Cross-Sections, Angles and Distances in Space: Difficult Problems with Solutions

Solution:
Let [tex]D[/tex] be the orthogonal projection of [tex]C[/tex] onto [tex]\alpha[/tex] and [tex]CD=h[/tex]. Then [tex]AD[/tex] and [tex]BD[/tex] are the projections of [tex]AC[/tex] and [tex]BC[/tex], so [tex]\angle CAD=30^{\circ}[/tex] and [tex]\angle CBD=45^{\circ}[/tex]. From the right triangles [tex]ADC[/tex] and [tex]BDC[/tex]:
[tex]AC=\frac{h}{\sin 30^{\circ}}=2h,\quad BC=\frac{h}{\sin 45^{\circ}}=h\sqrt{2}[/tex]
In the triangle [tex]ABC[/tex] ([tex]\angle ACB=90^{\circ}[/tex]) we find the hypotenuse by the Pythagorean theorem and then the altitude [tex]CH[/tex] to it:
[tex]AB=\sqrt{4h^2+2h^2}=h\sqrt{6},\quad CH=\frac{AC\times BC}{AB}=\frac{2h\times h\sqrt{2}}{h\sqrt{6}}=\frac{2h}{\sqrt{3}}[/tex]
The segment [tex]DH[/tex] is the projection of [tex]CH[/tex] onto [tex]\alpha[/tex] and the line [tex]AB[/tex] of [tex]\alpha[/tex] is perpendicular to [tex]CH[/tex], so by the theorem of the three perpendiculars [tex]AB\perp DH[/tex]. Hence [tex]\angle CHD[/tex] is the linear angle of the dihedral angle with edge [tex]AB[/tex], i.e. the required angle. It is an acute angle of the right triangle [tex]CDH[/tex], and
[tex]\sin \angle CHD=\frac{CD}{CH}=\frac{h}{\frac{2h}{\sqrt{3}}}=\frac{\sqrt{3}}{2}\Rightarrow \angle CHD=60^{\circ}[/tex]

Solution:
First we show that the diagonal [tex]B_1D[/tex] of the cube is perpendicular to the plane [tex](ACD_1)[/tex]. Its projection onto [tex](ABCD)[/tex] is [tex]BD[/tex] (as [tex]B_1B\perp(ABCD)[/tex]) and [tex]AC\perp BD[/tex] (diagonals of a square), so by the theorem of the three perpendiculars [tex]AC\perp B_1D[/tex].
Similarly, its projection onto [tex](DCC_1D_1)[/tex] is [tex]C_1D[/tex] (as [tex]B_1C_1\perp(DCC_1D_1)[/tex]) and [tex]CD_1\perp C_1D[/tex], so [tex]CD_1\perp B_1D[/tex]. Thus [tex]B_1D[/tex] is perpendicular to two intersecting lines of the plane, i.e. [tex]B_1D\perp(ACD_1)[/tex].
Let [tex]G[/tex] be the common point of [tex]B_1D[/tex] and the plane; the required distance is [tex]B_1G[/tex]. Since [tex]DA=DC=DD_1=a[/tex], their projections [tex]GA[/tex], [tex]GC[/tex], [tex]GD_1[/tex] on the plane are also equal, so [tex]G[/tex] is the center of the equilateral triangle [tex]ACD_1[/tex] with side [tex]a\sqrt{2}[/tex]. From the right triangle [tex]ADG[/tex]:
[tex]AG=\frac{a\sqrt{2}}{\sqrt{3}}=\frac{a\sqrt{6}}{3},\quad DG=\sqrt{AD^2-AG^2}=\sqrt{a^2-\frac{2a^2}{3}}=\frac{a\sqrt{3}}{3}[/tex]
The point [tex]G[/tex] lies inside the cube, hence on the segment [tex]DB_1[/tex], whose length is [tex]a\sqrt{3}[/tex]. Therefore
[tex]B_1G=B_1D-DG=a\sqrt{3}-\frac{a\sqrt{3}}{3}=\frac{2a\sqrt{3}}{3}[/tex]

Solution:
Let [tex]M[/tex] be the midpoint of [tex]AB[/tex], so [tex]AM=2\sqrt{6}[/tex]. The triangles [tex]ABC[/tex] and [tex]ABD[/tex] are isosceles with base [tex]AB[/tex], hence [tex]CM\perp AB[/tex] and [tex]DM\perp AB[/tex]. From the right triangles [tex]AMC[/tex] and [tex]AMD[/tex]:
[tex]CM=\sqrt{49-24}=5,\qquad DM=\sqrt{40-24}=4[/tex]
The line [tex]AB[/tex] is perpendicular to two intersecting lines of the plane [tex](CDM)[/tex], so [tex]AB\perp (CDM)[/tex]. In triangle [tex]CDM[/tex]:
[tex]CD^2+DM^2=9+16=25=CM^2\Rightarrow \angle CDM=90^{\circ}[/tex]
Let [tex]DH[/tex] be the altitude of the right triangle [tex]CDM[/tex] ([tex]H\in CM[/tex]). Then [tex]DH\perp CM[/tex], and [tex]DH\perp AB[/tex] because [tex]DH[/tex] lies in the plane [tex](CDM)[/tex]. So [tex]DH[/tex] is perpendicular to two intersecting lines of [tex](ABC)[/tex], i.e. [tex]DH\perp (ABC)[/tex] and [tex]DH[/tex] is the required perpendicular:
[tex]DH=\frac{CD\times DM}{CM}=\frac{3\times 4}{5}=\frac{12}{5}[/tex]

Solution:
The lateral faces are equilateral, so all edges of the pyramid are equal; let their length be [tex]a[/tex]. In triangle [tex]AVC[/tex] we have [tex]VA=VC=a[/tex] and [tex]AC=a\sqrt{2}[/tex] (the diagonal of the square base), so [tex]VA^2+VC^2=AC^2[/tex] and [tex]\angle AVC=90^{\circ}[/tex].
Let [tex]Q[/tex] be the midpoint of [tex]AN[/tex]. Then [tex]PQ[/tex] is a midline of triangle [tex]ABN[/tex], so [tex]PQ\parallel BN[/tex] and [tex]PQ=\frac{BN}{2}[/tex]. Hence the required angle is the angle between the lines [tex]VP[/tex] and [tex]PQ[/tex]; we find the sides of triangle [tex]VPQ[/tex].
The segments [tex]VP[/tex] and [tex]BN[/tex] are medians of the equilateral triangles [tex]VAB[/tex] and [tex]VBC[/tex], so
[tex]VP=BN=\frac{a\sqrt{3}}{2},\qquad PQ=\frac{BN}{2}=\frac{a\sqrt{3}}{4}[/tex]
The triangle [tex]AVN[/tex] is right-angled at [tex]V[/tex] with [tex]VN=\frac{a}{2}[/tex], and [tex]VQ[/tex] is the median to its hypotenuse, so it equals half of the hypotenuse:
[tex]AN=\sqrt{a^2+\frac{a^2}{4}}=\frac{a\sqrt{5}}{2},\qquad VQ=\frac{AN}{2}=\frac{a\sqrt{5}}{4}[/tex]
By the law of cosines in triangle [tex]VPQ[/tex]:
[tex]\cos\angle VPQ=\frac{VP^2+PQ^2-VQ^2}{2\times VP\times PQ}=\frac{\frac{3a^2}{4}+\frac{3a^2}{16}-\frac{5a^2}{16}}{2\times \frac{a\sqrt{3}}{2}\times \frac{a\sqrt{3}}{4}}=\frac{\frac{5a^2}{8}}{\frac{3a^2}{4}}=\frac{5}{6}[/tex]
The cosine is positive, so [tex]\angle VPQ[/tex] is acute and it is the angle between the lines [tex]VP[/tex] and [tex]BN[/tex]. The answer is [tex]\frac{5}{6}[/tex].

Solution:
First, [tex]A[/tex] and [tex]B[/tex] lie on the same side of [tex]\alpha[/tex]. Otherwise the segment [tex]AB[/tex] would cross [tex]\alpha[/tex] and [tex]AB\ge 6+2=8[/tex], with equality only if [tex]AB\perp\alpha[/tex]; then [tex]A[/tex] and [tex]B[/tex] would have the same projection [tex]A'[/tex] on [tex]\alpha[/tex], and [tex]CA^2=CA'^2+36\ne CA'^2+4=CB^2[/tex], a contradiction.
Let [tex]H[/tex] be the midpoint of [tex]AB[/tex]; then [tex]CH[/tex] is the altitude, median and angle bisector of the isosceles triangle. Let [tex]A'[/tex], [tex]B'[/tex], [tex]H'[/tex] be the projections of [tex]A[/tex], [tex]B[/tex], [tex]H[/tex] on [tex]\alpha[/tex]. The quadrilateral [tex]AA'B'B[/tex] is a trapezoid ([tex]AA'\parallel BB'[/tex]) with midline [tex]HH'[/tex]:
[tex]HH'=\frac{AA'+BB'}{2}=\frac{6+2}{2}=4[/tex]
The incenter [tex]I[/tex] lies on the bisector [tex]CH[/tex]. In triangle [tex]ACH[/tex] the segment [tex]AI[/tex] bisects the angle [tex]CAH[/tex], so by the angle bisector theorem
[tex]\frac{CI}{IH}=\frac{AC}{AH}=\frac{12}{4}=3\Rightarrow CI=\frac{3}{4}CH[/tex]
Let [tex]I'[/tex] be the projection of [tex]I[/tex] on [tex]\alpha[/tex]. Since [tex]C\in\alpha[/tex] and [tex]II'\parallel HH'[/tex], the right triangles [tex]CII'[/tex] and [tex]CHH'[/tex] are similar, so
[tex]II'=\frac{CI}{CH}\times HH'=\frac{3}{4}\times 4=3[/tex]

Solution:
The pyramid is regular, so the foot of its height is the center [tex]G[/tex] of the equilateral triangle [tex]ABC[/tex]: [tex]MG\perp (ABC)[/tex], [tex]MG=28[/tex]. Let [tex]Q[/tex] be the midpoint of [tex]AC[/tex]; the median [tex]BQ[/tex] is also an altitude of [tex]\triangle ABC[/tex] and passes through [tex]G[/tex]:
[tex]BQ=\frac{7\sqrt{3}}{2},\quad BG=\frac{2}{3}BQ=\frac{7\sqrt{3}}{3}[/tex]
[tex]AC\perp BQ[/tex] and [tex]AC\perp MQ[/tex] ([tex]MQ[/tex] is a median of the isosceles triangle [tex]ACM[/tex]), so [tex]AC\perp (BQM)[/tex]. In the plane [tex](BQM)[/tex] draw [tex]QQ_1\perp BM[/tex], [tex]Q_1\in BM[/tex]. Then [tex]BM[/tex] is perpendicular to the intersecting lines [tex]AC[/tex] and [tex]QQ_1[/tex], so the section is the triangle [tex]ACQ_1[/tex], and its altitude is [tex]QQ_1[/tex] (since [tex]AC\perp (BQM)[/tex] gives [tex]AC\perp QQ_1[/tex]).
The point [tex]Q_1[/tex] lies on the segment [tex]BM[/tex], because the angles of [tex]\triangle BQM[/tex] at [tex]B[/tex] and [tex]M[/tex] are acute: [tex]\angle QBM[/tex] is an acute angle of the right triangle [tex]BGM[/tex], and [tex]\angle BMQ[/tex] lies opposite the shortest side [tex]BQ[/tex]. From the right triangle [tex]BGM[/tex]:
[tex]BM=\sqrt{BG^2+MG^2}=\sqrt{\frac{49}{3}+784}=\sqrt{\frac{2401}{3}}=\frac{49\sqrt{3}}{3}[/tex]
[tex]MG[/tex] is the altitude of [tex]\triangle BQM[/tex] from [tex]M[/tex] (as [tex]G\in BQ[/tex]), so computing the area of [tex]\triangle BQM[/tex] in two ways, [tex]\frac{1}{2}BQ\times MG=\frac{1}{2}BM\times QQ_1[/tex], we get
[tex]QQ_1=\frac{BQ\times MG}{BM}=\frac{\frac{7\sqrt{3}}{2}\times 28}{\frac{49\sqrt{3}}{3}}=6[/tex]
Hence the area of the section is [tex]S=\frac{1}{2}AC\times QQ_1=\frac{1}{2}\times 7\times 6=21[/tex] square inches.

Solution:
Since [tex]MC\perp (ABCD)[/tex], the triangles [tex]BCM[/tex] and [tex]ACM[/tex] are right-angled at [tex]C[/tex]. By the Pythagorean theorem
[tex]BC=\sqrt{BM^2-MC^2}=\sqrt{3},\quad AC=\sqrt{AM^2-MC^2}=2\sqrt{2}[/tex]
Let [tex]O=AC\cap BD[/tex]. The diagonals of a rhombus bisect each other and are perpendicular, so [tex]AO=\sqrt{2}[/tex] and [tex]BD\perp AC[/tex]. Also [tex]BD\perp MC[/tex], because [tex]MC[/tex] is perpendicular to the base. Hence [tex]BD\perp (ACM)[/tex], and [tex]BD[/tex] meets this plane at [tex]O[/tex].
In the plane [tex](ACM)[/tex] draw [tex]OP\perp AM[/tex], [tex]P\in AM[/tex] ([tex]P[/tex] lies on the segment [tex]AM[/tex], since [tex]\angle MAC[/tex] is acute). As [tex]OP[/tex] lies in the plane [tex](ACM)[/tex], also [tex]OP\perp BD[/tex]. Thus [tex]OP[/tex] is the common perpendicular of [tex]AM[/tex] and [tex]BD[/tex]. From the right triangles [tex]ACM[/tex] and [tex]AOP[/tex]:
[tex]\sin \angle MAC=\frac{MC}{AM}=\frac{1}{3},\quad OP=AO\sin \angle MAC=\frac{\sqrt{2}}{3}[/tex]

Solution:
Let [tex]P[/tex] be the midpoint of [tex]AB[/tex], so [tex]AP=BP=8[/tex]. The triangle [tex]ABC[/tex] is isosceles with base [tex]AB[/tex], so its median [tex]CP[/tex] is also an altitude: [tex]CP\perp AB[/tex]. From the right triangle [tex]APC[/tex]:
[tex]CP=\sqrt{17^2-8^2}=15[/tex]
The triangle [tex]ABM[/tex] is isosceles and right-angled at [tex]M[/tex], so its median [tex]MP[/tex] is also an altitude, [tex]MP\perp AB[/tex], and, as the median to the hypotenuse,
[tex]MP=\frac{AB}{2}=8[/tex]
Since [tex]CP\perp AB[/tex] and [tex]MP\perp AB[/tex], the angle [tex]CPM[/tex] is the linear angle of the dihedral angle with edge [tex]AB[/tex]. By the law of cosines in the triangle [tex]CPM[/tex]:
[tex]\cos\angle CPM=\frac{CP^2+MP^2-CM^2}{2\times CP\times MP}=\frac{225+64-169}{2\times 15\times 8}=\frac{120}{240}=\frac{1}{2}[/tex]
Hence [tex]\angle CPM=60^{\circ}[/tex]; this angle is acute, so the angle between the planes is [tex]60^{\circ}[/tex].

Solution:
The edge [tex]AB[/tex] is perpendicular to [tex]BC[/tex] and to [tex]BB_1[/tex], so [tex]AB\perp(BCC_1B_1)[/tex]. In the face [tex]BCC_1B_1[/tex] draw [tex]CH\perp BC_1[/tex], [tex]H\in BC_1[/tex]. Then also [tex]CH\perp AB[/tex], so [tex]CH[/tex] is perpendicular to the two intersecting lines [tex]AB[/tex] and [tex]BC_1[/tex] of the plane [tex](ABC_1D_1)[/tex], i.e. [tex]CH\perp(ABC_1D_1)[/tex].
Hence [tex]D_1H[/tex] is the orthogonal projection of [tex]D_1C[/tex] on this plane, the required angle is [tex]\varphi=\angle CD_1H[/tex] and [tex]\sin \varphi=\frac{CH}{CD_1}[/tex]. From the right triangles [tex]BCC_1[/tex] and [tex]CDD_1[/tex]:
[tex]BC_1=CD_1=\sqrt{\left(2\sqrt{2}\right)^2+\left(2\sqrt{7}\right)^2}=\sqrt{36}=6[/tex]
[tex]CH=\frac{BC\times CC_1}{BC_1}=\frac{2\sqrt{2}\times 2\sqrt{7}}{6}=\frac{2\sqrt{14}}{3}[/tex]
Therefore
[tex]\sin \varphi=\frac{CH}{CD_1}=\frac{2\sqrt{14}}{3\times 6}=\frac{\sqrt{14}}{9}[/tex]

Solution:
Since [tex]AB\perp AD[/tex] and [tex]AB\perp DQ[/tex], we get [tex]AB\perp(ADQ)[/tex], hence [tex]AB\perp AQ[/tex]; similarly [tex]CB\perp CD[/tex] and [tex]CB\perp DQ[/tex] give [tex]CB\perp CQ[/tex]. By the Pythagorean theorem:
[tex]AQ=\sqrt{AD^2+DQ^2}=\sqrt{3}, \quad CQ=\sqrt{CD^2+DQ^2}=2, \quad QB=\sqrt{BC^2+CQ^2}=\sqrt{6}[/tex]
In the right triangle [tex]ABQ[/tex] we have [tex]\tan \angle ABQ=\frac{AQ}{AB}=1[/tex], so [tex]\angle ABQ=45^{\circ}[/tex]. In the right triangle [tex]BCQ[/tex] draw the height [tex]CL[/tex] to the hypotenuse [tex]QB[/tex]:
[tex]CL=\frac{BC\times CQ}{QB}=\frac{2\sqrt{2}}{\sqrt{6}}=\frac{2\sqrt{3}}{3}, \quad BL=\frac{BC^2}{QB}=\frac{2}{\sqrt{6}}=\frac{\sqrt{6}}{3}[/tex]
In the plane [tex](ABQ)[/tex] draw through [tex]L[/tex] the perpendicular to [tex]QB[/tex] and let it meet [tex]AB[/tex] at [tex]N[/tex]. The triangle [tex]BLN[/tex] is right-angled at [tex]L[/tex] with [tex]\angle B=45^{\circ}[/tex], so [tex]LN=BL=\frac{\sqrt{6}}{3}[/tex] and [tex]BN=BL\sqrt{2}=\frac{2\sqrt{3}}{3}\lt AB[/tex], i.e. [tex]N[/tex] lies on the edge [tex]AB[/tex].
Since [tex]CL\perp QB[/tex] and [tex]NL\perp QB[/tex], the angle [tex]CLN[/tex] is the linear angle of the dihedral angle at the edge [tex]QB[/tex]. As [tex]CB\perp AB[/tex], from the right triangle [tex]CBN[/tex]:
[tex]CN^2=BC^2+BN^2=2+\frac{4}{3}=\frac{10}{3}[/tex]
By the law of cosines in the triangle [tex]CLN[/tex]:
[tex]\cos\angle CLN=\frac{CL^2+LN^2-CN^2}{2\times CL\times LN}=\frac{\frac{4}{3}+\frac{2}{3}-\frac{10}{3}}{\frac{4\sqrt{2}}{3}}=-\frac{\sqrt{2}}{2}[/tex]
Hence the dihedral angle is [tex]135^{\circ}[/tex].

Solution:
Let the triangle be [tex]ABC[/tex] with [tex]BC=14[/tex], [tex]CA=30[/tex], [tex]AB=40[/tex], and let [tex]O[/tex] be the orthogonal projection of [tex]M[/tex] on the plane [tex](ABC)[/tex]; the required distance is [tex]MO[/tex]. The right triangles [tex]MOA[/tex], [tex]MOB[/tex], [tex]MOC[/tex] have the common leg [tex]MO[/tex] and equal hypotenuses [tex]MA=MB=MC=65[/tex], so they are congruent and [tex]OA=OB=OC[/tex]. Hence [tex]O[/tex] is the circumcenter of [tex]\triangle ABC[/tex] and [tex]OA=R[/tex] is its circumradius.
By Heron's formula with semiperimeter [tex]p=42[/tex], and then by the formula for the circumradius:
[tex]S=\sqrt{42\times 28\times 12\times 2}=\sqrt{28224}=168[/tex]
[tex]R=\frac{BC\times CA\times AB}{4S}=\frac{14\times 30\times 40}{4\times 168}=25[/tex]
(The triangle is obtuse, since [tex]40^2\gt 14^2+30^2[/tex], so [tex]O[/tex] lies outside it, but this does not change the computation.) From the right triangle [tex]MOA[/tex]:
[tex]MO=\sqrt{MA^2-OA^2}=\sqrt{65^2-25^2}=\sqrt{3600}=60[/tex]

Solution:
Let the edge of the pyramid be [tex]a[/tex]. A plane parallel to [tex](BCE)[/tex] cuts every face along a line parallel to the common line of that face with [tex](BCE)[/tex]. In the base it cuts along [tex]QR\parallel BC[/tex], [tex]R\in CD[/tex], so [tex]QR=a[/tex] and [tex]DR=AQ=3[/tex].
In the face [tex]ABE[/tex] it cuts along [tex]QS\parallel BE[/tex], [tex]S\in AE[/tex]. The triangle [tex]ABE[/tex] is equilateral, hence so is [tex]AQS[/tex], and [tex]AS=QS=3[/tex]. In the same way, in the face [tex]CDE[/tex] it cuts along [tex]RT\parallel CE[/tex], [tex]T\in DE[/tex], with [tex]DT=RT=3[/tex].
Since [tex]ES=ET=a-3[/tex], the triangle [tex]EST[/tex] is similar to the equilateral triangle [tex]EAD[/tex], so [tex]ST\parallel AD[/tex] and [tex]ST=a-3[/tex]. Thus the section [tex]QRTS[/tex] is an isosceles trapezoid with bases [tex]QR=a[/tex], [tex]ST=a-3[/tex] and legs [tex]QS=RT=3[/tex].
The projection of a leg on the longer base is [tex]\frac{a-(a-3)}{2}=\frac{3}{2}[/tex], so the height of the trapezoid is [tex]\sqrt{9-\frac{9}{4}}=\frac{3\sqrt{3}}{2}[/tex], and the projection of the diagonal [tex]QT[/tex] on [tex]QR[/tex] is [tex]a-\frac{3}{2}[/tex]. By the Pythagorean theorem:
[tex]QT^2=\left(a-\frac{3}{2}\right)^2+\frac{27}{4}=49\Rightarrow \left(a-\frac{3}{2}\right)^2=\frac{169}{4}\Rightarrow a-\frac{3}{2}=\frac{13}{2}\Rightarrow a=8[/tex]
The perimeter of the section is
[tex]QR+RT+TS+SQ=8+3+5+3=19[/tex]

Solution:
Let [tex]P[/tex] be the center of the box, the common midpoint of its space diagonals. The points [tex]M[/tex] and [tex]N[/tex] are symmetric with respect to [tex]P[/tex], so [tex]P[/tex] is the midpoint of [tex]MN[/tex]. Hence the line [tex]D_1P[/tex] lies in the plane [tex](MND_1)[/tex], and so does the other end [tex]B[/tex] of the diagonal [tex]D_1B[/tex].
So the section is the quadrilateral [tex]MBND_1[/tex]. Its opposite sides lie in pairs of parallel faces, so [tex]MB\parallel D_1N[/tex] and [tex]BN\parallel MD_1[/tex], i.e. it is a parallelogram. Its orthogonal projection on the base plane is the rectangle [tex]ABCD[/tex] with area [tex]8\times 4=32[/tex].
To find the angle [tex]\varphi[/tex] between the section plane and the base, let the line [tex]D_1M[/tex] meet the line [tex]DA[/tex] at [tex]X[/tex]. Since [tex]AM\parallel DD_1[/tex] and [tex]AM=\frac{1}{2}DD_1[/tex], the segment [tex]AM[/tex] is a midline of the triangle [tex]XDD_1[/tex], so [tex]XD=2AD=8[/tex]. In the same way the line [tex]D_1N[/tex] meets the line [tex]DC[/tex] at [tex]Y[/tex] with [tex]DY=2DC=16[/tex].
The line [tex]XY[/tex] is the common line of the section plane and the base plane. Let [tex]DH[/tex] be the altitude of the right triangle [tex]XDY[/tex] to the hypotenuse [tex]XY[/tex]:
[tex]XY=\sqrt{8^2+16^2}=8\sqrt{5},\qquad DH=\frac{XD\times DY}{XY}=\frac{128}{8\sqrt{5}}=\frac{16\sqrt{5}}{5}[/tex]
By the theorem of the three perpendiculars [tex]D_1H\perp XY[/tex], so [tex]\varphi=\angle D_1HD[/tex]. From the right triangle [tex]D_1DH[/tex]:
[tex]\tan \varphi=\frac{DD_1}{DH}=\frac{12\sqrt{5}/5}{16\sqrt{5}/5}=\frac{3}{4}\Rightarrow \cos\varphi=\frac{1}{\sqrt{1+\tan ^2\varphi}}=\frac{4}{5}[/tex]
By the theorem for the area of an orthogonal projection:
[tex]S=\frac{S_{ABCD}}{\cos\varphi}=\frac{32}{\frac{4}{5}}=40[/tex]

Solution:
Let [tex]O[/tex] be the center of the square [tex]ABCD[/tex], so [tex]MO\perp(ABCD)[/tex], and let [tex]L[/tex] be the midpoint of [tex]PQ[/tex]. The lateral faces [tex]ABM[/tex] and [tex]BCM[/tex] are congruent isosceles triangles, so their medians [tex]MP[/tex] and [tex]MQ[/tex] are equal. Hence the triangle [tex]PQM[/tex] is isosceles and its median [tex]ML[/tex] is the altitude through [tex]M[/tex]:
[tex]ML\perp PQ[/tex]
The segment [tex]PQ[/tex] is a midline of the triangle [tex]ABC[/tex], so [tex]PQ\parallel AC[/tex], and the median [tex]BO[/tex] of this triangle bisects [tex]PQ[/tex]; thus [tex]L[/tex] lies on [tex]BO[/tex], i.e. on the diagonal [tex]BD[/tex].
Moreover [tex]AC\perp BD[/tex] (diagonals of a square) and [tex]AC\perp MO[/tex], so [tex]AC\perp(BDM)[/tex] and therefore
[tex]PQ\perp(BDM)[/tex]
The points [tex]M[/tex] and [tex]L[/tex] lie in both planes [tex](PQM)[/tex] and [tex](BDM)[/tex], so [tex]ML[/tex] is their common line. In the plane [tex](BDM)[/tex] draw [tex]DF\perp ML[/tex], [tex]F\in ML[/tex]. Since [tex]DF[/tex] lies in [tex](BDM)[/tex], also [tex]DF\perp PQ[/tex].
Thus [tex]DF[/tex] is perpendicular to the two intersecting lines [tex]ML[/tex] and [tex]PQ[/tex] of the plane [tex](PQM)[/tex], so [tex]DF\perp(PQM)[/tex]. Hence the foot [tex]F[/tex] of the perpendicular from [tex]D[/tex] to [tex](PQM)[/tex] lies on the line [tex]ML[/tex], which contains the altitude through [tex]M[/tex] of the triangle [tex]PQM[/tex].

Solution:
We have [tex]AB\perp BC[/tex] and [tex]AB\perp BB_1[/tex], and these two lines intersect, so [tex]AB\perp (BCC_1B_1)[/tex]. Hence [tex]AB[/tex] is perpendicular to every line of this face; in particular [tex]AB\perp B_1C[/tex].
Also [tex]B_1C\perp BC_1[/tex], because they are the diagonals of the square [tex]BCC_1B_1[/tex].
Thus [tex]B_1C[/tex] is perpendicular to the two intersecting lines [tex]AB[/tex] and [tex]BC_1[/tex] of the plane [tex](ABC_1)[/tex], so [tex]B_1C\perp (ABC_1)[/tex].
The plane [tex](AB_1C)[/tex] contains the line [tex]B_1C[/tex], which is perpendicular to the plane [tex](ABC_1)[/tex], and a plane through a line perpendicular to another plane is perpendicular to that plane. Hence [tex](AB_1C)\perp (ABC_1)[/tex].

Solution:
All three angles are angles between lines, so they lie in [tex][0^{\circ}, 90^{\circ}][/tex]. Take a point [tex]P\ne O[/tex] on [tex]p[/tex] and let [tex]Q[/tex] be its orthogonal projection on [tex]\delta[/tex]. Then [tex]Q\in p'[/tex], and [tex]Q\ne O[/tex] because [tex]\alpha \ne 90^{\circ}[/tex]. In the right triangle [tex]OQP[/tex] ([tex]\angle OQP=90^{\circ}[/tex]) we have [tex]\angle POQ=\alpha[/tex], so
[tex]\cos \alpha =\frac{OQ}{OP}[/tex]
If [tex]\beta =0^{\circ}[/tex], then [tex]q=p'[/tex], [tex]\gamma =\alpha[/tex] and the equality is obvious. If [tex]\beta =90^{\circ}[/tex], then [tex]q\perp OQ[/tex], so by the theorem of three perpendiculars [tex]q\perp OP[/tex], i.e. [tex]\gamma =90^{\circ}[/tex] and both sides equal [tex]0[/tex].
Now let [tex]0^{\circ}\lt \beta \lt 90^{\circ}[/tex]. Draw [tex]QM\perp q[/tex], [tex]M\in q[/tex]. Then [tex]M\ne O[/tex] (as [tex]\beta \ne 90^{\circ}[/tex]), and the angle [tex]QOM[/tex] of the right triangle [tex]OMQ[/tex] is acute, so [tex]\angle QOM=\beta[/tex] and
[tex]\cos \beta =\frac{OM}{OQ}[/tex]
Since [tex]PQ\perp \delta[/tex] and [tex]QM\perp q[/tex], by the theorem of three perpendiculars [tex]PM\perp q[/tex]. Hence the triangle [tex]OMP[/tex] is right-angled at [tex]M[/tex]; its angle [tex]POM[/tex] is acute and therefore equals the angle [tex]\gamma[/tex] between [tex]p[/tex] and [tex]q[/tex], so [tex]\cos \gamma =\frac{OM}{OP}[/tex]. Multiplying the two equalities above:
[tex]\cos \alpha \times \cos \beta =\frac{OQ}{OP}\times \frac{OM}{OQ}=\frac{OM}{OP}=\cos \gamma[/tex]

Solution:
[tex]MN[/tex] is a midline of [tex]\triangle ABC[/tex] and [tex]PQ[/tex] is a midline of [tex]\triangle ACD[/tex], so
[tex]MN\parallel AC\parallel PQ[/tex]
[tex]Q[/tex] and [tex]N[/tex] are the midpoints of the opposite sides [tex]AD[/tex] and [tex]BC[/tex] of the square [tex]ABCD[/tex], so [tex]AQ\parallel BN[/tex], [tex]AQ=BN[/tex], [tex]ABNQ[/tex] is a parallelogram, and [tex]QN\parallel AB[/tex], [tex]QN=AB[/tex]. Since also [tex]A_1B_1\parallel AB[/tex] and [tex]A_1B_1=AB[/tex], we get [tex]QN\parallel A_1B_1[/tex] and [tex]QN=A_1B_1[/tex]. Hence [tex]QNB_1A_1[/tex] is a parallelogram and
[tex]NB_1\parallel QA_1[/tex]
Thus the intersecting lines [tex]MN[/tex] and [tex]NB_1[/tex] of the plane [tex](MNB_1)[/tex] are parallel to the intersecting lines [tex]PQ[/tex] and [tex]QA_1[/tex] of the plane [tex](PQA_1)[/tex], so by the criterion for parallel planes the two planes are parallel or coincide.
They do not coincide: since [tex]A_1\notin (ABCD)[/tex], the plane [tex](PQA_1)[/tex] meets the base plane [tex](ABCD)[/tex] only along the line [tex]PQ[/tex], while the point [tex]N\in (ABCD)[/tex] is not on [tex]PQ[/tex], because [tex]MN[/tex] and [tex]PQ[/tex] are different parallel lines (they lie on different sides of [tex]AC[/tex]). Therefore [tex](MNB_1)\parallel (PQA_1)[/tex].

Solution:
Let [tex]\pi[/tex] be the cutting plane. Since [tex]AB\parallel\pi[/tex] and [tex]CD\parallel\pi[/tex], the points [tex]A[/tex] and [tex]B[/tex] lie on one side of [tex]\pi[/tex], and so do [tex]C[/tex] and [tex]D[/tex]; as [tex]\pi[/tex] passes through interior points, these sides are different. Hence [tex]\pi[/tex] meets the edges [tex]AC[/tex], [tex]BC[/tex], [tex]BD[/tex], [tex]AD[/tex] at points [tex]K[/tex], [tex]L[/tex], [tex]M[/tex], [tex]N[/tex].
Since [tex]AB\parallel\pi[/tex], the plane [tex](ABC)[/tex] meets [tex]\pi[/tex] along a line parallel to [tex]AB[/tex], i.e. [tex]KL\parallel AB[/tex]; in the same way the plane [tex](ABD)[/tex] gives [tex]NM\parallel AB[/tex]. Since [tex]CD\parallel\pi[/tex], the planes [tex](ACD)[/tex] and [tex](BCD)[/tex] give [tex]KN\parallel CD[/tex] and [tex]LM\parallel CD[/tex]. Hence [tex]KLMN[/tex] is a parallelogram.
Let [tex]E[/tex] be the midpoint of [tex]CD[/tex]. The medians [tex]AE[/tex] and [tex]BE[/tex] of the equilateral triangles [tex]ACD[/tex] and [tex]BCD[/tex] are perpendicular to [tex]CD[/tex], so [tex]CD\perp(ABE)[/tex] and therefore [tex]CD\perp AB[/tex]. Since [tex]KL\parallel AB[/tex] and [tex]KN\parallel CD[/tex], the angle [tex]LKN[/tex] is right, so [tex]KLMN[/tex] is a rectangle.

Solution:
Let [tex]\pi[/tex] be the cutting plane and [tex]a[/tex] the edge of the tetrahedron. Since [tex]AB\parallel\pi[/tex] and [tex]CD\parallel\pi[/tex], the points [tex]A[/tex] and [tex]B[/tex] lie on one side of [tex]\pi[/tex], and so do [tex]C[/tex] and [tex]D[/tex]; as [tex]\pi[/tex] passes through interior points, these sides are different. Hence [tex]\pi[/tex] meets the edges [tex]AC[/tex], [tex]BC[/tex], [tex]BD[/tex], [tex]AD[/tex] at points [tex]K[/tex], [tex]L[/tex], [tex]M[/tex], [tex]N[/tex].
Since [tex]AB\parallel\pi[/tex], the plane [tex](ABC)[/tex] meets [tex]\pi[/tex] along a line parallel to [tex]AB[/tex], i.e. [tex]KL\parallel AB[/tex]; in the same way the plane [tex](ABD)[/tex] gives [tex]NM\parallel AB[/tex]. Since [tex]CD\parallel\pi[/tex], the planes [tex](ACD)[/tex] and [tex](BCD)[/tex] give [tex]KN\parallel CD[/tex] and [tex]LM\parallel CD[/tex]. Hence the cross-section [tex]KLMN[/tex] is a parallelogram, so its opposite sides are equal.
Let [tex]\frac{AK}{AC}=t[/tex]. From the similar triangles [tex]CKL[/tex] and [tex]CAB[/tex], and [tex]AKN[/tex] and [tex]ACD[/tex], we get
[tex]KL=a\times \frac{CK}{CA}=a(1-t), \quad KN=a\times \frac{AK}{AC}=at[/tex]
Hence the perimeter of the cross-section is
[tex]2(KL+KN)=2\left(a(1-t)+at\right)=2a[/tex]
which does not depend on the position of the cutting plane.

Solution:
Let [tex]AA_1=h[/tex]. Then [tex]AB=h\sqrt{2}[/tex], [tex]AC=AB\sqrt{2}=2h[/tex] and [tex]AK=\frac{AC}{4}=\frac{h}{2}[/tex].
First, [tex]BD\perp AC[/tex] (diagonals of a square) and [tex]BD\perp AA_1[/tex] (the prism is right), so [tex]BD\perp(ACC_1A_1)[/tex]. The line [tex]A_1K[/tex] lies in this plane, hence [tex]A_1K\perp BD[/tex].
Let [tex]O[/tex] be the midpoint of [tex]AC_1[/tex] (the center of the prism) and [tex]O'[/tex] the center of the base. The points [tex]O[/tex] and [tex]M[/tex] lie at height [tex]\frac{h}{2}[/tex] above [tex]O'[/tex] and [tex]D[/tex], respectively, so [tex]DO'OM[/tex] is a rectangle and [tex]MO\parallel DO'[/tex], i.e. [tex]MO\parallel BD[/tex]. The line [tex]MO[/tex] lies in the plane [tex](AMC_1)[/tex], and [tex]A_1K\perp MO[/tex].
Second, in the rectangle [tex]ACC_1A_1[/tex] ([tex]AC=2h[/tex], [tex]CC_1=h[/tex]) consider the right triangles [tex]AA_1K[/tex] and [tex]ACC_1[/tex]:
[tex]\tan \angle AA_1K=\frac{AK}{AA_1}=\frac{1}{2}=\frac{CC_1}{AC}=\tan \angle CAC_1\Rightarrow \angle AA_1K=\angle CAC_1[/tex]
Let [tex]P=A_1K\cap AC_1[/tex]. Since [tex]\angle A_1AC=90^{\circ}[/tex], we have [tex]\angle PAA_1=90^{\circ}-\angle CAC_1[/tex], so in the triangle [tex]AA_1P[/tex]:
[tex]\angle PA_1A+\angle PAA_1=\angle CAC_1+\left(90^{\circ}-\angle CAC_1\right)=90^{\circ}\Rightarrow \angle A_1PA=90^{\circ}[/tex]
Hence [tex]A_1K\perp AC_1[/tex]. The lines [tex]MO[/tex] and [tex]AC_1[/tex] lie in the plane [tex](AMC_1)[/tex] and intersect at [tex]O[/tex], and the line [tex]A_1K[/tex] is perpendicular to both of them; therefore [tex]A_1K\perp(AMC_1)[/tex].