1. Pappas Question of the Day
2. Van Brummelen Chapter 8: Stereographic Projection
3. Exercise 8.1
4. Exercise 8.3
5. Exercise 8.5
6. Exercise 8.6
7. Exercise 8.7
8. Exercise 8.8
9. Is Blaugust Back?
10. Conclusion
Pappas Question of the Day
Before we start with today's Pappas question, let me make an announcement about summer posts. I point out that the district whose calendar the blog follows will start next week, on August 15th. We have two chapters left in Van Brummelen's book, and I don't want our reading of this book to extend into a third summer.
I know -- it's hard to believe that the summer is almost over. But just today, the district sent out the email reminding us to get ready to return to substitute teaching.
Therefore my plan is to cover both Chapters 8 and 9 this week. I can't help but wish to follow the Pappas date pattern -- Chapter 8 on August 8th and Chapter 9 on the ninth. I'm actually posting this at 11PM Pacific Time on the seventh, so that I'll have all day on the eighth to read and answer some of the exercises for Chapter 9. But in all other time zones, today's post will be dated the eighth, and so I'm counting this as the August 8th post.
That extends to today's Pappas problem. And besides -- her problem for the seventh is on matrices, so I'd rather do her problem for the eighth, which is a Geometry problem.
And so today on her Mathematics Calendar 2018, Theoni Pappas writes:
For this cube find PQ = ?
(Here is the given information from the diagram: P and Q are opposite vertices of the cube, and the side length of the cube is sqrt(64/3).)
This is what we call a body diagonal of the cube, as opposed to a diagonal of just a face. Now it turns out that to calculate a body diagonal, we just use the "three-dimensional Pythagorean Theorem":
a^2 + b^2 + c^2 = d^2
This theorem usually appears in Geometry texts only to explain the 3D Distance Formula -- including the U of Chicago text, in Lesson 11-6. It's based on two applications of the Pythagorean Theorem -- the first to find a face diagonal, the second because the face diagonal and the side representing the third dimension form a right triangle whose hypotenuse is the body diagonal.
For a cube, all three dimensions are equal, so we can let them all equal s, the side of the cube:
s^2 + s^2 + s^2 = d^2
3s^2 = d^2
d = s sqrt(3)
Finally, we plug in the given value s = sqrt(64/3):
d = sqrt(64/3)sqrt(3)
d = sqrt(64)
d = 8
Therefore the body diagonal PQ = 8 -- and of course, this post is officially dated the eighth.
I prefer using that trick of letting s be a side, simplifying, and then plugging in s = sqrt(64/3), rather than having radicals all over the place. We know that Pappas had to use a radical for the side length so that the body diagonal winds up being a natural number (the date). If she had chosen an integer for the side, then the diagonal isn't an integer (the date) -- instead it would be that integer times sqrt(3).
Van Brummelen Chapter 8: Stereographic Projection
Chapter 8 of Glen Van Brummelen's Heavenly Mathematics is called "Stereographic Projection." It begins as follows:
"Astronomers needed to compute and observe long before the computer and the telescope. Before time-saving devices like logarithms and slide rules rescued astronomers from hours of drudgery, calculations were done by hand and were simply part of the job description. In fact, the word 'computer' referred originally to a person, not a machine."
In fact, one of those human computers is still alive and even had a movie made after her -- of course, I'm referring to Katherine Johnson.
The author continues by describing a 3D representation of the universe -- an astrolabe:
"The first component, the latitude plate (figure 8.1), resides inside a circular frame (the mater), and contains all celestial objects that do not move with respect to time: the horizon, the zenith, the North Pole, the celestial equator, and various other curves. The second component, the rete (figure 8.2), contains all the objects that move with the daily rotation of the celestial sphere, such as the ecliptic and the stars."
But how exactly can we build an astrolabe? Van Brummelen writes:
"We need a projection of the celestial sphere onto a flat surface so that objects stay in their proper places with respect to each other. It would be a significant bonus if the projection were to render the transformed curves so that an instrument maker could construct the device easily."
As it turns out, the transformation we need is called stereographic projection. This is a transformation much like the Common Core transformations (reflections, translations, and so on). But projections are unlike Common Core transformations. The latter map the plane to itself, and these are one-to-one correspondences, so they have inverses (and we used this fact in the extra proofs that I wrote back in my July 12th post). Projections, on the other hand, reduce the number of dimensions -- for example, the stereographic projection map a 3D sphere to a 2D plane. (On the other hand, the surface of the sphere maps uniquely to the plane, and so by restricting the domain, we have an inverse mapping the plane to the sphere's surface.)
The author describes stereographic projection more fully:
"Imagine a transparent celestial globe, with all points and curves of interest drawn onto it (figure 8.3). We construct a horizontal plane cutting through the sphere at the celestial equator, and we place a light source at the South Pole. Curves on the southern hemisphere, such as the Tropic of Capricorn, cast shadows on the part of the plane outside the equator. For curves on the northern hemisphere such as the Tropic of Cancer, imagine shadows being cast backwards; the shadows land on the part of the plane within the equator."
And these "shadows," of course, are the images of this projection. The domain is the sphere minus the South Pole, and the range is the plane:
"Of course, in practice we cannot build an infinitely large plane, so most astrolabes extended their representations of the celestial sphere southward only as far as the Tropic of Capricorn (see figure 8.1). Stereographic projection distorts areas dramatically; if the sphere to be projected is the Earth's surface, Antarctica would be an infinitely vast land mass surrounding the rest of the planet. Incidentally, this is just how many members of the Flat Earth Society consider Antarctica to situate itself in real life."
Which of the A-B-C-D properties does stereographic projection preserve? Clearly, these projections don't preserve D (distance). Neither do they preserve C (collinearity), but "cocircularity" is preserved in that circles are mapped to circles. (Here lines in the plane count as circles -- these correspond to circles on the sphere through the South Pole.) And B (betweenness) is not preserved (since this is directly related to colinearity). As it turns out, A (angle measure) is preserved -- the author gives the name conformal map to transformations preserving A. (Thus by definition, isometries and similarity transformations are also conformal maps.)
Back in my October 17th post, I wrote about a transformation called a "circle reflection." As it turns out, a circle reflection is the composite of an inverse stereographic projection (mapping the plane to the sphere minus one point) and a forward stereographic projection, except with the light source at the North Pole rather than the South Pole. And recall that the composite of two circle reflections in concentric circles is a dilation. Therefore, it follows that -- like stereographic projections -- circle reflections and dilations also preserve circles and angle measure.
Van Brummelen continues:
"One might wonder why we are discussing stereographic projection at all, since it seems to share little with spherical trigonometry beyond the use of the sphere. But projections have been at the heart of geometry and trigonometry for many centuries."
At this point Van Brummelen writes about "our favorite author Benjamin Martin," who wrote about this method in his 18th century book Young Trigonometer's Compleat Guide.
First, let's describe Van Brummelen's figure 8.4. Here C is the center of the unit sphere, and our goal is to project the great circle arc AG onto the plane, where DEFG is the circle where the sphere meets the plane -- the Equator/primitive circle. The points E and G is where our great circle meets the primitive circle, and
Here is Van Brummelen's derivation, which is attributable to Martin:
Derivation:
Supposing that we know the angle of inclination, how do we know where to draw A'? Consider figure 8.5, a view of the vertical cross-section of figure 8.4 through the center of the sphere, parallel to the page [that is, through
CA' = tan (1/2)(90 - angle of inclination).
Since we already know that the projected circle passes through E and G, we have located three points on this circle, which is enough to determine its position. [U of Chicago Lesson 4-5 -- dw]
Our next task is to locate the projection of a pole of great circle AG. Although it would be easy enough to add the pole to figure 8.4 and simply connect it with S, it was preferred that constructions remained on the primitive circle, to avoid having to draw in three dimensions. The portion of the projected circle that is within the primitive circle is EA'G (figure 8.6). Now imagine rotating the primitive circle out of the page, holding DCF in place but bringing E upward so that it is directly above C. Some of the points no longer refer to the same thing; for instance, G is now the point of projection S, at the South Pole. But all points on the line of measures DCA'F remain the same. [Recall that rotations in 3D don't have just a fixed point, but an entire fixed line -- the rotation's axis -- dw]
Now that we have rotated our circle, we can draw a line from the point of projection, now at G, through A' to reconstruct A. But we know our pole a is 90 degrees removed from A along this circle, so we trace out a 90-degree arc to locate a. Finally, connect a with G to determine the position of the projected pole a'. Since a' is on the line of measures, rotating our circle back to its original position at the primitive circle does not move a', and we have successfully constructed the projected pole.
Now Van Brummelen proceeds to use these projections to solve a triangle:
"Interested readers may follow along with Martin by looking up pages 150 to 152 in volume 2 of The Young Trigonometer's Compleat Guide, available online."
Hey, we're interested readers, aren't we? So here's the link:
https://archive.org/stream/youngtrigonomet01martgoog#page/n169
(I tried to use Google books, but I could only find Volume 1, not Volume 2.) I won't bother to repeat Van Brummelen's solution here, since it's so long. All he does is take Martin's solution and convert it to modern notation, which is helpful, but it's not worth it for me to type it all here.
So let's instead skip to the next section in Van Brummelen:
"Mathematics often advances in fits and starts with intervening periods of stability. Some new insight comes along and the field leaps forward, causing some of the existing ground to be disturbed."
You might have heard of the strange statement that 1 + 2 + 3 + 4 + 5 + 6 + ... = -1/12. This is due to something called "Cesaro summation," named for Italian mathematician Ernesto Cesaro. But now Van Brummelen describes a discovery of Ernesto's old brother Giuseppe, who wasn't even a mathematician, but a crystallographer. The author says of Giuseppe:
"Cesaro's idea, like all the great ones, is simple: project an arbitrary triangle ABC onto a plane using stereographic projection. Apply some identity from plane trigonometry to the projected triangle, and map the identity backward to the original spherical triangle ABC."
Van Brummelen now describes his figure 8.8. A spherical triangle ABC is placed with the vertex A at the North Pole. Recall that great circles through the South Pole (and thus the North Pole) are mapped to straight lines, and the North Pole is mapped to the center of the primitive circle. Thus sides AB and AC are mapped to segments
Derivation:
In the projection, extend tangent lines to the arc from B' and C', meeting at D. Since stereographic projection preserves angles, A'B'D is equal to Angle B on the original triangle, and A'C'D is equal to Angle C. Since the angles of quadrilateral A'B'DC sum to 360, we have:
Angle B'D'C' = 360 - (angle sum of spherical Triangle ABC)
= 360 - (180 + 2E) = 180 - 2E,
recalling that 2E is the triangle's spherical excess. By symmetry Angle C'B'D = B'C'D = E, which leads us to Angle A'B'C' = B - E and Angle A'C'B' = E.
The author adds:
"Here, finally, we see why in the previous chapter we defined the spherical excess to be 2E, rather than just E."
According to Van Brummelen, Cesaro refers to A'B'C' as the triangle of elements. In figure 8.9., the author shows us that we can apply the figure 8.5 result to cross-sections AA'SB'B and AA'SC'C to find two sides of the triangle of elements:
b' = tan(b/2) and c' = tan(c/2).
Derivation:
Finding a' is a nice geometrical exercise, which we solve differently from [Cesaro's colleague J.D.H.] Donnay. Since Triangle ABS is inscribed in a semicircle, it has a right angle at B. Thus Triangle ABS is similar to B'A'S, since they share two angles [AA~] (although we must be careful, since the similarity does not relate the vertices in the way suggested by the letter names). Thus SB/SA = SA'/SB'
or SA * SA' = SB * SB'. Likewise, on the right side of the figure we arrive at SA * SA' = SC * SC'. So SB * SB' = SC * SC' or SB/SC = SC'/SB', which implies that Triangles SBC and SCB' are similar since they share an angle at S [SAS~]. Combining these results, we have:
B'C'/BC = SC'/SB.
Now we must interpret each of the quantities in these ratios. The first is easy: B'C' = a'. To find BC, connect both B and C to A' and drop a perpendicular from A' to BC to discover that BC = 2sin(a/2). For SC', consider right triangle SA'C', which has SA' = 1 and Angle S = b/2, from which we have SC' = 1/cos(b/2). Finally, for SB consider right triangle SAB; we leave to the reader the conclusion that SB = cos(c/2).
Van Brummelen puts everything together to give us the cumbersome expression:
a' = sin(a/2)/(cos(b/2)cos(c/2)).
The author tells us that Cesaro multiplies all the sides by cos(b/2)cos(c/2) to make it easier. There are three more Cesaro triangles -- the derived triangle (or colunar triangle) formed by extending BA and BC to the antipodal point of B, the triangle of elements of the polar triangle, and the derived triangle of the polar triangle. Van Brummelen summarizes Cesaro's findings in figure 8.10:
Triangle of elements:
Angles: A, B - E, C - E
Sides: sin(a/2), sin(b/2)cos(c/2), sin(c/2)cos(b/2)
Derived triangle:
Angles: 180 - A, E, 180 - E
Sides: cos(a/2), sin(b/2)sin(c/2), cos(b/2)cos(c/2)
Triangle of elements of polar triangle:
Angles: 180 - a, s - b, s - c (where s = semiperimeter)
Sides: cos(A/2), cos(B/2)sin(C/2), cos(C/2)sin(B/2)
Derived triangle of polar triangle:
Angles: a, 180 - s, s - a
Sides: sin(A/2), cos(B/2)cos(C/2), sin(B/2)sin(C/2)
The point of all of this is that we can apply Euclidean trig identities to these four triangles to derive all of the spherical trig identities. Let's follow Van Brummelen's derivations:
Law of Cosines: Apply the planar Law of Cosines to the triangle of elements:
sin^2(a/2) = sin^2(c/2)cos^2(b^2) + sin^2(b/2)cos^2(c/2) - 2sin(b/2)cos(b/2)sin(c/2)cos(c/2)cos A.
Using the half-angle identities sin^2(theta/2) = (1/2)(1-cos theta), cos^2(theta/2) = (1/2)(1+cos theta):
(1/2)(1 - cos a) = (1/4)(1 - cos c)(1 - cos b) + (1/4)(1 - cos b)(1 - cos c)
- (1/2)sqrt((1 - cos b)(1 + cos b)(1 - cos c)(1 + cos c))cos A.
This can be simplified to:
cos a = cos b cos c + sin b sin c cos A.
Hint: (1 - cos b)(1 + cos b) = 1 - cos^2 b = sin^2 b.
Law of Sines: Apply the planar Law of Sines to the derived triangle:
cos(a/2)/sin A = sin(b/2)sin(c/2)/sin E.
Multiplying both sides by 2sin(a/2):
sin a/sin A = 2sin(a/2)sin(b/2)sin(c/2)/sin E.
The author explains that the RHS is symmetric wrt a, b, and c. This means that we can substitute b and c for a on the LHS to obtain:
sin a/sin A = sin b/sin B = sin c/sin C
as all three of these ratios are equal to the RHS.
Napier's Second Analogy: Apply the planar Law of Tangents to the triangle of elements. Yes -- I said tangents. Apparently, there's a Law of Tangents, but this isn't taught in high school trig as much as the Laws of Sines and Cosines are. The Law of Tangents is:
tan(1/2)(alpha - beta)/tan(1/2)(alpha + beta) = (a - b)/(a + b).
Set alpha and beta equal to the two angles at the bottom of the triangle:
tan(1/2)(B-C)/tan(1/2)(B+C-2E)=(sin(b/2)cos(c/2)-sin(c/2)cos(b/2))/(sin(b/2)cos(c/2)+sin(c/2)cos(b/2
But B + C - 2E = 180 - A, so:
tan(1/2)(B - C)/cot(1/2)A = sin(1/2)(b - c)/sin(1/2)(b + c),
which is Napier's second analogy. The other three of Napier's analogies may be obtained similarly.
Delambre's 2nd and 4th Analogies: Apply the planar Law of Sines to the triangle of elements:
sin(a/2)/sin A = sin(b/2)cos(c/2)/sin(B - E) = sin(c/2)cos(b/2)/sin(C - E).
At this point Van Brummelen uses two properties of ratios:
If w/x = y/z, then both ratios also equal (w + y)/(x + z) and (w - y)/(x - z).
I've seen something similar to this in other texts (often in the lessons on proportions or similarity), but not the U of Chicago text. Anyway, the author uses this to set up the sine sum/difference formulas:
sin(a/2)/sin A = sin(1/2)(b + c)/(sin(B - E) + sin(C - E)) = sin(1/2)(b - c)/(sin(B - E) - sin(C - E)).
We now apply the sum-to-product identities in the denominators. Here are the identities:
sin alpha + sin beta = 2sin(1/2)(alpha + beta)cos(1/2)(alpha - beta)
sin alpha - sin beta = 2cos(1/2)(alpha + beta)sin(1/2)(alpha - beta).
The author remarks that these are "identities largely forgotten by today's students." Anyway, we get:
sin(a/2)/(2sin(1/2)A cos(1/2)A) = sin(1/2)(b + c)/(2cos(1/2)A cos(1/2)(B - C))
= sin(1/2)(b - c)/(2sin(1/2)A sin(1/2)(B - C))
From this step Delambre's second and fourth analogies follow immediately:
sin(1/2)(A - B)/cos(1/2)C = sin(1/2)(a - b)/sin(1/2)c,
cos(1/2)(A - B)/sin(1/2)C = sin(1/2)(a + b)/sin(1/2)c.
Euler's Formula: No, not that Euler's formula (by which I assume Van Brummelen means the polyhedral formula V + F - E = 2 -- Euler had so many formulas). Instead, this is a formula that determines a triangle's spherical excess from its side lengths:
cos E = (1 + cos a + cos b + cos c)/(4cos(a/2)cos(b/2)cos(c/2)).
Proof:
Apply the planar Law of Cosines to the derived triangle:
sin^2(b/2)sin^2(c/2) = cos^2(a/2) + cos^2(b/2)cos^2(c/2) - 2cos(a/2)cos(b/2)cos(c/2)cos E
Double this expression and solve for the rightmost term; then factor the rest:
cos(a/2)cos(b/2)cos(c/2)cos E = cos^2(a/2) + (cos(b/2)cos(c/2) + sin(b/2)sin(c/2))
(cos(b/2)cos(c/2) - sin(b/2)sin(c/2))
or:
2cos(a/2)cos(b/2)cos(c/2)cos E = cos^2(a/2) + cos((b - c)/2)cos((b + c)/2).
Apply the identities (half-angle formula) cos^2(theta/2) = (1 + cos theta)/2 and (product-to sum f.) cos alpha cos beta = (cos(alpha + beta) + cos(alpha - beta))/2. Shuffle the terms, and we're done. QED
Van Brummelen writes:
"The divergent reactions to Donnay's book in the 1940's American mathematical community strike a familiar chord today, split between commitment to the practical payoff of the subject and appreciation for its intellectual elegance."
Indeed, it definitely strikes a familiar chord because it's just the traditionalist debate. For example, the author quotes B.M. Brown, who was more traditi -- um, pragmatic:
"This approach does not comment itself for the purpose of introducing students to spherical trigonometry...the total amount of preliminary work to be done more than offsets any subsequent advantage over the normal method."
And Van Brummelen summarizes Brown's position:
"Indeed, if the point of the study of mathematics is to generate answers to problems for engineers and scientists, then the overhead required by the theorems of stereographic projection is simply extra baggage."
The author contrasts this by quoting H.V. Craig:
"Among the sundry ills of the teaching of elementary mathematics, there are two in which the reviewer's opinions are serious, widespread, and chronic. One is the occurrence of rote methods including, of course, emphasis on the mere acquisition of manipulative techniques. The other is essentially a frame of mind -- a rigid and reactionary orthodoxy that insists on the strict segregation of mathematical concepts into compartments in accordance with well established custom."
If this had been the start of a modern article, I can already imagine Barry Garelick linking to it on his own blog and dismissing Craig as just another "reformer." Indeed, Van Brummelen concludes:
"Today's debates on the value of mathematics in education -- whether it is primarily a tool for science and commerce, or rather a journey of mental and conceptual cultivation -- have long and deep roots."
Wow -- so this is an actual traditionalist debate from 70 years ago! Oh, and which side of this debate would Van Brummelen be on? Well, the very fact that he even includes this chapter at all should speak for itself.
Exercise 8.1
Show that the stereographic image of an angle on the sphere is the same angle on the primitive plane. (In figure E-8.1, the angle in question is on the surface of the sphere at M. MtT and MrR are tangents to the great circles that form the angle at M; these tangents are drawn to meet the tangent plane to the sphere through S. Triangle mrt is on the plane of projection. Prove that Angle RMT = rmt. Hint: Notice that TM and TS are both tangents to the sphere through T, and are therefore equal.) [Brown 1913, 105-106]
Proof:
First, let's look at Van Brummelen's hint here. This is a theorem that often appears on the Pappas calendar and ought to appear in the U of Chicago text, yet it doesn't -- if two tangent segments to a circle (or sphere) are drawn from the same point, then the segments are congruent. This statement is proved using the HL Congruence Theorem. This is why TM = TS, as well as RM = RS.
A theorem that does appear in Lesson 13-5 of the U of Chicago text is that a tangent to a circle is perpendicular to the radius drawn at the point of tangency. Actually, the theorem for circles for stated, but not the analogous theorem for spheres -- a tangent line or plane to a sphere is perpendicular to the radius of the sphere drawn at the point of tangency. Thus Plane SRT, which is given to be tangent to the sphere at point S, must be perpendicular to the radius
As it turns out, the plane of projection mrt is also perpendicular to
So by the Two Perpendiculars Theorem, since both planes mrt and SRT are perpendicular to the same segment
You can't see the drawing (figure E-8.1), but I can tell you that M-SRT appears as a pyramid, and the plane mrt cuts this pyramid. The plane is parallel to the base of the pyramid. This implies that the corresponding segments are also parallel --
The rest of the proof fits the two-column format:
Statements Reasons
1. TM = TS, RM = RS 1. Van Brummelen's hint/discussion from above
2. Angle SMT = STM, 2. Isosceles Triangle Theorem
Angle SMR = SRM
3.
4. Angle STM = mtM, 4. Corresponding Angles Consequence
Angle STR = mtR
5. Angle SMT = mtM, 5. Transitive Property of Congruence
Angle SMR = mtR
6. tM = tm, rM = rm 6. Converse Isosceles Triangle Theorem
(Notice that SMT and mMt are the same angle, as are SMR and mMr.)
7. tr = tr 7. Reflexive Property of Congruence
8. Triangle rMt = rmt 8. SSS Congruence Theorem [steps 6,6,7]
9. Angle rMt = rmt 9. CPCTC
And notice that angle rMt is the same as RMT, so this is exactly what we set out to prove. QED
(I know, it's a bit confusing. From vertex M, we have rays MmS, MtT, and MrR, so that in the names of angles with vertex M, we have that t and T are interchangeable, as are r and R, but m interchanges with S, not M. Again, this is much easier if you can see the diagram.)
Exercise 8.3
Draw a chord within a circle. Connect the endpoints of that chord to the center of the circle, forming angle A at the center; then connect the endpoints of the chord to some other point on the far side of the circle, forming angle B at that far point. Show that Angle A = 2B.
In other words, prove the Inscribed Angle Theorem. Hey, we already proved the theorem back in Lesson 15-3 of the U of Chicago text, in my April 26th post. And that day, I also mentioned another book, Wayne Wickelgren's How to Solve Problems, which also proves this theorem. Well, it is an important theorem that shows up in many places (including the SBAC).
So let's cut and paste from my April 26th post, except I'll change the points around so that A is the center of the circle. In that proof, B is already the vertex of the inscribed angle, so all we need to do is change O to A at the center and A to, say, Z on the circle:
Given: Angle ZBC inscribed in Circle A
Prove: Angle ZBC = 1/2 * Arc ZC
Proof:
Case I: The auxiliary segment
Notice that the trick here was that between the central angle (whose measure equals that of the arc) and the inscribed angle is an isosceles triangle.
Let's move onto Case II. Well, the U of Chicago almost gives us a two-column proof here, so why don't we complete it into a full two-column proof. For Case II, A is in the interior of Angle ZBC.
Statements Reasons
1. A interior ZBC 1. Given
2. Draw ray BA inside ZBC 2. Definition of interior of angle
3. Angle ZBC = Angle ZBD + Angle DBC 3. Angle Addition Postulate
4. Angle ZBC = 1/2 * Arc ZD + 1/2 * Arc DC 4. Case I and Substitution
5. Angle ZBC = 1/2(Arc ZD + Arc DC) 5. Distributive Property
6. Angle ZBC = 1/2 * Arc ZC 6. Arc Addition Property and Substitution
The proof of Case III isn't fully given, but it's hinted that we use subtraction rather than addition as we did in Case II.
Returning to Van Brummelen, let's include the proof of Case III today then. It's basically cutting and pasting the same proof again but changing the + sign to -:
Statements Reasons
1. A exterior ZBC 1. Given
2. Draw ray BA outside ZBC 2. Definition of exterior of angle
3. Angle ZBC = Angle ZBD - Angle CBD 3. Angle Addition Postulate
4. Angle ZBC = 1/2 * Arc ZD - 1/2 * Arc CD 4. Case I and Substitution
5. Angle ZBC = 1/2(Arc ZD - Arc CD) 5. Distributive Property
6. Angle ZBC = 1/2 * Arc ZC 6. Arc Addition Property and Substitution
In Cases II and III, we proved that the inscribed angle is half of the intercepted arc. But the central angle equals the intercepted arc, so the inscribed angle B is half of the central angle A. In other words, we prove A = 2B. QED
All we need to do now is wait for me to get my next side-along reading book, which will probably ask the reader to prove the Inscribed Angle Theorem yet again....
Exercise 8.5
(The following three exercises work toward understanding Benjamin Martin's second right triangle construction in pp. 150-152 of The Young Trigonometer's Complete Guide, vol. 2.)
Let C be a point on the edge of the primitive circle. Show how to construct the stereographic projection of a circle with center C and a given radius.
Notice that even though circles map to circles, and points on the primitive circle such as C are fixed points, C will not be the center of the image circle! This is because stereographic projections do not preserve distance.
The preimage circle on the sphere will intersect the primitive circle in two points. These are also fixed points of the projection, and so they lie on the image circle as well. If r is the radius of the circle on the sphere, then the arc length between C and the two image points will also be r. Let's call these points H and I -- you'll find out why soon enough.
Now we must find the image of a third point. Two natural points to consider are the points on the same meridian as C, r units to the North and South of C.
Let's consider the cross section of the sphere containing C and both poles. This looks exactly like a unit circle, with the center of the sphere at (0, 0), C at (1, 0), the poles at (0, 1) and (0, -1), and the North and South preimage points are at (cos r, sin r) and (cos r, -sin r).
To find the image points, we draw a line from the light source at the South Pole (0, -1) to each of the preimage points. The x-intercept of each line is the image point, since the x-axis is where this plane (containing C, the poles, and preimage points) intersects the plane of the primitive circle.
Let's try the preimage point in the Northern Hemisphere first. It's easy to find the equation of the line passing through (0, -1) and (cos r, sin r) -- its slope is (1 + sin r)/cos r and its y-intercept is -1:
y = x(1 + sin r)/cos r - 1
To find the x-intercept, we set y = 0:
0 = x(1 + sin r)/cos r - 1
1 = x(1 + sin r)/cos r
x = cos r/(1 + sin r)
So the Northern image point lies on the ray from the center of the sphere to C, cos r/(1 + sin r) units from the center. It turns out that the Southern image point is:
x = cos r/(1 - sin r)
Now the point halfway between these two image points is the center of the image circle. The proof of this is a bit tricky. We start with the two fixed points, where the image and primitive circles meet (that is, H and I). So these two points are clearly equidistant from the center of the sphere, and they are also equidistant from C, since the arc length between each of them and C is r (congruent arcs, congruent chords).
Thus the line joining the center to C is the perpendicular bisector of
Now for any quadrilateral inscribed in a circle, its opposite angles are supplementary. (This isn't proved in the U of Chicago text, but it follows from the Inscribed Angle Theorem.) By the Kite Symmetry Theorem, we know that angles H and I are congruent -- and since they are both congruent and supplementary, the measure of each must be 90.
So by the Inscribed Angle Theorem, the arc between the Northern and Southern image points must be a semicircle. Thus the chord joining them is a diameter, and therefore its midpoint must be the center of the circle. QED
Now we can calculate this midpoint to find the center of the image circle:
center = (1/2)(cos r/(1 - sin r) + cos r/(1 + sin r))
= (1/2)((cos r + cos r sin r + cos r - cos r sin r)/(1 - sin^2 r))
= (1/2)(2 cos r/cos^2 r)
= 1/cos r
= sec r
The radius of this circle is also easily found:
radius = (1/2)(cos r/(1 - sin r) - cos r/(1 + sin r))
= (1/2)((cos r + cos r sin r - cos r + cos r sin r)/(1 - sin^2 r))
= (1/2)(2 cos r sin r/cos^2 r)
= sin r/cos r
= tan r
Notice that the center can never be the same as C (which is at 1) unless r = 0. In most cases, sec r > 1 and so the center lies outside the primitive circle.
A special case appears when r = 90. This means that the radius is a quadrant, and the preimage circle is a great circle (with C as one of its poles). This great circle passes through the North and South Poles, and so its image is a line. It passes through the center of the primitive circle and crosses it exactly 90 degrees away from C.
Exercise 8.6
Martin's goal in the second construction is to draw the projection of a triangle with the same elements as before (right angle at A, angle C = 56d 57', and BC = 44d 52'), but this time at the edge of the primitive circle rather than at the center. See figure E-8.6, taken from Martin's text [or just follow the Martin link above -- dw].
(a) First draw primitive circle DFCE and [perpendicular] diameters
By the way, if you can't see the diagram, at least draw a circle with C at the top, D at the bottom, E on the right, and F on the left.
I assume that this is the calculation from earlier, where we found:
distance from center = tan(1/2)(90 - angle of inclination)
The angle of inclination is given as 56d 57', so we write:
distance = tan(1/2)(90d - 56d 57')
= tan(1/2)(33d 03')
= tan(16d 31.5')
= 0.2967 (Press 2nd+MATRIX on TI graphing to find degrees and minutes.)
So G is placed this many units to the right on the diagram, towards E. We'll worry about B later.
(b) Use the construction of question 5 to draw the image of IBH, a circle with center C and radius 44d 52'. Finally, draw a line through the center and B, defining A and K.
OK, so from the previous problem, we know that this is a circle centered at (0, sec 44d 52') -- which is on the y-axis as C is placed on that axis -- with the other points H(sin 44d 52', cos 44d 52') as well as I(-sin 44d 52', cos 44d 52').
How do we find B? To me, it's easier to note that B lies on IBH, the circle with center (0, sec 44d 52') and radius tan 44d 52'. We also know that B lies on CBGD, but we're not given the radius or center of that circle -- all we know is that C(0, 1), D(0, -1), and G(tan 16d 31.5', 0) lie on the circle.
For simplicity, let's call the x-intercept of the circle (x, 0) for now. We wish to find the center of the circle that passes through the points (0, 1), (0, -1), and (x, 0). It's easy to see that the center must lie on the negative x-axis, so let's call it (-h, 0). The radius of this circle is then equal to x + h. Then the distance from (-h, 0) to (0, 1) must equal x + h:
sqrt(h^2 + 1) = x + h
h^2 + 1 = x^2 + 2xh + h^2
1 = x^2 + 2xh
1 - x^2 = 2xh
h = (1 - x^2)/(2x)
Now let's partially substitute in x = tan theta (since we already found x to be the tan something):
h = (1 - tan^2 theta)/(2theta)
h = (cot theta - tan theta)/2
R (radius of CBGD) = tan theta + h
R = (cot theta + tan theta)/2
Let's write these in terms of sines and cosines:
h = (cot theta - tan theta)/2
= (1/2)(cos theta/sin theta - sin theta/cos theta)
= (cos^2 theta - sin^2 theta)/(2sin theta cos theta)
= cos 2theta/sin 2theta
= cot 2theta
= tan(90 - 2theta)
R = (cot theta + tan theta)/2
= (1/2)(cos theta/sin theta + sin theta/cos theta)
= (cos^2 theta + sin^2 theta)/(2 sin theta cos theta)
= csc 2theta
= sec(90 - 2theta)
But notice that 90 - 2theta is just the original angle of inclination, 56d 57'. In other words:
Circle CBGD has center (-tan 56d 57', 0) and radius sec 56d 57'.
Circle IBH has center (0, sec 44d 52') and radius tan 44d 52'.
This is so simple, I'm surprised that neither Martin nor Van Brummelen ever brings this up!
So now what remains is to determine where these two circles intersect. This requires us to solve a system of equations -- not linear equations, but the equations of circles. To make it easier, let's write 56d 57' as C (for angle C) and 44d 52' as BC (for arc BC):
(x + tan C)^2 + y^2 = sec^2 C
x^2 + (y - sec BC) = tan^2 BC
We can expand the first equation and use a Pythagorean identity:
(x + tan C)^2 + y^2 = sec^2 C
x^2 + 2x tan C + tan^2 C + y^2 = sec^2 C
x^2 + y^2 + 2x tan C = sec^2 C - tan^2 C
x^2 + y^2 + 2x tan C = 1
We can do the same to the second equation. Here is the resulting system:
x^2 + y^2 + 2x tan C = 1
x^2 + y^2 - 2y sec BC = -1
2x tan C + 2y sec BC = 2
x tan C + y sec BC = 1
But this equation, while linear, still contains both x and y. Notice that the two circles will intersect in two points, and thus this is the equation of the line containing both.
But at this point, we can just use substitution. We solve the linear equation for y and then substitute it into the first equation, since it contains only one y:
x tan C + y sec BC = 1
y sec BC = 1 - x tan C
y = (1 - x tan C)/sec BC
x^2 + y^2 + 2x tan C = 1
x^2 + (1 - x tan C)^2/sec^2 BC + 2x tan C = 1
This is just a quadratic equation in x, so we should be able to use the Quadratic Formula to solve it. It turns out that the work required is very tedious. Fortunately at least the square root in the formula disappears, since the discriminant is a perfect square (check out the plus-or-minus +/- sign below):
x = (tan C cos^2 BC - tan C +/- sec C sin BC)/(tan^2 C cos^2 BC + 1)
Normally we'd plug this in to one of the equations and solve for y. Instead, I decided to solve the linear equation above for x and then substitute it into the second circle equation:
y = (cot^2 C sec BC + sec BC +/- csc C tan BC)/(cot^2 C sec^2 BC + 1)
We look at Van Brummelen's/Martin's diagram and see that point B, the desired point of intersection, lies to the lower-right on the line joining the two intersections (that is, it lies inside the primitive or unit circle). This suggests that at the +/- signs, we should choose the + sign for x and - for y. Plugging in the angles for C and BC gives:
x = 0.2418 or x = -0.9414
x = 1.734 or y = 0.4454
and the desired point is B(0.2418, 0.4454).
In solving this problem, I retained the exact angles C and BC as long as possible because we were able to obtain simple values for the center and radius of the circles -- for example, we found that Circle CBGD has center (-tan C, 0) and radius sec C (something that might help us solve other triangles later on). But the coordinates of B isn't just a simple trig function of C or BC, and so by this point it's better just to use the decimals.
After all, we still haven't found A or K yet. This line passes through the origin, and so to find it's equation, we just divide the values of x and y we found earlier:
y = 1.842x
Notice that this is the value of tan theta (where "theta" here refers to the angle made between x-axis and line ABK). Then A has coordinates (cos theta, sin theta), since it lies on the unit circle:
tan theta = 1.842
theta = 61.5
cos theta = 0.4771
sin theta = 0.8788
Therefore A has coordinates (0.4771, 0.8788), and K is its antipodal point, (-0.4771, -0.8788).
Exercise 8.7
(Originally I wasn't going to solve 8.7 -- ordinarily I choose random questions to answer, and I'd chosen 8.5 and 8.6, but not 8.7. But Exercises 8.5-8.7 go together -- and besides, the other question I'd randomly selected, 8.4, is just mindless substituting values into Martin's method. This question, which is Martin from another perspective, requires more thought -- thus it's more interesting.)
(a) Measure AB on the diagram from question 6, and use this value to determine Arc AB.
Well, we have points A(0.4771, 0.8788) and B(0.2418, 0.4454) found earlier, and so we only need the Distance Formula to find AB:
AB = sqrt((0.4771 - 0.2418)^2 + (0.8788 - 0.4454)^2)
AB = 0.4932
But how do we determine Arc AB from this? Notice that back in Exercise 8.5, we found the equation:
x = cos r/(1 + sin r)
to find the coordinate of the image of a point in the Northern Hemisphere on the meridian of C (well, technically we should say y =, not x =, since we eventually placed C on the y-axis).
But we can rotate the sphere easily, so this same formula should help us find the image of a point on the meridian of any point on the primitive circle, including A. We find the distance between the image of B and the center of the circle, which is 1 - AB = 0.5068. We then place this on the LHS:
0.5068 = cos r/(1 + sin r)
Solving this on a calculator (since I don't know an analytic way of solving this), we obtain a value of 36.25 degrees for AB. Notice that Martin has already found a value for AB earlier (in the part of his book that I didn't feel like quoting) -- 36d 15'. This means that our value is correct -- after all, there are 60 minutes in a degree, so 0.25 degree = 1/4 degree = 15 minutes.
(b) Determine AC. (This is a simple measurement.)
Actually, for us that is very simple. In Exercise 8.6, we already found theta = 61.5, where theta is the angle between A and the positive x-axis (that is, AE). Then AC is just the angle between A and the positive y-axis, so we find AC = 90 - 61.5 = 28.5.
Again, Martin already found a value for AC -- 28d 30'. It's clear that we've found the correct answer, since 0.5 degree = 1/2 degree = 30 minutes.
(c) Find Angle B, in a manner similar to how Martin found the same angle in the problem solved in this chapter.
Once again, Van Brummelen directs us to Martin's previous work. The confusing part is that Martin uses the same letters to stand for different points in the two diagrams.
Anyway, Martin's idea is to find the poles of the two great circles intersecting at B -- and these are known as ABK and CBGD. The pole of ABK is easy to find -- the line passes through the North and South Poles, hence its pole lies on the Equator (primitive circle), 90 degrees away from A. On Martin's diagram, this point is labeled M, and it lies 28d 30' clockwise from the x-axis. (That is the negative x-axis. Of course, all great circles have two poles, but only M is labeled.)
Now we must find the pole of CBGD. It crosses the primitive circle at C and D with an angle of inclination of 56d 57'. It follows that its pole lies on the meridian from E to F, with 90 - 56d 57', or 33d 03', as the distance from this pole to the primitive circle. Martin labels this point a.
Van Brummelen explains what Martin does next:
"Angle B is the angle of inclination between great circles, and the angle of inclination between great circles is equal to the distance between their poles. So we are really after Ma. But B, lying on both great circles, is 90 degrees removed from both M and a, so it is a pole of Ma. To determine Ma, extend lines from the projected pole B through the endpoints of Ma to the edge of the primitive circle. (The fact that M is already on the edge is a convenience, and is no obstacle to the argument.) Then, as we saw above, Mm will measure the length of Ma."
So far, we've been calculating all of our values, since you don't have the picture in front of you. But this is, unfortunately, where our calculations must end. There's no easy way to determine what the coordinates of m actually are. The arc maB does not map to a straight line, since only arcs through the North and South Poles do that (and FaE passes through the N/S Poles, not maB). Thus the image of maB must be a circle, and there's no easy way to find its center and radius. Otherwise we would find its equation and determine where it intersects the primitive circle (x^2 + y^2 = 1) to find m.
But based on the picture, maB appears to be very close to a line. So maybe we can at least estimate m by finding where line aB intersects the x-axis.
The coordinate of a is easy to find:
a = -cos r/(1 + sin r)
a = -cos 33d 03'(1 + sin 33d 03')
a = (-0.5424, 0)
So line aB passes through (-0.5424, 0) and (0.2418, 0.4454). Its slope is 0.568:
y = 0.568(x + 0.5424)
y = 0.568x + 0.3081
x^2 + y^2 = 1
x^2 + (0.568x + 0.3081)^2 = 1
1.323x^2 + 0.35x - 0.9051 = 0
x = -0.97 or x = 0.7054
We clearly want the negative value on the chart, -0.97. Instead of finding y, we just find its inverse cosine to determine that m is 14.07 degrees (14d 04') counterclockwise from the negative x-axis.
Since M is 28d 30' clockwise from the x-axis, we find that Mm (and hence B) is 42d 34'. Let's check to see what Martin obtained for Mm -- hmm, it's 42d 34' after all. So maybe Martin really did just draw maB as a line after all. Even Van Brummelen points out that Martin isn't always forthcoming with his justifications -- I drew the line and it matches Martin's answers, which is all that Van Brummelen asks us to do in this problem.
Exercise 8.8
When working out the side lengths of the triangle of elements and the derived triangle, Cesaro chooses A to be at the North Pole. What would happen if we chose B or C to be at the North Pole instead?
Well, for this problem, I see no reason why we wouldn't just obtain the same expressions but with the variables switched. In other words:
Triangle of elements:
A at North Pole: angles A, B - E, C - E, sides sin(a/2), sin(b/2)cos(c/2), sin(c/2)cos(b/2)
B at North Pole: angles B, C - E, A - E, sides sin(b/2), sin(c/2)cos(a/2), sin(a/2)cos(c/2)
C at North Pole: angles C, A - E, B - E, sides sin(c/2), sin(a/2)cos(b/2), sin(b/2)cos(a/2)
As for the derived triangle, recall that this is the triangle of elements of the colunar triangle formed by extending two sides of the original triangle to the antipodes of B. A simple substitution won't work if we place B at the North Pole. So instead, we assume that if the Pole is at B, then we extend the colunar triangle to the antipodes of C, and if the Pole is at C, then we extend the colunar triangle to the antipodes of A:
Derived triangle:
A at North Pole: angles 180 - A, E, A - E, sides cos(a/2), sin(b/2)sin(c/2), cos(b/2)cos(c/2)
B at North Pole: angles 180 - B, E, B - E, sides cos(b/2), sin(c/2)sin(a/2), cos(c/2)cos(a/2)
C at North Pole: angles 180 - C, E, C - E, sides cos(c/2), sin(a/2)sin(b/2), cos(a/2)cos(b/2)
And if we try to derive, say, the Law of Cosines from the triangle of elements with B at the North Pole, then we obtain the version of the Law with cos b isolated:
cos b = cos c cos a + sin c sin a cos B.
Is Blaugust Back?
Two years ago, a certain blogger -- a math teacher named Shelli -- set up a Blaugust challenge as part of the MTBoS. And I participated in Blaugust in preparation for my first (and only so far) year of teaching math.
Last year, Shelli didn't post the Blaugust challenge -- but this year, Blaugust is back!
http://statteacher.blogspot.com/
http://statteacher.blogspot.com/2018/07/all-about-mtbosblaugust-2018.html
I like participating in MTBoS challenges, but I don't think it's right to for me to do so this year. To me, Blaugust is for real math teachers -- and right now I'm not in the classroom, so therefore I'm not a real math teacher.
In fact, even though Shelli doesn't require it of all participants, I like to follow the list of prompts that she provides for Blaugust. Since today's post is dated the eighth, I'd respond to her eighth prompt:
Answer the prompts from Julie R’s #TMC18 keynote: I am a great teacher because… I am a teacher leader because… I want to grow as a #teacherleader this year by…
Again that's just it. I'm not a great teacher because if I were, I'd be in the classroom now -- and I wouldn't have had to leave the classroom my first year. For me, full participation in Blaugust is sad because it just reminds me how much I miss the classroom.
I will point out that the "Julie R" mentioned in this prompt refers to Julie Reulbach:
https://ispeakmath.org/
https://ispeakmath.org/2018/07/26/teacher-leader-you-are-enough-keynote-tmc18/
Notice that Reulbach's most recent post, as well as Shelli's seventh prompt, both refer to yet another math teacher, Sam Shah:
https://samjshah.com/
Now Shelli, Reulbach, and Shah -- those are great math teachers. They have classrooms that they're returning to by the end of this month. Even Barry Garelick, who just completed his second year of teaching, is a great math teacher in his own way. His traditionalist teaching style is at odds with that of most MTBoS members, and even though he has a blog, he won't make any Blaugust posts any time soon. Yet he has his own classroom to go to this month.
But I don't have a classroom to go to -- except, of course, just for one day as a sub. Therefore I am not a great math teacher -- at least not yet.
Two years ago, Shelli welcomed me to Blaugust by leaving the following comment:
Shelli:
If you are on twitter, there are some active science teachers there. If you are the only math/science teacher, I would highly recommend joining twitter because it provides you an online math/science department at your fingertips. I'm @druinok on there - come say hi! :)
And I told her that I was considering creating my own Twitter account. But that fell by the wayside when I joined Tina Cardone's "Day in the Life" challenge instead. At the time, it made more sense for me to join Cardone than Twitter -- after all, Twitter entailed creating a new account, while I already had this blog, which was all I needed for Cardone's challenge.
But now, in hindsight, I realize that I should have taken up Shelli on her offer and joined Twitter. I struggled to teach science that year, yet, as she pointed out, I might have met someone on Twitter who could help me with science. And becoming a better science teacher would have immediately helped me out with classroom management. After all, one reason I struggled with management was that the students didn't respect me -- and much of that was because I was supposed to be their science teacher and I failed to teach science.
Recall that I didn't blog all 180 days that school year. Each day that I didn't blog, I should have tweeted instead. After all, I was too busy to blog everyday, but it takes only a few minutes to type 140 characters (now 280, but 140 at the time). I now believe that if I had tweeted and communicated with more experienced teachers, I would have survived my first year -- and today I'd be preparing for my third year of teaching. I'd be on my way to becoming a great teacher.
Instead, I'm stuck as a sub. If, by chance, I happen to get hired as a regular teacher by the end of this month, then I'll sign my name on Shelli's page and formally join Blaugust. But most likely, I won't be hired at all this year -- and my name doesn't belong on Shelli's list of great math teachers. I'm not fit to rejoin the MTBoS.
Conclusion
This isn't a traditionalist post, but since this is one of my last summer vacation posts, it's my last chance to mention politics before the school year, when I try (if possible) to avoid political posts.
Here is an analogy to consider all of the recent political tracking discussions out there -- especially the ones where the discussion about tracking turns into a discussion about race (or gender).
Consider dividing up all Americans into two groups -- the idea is that the groups don't correspond to race or gender, but are more or less arbitrary.
To determine which group you're in, count all the letters in your first, middle, and last name. Then one group includes all people with an odd number of letters, and the other group includes all people with an even number of letters. For example, my first name has five letters, my middle name has three letters, and my last name has six letters, for a total of 14. Therefore I'm an "even." We refer to the even/odd division as "parity."
I decided upon an even/odd split since, for example, a long/short name split might unwittingly correlate very slightly with race. Some languages (such as Eastern European) have longer names, while others (such as Chinese and Korean) have shorter names. Certain Eastern European languages also add the letter "a" to the end of women's last names -- and that single letter might be enough to push more women into the longer name group.
The even/odd split also worries me slightly. Some languages, such as Japanese, tend to alternate between consonants and vowels, so names from this language might tilt towards even lengths. But I assume that there are enough exceptions (names starting with a vowel, the clusters "ch" or "sh") to make the split race-independent. If the even/odd split becomes a problem, we might try dividing into three groups -- multiples of three, one more than a multiple, one less than a multiple.
Anyway, suppose we divide everyone into the even/odd groups. Now suppose that we're at a school where there are two math tracks, a high track and a low track. My question is, do we expect there to be more evens on the high track, or more odds?
The only reasonable answer is that the high track is to be equally split between evens/odds. There's absolutely no reason to expect a correlation between parity and track. I'd be surprised if all the evens were on the low track and all the odds were on the high track. After all, parity of letters in the name is just an arbitrary division.
My wish, therefore, is for any correlation between race and track -- or even gender and track -- to be as surprising (and therefore as rare) as one between parity and track.
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