1}{2}\times 2\times 4\times \frac{3\sqrt{15}}{16}=\frac{3\sqrt{15}}{4}[/tex]
[tex]V=\frac{1}{3}S_{ABC}\times DL=\frac{1}{3}\times \frac{3\sqrt{15}}{4}\times \frac{3\sqrt{3}}{5}=\frac{9\sqrt{45}}{60}=\frac{9\sqrt{5}}{20}[/tex]

Solution:
Let the pyramid be [tex]ABCDM[/tex] with square base [tex]ABCD[/tex] of side [tex]a[/tex] and height [tex]MH=h[/tex], and let the dihedral angles be [tex]\alpha[/tex] at [tex]AB[/tex], [tex]2\alpha[/tex] at [tex]BC[/tex] and [tex]DA[/tex], and [tex]4\alpha[/tex] at [tex]CD[/tex]. Every dihedral angle of the pyramid is less than [tex]180^{\circ}[/tex], so [tex]4\alpha\lt 180^{\circ}[/tex], i.e. [tex]\alpha\lt 45^{\circ}[/tex] and [tex]2\alpha\lt 90^{\circ}[/tex].
If [tex]N[/tex] is the foot of the perpendicular from [tex]H[/tex] to the line of a base edge, then [tex]MN[/tex] is perpendicular to that edge (theorem of the three perpendiculars). If [tex]H[/tex] is on the same side of the edge as the base, the dihedral angle at this edge equals [tex]\angle MNH[/tex] and is acute; if [tex]H[/tex] is on the other side, it equals [tex]180^{\circ}-\angle MNH[/tex] and is obtuse. In both cases [tex]HN=h\cot \angle MNH[/tex].
The angles at [tex]BC[/tex] and [tex]DA[/tex] are equal and acute, so [tex]H[/tex] lies between the lines [tex]BC[/tex] and [tex]DA[/tex], at distance [tex]h\cot 2\alpha[/tex] from each of them. Hence [tex]H[/tex] lies on the line [tex]KL[/tex] through the midpoints [tex]K[/tex] of [tex]AB[/tex] and [tex]L[/tex] of [tex]CD[/tex], and [tex]a=2h\cot 2\alpha[/tex]. The angle at [tex]AB[/tex] is acute, so [tex]HK=h\cot \alpha[/tex].
At the edge [tex]CD[/tex]: if [tex]4\alpha\lt 90^{\circ}[/tex], then [tex]H[/tex] is between [tex]AB[/tex] and [tex]CD[/tex], [tex]HL=h\cot 4\alpha[/tex] and [tex]HK+HL=a[/tex]; if [tex]4\alpha\gt 90^{\circ}[/tex], then [tex]H[/tex] is beyond [tex]CD[/tex], [tex]HL=-h\cot 4\alpha[/tex] and [tex]HK-HL=a[/tex]; if [tex]4\alpha=90^{\circ}[/tex], then [tex]H=L[/tex] and [tex]HK=a[/tex]. In every case
[tex]h\cot \alpha+h\cot 4\alpha=2h\cot 2\alpha\iff\cot \alpha-\cot 2\alpha=\cot 2\alpha-\cot 4\alpha[/tex]
Since
[tex]\cot x-\cot 2x=\frac{\sin 2x\cos x-\cos 2x\sin x}{\sin x\sin 2x}=\frac{\sin x}{\sin x\sin 2x}=\frac{1}{\sin 2x}[/tex]
the equation becomes [tex]\frac{1}{\sin 2\alpha}=\frac{1}{\sin 4\alpha}[/tex], i.e. [tex]2\sin 2\alpha\cos 2\alpha=\sin 2\alpha[/tex]. As [tex]\sin 2\alpha\ne 0[/tex], we get [tex]\cos 2\alpha=\frac{1}{2}[/tex], and since [tex]0^{\circ}\lt 2\alpha\lt 90^{\circ}[/tex], [tex]2\alpha=60^{\circ}[/tex] and [tex]\alpha=30^{\circ}[/tex].
The four angles are [tex]30^{\circ}[/tex], [tex]60^{\circ}[/tex], [tex]120^{\circ}[/tex], [tex]60^{\circ}[/tex] (the foot [tex]H[/tex] lies outside the base, beyond [tex]CD[/tex]), so the smallest of them is [tex]30^{\circ}[/tex].

Solution:
The heights of the parallelogram to the sides [tex]AB[/tex] and [tex]AD[/tex] are [tex]\frac{90}{18}=5[/tex] and [tex]\frac{90}{10}=9[/tex]. The point [tex]O[/tex] is the center of symmetry of the parallelogram, so it is equidistant from opposite sides: its distance to the lines [tex]AB[/tex] and [tex]CD[/tex] is [tex]\frac{5}{2}[/tex], and to the lines [tex]AD[/tex] and [tex]BC[/tex] it is [tex]\frac{9}{2}[/tex].
Let [tex]OK\perp AB[/tex] and [tex]OL\perp AD[/tex] ([tex]K[/tex] on the line [tex]AB[/tex], [tex]L[/tex] on the line [tex]AD[/tex], possibly on the extension of the side). Since [tex]MO\perp(ABCD)[/tex], by the theorem of three perpendiculars [tex]MK\perp AB[/tex] and [tex]ML\perp AD[/tex], i.e. [tex]MK[/tex] and [tex]ML[/tex] are the heights of the faces [tex]ABM[/tex] and [tex]ADM[/tex]. From the right triangles [tex]MOK[/tex] and [tex]MOL[/tex]:
[tex]MK=\sqrt{6^2+\left(\frac{5}{2}\right)^2}=\frac{13}{2}, \quad ML=\sqrt{6^2+\left(\frac{9}{2}\right)^2}=\frac{15}{2}[/tex]
Since [tex]OA=OC[/tex] and [tex]OB=OD[/tex], the right triangles [tex]MOA[/tex] and [tex]MOC[/tex] are congruent, and so are [tex]MOB[/tex] and [tex]MOD[/tex]; hence [tex]MA=MC[/tex], [tex]MB=MD[/tex], and [tex]\triangle ABM\cong\triangle CDM[/tex], [tex]\triangle ADM\cong\triangle CBM[/tex]. Therefore
[tex]S=2\times \frac{1}{2}\times 18\times \frac{13}{2}+2\times \frac{1}{2}\times 10\times \frac{15}{2}=117+75=192[/tex]

Solution:
Let [tex]EO[/tex] be the height of the pyramid. The right triangles [tex]EOA[/tex], [tex]EOB[/tex], [tex]EOC[/tex], [tex]EOD[/tex] have the common leg [tex]EO[/tex] and equal hypotenuses, so [tex]OA=OB=OC=OD=R[/tex]. Thus [tex]O[/tex] is the center of a circle circumscribed about [tex]ABCD[/tex], the quadrilateral is cyclic and [tex]\angle BAD+\angle BCD=180^{\circ}[/tex].
Let [tex]\angle BCD=\varphi[/tex]; then [tex]\cos\angle BAD=-\cos\varphi[/tex]. By the law of cosines in [tex]\triangle BCD[/tex] and [tex]\triangle BAD[/tex]:
[tex]BD^2=25^2+35^2-2\times 25\times 35\cos\varphi=1850-1750\cos\varphi[/tex]
[tex]BD^2=5^2+17^2+2\times 5\times 17\cos\varphi=314+170\cos\varphi[/tex]
Equating, [tex]1920\cos\varphi=1536[/tex], hence
[tex]\cos\varphi=\frac{4}{5}, \quad \sin \varphi=\frac{3}{5}, \quad BD^2=314+136=450, \quad BD=15\sqrt{2}[/tex]
The circle through [tex]A[/tex], [tex]B[/tex], [tex]C[/tex], [tex]D[/tex] is the circumcircle of [tex]\triangle BCD[/tex], so by the law of sines
[tex]2R=\frac{BD}{\sin \varphi}=\frac{15\sqrt{2}\times 5}{3}=25\sqrt{2}\Rightarrow R^2=\frac{625}{2}[/tex]
The triangle [tex]CDE[/tex] is isosceles ([tex]EC=ED[/tex]), so its base angles are acute and the right angle is at [tex]E[/tex]. Hence [tex]CD=EC\sqrt{2}[/tex] and [tex]EC^2=\frac{35^2}{2}=\frac{1225}{2}[/tex]. From the right triangle [tex]EOC[/tex]:
[tex]EO^2=EC^2-R^2=\frac{1225}{2}-\frac{625}{2}=300\Rightarrow EO=10\sqrt{3}[/tex]

Solution:
Put [tex]AB=a[/tex]; then [tex]AA_1=a-1[/tex] and [tex]AC=a+1[/tex], where [tex]a\gt 1[/tex]. Let [tex]\angle BAC=\varphi[/tex].
Complete the prism to the right prism [tex]ABCDA_1B_1C_1D_1[/tex] whose base [tex]ABCD[/tex] is a parallelogram. Then [tex]ABC_1D_1[/tex] is a parallelogram, so [tex]AD_1\parallel BC_1[/tex] and hence [tex]AB_1\perp AD_1[/tex]; from the right triangle [tex]AB_1D_1[/tex] we get [tex]B_1D_1^2=AB_1^2+AD_1^2[/tex].
Since the prism is right, [tex]AB_1^2=AB^2+AA_1^2[/tex], [tex]AD_1^2=AD^2+DD_1^2=BC^2+AA_1^2[/tex] and [tex]B_1D_1=BD[/tex]. In the parallelogram [tex]ABCD[/tex] the sum of the squares of the diagonals equals the sum of the squares of the sides, so [tex]BD^2=2AB^2+2BC^2-AC^2[/tex]. Substituting:
[tex]2AB^2+2BC^2-AC^2=AB^2+BC^2+2AA_1^2\Rightarrow BC^2=AC^2-AB^2+2AA_1^2[/tex]
On the other hand, by the law of cosines [tex]BC^2=AB^2+AC^2-2AB\times AC\cos\varphi[/tex]. Comparing the two expressions:
[tex]\cos\varphi=\frac{AB^2-AA_1^2}{AB\times AC}=\frac{a^2-(a-1)^2}{a(a+1)}=\frac{2a-1}{a(a+1)}[/tex]
Put [tex]t=2a-1\gt 1[/tex]; then [tex]a=\frac{t+1}{2}[/tex] and [tex]a(a+1)=\frac{t^2+4t+3}{4}[/tex]. By the AM-GM inequality [tex]t+\frac{3}{t}\ge 2\sqrt{3}[/tex], so
[tex]\cos\varphi=\frac{4t}{t^2+4t+3}=\frac{4}{t+\frac{3}{t}+4}\le\frac{4}{2\sqrt{3}+4}=4-2\sqrt{3}[/tex]
Equality holds exactly when [tex]t=\sqrt{3}[/tex], i.e. [tex]a=\frac{1+\sqrt{3}}{2}\gt 1[/tex]. Then [tex]2a^2=2+\sqrt{3}[/tex], [tex]2a=1+\sqrt{3}[/tex] and
[tex]BC^2=(a+1)^2-a^2+2(a-1)^2=2a^2-2a+3=(2+\sqrt{3})-(1+\sqrt{3})+3=4\Rightarrow BC=2[/tex]

Solution:
Let [tex]AB=a[/tex], let [tex]O[/tex] be the center of the square [tex]ABCD[/tex] and [tex]K[/tex], [tex]N[/tex] the midpoints of [tex]AB[/tex], [tex]CD[/tex]. Since the pyramid is regular, [tex]MO\perp(ABCD)[/tex], all lateral faces are congruent isosceles triangles of the same area [tex]S[/tex], and the slant heights are [tex]MK=MN=k[/tex].
Since [tex]MK\perp AB[/tex] and [tex]OK\perp AB[/tex], the angle [tex]\varphi=\angle MKO=\angle MKN[/tex] is the linear angle of the dihedral angle at [tex]AB[/tex]. From the right triangle [tex]MOK[/tex]:
[tex]\cos\varphi=\frac{OK}{MK}=\frac{a}{2k}\Rightarrow k=\frac{a}{2\cos\varphi}[/tex]
The bisector plane contains [tex]AB\parallel CD[/tex], so it cuts the face [tex]MCD[/tex] along a segment [tex]PQ\parallel CD[/tex] ([tex]P\in MD[/tex], [tex]Q\in MC[/tex]). The plane [tex]MKN[/tex] is perpendicular to [tex]AB[/tex], so it cuts the bisector plane along the bisector [tex]KR[/tex] of [tex]\angle MKN[/tex], where [tex]R[/tex] is the common point of [tex]PQ[/tex] and [tex]MN[/tex]. By the angle bisector theorem in [tex]\triangle KMN[/tex] (with [tex]KN=a[/tex]):
[tex]\frac{MR}{RN}=\frac{KM}{KN}=\frac{k}{a}\Rightarrow \lambda=\frac{MR}{MN}=\frac{k}{k+a}[/tex]
Since [tex]PQ\parallel CD[/tex], we have [tex]\frac{MP}{MD}=\frac{MQ}{MC}=\lambda[/tex], so [tex]\triangle MPQ\sim\triangle MDC[/tex] with ratio [tex]\lambda[/tex] and [tex]S_{MPQ}=\lambda^2S[/tex]. The triangles [tex]MAP[/tex] and [tex]MAD[/tex] have the same height from [tex]A[/tex], so [tex]S_{MAP}=\lambda S[/tex]; similarly [tex]S_{MBQ}=\lambda S[/tex].
The part of the lateral surface that contains the apex [tex]M[/tex] has area [tex]S+2\lambda S+\lambda^2S=(1+\lambda)^2S[/tex], and the whole lateral surface has area [tex]4S[/tex]. The two parts are equal, so
[tex](1+\lambda)^2S=2S\Rightarrow 1+\lambda=\sqrt{2}\Rightarrow \lambda=\sqrt{2}-1[/tex]
Substituting [tex]k=\frac{a}{2\cos\varphi}[/tex]:
[tex]\lambda=\frac{k}{k+a}=\frac{1}{1+2\cos\varphi}=\sqrt{2}-1\Rightarrow 1+2\cos\varphi=\sqrt{2}+1\Rightarrow \cos\varphi=\frac{\sqrt{2}}{2}\Rightarrow \varphi=45^{\circ}[/tex]

Solution:
Let [tex]O[/tex] be the orthogonal projection of [tex]M[/tex] on the plane [tex]ABC[/tex], so [tex]MO[/tex] is the height. The right triangles [tex]MOB[/tex] and [tex]MOC[/tex] have the common leg [tex]MO[/tex] and equal hypotenuses [tex]MB=MC[/tex], so [tex]OB=OC[/tex] and [tex]O[/tex] lies on the perpendicular bisector of [tex]BC[/tex].
Let [tex]OP\perp AB[/tex] and [tex]OQ\perp AC[/tex] ([tex]P[/tex] on the line [tex]AB[/tex], [tex]Q[/tex] on the line [tex]AC[/tex]). By the theorem of three perpendiculars [tex]MP\perp AB[/tex] and [tex]MQ\perp AC[/tex], so [tex]\angle MPO[/tex] and [tex]\angle MQO[/tex] are the angles between the planes [tex]ABM[/tex], [tex]ACM[/tex] and the base. They are equal, so the right triangles [tex]MOP[/tex] and [tex]MOQ[/tex] (common leg [tex]MO[/tex]) are congruent and [tex]OP=OQ[/tex].
A dihedral angle at a base edge is acute when [tex]O[/tex] lies on the same side of that edge as the base, and obtuse when [tex]O[/tex] lies on the other side. Since the dihedral angles at [tex]AB[/tex] and [tex]AC[/tex] are equal and [tex]OP=OQ[/tex], the point [tex]O[/tex] lies in [tex]\angle BAC[/tex] or in its vertical angle, i.e. on the line [tex]AL[/tex] of the bisector of [tex]\angle BAC[/tex] ([tex]L\in BC[/tex]). As [tex]AB\ne AC[/tex], this line meets the perpendicular bisector of [tex]BC[/tex] at exactly one point, and that point is [tex]O[/tex].
Now we work in the base. Let [tex]H[/tex] be the foot of the altitude from [tex]A[/tex] and [tex]N[/tex] the midpoint of [tex]BC[/tex]. From [tex]AB^2-BH^2=AC^2-CH^2[/tex]:
[tex]225-BH^2=169-(14-BH)^2\Rightarrow BH=9,\quad AH=\sqrt{225-81}=12[/tex]
By the angle bisector theorem [tex]\frac{BL}{LC}=\frac{15}{13}[/tex], so [tex]BL=\frac{15}{28}\times 14=\frac{15}{2}[/tex]; also [tex]BN=7[/tex]. Hence the points on [tex]BC[/tex] are in the order [tex]B[/tex], [tex]N[/tex], [tex]L[/tex], [tex]H[/tex] with [tex]NL=\frac{1}{2}[/tex] and [tex]LH=\frac{3}{2}[/tex], and the line [tex]AL[/tex] meets the perpendicular to [tex]BC[/tex] through [tex]N[/tex] at a point [tex]O[/tex] on the other side of [tex]BC[/tex] from [tex]A[/tex]. The right triangles [tex]LHA[/tex] and [tex]LNO[/tex] are similar (vertical angles at [tex]L[/tex]), so
[tex]ON=AH\times \frac{LN}{LH}=12\times \frac{1}{3}=4[/tex]
Let [tex]T[/tex] be the foot of the perpendicular from [tex]A[/tex] to the line [tex]ON[/tex]. Then [tex]AHNT[/tex] is a rectangle, so [tex]AT=HN=2[/tex], [tex]TN=AH=12[/tex] and [tex]TO=TN+NO=12+4=16[/tex], hence
[tex]AO^2=AT^2+TO^2=2^2+16^2=260[/tex]
From the right triangle [tex]AOM[/tex] we find the height, and the base area is [tex]S_{ABC}=\frac{BC\times AH}{2}=\frac{14\times 12}{2}=84[/tex]:
[tex]MO=\sqrt{AM^2-AO^2}=\sqrt{324-260}=8,\quad V=\frac{1}{3}S_{ABC}\times MO=\frac{1}{3}\times 84\times 8=224[/tex]

Solution:
Let [tex]O[/tex] be the foot of the height from [tex]M[/tex]. Since the plane [tex](ABM)[/tex] is perpendicular to the base, this height lies in the plane [tex](ABM)[/tex], so [tex]O[/tex] lies on the line [tex]AB[/tex].
The right triangles [tex]MOA[/tex], [tex]MOB[/tex], [tex]MOC[/tex], [tex]MOD[/tex] have the common leg [tex]MO[/tex] and equal hypotenuses [tex]\sqrt{5}[/tex], so [tex]OA=OB=OC=OD=R[/tex]. Thus [tex]O[/tex] is the midpoint of [tex]AB[/tex], the base is inscribed in a circle with center [tex]O[/tex], [tex]AB=2R[/tex] is its diameter and [tex]MO=\sqrt{5-R^2}[/tex].
Each lateral face is an isosceles triangle with legs [tex]\sqrt{5}[/tex]. If its base is [tex]t[/tex], its height to the base is [tex]\sqrt{5-\frac{t^2}{4}}[/tex], so its area [tex]f(t)[/tex] satisfies
[tex]f(t)^2=\frac{t^2}{4}\left(5-\frac{t^2}{4}\right)=\frac{t^2(20-t^2)}{16}[/tex]
Since [tex]u(20-u)=v(20-v)[/tex] exactly when [tex]u=v[/tex] or [tex]u+v=20[/tex], two faces with bases [tex]t[/tex] and [tex]s[/tex] have equal areas if and only if [tex]t=s[/tex] or [tex]t^2+s^2=20[/tex]. The chords [tex]BC[/tex], [tex]CD[/tex], [tex]DA[/tex] are shorter than the diameter [tex]AB[/tex], so comparing the faces [tex]BCM[/tex], [tex]CDM[/tex], [tex]DAM[/tex] with the face [tex]ABM[/tex] gives
[tex]BC^2=CD^2=DA^2=20-AB^2=20-4R^2[/tex]
Thus [tex]BC=CD=DA[/tex]. These equal chords divide the semicircle containing [tex]C[/tex] and [tex]D[/tex] into three equal arcs of [tex]60^{\circ}[/tex], so the triangles [tex]OBC[/tex], [tex]OCD[/tex], [tex]ODA[/tex] are equilateral and [tex]BC=CD=DA=R[/tex]. Then
[tex]R^2=20-4R^2\Rightarrow R=2, \quad MO=\sqrt{5-R^2}=1[/tex]
The base consists of three equilateral triangles with side [tex]2[/tex], so its area is [tex]3\times \frac{2^2\sqrt{3}}{4}=3\sqrt{3}[/tex], and
[tex]V=\frac{1}{3}\times 3\sqrt{3}\times 1=\sqrt{3}[/tex]
(Check: every lateral face then has area [tex]2[/tex].)

Solution:
Let the pyramid be [tex]ABCDM[/tex] with square base [tex]ABCD[/tex], base center [tex]O[/tex] and height [tex]MO[/tex], and let [tex]K[/tex] be the midpoint of [tex]AB[/tex], so [tex]MK[/tex] is an apothem and [tex]OK=\frac{a}{2}[/tex]. Since [tex]OA=\frac{a\sqrt{2}}{2}[/tex], from the right triangle [tex]AOM[/tex]
[tex]MO=\sqrt{MA^2-OA^2}=\sqrt{a^2-\frac{a^2}{2}}=\frac{a\sqrt{2}}{2}[/tex]
Let [tex]x[/tex] be the edge of the cube. Its lower face lies in the base and its lateral edges are perpendicular to the base, so its four upper vertices lie in the plane parallel to the base at distance [tex]x[/tex] from it. This plane meets [tex]MO[/tex] at a point [tex]Q[/tex] with [tex]OQ=x[/tex], [tex]MQ=\frac{a\sqrt{2}}{2}-x[/tex], and meets each apothem at one point, which must be the upper vertex of the cube on it. By symmetry these four points form the upper face of the cube, a square with center [tex]Q[/tex].
Let [tex]P[/tex] be the upper vertex on [tex]MK[/tex]. Then [tex]QP[/tex] is half of the diagonal of a face of the cube, i.e. [tex]QP=\frac{x\sqrt{2}}{2}[/tex]. Since [tex]QP\parallel OK[/tex], the triangles [tex]MQP[/tex] and [tex]MOK[/tex] are similar, so
[tex]\frac{QP}{OK}=\frac{MQ}{MO}\Rightarrow \frac{x\sqrt{2}}{a}=1-\frac{x\sqrt{2}}{a}\Rightarrow \frac{x\sqrt{2}}{a}=\frac{1}{2}\Rightarrow x=\frac{a\sqrt{2}}{4}[/tex]
Therefore the surface area and the volume of the cube are
[tex]S=6x^2=6\times \frac{a^2}{8}=\frac{3a^2}{4}, \quad V=x^3=\frac{2\sqrt{2}a^3}{64}=\frac{\sqrt{2}a^3}{32}[/tex]

Solution:
All base edges equal [tex]3[/tex], so [tex]ABCD[/tex] is a rhombus; let [tex]O[/tex] be the intersection point of its diagonals, so [tex]AC\perp BD[/tex] and [tex]O[/tex] is the midpoint of both. The triangle [tex]SAC[/tex] is isosceles ([tex]SA=SC=3[/tex]), so its median [tex]SO[/tex] is perpendicular to [tex]AC[/tex]. Together with [tex]BO\perp AC[/tex] this gives [tex]AC\perp (SBD)[/tex], so [tex]AO[/tex] and [tex]CO[/tex] are the distances from [tex]A[/tex] and [tex]C[/tex] to the plane [tex](SBD)[/tex]:
[tex]V=V_{ASBD}+V_{CSBD}=\frac{1}{3}S_{SBD}(AO+CO)=\frac{1}{3}AC\times S_{SBD}[/tex]
The right triangles [tex]AOS[/tex] and [tex]AOB[/tex] have a common leg [tex]AO[/tex] and equal hypotenuses [tex]AS=AB=3[/tex], so [tex]SO=BO=DO[/tex]. Thus the median [tex]SO[/tex] of triangle [tex]SBD[/tex] equals half of [tex]BD[/tex], hence [tex]\angle BSD=90^{\circ}[/tex] and
[tex]S_{SBD}=\frac{1}{2}SB\times SD=\frac{3x}{2},\quad BD^2=9+x^2[/tex]
From the right triangle [tex]AOB[/tex]:
[tex]AO^2=AB^2-\frac{BD^2}{4}=9-\frac{9+x^2}{4}=\frac{27-x^2}{4}\Rightarrow AC=\sqrt{27-x^2}[/tex]
This requires [tex]0\lt x\lt 3\sqrt{3}[/tex], and the volume as a function of [tex]x[/tex] is
[tex]V(x)=\frac{1}{3}\sqrt{27-x^2}\times \frac{3x}{2}=\frac{x\sqrt{27-x^2}}{2}[/tex]
By the AM-GM inequality,
[tex]x\sqrt{27-x^2}=\sqrt{x^2(27-x^2)}\le\frac{x^2+(27-x^2)}{2}=\frac{27}{2}[/tex]
with equality for [tex]x^2=27-x^2[/tex], i.e. [tex]x=\frac{3\sqrt{6}}{2}[/tex]. Hence the greatest volume is
[tex]V_{\max}=\frac{1}{2}\times \frac{27}{2}=\frac{27}{4}[/tex]

Solution:
Take [tex]ABC[/tex] as the base:
[tex]S_{ABC}=\frac{1}{2}\times 6\times 5\times \sin 60^{\circ}=\frac{15\sqrt{3}}{2}[/tex]
Let [tex]DH[/tex] be the height of the pyramid ([tex]H[/tex] in the plane [tex]ABC[/tex]) and let [tex]P[/tex] and [tex]Q[/tex] be the feet of the perpendiculars from [tex]D[/tex] to [tex]CA[/tex] and [tex]CB[/tex]. The right triangles [tex]CPD[/tex] and [tex]CQD[/tex] have hypotenuse [tex]CD=4[/tex] and an angle of [tex]45^{\circ}[/tex] at [tex]C[/tex], so they are isosceles and
[tex]CP=CQ=DP=DQ=\frac{4}{\sqrt{2}}=2\sqrt{2}[/tex]
Since [tex]2\sqrt{2}\lt 5[/tex], the point [tex]P[/tex] lies on the segment [tex]CA[/tex] and [tex]Q[/tex] on [tex]CB[/tex]. The segment [tex]HP[/tex] is the orthogonal projection of [tex]DP[/tex] on the plane [tex]ABC[/tex] and [tex]DP\perp CA[/tex], so by the theorem of three perpendiculars [tex]HP\perp CA[/tex]; similarly [tex]HQ\perp CB[/tex].
The right triangles [tex]CPH[/tex] and [tex]CQH[/tex] have the common hypotenuse [tex]CH[/tex] and equal legs [tex]CP=CQ[/tex], so they are congruent and [tex]CH[/tex] is the bisector of [tex]\angle ACB[/tex], i.e. [tex]\angle PCH=30^{\circ}[/tex]. From the right triangle [tex]CPH[/tex]:
[tex]HP=CP\tan 30^{\circ}=\frac{2\sqrt{2}}{\sqrt{3}}[/tex]
From the right triangle [tex]DHP[/tex] (right angle at [tex]H[/tex]):
[tex]DH^2=DP^2-HP^2=8-\frac{8}{3}=\frac{16}{3}\Rightarrow DH=\frac{4}{\sqrt{3}}[/tex]
Therefore the volume is
[tex]V=\frac{1}{3}\times \frac{15\sqrt{3}}{2}\times \frac{4}{\sqrt{3}}=10[/tex]

Solution:
Let [tex]DO[/tex] be the height of the pyramid and [tex]DT\perp AB[/tex] ([tex]T[/tex] on [tex]AB[/tex]). By the theorem of three perpendiculars [tex]OT\perp AB[/tex], so [tex]\angle DTO[/tex] is the angle between the face [tex]ABD[/tex] and the base; the same holds for the other two faces. These three right triangles have the common leg [tex]DO[/tex] and equal acute angles opposite to it, so they are congruent and [tex]O[/tex] is equidistant from the sides of [tex]\triangle ABC[/tex]. Hence [tex]O[/tex] is the incenter of [tex]ABC[/tex] and [tex]T[/tex] is the point where the incircle touches [tex]AB[/tex].
In the right triangle [tex]ABD[/tex] ([tex]\angle ADB=90^{\circ}[/tex]):
[tex]AB=\sqrt{15^2+20^2}=25,\quad DT=\frac{15\times 20}{25}=12,\quad AT=\frac{15^2}{25}=9,\quad BT=\frac{20^2}{25}=16[/tex]
Let the incircle touch [tex]BC[/tex] at [tex]P[/tex] and [tex]CA[/tex] at [tex]Q[/tex]. Tangent segments from a point to a circle are equal, so [tex]AQ=AT=9[/tex], [tex]BP=BT=16[/tex] and
[tex]CP=CQ=\frac{60-2\times 25}{2}=5\Rightarrow BC=21,\quad CA=14[/tex]
With semiperimeter [tex]p=30[/tex], Heron's formula gives the area and then the inradius:
[tex]S_{ABC}=\sqrt{30\times 5\times 9\times 16}=60\sqrt{6},\quad r=\frac{S_{ABC}}{p}=2\sqrt{6}[/tex]
From the right triangles [tex]DOT[/tex], [tex]OQC[/tex] and [tex]DOC[/tex]:
[tex]DO^2=DT^2-OT^2=144-24=120,\quad OC^2=OQ^2+CQ^2=24+25=49[/tex]
[tex]DC^2=DO^2+OC^2=120+49=169\Rightarrow DC=13[/tex]

Solution:
Consider a lateral face [tex]A_iA_{i+1}B_{i+1}B_i[/tex] (indices are taken modulo [tex]n[/tex]). The planes of the bases are parallel, and a third plane cuts two parallel planes along parallel lines, so [tex]A_iA_{i+1}\parallel B_iB_{i+1}[/tex] and the face is a trapezoid.
The four vertices of this face lie on the sphere and in the plane of the face, and a plane cuts a sphere along a circle, so the trapezoid is inscribed in a circle. Since [tex]A_iA_{i+1}\parallel B_iB_{i+1}[/tex] and opposite angles of a cyclic quadrilateral add up to [tex]180^{\circ}[/tex], we get
[tex]\angle A_iB_iB_{i+1}=180^{\circ}-\angle B_iA_iA_{i+1}=\angle A_{i+1}B_{i+1}B_i[/tex]
So the angles at the base [tex]B_iB_{i+1}[/tex] are equal, the trapezoid is isosceles and [tex]A_iB_i=A_{i+1}B_{i+1}[/tex]. Going around all lateral faces, we obtain
[tex]A_1B_1=A_2B_2=\ldots=A_nB_n=l[/tex]
Let [tex]h[/tex] be the distance between the planes of the bases (the height of the frustum). For every [tex]i[/tex] drop [tex]B_iH_i\perp(A_1A_2\ldots A_n)[/tex]; then [tex]B_iH_i=h[/tex] and [tex]A_iH_i[/tex] is the orthogonal projection of [tex]A_iB_i[/tex] on the base, so the angle between the lateral edge [tex]A_iB_i[/tex] and the base is [tex]\varphi_i=\angle B_iA_iH_i[/tex]. From the right triangle [tex]A_iH_iB_i[/tex]
[tex]\sin \varphi_i=\frac{B_iH_i}{A_iB_i}=\frac{h}{l}[/tex]
This number does not depend on [tex]i[/tex], and the angle between a line and a plane is at most [tex]90^{\circ}[/tex], so [tex]\varphi_1=\varphi_2=\ldots=\varphi_n[/tex], which proves the statement.