/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 34 Imagine a triangle \(A B C\) on ... [FREE SOLUTION] | 91影视

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Chapter 1: Problem 34

Imagine a triangle \(A B C\) on the unit sphere (with radius \(r=\) 1), with angle \(\alpha\) between \(A B\) and \(A C,\) angle \(\beta\) between \(B C\) and \(B A,\) and angle \(\gamma\) between \(C A\) and \(C B\). You are now in a position to derive the remarkable formula for the area of such a spherical triangle. (a) Let the two great circles containing the sides \(A B\) and \(A C\) meet again at \(A^{\prime} .\) If we imagine \(A\) as being at the North pole, then \(A^{\prime}\) will be at the South pole, and the angle between the two great circles at \(A^{\prime}\) will also be \(\alpha .\) The slice contained between these two great circles is called a lune with angle \(\alpha\) (i) What fraction of the surface area of the whole sphere is contained in this lune of angle \(\alpha ?\) Write an expression for the actual area of this lune. (ii) If the sides \(A B\) and \(A C\) are extended backwards through \(A,\) these backward extensions define another lune with the same angle \(\alpha,\) and the same surface area. Write down the total area of these two lunes with angle \(\alpha\) (b)(i) Repeat part (a) for the two sides \(B A, B C\) meeting at the vertex \(B,\) to find the total area of the two lunes meeting at \(B\) and \(B^{\prime}\) with angle \(\beta\). (ii) Do the same for the two sides \(C A, C B\) meeting at the vertex \(C,\) to find the total area of the two lunes meeting at \(C\) and \(C^{\prime}\) with angle \(\gamma\). (c)(i) Add up the areas of these six lunes (two with angle \(\alpha,\) two with angle \(\beta,\) and two with angle \(\gamma\) ). Check that this total includes every part of the sphere at least once. (ii) Which parts of the sphere have been covered more than once? How many times have you covered the area of the original triangle \(A B C ?\) And how many times have you covered the area of its sister triangle \(A^{\prime} B^{\prime} C^{\prime} ?\) (iii) Hence find a formula for the area of the triangle \(A B C\) in terms of its angles \(-\alpha\) at \(A, \beta\) at \(B,\) and \(\gamma\) at \(C\)

Short Answer

Expert verified
The area of triangle \( ABC \) is \( (\alpha + \beta + \gamma - \pi) \).

Step by step solution

01

Understanding the Lune's Fraction

The lune, defined by two great circles meeting at angle \( \alpha \), represents a fraction \( \frac{\alpha}{2\pi} \) of the total surface area of the sphere, since the whole sphere corresponds to an angle \( 2\pi \) radians.
02

Calculating Lune Area

To find the actual area of the lune, multiply the sphere's total surface area \( 4\pi r^2 \) (with radius \( r = 1 \)) by the fraction \( \frac{\alpha}{2\pi} \). Therefore, the area of one lune is \( 2\alpha \).
03

Total Area of Two Lunes with Angle \( \alpha \)

The area of two lunes, each with angle \( \alpha \), is simply twice the area of one lune. Therefore, the total area is \( 4\alpha \).
04

Repeating for Angles \( \beta \) and \( \gamma \)

Similarly, for angle \( \beta \), the total area of its two lunes is \( 4\beta \). For angle \( \gamma \), the total area of the two lunes is \( 4\gamma \).
05

Sum of All Lune Areas

Add the areas of all six lunes together: \( 4\alpha + 4\beta + 4\gamma \). This results in the total area being \( 4(\alpha + \beta + \gamma) \).
06

Overlapping Coverage Analysis

The union of these areas covers the sphere, but overlaps the spherical triangle \( ABC \) three times: once directly, and once for each of the two pairs of opposite triangles (including \( A'B'C' \)).
07

Deriving the Area Formula

The total of \( 4(\alpha + \beta + \gamma) \) includes the area of the sphere \( 4\pi \) plus the two times the spherical excess. The area of the triangle \( ABC \) is the spherical excess: \( (\alpha + \beta + \gamma) - \pi \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Spherical Triangles
A spherical triangle is a fascinating concept from spherical geometry, which is the geometry of shapes on a sphere's surface. Unlike traditional triangles on a flat plane, spherical triangles have sides formed by arcs of great circles. Great circles are circles on a sphere's surface whose center aligns with the sphere's center, much like the equator on Earth. These triangles help us understand complex geometrical relationships on curved surfaces. They are defined by three vertices and three angles. The angles of a spherical triangle are measured in radians and always add up to more than 180 degrees, distinguishing them from their planar counterparts.
Great Circles
Great circles are the largest circles that can be drawn on a sphere's surface. They are significant in spherical geometry because they act like straight lines do in plane geometry. A great circle divides the sphere into two equal hemispheres.
  • For navigators and pilots, great circles are vital because they represent the shortest path between any two points on the sphere.
  • Any two great circles intersect at two points, which is crucial in forming the sides of a spherical triangle.
Great circles are the building blocks for understanding spherical shapes, being equivalent to geodesics, or the shortest paths, on the sphere.
Lune
A lune in spherical geometry is a two-dimensional surface on a sphere, shaped like a crescent. It is formed by two great circles that intersect at the sphere's center, creating a wedge-like shape.
  • Each lune is defined by its angle, known as the lune angle, measured between the two intersecting great circle arcs.
  • The area of a lune on a unit sphere (radius 1) is computed using the formula: \(2\text{(lune angle)}\), meaning if the lune angle is \(\alpha\), the lune has an area of \(2\alpha\).
  • In problems like the one provided, lunes help parcel up the sphere's surface for further calculations, such as determining spherical triangle areas.
Understanding lunes is instrumental in comprehending how areas can be dissected and quantified on a curved surface.
Spherical Excess
Spherical excess is a unique property of spherical triangles where the sum of their interior angles exceeds 180 degrees. This excess is a measure of the triangle's area on the sphere.The concept of spherical excess helps us derive the area of spherical triangles:
  • The spherical excess is calculated as \(E = \alpha + \beta + \gamma - \pi\), where \(\alpha, \beta,\text{ and }\gamma\) are the angles of the spherical triangle.
  • The area of a spherical triangle is directly proportional to its spherical excess on a unit sphere. Hence, the area is precisely equal to the spherical excess itself.
  • This relationship highlights how spherical geometry deviates from planar geometry, where angles are bounded within 180 degrees.
Mastering the concept of spherical excess is crucial for solving problems related to spherical triangles.

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Most popular questions from this chapter

(a) Compute by mental arithmetic (using pencil only to record results), then learn by heart: (i) the squares of positive integers: first up to \(12^{2}\); then to \(31^{2}\) (ii) the cubes of positive integers up to \(11^{3}\) (iii) the powers of 2 up to \(2^{10}\). (b) How many squares are there: (i) \(<1000 ?\) (ii) \(<10000 ?\) (iii) \(<100000 ?\) (c) How many cubes are there: (i) (ii) (iii) \(<1000000 ?\) (d) (i) Which powers of 2 are squares? (ii) Which powers of 2 are cubes? (e) Find the smallest square greater than 1 that is also a cube. Find the next smallest. Evaluating powers, and the associated index laws, constitute an example of a direct operation. For each direct operation, we need to think carefully about the corresponding inverse operation - here "extracting roots". In particular, we need to be clear about the distinction between the fact that the equation \(x^{2}=4\) has two different solutions, while \(\sqrt{4}\) has just one value (namely 2).

(Pythagoras' Theorem) Let \(\triangle A B C\) be a right angled triangle, with a right angle at \(C .\) Draw the squares \(A C Q P, C B S R,\) and \(B A U T\) on the three sides, external to \(\triangle A B C\). Use the resulting diagram to prove in your head that the square \(B A U T\) on \(B A\) is equal to the sum of the other two squares by: \- drawing the line through \(C\) perpendicular to \(A B\), to meet \(A B\) at \(X\) and \(U T\) at \(Y\) \- observing that \(P A\) is parallel to \(Q C B\), so that \(\triangle A C P\) (half of the square \(A C Q P,\) with base \(A P\) and perpendicular height \(A C)\) is equal in area to \(\triangle A B P\) (with base \(A P\) and the same perpendicular height) \- noting that \(\triangle A B P\) is SAS-congruent to \(\triangle A U C,\) and that \(\triangle A U C\) is equal in area to \(\triangle A U X\) (half of rectangle \(A U Y X,\) with base \(A U\) and height \(A X)\) \- whence \(A C Q P\) is equal in area to rectangle \(A U Y X\) \- similarly \(B C R S\) is equal in area to \(B T Y X\). The proof in Problem 18 is the proof to be found in Euclid's Elements Book 1, Proposition 47. Unlike many proofs, \- it is clear what the proof depends on (namely SAS triangle congruence, and the area of a triangle), and \- it reveals exactly how the square on the hypotenuse \(A B\) divides into two summands \(-\) one equal to the square on \(A C\) and one equal to the square on \(B C\).

Write out the first 12 or so powers of 4 : $$ 4,16,64,256,1024,4096,16384,65536, \ldots $$ Now create two sequences: the sequence of final digits: \(4,6,4,6,4,6, \ldots\) the sequence of leading digits: \(4,1,6,2,1,4,1,6, \ldots\) Both sequences seem to consist of a single "block", which repeats over and over for ever. (a) How long is the apparent repeating block for the first sequence? How long is the apparent repeating block for the second sequence? (b) It may not be immediately clear whether either of these sequences really repeats forever. Nor may it be clear whether the two sequences are alike, or whether one is quite different from the other. Can you give a simple proof that one of these sequences recurs, that is, repeats forever? (c) Can you explain why the other sequence seems to recur, and decide whether it really does recur forever?

Reveals the triple (3,4,5) as the first instance \((m=1)\) of a one-parameter infinite family of triples, which continues $$ (5,12,13)(m=2),(7,24,25)(m=3),(9,40,41)(m=4), \ldots $$ whose general term is $$ (2 m+1,2 m(m+1), 2 m(m+1)+1) $$ The triple (3,4,5) is also the first member of a quite different "one- parameter infinite family" of triples, which continues $$ (6,8,10),(9,12,15), \ldots $$ Here the triples are scaled-up versions of the first triple (3,4,5) . In general, common factors simply get in the way: If \(a^{2}+b^{2}=c^{2}\) and \(H C F(a, b)=s,\) then \(s^{2}\) divides \(a^{2}+b^{2},\) and \(a^{2}+b^{2}=c^{2} ;\) so \(s\) divides \(c .\) And if \(a^{2}+b^{2}=c^{2}\) and \(H C F(b, c)=s,\) then \(s^{2}\) divides \(c^{2}-b^{2}=a^{2},\) so \(s\) divides \(a .\) Hence a typical Pythagorean triple has the form \((s a, s b, s c)\) for some scale factor \(s,\) where \((a, b, c)\) is a triple of integers, no two of which have a common factor: any such triple is said to be primitive (that is, basic - like prime numbers). Every Pythagorean triple is an integer multiple of some primitive Pythagorean triple. The next problem invites you to find a simple formula for all primitive Pythagorean triples.

The "DIN A" series of paper sizes is determined by two conditions. The basic requirement is that all the DIN A rectangles are similar; the second condition is that when we fold a given size exactly in half, we get the next smaller size. Hence \- a sheet of paper of size \(\mathrm{A} 3\) folds in half to give a sheet of size \(\mathrm{A} 4-\) which is similar to \(\mathrm{A} 3 ;\) and \- a sheet of size \(A 4\) folds in half to give a sheet of size \(A 5 ;\) etc.. (a) Find the constant ratio \(r=\) "(longer side length) : (shorter side length)" for all DIN A paper sizes. (b)(i) To enlarge A4 size to A3 size (e.g. on a photocopier), each length is enlarged by a factor of \(r\). What is the "enlargement factor" to get from A3 size back to A4 size? (ii) To "enlarge" A4 size to A5 size (e.g. on a photocopier), each length is "enlarged" by a factor of \(\frac{1}{r} .\) What is the enlargement factor to get from A5 size back to A4 size?
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