D. Polyhedral Angles

If you try to make a model of a three dimensional figure whose faces are all equilateral triangles except one which is a regular hexagon, you will probably find that it is impossible to build such a model. In your model the sides of the triangles should fit to form edges of three dimensional figure, but they will not, since there seems to be something wrong with the angles.

Hence, we need to investigate the properties of the angles at a vertex of three dimensional figures.

Definition: (polyhedral angle)

A figure formed by three or more planes that meet in a point and are so situated that they may be intersected by another plane to form a polygon is called a polyhedral angle.

In Figure 2.22, the angles at vertex A of the three dimensional figure are ∠DAB, ∠DAC and ∠CAB. The union

of these angles and their interiors

form a polyhedral angle.

The portions of the planes which form the polyhedral angle are called its faces.

The common point of meeting of the planes is called the vertex of the polyhedral angle.

The dihedral angles formed by the faces are called the dihedral angles of the polyhedral angle.

The edges of the dihedral angles are the edges of the polyhedral angle.

A face angle of the polyhedral angle is formed by the edges of any face.

A polyhedral angle may be read by naming the vertex or by naming the vertex and a point on each edge. Thus, the polyhedral angle in Figure 2.23 may be read as “polyhedral angle P” or “polyhedral angle P-ABCDE”.

In the same figure, the surfaces containing P, A, B or P, B, C are some of the faces of the polyhedral angle.

A polyhedral angle may have many face angles. The least it may have is three. The polyhedral angle having three faces is called a trihedral angle.

Polyhedral angles of four, five, six or eight faces are called respectively tetrahedral, pentahedral, hexahedral, octahedral angles.

A polyhedral angle is convex if the section formed by a plane intersecting all the faces is a convex polygon.

Polyhedral angle V in Figure 2.24 (a) is not convex, since polygon ABCDEF in plane m is not convex. On the other hand, polyhedral angle W in part (b) is convex, since the section in plane n is a convex polygon KLMNP.

Two polyhedral angles are congruent if the face angles and dihedral angles of one are equal, each to each, to the face angles and dihedral angles of the other, and arranged in the same order.

Two polyhedral angles are symmetric, if the face angles and dihedral angles of one are equal, each to each, to the face angles and dihedral angles of the other, but arranged in opposite order.

Congruent polyhedral angles can be made to coincide, but symmetric polyhedral angles cannot. In Figure 2.25, polyhedral angles V – ABC and V – A'B'C' are symmetric and they can not be made to coincide.

We will now investigate the properties of the face angles of a convex polyhedral angle.

Theorems: 1. The sum of the measures of any two face angles of a trihedral angle is greater than the measure of the third face angle.

2. The sum of the measures of the face angles of a convex polyhedral angle is less than 360°.

3. In any trihedral angle;

a. each dihedral angle is less than 180°.

b. the sum of the dihedral angles is less than 540°.

c. the sum of the dihedral angles is greater than 180°.

4. Two trihedral angles are either congruent or symmetric if the three face angles of the one are equal respectively to the three face angles of the other.

Proofs:

1. Referring to Figure 2.26, polyhedral angle A – FGH is a trihedral angle whose largest face angle is angle FAH.

Let us draw ray AR in plane FAH so that ∠FAG = ∠FAR.

Choose any points B and C on rays AF and AH, respectively.

Let ray AR intersect BC at a point E.

Then select D on ray AG so that AD = AE.

Then ΔBAD ≅ ΔBAE by S.A.S. and so BD = BE.

Now, in DBDC, BD + DC > BC and BC = BE + EC ⇒ DC > EC. (BD = BE)

And in DACD and DACE, AC = AC, AD = AE and DC > EC, so ∠DAC > ∠EAC.

Hence, ∠DAC + ∠BAD > ∠EAC + ∠BAE (∠BAD = ∠BAE)

∠DAC + ∠BAD > ∠BAC.

Since angle BAC is the largest of the three face angles, the two following inequalities can be written,

∠BAC + ∠BAD > ∠DAC and ∠BAC + ∠DAC > ∠BAD.

The proof for the case where there is no largest face angle is similar.

2. Consider a convex polyhedral angle formed by point F and convex polygon ABCD, as illustrated in Figure 2.27.

Suppose E is any point in the interior of ABCD.

In the drawing, there are as many triangles having E as a vertex as there are triangles having F as a

vertex. (why?)

The sum of the measures of all the angles in the triangles having E as a vertex is the same as the sum of the measures of all the angles in the triangles having F as a vertex. (why?)

Hence, we may generalize the proof to all convex polyhedral angles. Namely, the number of faces of polyhedral angle F is not important and if we can prove that the sum of the measures of the face angles of polyhedral angle F is smaller than 360° then we may accept that it holds for all convex polyhedral angles.

Thus, it is enough to prove 360°>∠AFD+∠AFB+∠BFC+∠CFD.

By Theorem 2.7, for the trihedral angles at A, B, C and D, we can write

If we add both sides of the inequalities side by side, we get:

∠ADE + ∠EAD + ∠EBA + ∠BAE + ∠ECB + ∠CBE + ∠EDC + ∠DCE is smaller than ∠FAD + ∠FDA + ∠FAB + ∠FBA + ∠FCB + ∠FBC + ∠FCD + ∠FDC.

In DFAD, DFAB, DFDC and DFCB, we can write

∠FAD + ∠FDA = 180° – ∠AFD

∠FAB + ∠FBA = 180° – ∠AFB

∠FCB + ∠FBC = 180° – ∠BFC

∠FCD + ∠FDC = 180° – ∠CFD.

Similarly, in DEAD, DEDC, DECB and DEBA, we get

∠ADE + ∠EAD = 180° – ∠AED

∠EBA + ∠BAE = 180° – ∠AEB

∠ECB + ∠CBE = 180° – ∠BEC

∠EDC + ∠DCE = 180° – ∠CED.

If we substitute these values into the above inequality, we obtain

∠AED+∠AEB+∠BEC+∠CED > ∠AFD+∠AFB+∠BFC+∠CFD.

In quadrilateral ABCD, the sum of the angles around point E is 360°.

Thus, we conclude that 360° > ∠AFD + ∠AFB + ∠BFC + ∠CFD.

3. a. In trihedral angle V-PQR, let the measures of dihedral angles VP, VQ and VR be respectively

x, y and z.

On the edges of the trihedral angle, let us choose A, B, C so that VA = VB = VC and draw AB, BC

and AC as in Figure 2.28.

Through any point D on VA, if we draw a plane perpendicular to VA, cutting face VAB along DK, face VAC a long DL and the plane of ABC along KL, then ∠KDA = ∠LDA = 90°.

Therefore, ∠KDL is the plane angle of dihedral angle VP. In other words, ∠KDL = x.

On the other hand, D-AKL is a trihedral angle.

By Theorem 2, ∠KDA + ∠LDA + ∠KDL < 360°, 90° + 90° + x < 360°.

So x < 180°.

Similarly, we can obtain y < 180° and z < 180°.

B.

So x + y + z < 540°.

c. In DKDL, if we draw DN and AN so that DN ⊥ KL then by the three perpendiculars theorem, AN ⊥ KL.

Therefore, in right triangles ANK and DNK, each of angles NDK and NAK is acute.

.

Since AK is the hypotenuse of right triangle ADK, DK < AK and therefore,

sin (∠NAK) < sin (∠NDK).

In other words, m(∠NAK) < m(∠NDK) because they are both acute angles.

Similarly, we can obtain m(∠NDL) > m(∠NAL).

So, we can write, m(∠KDL) > m(∠KAL).

Since x = m(∠KDL) and m(∠BAC) = m(∠KAL), x > m(∠BAC).

In the same manner, we can prove that, y > m(∠ABC) and z > m(∠BCA).

On the other hand, in DABC, m(∠BAC) + m(∠ABC) + m(∠BCA) = 180°.

Therefore, x + y + z > m(∠BAC) + m(∠ABC) + m(∠BCA),

x + y + z > 180°.

4. Let us choose points A, B, C, E, F and G on the edges of the trihedral angles so that

VA = VB = VC = WE = WF = WG and draw AB, BC, AC, EF, FG and EG as in Figure 2.29.

Given that angles AVB, BVC, CVA are equal respectively to angles EWF, FWG, GWE.

Therefore, by S.A.S., we can write

ΔAVB ≅ ΔEWF, ΔBVC ≅ ΔFWG and ΔCVA ≅ ΔGWE.

Hence, AB = EF, BC = FG and CA = GE.

By S.S.S., ΔABC ≅ ΔEFG.

So, ∠BAC = ∠FEG.

Through any point D on VA, let us draw DK in face AVB and DL in face AVC so that DK ⊥ VA and DL ⊥ VA and draw KL.

Then let us take point H on WE so that AD = EH and draw HM, HN and MN in the same manner.

Therefore HM ⊥ WE and HN ⊥ WE.

Since, ∠BAV = ∠FEW, AD = EH and ∠ADK = ∠EHM = 90°, ΔADK ≅ ΔEHM.

Therefore, AK = EM and DK = HM.

Similarly, we can get AL = EN and DL = HN.

So, by S.A.S., ΔKAL ≅ ΔMEN (∠BAC = ∠FEG)

Then, KL = MN and therefore, by S.S.S., ΔKDL ≅ ΔMHN.

Hence, we can conclude that ∠KDL = ∠MHN.

Namely, dihedral angles VA and WE are equal.

In the same manner, we can prove that dihedral angles VB and VC are equal to dihedral angles WF and WG respectively.

So, the corresponding parts (face angles and dihedral angles) of trihedral angles V-ABC and W-EFG are equal to each other, respectively.

By the definition, if their corresponding parts are arranged in the same order then trihedral angles V-ABC and W-EFG are congruent.

Otherwise, they will be symmetric.

Example 57: Decide whether a trihedral angle can be constructed in which the face angles are respectively

a) 60°, 40°, and 110° b) 65°, 150°, and 155°

c) 60°, 80°, and 100° d) 120°, 120°, and 120°

Solution:

a) Since the sum of the measures of any two face angles must be greater than the measure of the third face angle, the trihedral angle can not be constructed (60°+40° is not greater than 110°).

b) The sum of the measures of the face angles is greater than 360°. But by Theorem 2, it can not be. So, the trihedral angle can not be constructed with the given face angles.

c) Sum of any two face angles is always greater than the third one and sum of all face angles is smaller than 360° so it is possible to construct a trihedral angle.

d) Sum of the face angles must be smaller than 360° but 120° + 120° + 120° = 360°. So it is not possible to construct. When we add them we will get a plane.

Example 58: Can a trihedral angle be constructed with the following dihedral

angles?

a) 70°, 45°, 45° b) 60°, 185°, 35°

c) 90°, 90°, 90° d) 20°, 30°, 120°

Solution: a) Since the sum of the dihedral angles is less than 180°, the trihedral angle cannot be constructed.

b) The trihedral angle cannot be constructed because one of the dihedral angles is greater than 180°.

c) Each dihedral angle is smaller than 180°, sum of the angles is between 180° and 540° so it is possible to construct a trihedral angle.

d) Since the sum of the dihedral angles is 20° + 30° + 120° = 170 < 180°, the trihedral angle cannot be constructed.

Дата добавления: 2015-07-20