THEOREMS FOR TRIHEDRAL ANGLES
§1. The polar trihedral angle
5.1. Given a trihedral angle with plane angles α, β, γ and the dihedral angles A, B and C, respectively, opposite to them, prove that there exists a trihedral angle with plane angles π − A, π − B and π − C and dihedral angles π − α, π − β and π − γ.
5.2. Prove that if dihedral angles of a trihedral angle are right ones, then its plane angles are also right ones.
5.3. Prove that trihedral angles are equal if the corresponding dihedral angles are equal.
§2. Inequalities with trihedral angles
5.4. Prove that the sum of two plane angles of a trihedral angle is greater than the third plane angle.
5.5. Prove that the sum of plane angles of a trihedral angle is smaller than 2π and the sum of its dihedral angles is greater than π.
5.6. A ray SC′ lies inside the trihedral angle SABC with vertex S. Prove that the sum of plane angles of a trihedral angle SABC is greater than the sum of plane angles of the trihedral angle SABC′.
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§3. Laws of sines and cosines for trihedral angles
5.7. Let α, β and γ be plane angles of a trihedral angle, A, B and C the dihedral angles opposite to them. Prove that
(The law of sines for a trihedral angle) sin α : sin A = sin β : sin B = sin γ : sin C.
5.8. Let α, β and γ be plane angles of a trihedral angle A, B and C the dihedral angles opposite to them.
a) Prove that
(The first law of cosines for a trihedral angle)
cos α = cos β cos γ + sin β sin γ cos A.
b) Prove that
(The second law of cosines for a trihedral angle)
cos A = − cos B cos C + sin B sin C cos α.
5.9. Plane angles of a trihedral angle are equal to α, β and γ; the edges opposite to them form angles a, b and c with the planes of the faces. Prove that
sin α sin a = sin β sin b = sin γ sin c.
5.10. a) Prove that if all the plane angles of a trihedral angle are obtuse ones, then all its dihedral angles are also obtuse ones.
b) Prove that if all the dihedral angles of a trihedral angle are acute ones, then all its plane angles are also acute ones.
§4. Miscellaneous problems
5.11. Prove that in an arbitrary trihedral angle the bisectors of two plane angles and the angle adjacent to the third plane angle lie in one plane.
5.12. Prove that the pairwise angles between the bisectors of plane angles of a trihedral angle are either simultaneously acute, or simultaneously obtuse, or simul-taneously right ones.
5.13. a) A sphere tangent to faces SBC, SCA and SAB at points A1, B1 and C1 is inscribed in trihedral angle SABC. Express the value of the angle ASB1 in terms of the plane angles of the given trihedral angle.
b) The inscribed and escribed spheres of tetrahedron ABCD are tangent to the face ABC at points P and P′, respectively. Prove that lines AP and AP′ are symmetric through the bisector of angle BAC.
5.14. The plane angles of a trihedral angle are not right ones. Through the vertices of tetrahedral angle planes perpendicular to the opposite faces are drawn.
Prove that these planes intersect along one line.
5.15. a) The plane angles of a trihedral angle are not right ones. In the planes of the trihedral angle’s faces there are drawn lines perpendicular to the respective opposite edges. Prove that all three lines are parallel to one plane.
b) Two trihedral angles with common vertex S are placed so that the edges of the second angle lie in the planes of the corresponding faces of the first angle and are perpendicular to its opposite edges. Find the plane angles of the first trihedral angle.
§5. Polyhedral angles
5.16. a) Prove that for any convex tetrahedral angle there exists a section which is a parallelogram and all such sections are parallel to each other.
b) Prove that there exists a section of a convex four-hedral angle with equal plane angles which is a rhombus.
5.17. Prove that any plane angle of a polyhedral angle is smaller than the sum of all the other plane angles.
5.18. One of two convex polyhedral angles with common vertex lies inside the other one. Prove that the sum of the plane angles of the inner polyhedral angle is smaller than the sum of the plane angles of the outer polyhedral angle.
5.19. a) Prove that the sum of dihedral angles of a convex n-hedral angle is greater than (n − 2)π.
b) Prove that the sum of plane angles of a convex n-hedral angle is smaller than 2π.
5.20. The sum of plane angles of a convex n-hedral angle is equal to the sum of its dihedral angles. Prove that n = 3.
5.21. A sphere is inscribed in a convex four-hedral angle. Prove that the sums of its opposite plane angles are equal.
5.22. Prove that a convex four-hedral angle can be inscribed in a cone if and only if the sums of its opposite dihedral angles are equal.
§6. Ceva’s and Menelaus’s theorems for trihedral angles
Before we pass to Ceva’s and Menelaus’s theorems for trihedral angles we have to prove (and formulate) Ceva’s and Menelaus’s theorems for triangles. To formulate these theorems, we need the notion of the ratio of oriented segments that lie on the same line.
Let points A, B, C and D lie on one line. By the ratio of oriented segments AB and CD we mean the number CDAB whose absolute value is equal to ABCD and which is positive if vectors {AB} and {CD} are similarly directed and negative if the directions of these vectors are opposite.
5.23. On sides AB, BC and CA of triangle ABC (or on their extensions), points C1, A1 and B1,respectively, are taken.
a) Prove that points A1, B1 and C1 lie on one line if and only if
(Menelaus’s theorem) A1B
A1C ·B1C B1A · C1A
C1B = 1.
b) Prove that if lines AA1, BB1 and CC1 are not pairwise parallel, then they intersect at one point if and only if
(Ceva’s theorem) A1B
A1C · B1C B1A ·C1A
C1B = −1.
Let rays l, m and n with a common origin lie in one plane. In this plane, select a positive direction of rotation. In this section we will denote by sin(n,m)sin(l,m) the ratio of sines of the angles through which one has to rotate in the positive direction rays l and n in order for them to coincide with ray m. Clearly, this ratio does not depend
on the choice of the positive direction of the rotation in plane: as we vary this direction both the numerator and the denominator adjust accordingly.
Let half-planes α, β and γ have a common boundary. Select one of the positive directions of rotation about this line (the boundary) as the positive one. In this section we will denote by sin(α,β)sin(γ,β) the ratio of the sines of the angles through which one has to turn in the positive direction the half-planes α and γ in order for them to coincide with β. Clearly, this quantity does not depend on the choice of the positive direction of rotation.
5.24. Given a trihedral angle with vertex S and edges a, b and c. Rays α, β and γ starting from S lie in the planes of the faces opposite to edges a, b and c, respectively. intersect along one line if and only if
(First Ceva’s theorem) sin(a, γ)
sin(b, γ) ·sin(b, α)
sin(c, α)· sin(c, β) sin(a, β) = −1.
5.25. Given are a trihedral angle with vertex S and edges a, b, c and rays α, β and γ, respectivly, starting from S and lying in the planes of the faces opposite to these edges. Let l and m be two rays with a common vertex. Denote by lm the plane determined by these rays.
c) Prove that the planes passing through pairs of rays a and α, b and β, c and γ intersect along one line if and only if
(Second Ceva’s theorem) sin(ab, aα)
sin(ac, aα)· sin(bc, bβ)
sin(ba, bβ)·sin(ca, cγ) sin(cb, cγ) = −1.
5.26. In trihedral angle SABC, a sphere tangent to faces SBC, SCA and SAB at points A1, B1and C1, respectively, is inscribed. Prove that planes SAA1, SBB1
and SCC1 intersect along one line.
5.27. Given a trihedral angle with vertex S and edges a, b and c. R are placed in planes of the faces opposite to edges a, b and c. Let rays α′, β′ and γ′ be symmetric to rays α, β and γ, respectively, through the bisectors of the corresponding faces.
a) Prove that rays α, β and γ lie in one plane if and only if rays α′, β′ and γ′ lie in one plane.
b) Prove that the planes passing through pairs of rays a and α, b and β, c and γ intersect along one line if and only if the planes passing through the pairs of rays a and α′, b and β′, c and γ′ intersect along one line.
5.28. Given a trihedral angle with vertex S and edges a, b and c. Lines α, β and γ lie in the planes of the faces opposite to edges a, b and c, respectively. Let α′ be the line along which the plane symmetric to the plane aα through the bisector plane of the dihedral angle at edge a intersects the plane of face bc; lines β′ and γ′ are similarly defined.
a) Prove that lines α, β and γ lie in one plane if and only if the lines α′, β′ and γ′ lie in one plane.
b) Prove that the planes passing through pairs of lines a and α, b and β, c and γ intersect along one line if and only if the planes passing through the pairs of lines a and α′, b and β′, c and γ′ intersect along one line.
5.29. Given tetrahedron A1A2A3A4 and a point P . For every edge AiAj con-sider the plane symmetric to plane P AiAjthrough the bisector plane of the dihedral angle at edge AiAj. Prove that either all these 6 planes intersect at one point or all of them are parallel to one line.
5.30. Given trihedral angle SABC such that ∠ASB = ∠ASC = 90◦. Planes πb and πc pass through edges SB and SC and planes πb′ and πc′ are symmetric to πb and πc, respectively, through the bisector planes of the dihedral angles at these edges. Prove that the projections of the intersection lines of planes πb and πc, π′b and πc′ to plane BSC are symmetric through the bisector of angle ∠BSC.
5.31. Let the Monge’s point of tetrahedron ABCD (see Problem 7.32) lie in the plane of face ABC. Prove that through point D planes pass in which there lie:
a) intersection points of the heights of faces DAB, DBC and DAC;
b) the centers of the circumscribed circles of faces DAB, DBC and DAC.
Problems for independent study
5.32. A sphere with center O is inscribed in the trihedral angle with vertex S.
Prove that the plane passing through the three tangent points is perpendicular to line OS.
5.33. Given trihedral angle SABC with vertex S; the dihedral angles ∠A, ∠B and ∠C at edges SA, SB and SC; the plane angles α, β and γ opposite to them.
a) The bisector plane of the dihedral angle at edge SA intersects face SBC along ray SA1. Prove that
sin A1SB : sin A1SC = sin ASB : sin ASC.
b) The plane passing through edge SA perpendicularly to face SBC intersects this face along ray SA1. Prove that
sin A1SB : sin A1SC = (sin β cos C) : (sin γ cos B).
We assume here that all the plane angles of the given trihedral angle are acute ones; consider on your own the case when among the plane angles of the trihedral angle obtuse angles are encountered.
5.34. Let a, b and c be the unit vectors directed along the edges of trihedral angle SABC.
a) Prove that the planes passing through the edges of the trihedral angle and the bisectors of the opposite faces intersect along one line and this line is given by vector a + b + c.
b) Prove that the bisector planes of the dihedral angles of the trihedral angle intersect along one line and this line is given by the vector
a sin α + b sin β + c sin γ.
c) Prove that the planes passing through the edges of the trihedral angle per-pendicularly to their opposite faces intersect along one line and this line is given by the vector
a sin α cos B cos C + b sin β cos A cos C + c sin γ cos A cos B.
d) Prove that the planes passing through the bisectors of the faces perpendicu-larly to the planes of these faces intersect along one line and this line is determined by the vector
[a, b] + [b, c] + [c, a]
(Recall the definition of the vector product [a, b] of vectors a and b.)
5.35. In a convex tetrahedral angle the sums of the opposite plane angles are equal. Prove that a sphere can be inscribed in this tetrahedral angle.
5.36. Projections SA′, SB′ and SC′ of edges SA, SB and SC of a trihedral angle to the faces opposite to them form the edges of a new trihedral angle. Prove that the bisector planes of the new angle are SAA′, SBB′ and SCC′.
Solutions
5.1. Inside the given trihedral angle with vertex S take an arbitrary point S′and from it drop perpendiculars S′A′, S′B′ and S′C′ to faces SBC, SAC and SAB, respectively. Clearly, the plane angles of trihedral angle S′A′B′C′ complement the dihedral angles of trihedral angle SABC to π. To complete the proof it remains to notice that edges SA, SB and SC are perpendicular to faces S′B′C′, S′A′C′ and S′A′B′, respectively.
Angle S′A′B′C′ is called the complementary or polar one to angle SABC.
5.2. Consider the trihedral angle polar to the given one (see Problem 5.1). Its plane angles are right ones; hence, its dihedral angles are also right ones. Therefore, the plane angles of the initial trihedral angle are also right ones.
5.3. The angles polar to the given trihedral angles have equal plane angles;
hence, they are equal themselves.
5.4. Consider trihedral angle SABC with vertex S. The inequality ∠ASC <
∠ASB + ∠BSC is obvious if ∠ASC ≤ ∠ASB. Therefore, let us assume that
∠ASC≥ ∠ASB. Then, inside face ASC, we can select a point B′so that ∠ASB′=
∠ASB and SB′ = SB, i.e., ∠ASB = ∠ASB′. We may assume that point C lies in plane ABB′. Since
AB′+ B′C = AC < AB + BC = A′B + BC,
it follows that B′C < BC. Hence, ∠B′SC < ∠BSC. It remains to notice that
∠B′SC = ∠ASC − ∠ASB.
5.5. First solution. On the edges of the trihedral angle draw equal segments SA, SB and SC starting from vertex S. Let O be the projection of S to plane ABC.
The isosceles triangles ASB and AOB have a common base AB and AS > AO.
Hence, ∠ASB < ∠AOB. By writing similar inequalities for the two other angles and taking their sum we get
∠ASB + ∠BSC + ∠CSA < ∠AOB + ∠BOC + ∠COA≤ 2π.
The latter inequality becomes a strict one only if point O lies outside triangle ABC.
To prove the second part, it suffices to apply the already proved inequality to the angle polar to the given one (see Problem 5.1). Indeed, if α, β and γ are dihedral angles of the given trihedral angle, then
(π − α) + (π − β)(π − γ) < 2π, i.e., α + β + γ > π.
Second solution. Let point A′ lie on the extension of edge SA beyond vertex S. By Problem 5.4
∠A′SB + ∠A′SC > ∠BSC, i.e., (π − ∠ASB) + (π − ∠ASC) > ∠BSC;
hence, 2π > ∠ASB + ∠BSC + ∠CSA.
Proof of the second part of the problem is performed as in the first solution.
5.6. Let K be the intersection point of face SCB with line AC′. By Problem 5.4 we have ∠C′SK + ∠KSB > ∠C′SB and
∠CSA + ∠CSK > ∠ASK = ∠ASC′+ ∠C′SK.
Adding these inequalities and taking into account that ∠CSK + ∠KSB = ∠CSB we get the desired statement.
5.7. On edge SA of trihedral angle SABC, take an arbitrary point M . Let M′ be the projection of M to plane SBC, let P and Q be the projections of M to lines SB and SC. By the theorem on three perpendiculars M′P ⊥ SB and M′Q ⊥ SC.
If SM = a, then M Q = a sin β and
M M′ = M Q sin C = a sin β sin C.
Similarly,
M M′= M P sin B = a sin γ sin B.
Therefore,
sin β : sin B = sin γ : sin C.
The second equality is similarly proved.
5.8. a) First solution. On segment SA take a point, M , and at it erect perpendiculars P M and QM to edge SA in planes SAB and SAC, respectively (points P and Q lie on lines SB and SC). By expressing the length of the side P Q in triangles P QM and P QS with the help of the law of cosines and equating these expressions we get the desired equality after simplifications.
Second solution. Let a, b and c be unit vectors directed along edges SA, SB and SC, respectively. Vector b lying in plane SAB can be represented in the form
b = a cos γ + u, where u ⊥ a and |u| = sin γ.
Similarly,
c = a cos β + v, where v ⊥ a and |v| = sin β.
It is also clear that the angle between vectors u and v is equal to ∠A.
On the one hand, the inner product of vectors b and c is equal to cos α. On the other hand, the product is equal to
(a cos γ + u, a cos β + v) = cos β cos γ + sin β sin γ cos ∠A.
b) To prove it, it suffices apply the first law of cosines to the angle polar to the given trihedral angle (cf. Problem 5.1).
5.9. Let us draw three planes parallel to the faces of the trihedral angle at distance 1 from them and intersecting the edges. Together with the planes of the faces they constitute a parallelepiped all the heights of which are equal to 1 and, therefore, the areas of all its faces are equal. Now, notice that the lengths of the edges of this parallelepiped are equal to sin a1 , sin b1 and sin c1 . Therefore, the areas of its faces are equal to
sin α
sin b sin c, sin β
sin a sin c, and sin γ sin a sin b. By equating these expressions we get the desired statement.
5.10. a) By the first theorem on cosines for a trihedral angle (Problem 5.8 a)) sin β sin γ cos A = cos α − cos β cos γ.
By the hypothesis cos α < 0 and cos β cos γ > 0; hence, cos A < 0.
b) To prove it, it suffices to make use of the second theorem on cosines (Problem 5.8 b)).
5.11. First solution. On the edges of the trihedral angle, draw equal segments SA, SB and SC beginning from vertex S. The bisectors of angles ASB and BSC pass through the midpoints of segments AB and BC, respectively, and the bisector of the angle adjacent to angle CSA is parallel to CA.
Second solution. On the segments of the trihedral angle draw equal vectors a, b and c beginning from vertex S. The bisectors of angles ASB and BSC are parallel to vectors a + b and b + c and the bisector of the angle adjacent to angle CSA is parallel to the vector c − a. It remains to notice that
(a + b) + (c − a) = b + c.
5.12. On the edges of the trihedral angle draw unit vectors a, b and c starting from its vertex. Vectors a + b, b + c and a + c determine the bisectors of the plane angles. It remains to verify that all the pairwise inner products of these sums are of the same sign. It is easy to see that the inner product of any pair of these vectors is equal to
1 + (a, b) + (b, c) + (c, a).
5.13. a) Let α, β and γ be the plane angles of trihedral angle SABC; let x = ∠ASB1= ∠ASC1, y = ∠BSA1= ∠BSC1 and z = ∠CSA1= ∠CSB1. Then
x + y = ∠ASC1+ ∠BSC1= ∠ASB = γ, y + z = α, z + x = β.
Hence,
x = 1
2(β + γ − α).
b) Let point D′ lie on the extension of edge AD beyond point A. Then the escribed sphere of the tetrahedron tangent to face ABC is inscribed in trihedral angle ABCD′ with vertex A. From the solution of heading a) it follows that
∠BAP = ∠BAC + ∠BAD− ∠CAD
2 ;
∠CAP′= ∠BAC + ∠CAD′− ∠BAD′
2 .
Since ∠BAD′= 180◦−∠BAD and ∠CAD′ = 180◦−∠CAD, we see that ∠BAP =
∠CAP′; hence, lines AP and AP′are symmetric through the bisector of angle BAC.
5.14. Let us select points A, B and C on the edges of the trihedral angle with vertex S so that SA ⊥ ABC (the plane that passes through point A of one edge perpendicularly to the edge intersects the other two edges because the plane angles are not right ones). Let AA1, BB1 and CC1 be the heights of triangle ABC. It suffices to verify that SAA1, SBB1 and SCC1are the planes spoken about in the formulation of the problem.
Since BC ⊥ AS and BC ⊥ AA1, it follows that BC ⊥ SAA1; hence, planes SBC and SAA1 are perpendicular to each other. Since BB1 ⊥ SA and BB1 ⊥ AS, we see that BB1 ⊥ SAC and, therefore, planes SBB1 and SAC are perpendicular.
We similarly prove that planes SCC1and SBC are perpendicular to each other.
5.15. a) Let a, b and c be vectors directed along the edges SA, SB and SC of the trihedral angle. The line lying in plane SBC and perpendicular to edge SA is parallel to vector (a, b)c − (a, c)b. Similarly, two other lines are parallel to vectors
5.15. a) Let a, b and c be vectors directed along the edges SA, SB and SC of the trihedral angle. The line lying in plane SBC and perpendicular to edge SA is parallel to vector (a, b)c − (a, c)b. Similarly, two other lines are parallel to vectors