§1. Lengths, perimeters
10.1. Let a, b and c be the lengths of sides of a parallelepiped, d that of its of its diagonals. Prove that
a2+ b2+ c2≥d2 3 .
10.2. Given a cube with edge 1, prove that the sum of distances from an arbitrary point to all its vertices is no less than 4√
3.
10.3. In tetrahedron ABCD the planar angles at vertex A are equal to 60◦. Prove that
AB + AC + AD ≤ BC + CD + DB.
10.4. From points A1, A2 and A3 that lie on line a perpendiculars AiBi are dropped to line b. Prove that if point A2 lies between A1 and A3 then the length of segment A2B2 is confined between the lengths of segments A1B1and A3B3.
10.5. A segment lies inside a convex polyhedron. Prove that the segment is not longer than the longest segment with the endpoints at vertices of the polyhedron.
10.6. Let P be the projection of point M to the plane that contains points A, B and C. Prove that if one can construct a triangle from segments P A, P B and P C, then from segments M A, M B and M C one can also construct a triangle.
10.7. Points P and Q are taken inside a convex polyhedron. Prove that one of the vertices of the polyhedron is closer to Q than to P .
10.8. Point O lies inside tetrahedron ABCD. Prove that the sum of the lengths of segments OA, OB, OC and OD does not exceed the sum of the lengths of tetrahedron’s edges.
10.9. Inside the cube with edge 1 several segments lie and any plane parallel to one of the cube’s faces does not intersect more than one segment. Prove that the sum of the lengths of these segments does not exceed 3.
10.10. A closed broken line passes along the surface of a cube with edge 1 and has common points with all the cube’s faces. Prove that its length is no less than 3√
2.
10.11. A tetrahedron inscribed in a sphere of radius R contains the center of the sphere. Prove that the sum of the lengths of the tetrahedron’s edges is greater than 6R.
10.12. The section of a regular tetrahedron is a quadrilateral. Prove that the perimeter of this quadrilateral is confined between 2a and 3a, where a is the length of the tetrahedron’s edge.
§2. Angles
10.13. Prove that the sum of the angles of a spatial quadrilateral does not exceed 360◦.
10.14. Prove that not more than 1 vertex of a tetrahedron has a property that the sum of any two of plane angles at this vertex is greater than 180◦.
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10.15. Point O lies on the base of triangular pyramid SABC. Prove that the sum of the angles between ray SO and the lateral edges is smaller than the sum of the plane angles at vertex S while being greater than half this sum.
10.16. a) Prove that the sum of the angles between the edges of a trihedral angle and the planes of the faces opposite to them does not exceed the sum of its plane angles.
b) Prove that if dihedral angles of a trihedral angle are acute ones then the sum of the angles between its edges and planes of faces opposite to them is not less than a half sum of its plane angles.
10.17. The diagonal of a rectangular parallelepiped constitutes angles α, β and γ with its edges. Prove that α + β + γ < π.
10.18. All the plane angles of a convex quadrangular angle are equal to 60◦. Prove that the angles between its opposite edges cannot be neither simultaneously acute nor simultaneously obtuse.
10.19. Prove that the sum of all the angles that have a common vertex inside a tetrahedron and subtend the edges of that tetrahedron is greater than 3π.
10.20. a) Prove that the sum of dihedral angles at edges AB, BC, CD and DA of tetrahedron ABCD is smaller than 2π.
b) Prove that the sum of dihedral angles of a tetrahedron is confined between 2π and 3π.
10.21. The space is completely covered by a finite set of (infinite one way) right circular coni with angles ϕ1, . . . , ϕn. Prove that
ϕ21+ · · · + ϕ2n ≥ 16.
§3. Areas
10.22. Prove that the area of any face of a tetrahedron is smaller(?) than the sum of the areas of its other three faces.
10.23. A convex polyhedron lies inside another polyhedron. Prove that the surface area of the outer polyhedron is greater than the surface area of the inner one.
10.24. Prove that for any tetrahedron there exist two planes such that the ratio of the areas of the tetrahedron’s projections to them is not less than√
2.
10.25. a) Prove that the area of any triangular section of a tetrahedron does not exceed the area of one of the tetrahedron’s faces.
b) Prove that the area of any quadrangular section of a tetrahedron does not exceed the area of one of the tetrahedron’s faces.
10.26. A plane tangent to the sphere inscribed in a cube cuts off it a triangular pyramid. Prove that the surface area of this pyramid does not exceed the area of the cube’s face.
§4. Volumes
10.27. On each edge of a tetrahedron a point is fixed. Consider four tetrahedrons one of the vertices of each of which is a vertex of the initial tetrahedron and the remaining vertices are fixed points belonging to the edges that go out of this vertex.
Prove that the volume of one of the tetrahedrons does not exceed 18 of the initial tetrahedron’s volume.
10.28. The lengths of each of the 5 edges of a tetrahedron do not exceed 1.
Prove that its volume does not exceed 18.
10.29. The volume of a convex polyhedron is equal to V and its surface area is equal to S.
a) Prove that if a sphere of radius r is placed inside the polyhedron, then VS ≥ r3. b) Prove that a sphere of radius VS can be placed inside the polyhedron.
c) A convex polyhedron is placed inside another one. Let V1 and S1 be the volume and the surface area of the outer polyhedron, V2 and S2 same of the outer one. Prove that
3V1
S1 ≥V2
S2
.
10.30. Inside a cube, a convex polyhedron is placed whose projection onto each face of the cube coincides with this face. Prove that the volume of the polyhedron is not less than 13 the volume of the cube.
10.31. The areas of the projections of the body to coordinate axes are equal to S1, S2 and S3. Prove that its volume does not exceed√
S1S2S3.
§5. Miscellaneous problems
10.32. Prove that the radius of the inscribed circle of any face of a tetrahedron is greater than the radius of the sphere inscribed in the tetrahedron.
10.33. On the base of a triangular pyramid OABC with vertex O point M is taken. Prove that
OM · SABC≤ OA · SMBC+ OB · SMAC+ OC · SMAB.
10.34. Let r and R be the radii of the inscribed and circumscribed spheres of a regular quadrangular pyramid. Prove that
R
r ≥ 1 +√ 2.
10.35. Is it possible to cut a hole in a cube through which another cube of the same size can be pulled?
10.36. Sections M1 and M2 of a convex centrally symmetric polyhedron are parallel and M1passes through the center of symmetry.
a) Is it true that the area of M1 is not less than the area of M2?
b) Is it true that the radius of the minimal circle that contains M1 is not less than the radius of the minimal circle that contains M2?
10.37. A convex polyhedron sits inside a sphere of radius R. The length of its i-th edge is equal to li and the dihedral angle at this edge is equal to ϕi. Prove that
Xli(π − ϕi) ≤ 8πR.
Problems for independent study
10.38. Triangle A′B′C′ is a projection of triangle ABC. Prove that the hights of triangle A′B′C′ are no longer than the corresponding hights of triangle ABC.
10.39. A sphere is inscribed into a truncated cone. Prove that the surface area of the ball is smaller than the area of the lateral surface of the cone.
10.40. The largest of the perimeters of tetrahedron’s faces is equal to d and the sum of the lengths of its edges is equal to D. Prove that
3d < 2D ≤ 4d.
10.41. Inside tetrahedron ABCD a point E is fixed. Prove that at least one of segments AE, BE and CE is shorter than the corresponding segment AD, BD and CD.
10.42. Is it possible to place 5 points inside a regular tetrahedron with edge 1 so that the pairwise distances between these points would be not less than 1?
10.43. The plane angles of a trihedral angle are α, β and γ. Prove that cos2α + cos2β + cos2γ ≤ 1 + 2 cos α cos β cos γ.
10.44. The base of pyramid ABCDE is a parallelogram ABCD. None of the lateral faces is an acute triangle. On edge DC, there is a point M such that line EM is perpendicular to BC. Moreover, diagonal AC of the base and lateral edges ED and EB are connected by relations AC ≥ 54EB ≥ 53ED. Through vertex B and the midpoint of one of lateral edges a section is drawn; the section is an isosceles trapezoid. Find the ratio of the area of the section to the area of the pyramid’s base.
Solutions 10.1. Since d ≤ a + b + c, it follows that
d2≤ a2+ b2+ c2+ 2ab + 2bc + 2ca ≤ 3(a2+ b2+ c2).
10.2. If P Q is the diagonal of cube with edge 1 and X is an arbitrary point, then P X + QX ≥ P Q =√
2. Since cube has 4 diagonals, the sum of the distances from X to all the vertices of the cube is not less than 4√
3.
10.3. First, let us prove that if ∠BAC = 60◦, then AB + AC ≤ 2BC. To this end let us consider points B′ and C′ symmetric to points B and C through the bisector of angle A. Since in any convex quadrilateral the sum of the lengths of diagonals is greater than the sum of the lengths of a pair of opposite sides,
BC + B′C′≥ CC′+ BB′
(the equality is attained if AB = AC). It remains to notice that B′C′ = BC, CC′ = AC and BB′= AB.
We similarly prove inequalities AC + AD ≤ 2CD and AD + AB ≤ 2DB. By adding up these inequalities we get the desired statement.
10.4. Let us draw through line b a plane Π parallel to a. Let Ci be the projec-tion of point Ai to plane Π. By the theorem on three perpendiculars, CiBi ⊥ b;
therefore, the length of segment B2C2is confined between the length of B1C1 and that of B3C3; the lengths of all three segments AiCi are equal.
10.5. In the proof we will several times make use of the following planimetric statement:
If point X lies on side BC of triangle ABC, then either AB ≥ AX or AC ≥ AX.
(Indeed, one of the angles BXA or CXA is not less than 90◦; if ∠BXA ≥ 90◦, then AB ≥ AX and if ∠CAX ≥ 90◦, then AC ≥ AX.)
Let us extend the given segment to its intersection with the polyhedron’s faces at certain points P and Q; this might only increase the length of the segment. Let M N be an arbitrary segment with the endpoints on the edges of the polyhedron;
let P belong to M N . Then either M Q ≥ P Q or NQ ≥ P Q.
Let, for definiteness, M Q ≥ P Q. Point M lies on an edge AB and either AQ ≥ MQ or BQ ≥ MQ. We have replaced segment P Q by a longer segment one of whose endpoints lies in a vertex of the polyhedron. Now, performe similar argument for the endpoint Q of the obtained segment. We can replace P Q by a longer segment with the endpoints in vertices of the polyhedron.
10.6. Let a = P A, b = P B and c = P C. We can assume that a ≤ b ≤ c. Then 10.7. Let us consider plane Π that passes through the midpoint of segment P Q perpendicularly to it. Suppose that all the vertices of the polyhedron are not closer to point Q than to point P . Then all the vertices of the polyhedron lie on the same side of plane Π as point P does. Therefore, point Q lies outside the polyhedron which contradicts the hypothesis.
Figure 75 (Sol. 10.8)
10.8. Let M and N be the intersection points of planes AOB and COD with edges CD and AB, respectively (Fig. 75). Since triangle AOB lies inside triangle AM B, it follows that
AO + BO ≤ AM + BM.
Similarly,
CO + DO ≤ CN + DN.
Therefore, it suffices to prove that the sum of the lengths of segments AM , BM , CN and DN does not exceed the sum of the lengths of the edges of tetrahedron ABCD.
First, let us prove that if X is a point on side A′B′ of triangle A′B′C′, then the length of segment C′X does not exceed a semi-perimeter of triangle A′B′C′. Indeed,
C′X ≤ C′B′+ B′X and C′X ≤ C′A + A′X.
Therefore,
2C′X ≤ A′B′+ B′C′+ C′A′. Thus,
2AM ≤ AC + CD + DA, 2BM ≤ BC + CD + DB, 2CN ≤ BA + AC + CB, 2DN ≤ BA + AD + DB.
By adding up all these inequalities we get the desired statement.
10.9. Let us enumerate the segments and consider the i-th segment. Let li be its length, xi, yi, zi the lengths of projections on the cube’s edges. It is easy to verify that li≤ xi+ yi+ zi.
On the other hand, if any plane parallel to the cube’s face intersects not more than 1 segment, then the projections of these segments to each edge of the cube do not have common points. Therefore,P xi≤ 1,P yi ≤ 1,P zi ≤ 1 and, finally, P li≤ 3.
10.10. Consider the projections on 3 nonparallel edges of the cube. The projec-tion of the given broken line on any edge contains both endpoints of the edge and, therefore, it coincides with the whole edge. Hence, the sum of the lengths of the projections of the broken line’s links on any edge is no less than 2 and the sum of the lengths of projections on all the three edges is not less than 6.
One of the three lengths of projections of any broken line’s link on the cube’s edges is zero; let two other lengths of projections be equal to a and b. Since (a + b)2 ≤ 2(a2+ b2), it follows that the sum of the lengths of the links of the broken line is no less than the sum of the lengths of these projections on the three edges of the cube divided by √
2; hence, it is no less than √62 = 3√ 2.
10.11. Let v1, v2, v3 and v4 be vectors that go from the center of the sphere to the vertices of the tetrahedron. Since the center of the sphere lies inside the tetrahedron, there exist positive numbers λ1, . . . , λ4 such that
λ1v1+ λ2v2+ λ3v3+ λ4v4= 0
(see Problem 7.16). We may assume that λ1+ · · · + λ4= 1. Let us prove that then λi≤12. Let, for example, λ1> 12. Then
R
2 < |λ1v1| = |λ2v2+ λ3v3+ λ4v4| ≤ (λ2+ λ3+ λ4)R = (1 − λ1)R <R 2. We have got a contradiction because λi≤ 12. Therefore,
|v1+ · · · + v4| = |(1 − 2λ1)v1+ · · · + (1 − 2λ4)v4|
≤ ((1 − 2λ1) + · · · + (1 − 2λ4))R = 2R.
Since
X|vi− vj|2= (4R)2− |X vi|2
(see the solution of Problem 14.15) and |P vi|2≤ 2R, it follows that X|vi− vj|2≥ (16 − 4)R2= 12R2.
And since 2R > |vi− vj|, it follows that 2RX
|vi− vj| >X
|vi− vj|2≥ 12R2.
10.12. Let us consider all the sections of the tetrahedron by the planes parallel to the given sections. Those of them that are quadrilaterals turn under the projection on the line perpendicular to the planes of the sections into the inner points of segment P Q, where points P and Q correspond to sections with planes passing through the vertices of the tetrahedron (Fig. 76 a)).
The length of the side of the section that belongs to a fixed face of the tetrahedron is a linear function on segment P Q. Therefore, the perimeter of the section being the sum of linear functions is a linear function on segment P Q. The value of a linear function at an arbitrary point of P Q is confined between its values at points P and Q.
Therefore, it suffices to verify that the perimeter of the section of a regular tetrahedron by a plane that passes through a vertex of the tetrahedron is confined between 2a and 3a (except for the cases when the section consists of one point; but such a section cannot correspond to neither P nor Q). If the section is an edge of the tetrahedron then the value of the considered linear function is equal to 2a for it.
Figure 76 (Sol. 10.12)
Since the length of any segment with the endpoints on sides of an equilateral tri-angle does not exceed the length of this tritri-angle’s side, the perimeter of a triangular section of the tetrahedron does not exceed 3a.
If the plane of the section passes through vertex D of tetrahedron ABCD and intersects edges AB and AC, then we will unfold faces ABD and ACD to plane ABC (Fig. 76 b)). The sides of the section connect points D′and D′′and, therefore, the sum of their lengths is no less than D′D′′= 2a.
10.13. If the vertices of a spatial quadrilateral ABCD are not in one plane, then
∠ABC < ∠ABD + ∠DBC and ∠ADC < ∠ADB + ∠BDC
(cf. Problem 5.4). Adding up these inequalities and adding further to both sides angles ∠BAD and ∠BCD we get the desired statement, because the sums of the angles of triangles ABD and DBC are equal to 180◦.
10.14. Suppose that vertices A and B of tetrahedron ABCD have the indicated property. Then
∠CAB + ∠DAB > 180◦and ∠CBA + ∠DBA > 180◦. On the other hand,
∠CAB + ∠CBA = 180◦− ∠ACB < 180◦and ∠DBA + ∠DAB < 180◦. Contradiction.
10.15. By Problem 5.4 ∠ASB < ∠ASO + ∠BSO. Since ray SO lies inside the trihedral angle SABC, it follows that
∠ASO + ∠BSO < ∠ASC + ∠BSC
(cf. Problem 5.6). By writing down two more pairs of such inequalities and taking their sum we get the desired statement.
10.16. a) Let α, β and γ be the angles between edges SA, SB and SC and the planes of the faces opposite to them, respectively. Since the angle between line l and plane Π does not exceed the angle between line l and any line in plane Π, it follows that
α ≤ ∠ASB, β ≤ ∠BSC and γ ≤ ∠CSA.
b) The dihedral angles of the trihedral angle SABC are all acute and, therefore, the projection SA1of ray SA to plane SBC lies inside angle BSC. Therefore, the inequalities
∠ASB≤ ∠BSA1+ ∠ASA1and ∠ASC ≤ ∠ASA1+ ∠CSA1
yield
∠ASB + ∠ASC− ∠BSC ≤ 2∠ASA1.
Write similar inequalities for edges SB and SC and take their sum. We get the desired statement.
10.17. Let O be the center of the rectangular parallelepiped ABCDA1B1C1D1. Height OH of an isosceles triangle AOC is parallel to edge AA1 and, therefore,
∠AOC = 2α, where α is the angle between edge AA1 and diagonal AC1. Similar arguments show that the plane angles of the trihedral angle OACD1 are equal to 2α, 2β and 2γ. Therefore, 2α + 2β + 2γ < 2π.
10.18. Let S be the vertex of the given angle. From solutions of Problem 5.16 b) it follows that it is possible to intersect this angle with a plane so that in the section we get rhombus ABCD, where SA = SC and SB = SD, and the projection of vertex S to the plane of the section coincides with the intersection point of the diagonals of the rhombus, O. Angle ASC is acute if AO < SO and obtuse if AO > SO. Since ∠ASB = 60◦, it follows that
AB2= AS2+ BS2− AS · BS.
Expressing, thanks to Pythagoras theorem, AB, AS and BS via AO, BO and SO we get after simplification and squaring
(1 + a2)(1 + b2) = 4, where a = AOSO and b = BOSO.
Therefore, the inequalities a > 1 and b > 1, as well as inequalities a < 1 and b < 1, cannot hold simultaneously.
10.19. Let O be a point inside tetrahedron ABCD; let α, β and γ be angles with vertex O that subtend the edges AD, BD and CD; let a, b and c be angles with vertex O that subtend the edges BC, CA and AB; P the intersection point of line DO with face ABC. Since ray OP lies inside the trihedral angle OABC, it follows that
∠AOP + ∠BOP < ∠AOC + ∠BOC (cf. Problem 5.6), i.e., π − α + π − β < b + a and, therefore,
α + β + a + b > 2π.
Similarly,
β + γ + b + c > 2π and α + γ + a + c > 2π.
Adding up these inequalities we get the desired statement.
10.20. a) Let us apply the statement of Problem 7.19 to tetrahedron ABCD.
Let a, b, c and d be normal vectors to faces BCD, ACD, ABD and ABC, respec-tively. The sum of these vectors is equal to 0 and, therefore, there exists a spatial quadrilateral the vectors of whose consecutive sides are a, b, c and d.
The angle between sides a and b of this quadrilateral is equal to the dihedral angle at edge CD (cf. Fig. 77). Similar arguments show that the considered sum of the dihedral angles is equal to the sum of plane angles of the obtained quadrilateral which is smaller than 2π (Problem 10.13).
Figure 77 (Sol. 10.20)
b) Let us express the inequality obtained in heading a) for each pair of the opposite edges of the tetrahedron and add up these three inequalities. Each dihedral angle of the tetrahedron enters two such inequalities and, therefore, the doubled
b) Let us express the inequality obtained in heading a) for each pair of the opposite edges of the tetrahedron and add up these three inequalities. Each dihedral angle of the tetrahedron enters two such inequalities and, therefore, the doubled