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

Solution:
The perpendiculars [tex]AA_1[/tex], [tex]MM_1[/tex] and [tex]BB_1[/tex] to [tex]\alpha[/tex] are parallel, so they lie in one plane, which meets [tex]\alpha[/tex] along the line [tex]A_1B_1[/tex]; hence [tex]M_1[/tex] lies on [tex]A_1B_1[/tex]. Since [tex]AB[/tex] does not meet [tex]\alpha[/tex], the points [tex]A[/tex] and [tex]B[/tex] are on the same side of [tex]\alpha[/tex], so [tex]AA_1B_1B[/tex] is a right trapezoid.
Through [tex]A[/tex] draw a line parallel to [tex]A_1B_1[/tex]; let it meet [tex]MM_1[/tex] at [tex]K[/tex] and [tex]BB_1[/tex] at [tex]L[/tex]. Then [tex]AA_1M_1K[/tex] and [tex]AA_1B_1L[/tex] are rectangles, so
[tex]KM_1=LB_1=AA_1=4,\quad BL=BB_1-LB_1=9-4=5[/tex]
Since [tex]KM\parallel LB[/tex], the triangles [tex]AKM[/tex] and [tex]ALB[/tex] are similar:
[tex]\frac{MK}{BL}=\frac{AM}{AB}=\frac{2}{5}\Rightarrow MK=\frac{2}{5}\times 5=2[/tex]
Therefore [tex]MM_1=MK+KM_1=2+4=6[/tex] in.

Solution:
In the equilateral triangles [tex]ABC[/tex] and [tex]DBC[/tex] the medians [tex]AN[/tex] and [tex]DN[/tex] are also altitudes, so [tex]BC\perp AN[/tex] and [tex]BC\perp DN[/tex]; hence [tex]BC\perp(AND)[/tex].
Therefore the angle [tex]\varphi[/tex] between [tex]BM[/tex] and the plane [tex](AND)[/tex] is complementary to the angle [tex]MBC[/tex] between [tex]BM[/tex] and the line [tex]BC[/tex], which is perpendicular to this plane:
[tex]\varphi=90^{\circ}-\angle MBC\Rightarrow \sin \varphi=\cos\angle MBC[/tex]
The tetrahedron is regular, so the foot of its height from [tex]B[/tex] is the center [tex]M[/tex] of the face [tex]ACD[/tex]: [tex]BM\perp(ACD)[/tex] and the triangle [tex]BMC[/tex] is right-angled at [tex]M[/tex]. Here [tex]CM[/tex] is the circumradius of the equilateral triangle [tex]ACD[/tex] with side [tex]1[/tex]:
[tex]CM=\frac{\sqrt{3}}{3},\quad BM=\sqrt{BC^2-CM^2}=\sqrt{1-\frac{1}{3}}=\frac{\sqrt{6}}{3}[/tex]
Hence
[tex]\sin \varphi=\cos\angle MBC=\frac{BM}{BC}=\frac{\sqrt{6}}{3}[/tex]

Solution:
Let [tex]M[/tex] be the midpoint of [tex]AB[/tex]. Triangle [tex]ABD[/tex] is isosceles ([tex]AD=BD[/tex]), so its median [tex]DM[/tex] is also an altitude: [tex]DM\perp AB[/tex]. From the right triangle [tex]AMD[/tex]:
[tex]DM=\sqrt{AD^2-AM^2}=\sqrt{10^2-6^2}=8[/tex]
Since [tex]CD\perp (ABC)[/tex], the segment [tex]CM[/tex] is the orthogonal projection of [tex]DM[/tex] on the plane [tex](ABC)[/tex], and by the theorem of the three perpendiculars [tex]CM\perp AB[/tex]. Both sides of [tex]\angle DMC[/tex] are perpendicular to the common line [tex]AB[/tex] of the two planes, so [tex]\angle DMC[/tex] is the angle between the planes [tex](ABD)[/tex] and [tex](ABC)[/tex].
In the right triangle [tex]DCM[/tex] ([tex]\angle DCM=90^{\circ}[/tex]):
[tex]\sin \angle DMC=\frac{CD}{DM}=\frac{4}{8}=\frac{1}{2}\Rightarrow \angle DMC=30^{\circ}[/tex]

Solution:
In the right triangle [tex]ABC[/tex] (right angle at [tex]C[/tex]) let [tex]CH[/tex] be the altitude to the hypotenuse, [tex]H\in AB[/tex]:
[tex]AB=\sqrt{15^2+20^2}=25,\qquad CH=\frac{AC\times BC}{AB}=\frac{15\times 20}{25}=12[/tex]
The prism is right, so [tex]CC_1\perp (ABC)[/tex] and [tex]CH[/tex] is the orthogonal projection of [tex]C_1H[/tex] on the base. Since [tex]CH\perp AB[/tex], by the theorem of the three perpendiculars [tex]C_1H\perp AB[/tex]. Hence [tex]C_1H[/tex] is the distance from [tex]C_1[/tex] to [tex]AB[/tex], i.e. [tex]C_1H=13[/tex].
From the right triangle [tex]CHC_1[/tex] (right angle at [tex]C[/tex]) we get the height of the prism:
[tex]CC_1=\sqrt{C_1H^2-CH^2}=\sqrt{13^2-12^2}=5[/tex]

Solution:
Let [tex]D[/tex] be the orthogonal projection of [tex]C[/tex] on [tex]\alpha[/tex]. Then [tex]CD\perp\alpha[/tex], [tex]AD[/tex] is the projection of [tex]CA[/tex], and the angle between [tex]CA[/tex] and [tex]\alpha[/tex] is [tex]\angle CAD=30^{\circ}[/tex]. From the right triangle [tex]ADC[/tex] ([tex]\angle ADC=90^{\circ}[/tex]):
[tex]CD=CA\sin 30^{\circ}=10\times \frac{1}{2}=5[/tex]
Since [tex]CB=5\sqrt{2}\ne CD[/tex], we have [tex]B\ne D[/tex], so [tex]BD[/tex] is the projection of [tex]CB[/tex] on [tex]\alpha[/tex] and the required angle is the acute angle [tex]\angle CBD[/tex] of the right triangle [tex]BDC[/tex] ([tex]\angle BDC=90^{\circ}[/tex]):
[tex]\sin \angle CBD=\frac{CD}{CB}=\frac{5}{5\sqrt{2}}=\frac{\sqrt{2}}{2}\Rightarrow \angle CBD=45^{\circ}[/tex]

Solution:
Let [tex]a[/tex] be the edge of the cube and [tex]O_1[/tex] the center of the top face [tex]A_1B_1C_1D_1[/tex]. The plane [tex](ACC_1)[/tex] is the plane of the diagonal section [tex]ACC_1A_1[/tex].
We have [tex]D_1O_1\perp A_1C_1[/tex] (the diagonals of a square are perpendicular) and [tex]D_1O_1\perp AA_1[/tex] (because [tex]AA_1\perp (A_1B_1C_1D_1)[/tex]). These two intersecting lines lie in the plane [tex](ACC_1)[/tex], hence [tex]D_1O_1\perp (ACC_1)[/tex].
So [tex]AO_1[/tex] is the orthogonal projection of [tex]AD_1[/tex] on the plane, and the required angle is [tex]\angle D_1AO_1[/tex]. In the right triangle [tex]AO_1D_1[/tex] ([tex]\angle AO_1D_1=90^{\circ}[/tex]) the leg [tex]D_1O_1[/tex] is half a face diagonal and the hypotenuse [tex]AD_1[/tex] is a face diagonal:
[tex]\sin \angle D_1AO_1=\frac{D_1O_1}{AD_1}=\frac{\frac{a\sqrt{2}}{2}}{a\sqrt{2}}=\frac{1}{2}\Rightarrow \angle D_1AO_1=30^{\circ}[/tex]

Solution:
Let [tex]B_1[/tex] be the orthogonal projection of [tex]B[/tex] onto [tex]\alpha[/tex]. Since [tex]C\in\alpha[/tex], the projection of the leg [tex]BC[/tex] is [tex]CB_1=\frac{a\sqrt{2}}{2}[/tex], and the triangle [tex]BB_1C[/tex] is right-angled at [tex]B_1[/tex]. The triangle [tex]ABC[/tex] is isosceles, so [tex]BC=a[/tex]:
[tex]BB_1=\sqrt{BC^2-CB_1^2}=\sqrt{a^2-\frac{a^2}{2}}=\frac{a\sqrt{2}}{2}[/tex]
Since [tex]A\in\alpha[/tex], the projection of [tex]AB[/tex] onto [tex]\alpha[/tex] is [tex]AB_1[/tex], so the required angle is [tex]\angle BAB_1[/tex]. In the right triangle [tex]ABB_1[/tex] ([tex]BB_1\perp\alpha[/tex]) the hypotenuse is [tex]AB=a\sqrt{2}[/tex]:
[tex]\sin \angle BAB_1=\frac{BB_1}{AB}=\frac{a\sqrt{2}/2}{a\sqrt{2}}=\frac{1}{2}\Rightarrow \angle BAB_1=30^{\circ}[/tex]

Solution:
Let [tex]ABCDM[/tex] be the pyramid with apex [tex]M[/tex] and base edge [tex]a[/tex], and let [tex]P[/tex] and [tex]Q[/tex] be the midpoints of [tex]AB[/tex] and [tex]BC[/tex]. The lateral edges are equal, so their projections on the base are equal too, and the foot [tex]O[/tex] of the height [tex]MO[/tex] is the center of the square (the intersection point of its diagonals).
The plane meets the base along [tex]PQ[/tex], so the section is the triangle [tex]MPQ[/tex]. Its orthogonal projection on the base is the triangle [tex]OPQ[/tex], in which [tex]OP\perp AB[/tex] and [tex]OQ\perp BC[/tex], so [tex]\angle POQ=90^{\circ}[/tex] and [tex]OP=OQ=\frac{a}{2}[/tex]:
[tex]S_{OPQ}=\frac{1}{2}\times \frac{a}{2}\times \frac{a}{2}=\frac{a^2}{8}[/tex]
The area of the orthogonal projection of a plane figure equals the area of the figure times the cosine of the angle between the two planes:
[tex]\frac{a^2}{8}=S\cos 60^{\circ}=\frac{S}{2}\Rightarrow a^2=4S\Rightarrow a=2\sqrt{S}[/tex]

Solution:
The pyramid is regular, so its apex is projected onto the center of the base: [tex]MO\perp(ABCD)[/tex]. The line [tex]AB[/tex] lies in the base plane, hence [tex]MO\perp AB[/tex].
The center [tex]O[/tex] is the midpoint of the diagonal [tex]BD[/tex], so [tex]ON[/tex] is a midline of the triangle [tex]ABD[/tex] and [tex]ON\parallel AD[/tex]. Since [tex]AD\perp AB[/tex], we get [tex]ON\perp AB[/tex].
Thus the line [tex]AB[/tex] is perpendicular to the two intersecting lines [tex]MO[/tex] and [tex]ON[/tex] of the plane [tex](OMN)[/tex], so
[tex]AB\perp(OMN)[/tex]
The plane [tex](AMB)[/tex] passes through the line [tex]AB[/tex], which is perpendicular to the plane [tex](OMN)[/tex]. By the criterion for perpendicular planes, [tex](AMB)\perp(OMN)[/tex].

Solution:
Since [tex]MD\perp(ABCD)[/tex], the line [tex]MD[/tex] is perpendicular to every line of the base through [tex]D[/tex], in particular to [tex]DA[/tex], [tex]DC[/tex] and [tex]DB[/tex]. Hence the triangles [tex]MDA[/tex], [tex]MDC[/tex] and [tex]MDB[/tex] are right-angled at [tex]D[/tex], and by the Pythagorean theorem
[tex]MA^2=MD^2+DA^2, \quad MC^2=MD^2+DC^2, \quad MB^2=MD^2+DB^2[/tex]
Adding the first two equalities, we get
[tex]MA^2+MC^2=2MD^2+DA^2+DC^2[/tex]
In the rectangle [tex]ABCD[/tex] the triangle [tex]DAB[/tex] is right-angled at [tex]A[/tex] and [tex]AB=DC[/tex], so
[tex]DB^2=DA^2+AB^2=DA^2+DC^2[/tex]
Therefore
[tex]MB^2+MD^2=2MD^2+DB^2=2MD^2+DA^2+DC^2=MA^2+MC^2[/tex]

Solution:
Since [tex]ABCD[/tex] is a square, [tex]BC\parallel AD[/tex], and [tex]AD[/tex] lies in the plane [tex](ADP)[/tex], hence [tex]BC\parallel(ADP)[/tex]. The plane [tex](BCM)[/tex] contains [tex]BC[/tex] and has the common point [tex]M[/tex] with [tex](ADP)[/tex], so it cuts [tex](ADP)[/tex] along a line through [tex]M[/tex] parallel to [tex]BC[/tex]. This line lies in [tex](ADP)[/tex] and meets [tex]DP[/tex] at [tex]N[/tex], so
[tex]MN\parallel AD\parallel BC[/tex]
In [tex]\triangle ADP[/tex] the segment [tex]MN[/tex] passes through the midpoint [tex]M[/tex] of [tex]AP[/tex] and is parallel to [tex]AD[/tex], so it is a midline: [tex]N[/tex] is the midpoint of [tex]DP[/tex] and
[tex]MN=\frac{AD}{2}=\frac{BC}{2}[/tex]
Thus [tex]MN\parallel BC[/tex] and [tex]MN\ne BC[/tex], so [tex]BMNC[/tex] is a trapezoid with bases [tex]BC[/tex] and [tex]MN[/tex].
The lateral faces [tex]ABP[/tex] and [tex]DCP[/tex] are congruent triangles ([tex]AB=DC[/tex], [tex]PA=PB=PC=PD[/tex]), and [tex]BM[/tex], [tex]CN[/tex] are their corresponding medians to the sides [tex]AP[/tex] and [tex]DP[/tex], so [tex]BM=CN[/tex]. Therefore [tex]BMNC[/tex] is an isosceles trapezoid.