Thursday, July 12, 2018

Van Brummelen Chapter 6 The Modern Approach: Oblique Triangles

Table of Contents

1. Van Brummelen Chapter 6 The Modern Approach: Oblique Triangles
2. Exercise 6.4
3. Exercise 6.5
4. Exercise 6.13
5. Exercise 6.16
6. Another Error on the Pappas Calendar?
7. A Line and Its Translation Image Are Parallel?
8. Conclusion

Van Brummelen Chapter 6 The Modern Approach: Oblique Triangles

Chapter 6 of Glen Van Brummelen's Heavenly Mathematics is called "The Modern Approach: Oblique Triangles." Here's how it begins:

"So far spherical trigonometry hasn't looked much like the plane theory we learned in high school. However, the parallels often lie just below the surface."

In this chapter, Van Brummelen writes about applying spherical trig to oblique triangles -- that is, triangles that aren't right triangles. In plane trig, we generally solve oblique triangles by using the Laws of Sines and Cosines. The author has already discussed the spherical Law of Sines, and so this chapter is all about the spherical Law of Cosines.

But first, he writes about the familiar planar cosine law. He gives the formula:

c^2 = a^2 + b^2 - 2ab cos C

and points out the apparent link to the Pythagorean Theorem. He then asks, who was the first person to discover the Law of Cosines:

"I've been asked this question before, and it sounds like the answer should be a simple fill-in-the-blank."

Van Brummelen makes the argument that the discoverer of the cosine law was none other than Euclid, despite his lacking a cosine ratio. He points out a certain proposition in Book II of The Elements, and since we've linked to Euclid before, we might as well do so again here.

Proposition 12.

In obtuse-angled triangles the square on the side opposite the obtuse angle is greater than the sum of the squares on the sides containing the obtuse angle by twice the rectangle contained by one of the sides about the obtuse angle, namely that on which the perpendicular falls, and the straight line cut off outside by the perpendicular towards the obtuse angle.
Proposition 13.
In acute-angled triangles the square on the side opposite the acute angle is less than the sum of the squares on the sides containing the acute angle by twice the rectangle contained by one of the sides about the acute angle, namely that on which the perpendicular falls, and the straight line cut off within by the perpendicular towards the acute angle.

Van Brummelen decides to focus on II.13, the acute angle case. By the way, he points out that some people refer to Book II as "geometric algebra," and indeed, David Joyce (the author of the above links) does refer to it as such.

Now Van Brummelen explains that II.13 really is the Law of Cosines:

"In Triangle ABC with an acute angle as C and a perpendicular dropped from A onto BC (defining D), c^2 = a^2 + b^2 - 2AD * BC. But BC = a and AD = b cos C, so what Euclid is 'really' saying is: c^2 = a^2 + b^2 - 2ab cos C."

And the author provides Euclid's full proof, written in modern algebraic notation:

Let x = CD, which we remember is equal to b cos C. Then Euclid asserts:

a^2 + x^2 = 2ax + (a - x)^2,

which we may verify with a little algebra, as long as we close our ears to the historians' howls of protest. (Euclid himself appeals to a previous geometric theorem at this point.) We add DA^2 to both sides:

a^2 + x^2 + DA^2 = 2ax + (a - x)^2 + DA^2.

Apply the Pythagorean Theorem to both sides; we get:

a^2 + b^2 = 2ax + c^2.

Finally, a bit of rearrangement takes us to:

c^2 = a^2 + b^2 - 2ax,

which is what we wanted to prove. QED

In fact, we see that Joyce writes something similar under the link for II.12, the obtuse angle case:

"This conclusion is very close to the law of cosines for oblique triangles, since AD equals –b cos A, the cosine of an obtuse angle being negative. Trigonometry was developed some time after the Elements was written, and the negative numbers needed here (for the cosine of an obtuse angle) were not accepted until long after most of trigonometry was developed. Nonetheless, this proposition and the next may be considered geometric versions of the law of cosines."

Notice that while most high school Geometry classes nowadays don't teach or prove the Law of Cosines, a few do. This version of Euclid's proof is indeed suitable for an Honors Geometry course.

Now Van Brummelen turns next to the spherical analog of this law. In his figure 6.2, he draws the same diagram on the spherical, adding a new variable h = AD, the altitude of the triangle that divides it into two right triangles. Here is the author's derivation:

This time Pythagoras looks a bit different:

cos b = cos h cos x and cos c = cos h cos (a - x)

Solving both expressions for cos h and setting them equal to each other has the advantage of removing the reference to the undesirable h:

cos b/cos x = cos c/cos (a - x).

From here we solve for cos c, apply the cosine subtraction law, and let the algebra run its course:

cos c cos x = cos b(cos a cos x + sin a sin x)
cos c = cos a cos b + sin a cos b tan x.

To get rid of the tan x term we apply identity four in the first column of the Napier's Rules theorems [from my Fourth of July post -- dw]. This takes us directly to the spherical Law of Cosines:

cos c = cos a cos b + sin a sin b cos C.

Van Brummelen points out that just as the planar Law of Cosines is a generalization of the planar Pythagorean Theorem, the spherical law generalizes its respective Pythagoras. The cos a cos b part comes from spherical Pythagoras, while by the locality principle, sin a and sin b approach a, b. The only thing from the planar law is the 2 -- make that the negative 2. But it's easy to see where the -2 went -- recall that cos c doesn't approach c^2, but 1 - (1/2)c^2. So the -1/2 factor swallows up the -2 we expect to see in the Law of Cosines.

Now Van Brummelen demonstrates applications of the Law of Cosines. He points out that many texts use the Law of Cosines to compute distances on common sea routes. In his first example, students are asked to find the distance traveled from Queenstown/Cobh, Ireland (51.78N, 8.18W) to Sandy Hook, New York Harbor (40.47N, 74.13W). Notice that this Sandy Hook is in New Jersey, and so it has nothing to do with the Sandy Hook school shooting in Connecticut (the two Sandy Hooks are about a hundred miles apart). This text is from the "Golden Age of Textbooks" in 1895, seventeen years before the the voyage of the Titanic along this route (which is itself a century before the tragedy at the other Sandy Hook).

As the author explains, the key to this problem is to join both New York (Y in figure 6.3) and Queenstown (Q) to the North Pole (N). Then YN = 90 - 40.47 = 49.53 and QN = 90 - 51.78 = 38.22, while the angle at N is the difference between the two longitudes, 65.95. We now apply the Law of Cosines, letting the C in the formula be the angle at the North Pole, and we find:

cos QY = cos 38.22 cos 49.53 + sin 38.22 sin 49.53 cos 65.95.

Thus QY = 45.43, which we multiply by 60 to get 2726 nautical miles, or (using the 1.15078 conversion factor) 3137 statute miles (that is, what we ordinarily call "miles"). The author points out that in practice, the ship would have traveled 2783 to 2889 miles, since Newfoundland and Nova Scotia get in the way of the true great circle route.

Van Brummelen's next example is the ASA case. He points out that in Euclidean ASA cases, we don't use the Law of Cosines -- instead, we use the fact that the angles of a triangle add up to 180 to find the third angle and then use the Law of Sines. The problem is that the angles of a triangle don't add up to 180, or any fixed value, in spherical geometry. Instead, the author uses:

Polar Duality Theorem: The sides of a polar triangle are the supplements of the angles of the original, and the angles of a polar triangle are the supplements of the sides of the original.

We discussed this theorem back in my Eid-al-Fitr post on June 15th, as well as in our discussion of spherical geometry from Legendre in years past. Anyway, Van Brummelen applies this to the Law of Cosines to obtain the "Law of Cosines for Angles."

cos(180 - C) = cos(180 - A)cos(180 - B) + sin(180 - A)sin(180 - B)cos(180 - c)
cos C = -cos A cos B + sin A sin B cos c.

Here's the author's example -- suppose we fly from Vancouver to Edmonton on a great circle route, measuring the distance we travel as well as the headings at departure and arrival. From this information alone, an application of the Law of Cosines for Angles gives us the difference in longitude between two cities, and the values of both latitudes follow immediately. The distance from Vancouver to Edmonton (figure 6.4) is VE = 507 statute miles, or 440.9 nautical miles, which corresponds to 7.35 degrees. We leave Vancouver with a heading 50.7 east of north, and arrive in Edmonton with a heading 58.22 east of north. Joining our two cities to the North Pole as before, we have Angle V = 50.7 and Angle E = 121.78. Apply the Law of Cosines for Angles, letting c be the journey from Edmonton:

cos N = -cos V cos E + sin V sin E cos VE.

This gives us Angle N = 9.6, the difference in longitude between Vancouver and Edmonton. Now that we have all three angles, we calculate the two latitudes using the Law of Sines; Vancouver works out to 49.3, and Edmonton to 53.6.

Van Brummelen tells us that with the Law of Cosines for Angles, we can even solve the AAA case -- which is impossible in plane trig. But, as warns us, not even on the sphere can we avoid the ambiguous case of SSA. Here's his example:

"A ship leaves Honolulu (latitude 21.31N) traveling towards Tokyo (latitude 35.7N) on a great circle route with a heading of 60.5 west of north. What will be the length of the voyage, in miles?"

From figure 6.5 we see that we are in an SSA situation. Unfortunately, there are two endpoints that satisfy the givens. The first is X, a spot in the middle of the Pacific Ocean. If you extend BX past X, it will eventually reach its northernmost point, and then start heading slightly southward, crossing the 35.7N latitude circle again at Tokyo.

The mathematical face of this ambiguity appears immediately when we use the Law of Sines to find Angle A: sin A = sin a sin B/sin b = 0.9985, so A = 86.84 or 180 - 86.84 = 93.16. We can tell from our diagram that we want the smaller angle 86.64, but if the ambiguity had passed without notice, we might have run out of fuel at X, with no land in sight.

We now know a, b, A, and B; but finding c and C proves surprisingly awkward. It is possible to use the Law of Cosines (rearranging the letters in the diagram appropriately), but only by changing sin c into sqrt(1 - cos^2 c), eventually requiring us to solve a quadratic equation for cos c. A cleaner approach would be nice.

It's at this point when Van Brummelen suddenly brings up the feud between Newton and Leibniz -- and I already mentioned this in my Fourth of July post. So let's just skip to the point he's making -- two mathematicians developed the "cleaner approach." One is Napier, and the other is the French astronomer Jean-Baptiste Delambre. Here the author derives the first of "Delambre's analogies":

Derivation (from Van Brummelen):
Delambre's analogies are usually demonstrated disappointingly through algebraic manipulation of various known identities. It is not easy to approach them geometrically, but at least one textbook (Isaac Todhunter's classic) gives it a go. The argument, based on Delambre's original demonstration, begins with Triangle ABC in figure 6.6. We begin by bisecting AB at M and drawing a perpendicular upward [towards V]. Next, bisect BCP to form CV. Drop arcs perpendicular to CP and CB, defining P and Q respectively. Finally, join AV and BV. We show first that Triangles AVP and BVQ are equal, by matching three elements of the triangles. They both contain a right angle; since MV is perpendicular to AB we know that AV = BV; finally, since Angle BCP was bisected by CV we know that PV = QV. So the two triangles are equal.

[In a footnote, the author realizes that he just "proved" the triangles "congruent" by SSA -- and it was his own students who told him that he was an, um, backwards-SSA, for making this mistake. The fact that these are right triangles is of no help, since HL is also invalid in spherical geometry. He leaves it to readers to figure out how to patch this flaw -- now I'm wondering, is the U of Chicago's "SsA" theorem valid in spherical geometry. This would require that the corresponding sides opposite the right angles be longer than the other given congruent sides -- that is, AV > PV and BV > QV. These inequalities certainly appear true from the diagram.]

This fact allows us to label the angles at A and B as we have done in the diagram. The angles at the base of the original triangle are Angle A = y + x and Angle B = y - x; a little algebra yields x = (A - B)/2 and y = (A + B)/2. We are in a similar algebraic situation with respect to sides a and b of the original triangle: a = BQ + CQ and b = AP - CP, but BQ = AP (since triangles AVP and BVQ are equal) and CQ = CP, so BQ = (a + b)/2 and CQ = (a - b)/2. Turning to the angles at the top of the diagram, we know that Angle AVB = PVQ since both are composed of AVQ and equal angles. Cutting these angles in half, we have Angle AVM = PVC.

We're in the home stretch of the proof. Consider the two right-angled triangles AVM and PVC, with equal angles at V. Apply Geber's Theorem [Fourth of July post] to both and bring the results together:

sin VAM cos AM = cos AVM = sin PCV cos CP.

Substituting each of the angles for the values from the elements of the original triangle and rearranging the terms, we finally arrive at Delambre's first analogy:

sin(1/2)(A + B)/cos(1/2)C = cos(1/2)(a - b)/cos(1/2)c.

Here are the other three analogies:

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 = cos(1/2)(a + b)/cos(1/2)c, and
cos(1/2)(A - B)/sin(1/2)C = sin(1/2)(a + b)/sin(1/2)c.

The remaining Napier's analogies are found by dividing Delambre's analogies by one another:

tan(1/2)(A + B)/cot(1/2)C = cos(1/2)(a - b)/cos(1/2)(a + b),
tan(1/2)(A - B)/cot(1/2)C = sin(1/2)(a - b)/sin(1/2)(a + b),
tan(1/2)(a + b)/tan(1/2)c = cos(1/2)(A - B)/cos(1/2)(A + B), and
tan(1/2)(a - b)/tan(1/2)c = sin(1/2)(A - B)/sin(1/2)(A + B).

The author uses Napier's third analogy to solve the Honolulu-to-Tokyo problem:

tan(1/2)(68.69 + b54.3/tan(1/2)c = cos(1/2)(86.84 - 60.5)/cos(1/2)(86.84 + 60.5)

from which we have tan(1/2)c = 0.5317, and so c = 28 degrees, equal to 1680 nautical miles or 1933 statute miles. We can find C easily using any relevant identity.

Van Brummelen ends the chapter as follows:

"There is plenty more to spherical trigonometry than we've seen so far, but this concludes the basic theory needed for solving triangles. In the remaining three chapters we'll see some special topics and applications."

Exercise 6.4

A ship sailing on a great circle from Ceylon to Madagascar crosses the meridian of 79E longitude bearing S50W. After sailing 2060 nautical miles farther, it crosses the meridian of 52E longitude. Find its latitude and the bearing of its course at this point and its latitude at the first point. [Brink 1942, 41].

(Yes, this is definitely a 1942 question, since "Ceylon" is now known as Sri Lanka.)

Let's set up the triangle. We'll use C for "Ceylon" and M for Madagascar, and as usual we'll label the North Pole N. Here are the givens: the angle at C is 130 (since that's 50 degrees west of south, not north) and CM is 2060/60 = 34+1/3 degrees. The angle at N is the difference between longitude, 27.

This is an AAS situation. We must first use the Law of Sines to find NM and then Napier's analogies to find the remaining parts:

sin NM/sin C = sin CM/sin N
sin NM/sin 130 = sin(34+1/3)/sin 27
sin NM = sin(34+1/3)sin130/sin 27
sin NM = 0.95168
NM = 72.1 or NM = 180 - 72.1 = 107.9

Hmm, so there are two possible answers. I suspect that the answer is 107.9, because Madagascar is in the southern hemisphere. The actual latitude is 90 - 72.1 = 17.9S.

Now we use Napier's analogies to find M and NC. We could use any of the analogies, but let's follow Van Brummelen and use the third analogy:

tan(1/2)(107.9 + 34.3)/tan(1/2)NC = cos(1/2)(130 - 27)/cos(1/2)(130 + 27)
tan(1/2)NC = tan 71.1 cos 78.5/cos 51.5
tan(1/2)NC = 0.93587
(1/2)NC = 43.1
NC = 86.2

So the latitude of Ceylon is 90 - 86.2 = 3.8N.

Now all that remains is to find the angle at M. Van Brummelen tells us that we can use any formula at this point, so let's use the Law of Cosines for Angles:

cos M = -cos N cos C + sin N sin C cos NC
cos M = -cos 27 cos 130 + sin 27 sin 130 cos 86.2
cos M = 0.59574
M = 53.4

This means that the bearing from Madagascar back to Ceylon is N53.4E. The ship is now heading away from Ceylon, so the bearing at Madagascar is S53.4W.

Therefore our answers are Madagascar 17.9S, Ceylon 3.8N, final bearing S53.4W.

I decided to check my answers by looking up the latitudes of Sri Lanka and Madagascar. I find that the latitude of Sri Lanka is about 7N and the latitude of Madagascar is about 19S. So our answers are fairly close, though I'd prefer to be a bit closer for Sri Lanka.

Also, I wonder whether there's a real way to assert that Madagascar is closer to 18S than 18N, which can't be determined from sin NM = 0.95168. If we assumed that Madagascar is at 17.9N, then we obtain 43.6N for the latitude of Ceylon (which is way off) and S35.6W for the final bearing. All of the identities are satisfied for this second triangle. After all, AAS is not a valid congruence theorem in spherical geometry -- therefore AAS, like SSA, leads to an ambiguous case.

Exercise 6.5

Charles A. Lindbergh flew his plane The Spirit of St. Louis on the great circle route from New York (40.75N, 73.97W) to Paris (48.83N, 2.33E). He left New York at 7:52 AM (Eastern Standard Time) on May 20th, 1927, and arrived at Paris the next day at 5:24 PM (Eastern Standard Time). What was his average speed for the flight? [Rosenbach/Whitman/Moskovitz 1937, 332]

Once again, we place N on top of the world, Y at the Big Apple, and P at the City of Light. The angle at N is 73.97 + 2.33 = 76.3, NY = 90 - 40.75 = 49.25, and NP = 90 - 48.83 = 41.17.

Let's use the Law of Cosines:

cos YP = cos NY cos NP + sin NY sin NP cos N
cos YP = cos 49.25 cos 41.17 + sin 49.25 sin 41.17 cos 76.5
cos YP = 0.60779
YP = 52.57 degrees

This equals 3154.2 nautical miles or 3629.8 statute miles. The famous transatlantic flight lasted for 33 hours and 32 minutes (33+32/60 hours), and so the average speed is 108.2 mph. A modern jet travels about four or five times as fast as Lindbergh's flight, but notice that this problem comes from a text written a mere decade after the famous flight.

There's no need to calculate either Angle Y or Angle P, since the question doesn't ask for the heading at either point.

Exercise 6.13

Napier's analogies were sometimes used as a method of solving geographical problems. Suppose that we know the latitudes of Edmonton (53.6N) and Vancouver (49.3N), as well as their difference in longitude (9.6). The Law of Cosines is the obvious choice here, but instead use Napier's analogies to determine the headings of a great circle path between the cities, and then use some other identity to determine the distance.

Well, let's look at the first two identities again:

tan(1/2)(A + B)/cot(1/2)C = cos(1/2)(a - b)/cos(1/2)(a + b),
tan(1/2)(A - B)/cot(1/2)C = sin(1/2)(a - b)/sin(1/2)(a + b).

The first equation gives the value of A + B, while the second provides the value of A - B. So we can use these two to find E + V and E - V, and then these two equations are solved easily for E and V:

tan(1/2)(E + V)/cot(1/2)9.6 = cos(1/2)(40.7 - 36.4)/cos(1/2)(40.7 + 36.4)
tan(1/2)(E + V) = cos 2.15/(cos 38.55*tan 4.8)
tan(1/2)(E + V) = 15.217
(1/2)(E + V) = 86.24

tan(1/2)(E - V)/cot(1/2)9.6 = sin(1/2)(40.7 - 36.4)/sin(1/2)(40.7 + 36.4)
tan(1/2)(E - V) = sin 2.15/(sin 38.55*tan 4.8)
tan(1/2)(E - V) = 0.71689
(1/2)(E - V) = 35.64

(1/2)(E + V) = 86.24
(1/2)(E - V) = 35.64
E = 121.88
V = 50.6

Since in the original problem, we're traveling from Vancouver to Edmonton, we read the heading at Vancouver as N50.6E, and the heading at Edmonton becomes 180 - 121.88 = N58.12E.

We must use some other identity to find the distance between the two cities -- and Van Brummelen specifically tells us not to use the Law of Cosines. So let's use the Law of Sines instead:

sin VE/sin 9.6 = sin 36.4/sin 50.6
sin VE = sin 9.6 sin 36.4/sin 50.6
sin VE = 0.12806
VE = 7.36 degrees = 441 nautical miles = 508 statute miles

We double-check the answers with figure 6.4 from the book. In that figure, V = 50.7, E = 121.78, and VE = 507 miles, so our answers are reasonable.

Exercise 6.16

(a) Show that cot a sin b = cos b cos C + cot A sin C.

We mimic Van Brummelen's derivation of the Law of Cosines. We start by dividing ABC into two triangles and then applying the ten Chapter 5 formulas to these two triangles.

Notice that the statement to be proved starts with cot a sin b. We don't know a difference formula for cotangent, but we do know one for sine. Along with the fact that B doesn't appear in the goal formula, this indicates that we should drop the perpendicular from B to form Triangles ABD and CBD. We'll let CD = x and AD = b - x and choose the formula involving sin(b - x) to use the difference formula (h, again, equals the height, which this time is BD):

sin(b - x) = tan h cot A
sin b cos x - cos b sin x = tan h cot A

Applying this same formula to Triangle CBD gives:

sin x = tan h cot C

Dividing these two formulas gives:

sin b/tan x - cos b = cot A/cot C

We apply another formula to CBD involving tan x:

cos C = tan x cot a
tan x = cos C/cot a
1/tan x = cot a/cos C

We substitute this in to eliminate tan x:

sin b cot a/cos C - cos b = cot A/cot C

Multiplying by cos C gives:

sin b cot a - cos b cos C = cot A sin C (since cos C/cot C = cos C tan C = sin C)
cot a sin b = cos b cos C + cot A sin C

which is exactly what we want to prove. QED

(b) Use the above result (with the variables rearranged appropriately) in conjunction with Triangles AVP and CVP in the derivation of Delambre's first analogy (figure 6.6) to demonstrate Napier's second analogy.

Let's begin by repeating the information from figure 6.6, which I can see, but you can't.

From earlier:
The argument, based on Delambre's original demonstration, begins with Triangle ABC in figure 6.6. We begin by bisecting AB at M and drawing a perpendicular upward [towards V]. Next, bisect BCP to form CV. Drop arcs perpendicular to CP and CB, defining P and Q respectively. Finally, join AV and BV. We show first that Triangles AVP and BVQ are equal, by matching three elements of the triangles. They both contain a right angle; since MV is perpendicular to AB we know that AV = BV; finally, since Angle BCP was bisected by CV we know that PV = QV. So the two triangles are equal. This fact allows us to label the angles at A and B as we have done in the diagram. The angles at the base of the original triangle are Angle A = y + x and Angle B = y - x; a little algebra yields x = (A - B)/2 and y = (A + B)/2. We are in a similar algebraic situation with respect to sides a and b of the original triangle: a = BQ + CQ and b = AP - CP, but BQ = AP (since triangles AVP and BVQ are equal) and CQ = CP, so BQ = (a + b)/2 and CQ = (a - b)/2.

(We'll cut it off there, since that's all the given information we need to solve the problem.)

We're instructed to apply the formula from (a) to Triangles AVP and CVP, so let's do so:

cot VP sin CP = cos CP cos CPV + cot VCP sin CPV
cot VP sin AP = cos AP cos APV + cot VAP sin APV

First of all, from the diagram, there's a right angle at P. So CPV and APV are right angles, which means cos CPV = cos APV = cos 90 = 0 and sin CPV = sin APV = sin 90 = 1:

cot VP sin CP = cot VCP
cot VP sin AP = cot VAP

Now let's plug in the values from the info above -- CP = CQ = (a - b)/2 and AP = BQ = (a + b)/2. In the diagram we see that Angle VAP = x, so we obtain VAP = x = (A - B)/2 from above.

That leaves us with Angle VCP. From the diagram, BCP and BCA (which is C from the original triangle ABC) form a linear pair, so BCP = 180 - C. And from the info above, ray CV bisects BCP, and so VCP = (180 - C)/2. Now we can fill in the equations:

cot VP sin (1/2)(a - b) = cot (1/2)(180 - C)
cot VP sin (1/2)(a + b) = cot (1/2)(A - B)

Of course, there's still one value we didn't substitute into the equation -- VP. But that doesn't matter, because we're just going to divide the equations:

sin(1/2)(a - b)/sin(1/2)(a + b) = cot(1/2)(180 - C)/cot(1/2)(A - B)
sin(1/2)(a - b)/sin(1/2)(a + b) = tan(1/2)(A - B)/tan(1/2)(180 - C)

But notice that tan(1/2)(180 - C) = tan(90 - (1/2)C) = cot (1/2)C, so we have:

sin(1/2)(a - b)/sin(1/2)(a + b) = tan(1/2)(A - B)/cot(1/2)C,

which, as you may have noticed, is exactly the formula we set out to prove, Napier's second analogy:

tan(1/2)(A - B)/cot(1/2)C = sin(1/2)(a - b)/sin(1/2)(a + b). QED

Another Error on the Pappas Calendar?

Tomorrow on her Mathematics Calendar 2018, Theoni Pappas writes:

1, 2, 3, 10, 11, 12, ____, 100, 101, 102, ...

It appears that Pappas intends this to be the sequence of natural numbers base 4 -- after all, the list begins 1, 2, 3, and then skips up to 10. And so the next value in the sequence after 12 should be 13 -- and of course, tomorrow's date is the thirteenth (albeit in decimal -- triskaidekaphobes beware on Friday the 13th!).

{4} (default quaternary)

But the next number after 13 isn't 100 -- it's 20. The sequence should have read:

1, 2, 3, 10, 11, 12, _13_, 20, 21, 22, 23, 30, 31, 32, 33, 100, 101, 102, ...

Thus assuming that quaternary is intended here, Pappas makes another error. Of course, it's possible that Pappas doesn't intend this sequence to be the quaternary naturals. But I performed a OEIS search and no sequence in the database begins with 1, 2, 3, 10, 11, 12, 13, 100.

{a} (default decimal)

The closest I found is:

This is the "square base" with a ones place, fours place, nines place, 16's place, and so on. But even then, we don't skip directly from 13 (seven) to 100 (nine) -- we still need 20 (eight).

Also, notice that any list of numbers can be the start of a sequence. We could take the Pappas list of ten numbers:

1, 2, 3, 10, 11, 12, _13_, 100, 101, 102, ...

and find an interpolating polynomial of degree nine that passes through these points. The Wolfram online calculator finds the following polynomial:

23x^9/2340 - 27x^8/80 + 12967x^7/1890 - 28081x^6/360 + 587903x^5/1080 - 1731947x^4/720
     + 5413807x^3/810 - 501803x^2/45 + 6300671x/630 - 3600

By the way, the next number in the sequence after 102 is 7543. But seriously, I doubt this is what Pappas has in mind.

As I wrote back in my May 25th post, number bases is one of my favorite topics. Base four, while superior to base two, is still considered a bit too small to be human scale.

But it's precisely because four is so small that certain SPD divisibility tests that are impractical in human scale bases become usable in quaternary. We see that 4^3 = 64, and so its neighbors 63 and 65 provide SPD tests for 7 and 9 (cube-omega) as well as 13 (cube-alpha), and all of these require memorizing at most ten multiples. We know that 11 resists being resolved by SPD, but on May 25th I mentioned how 11 can be resolved in binary -- and we can use that same test in quaternary by first converting it to binary (two digits at a time). This means that we have divisibility rules for all numbers up to the square of the base 4.

Even though base 4 isn't human scale, its square 16 is (albeit on the upper end of) human scale. So these divisibility tests are available in hexadecimal by first converting it two quaternary (two digits at a time). Of course, all of these tests are esoteric -- the three most common primes 2, 3, and 5 all have intuitive tests in both 4 and 16.

One interesting place where these divisibility rules might come into play is Mocha/Atari music, if the degree is written in hexadecimal (which is logical, since this is on a computer). Then knowing whether a number is divisible by 7, 11, 13, or 17 will help us determine whether the color of the note is red, amber, ocher, or umber. That's all I'm going to say about this -- don't worry, this won't turn into yet another music post.

A Line and Its Translation Image Are Parallel?

By now, I wanted to post a complete proof of "a line and its translation image are parallel" -- the proof of which I made the Fermat-like announcement last year ("I have the proof, but I can't fit it within the margins of this blog!").

Note: This is a correction of an earlier error.

To complete the proof, we begin with two important lemmas about reflections:

Line Perpendicular to Mirror Theorem:
If a line is perpendicular to the mirror, then it is invariant (that is, its image is itself).

Line Parallel to Mirror Theorem:
If a line is parallel to the mirror, then its image is also parallel to the mirror.

I've proved the first lemma in years past, so let's repeat the proof:

Proof (Line Perpendicular to Mirror):
We're given that l, the line, is perpendicular to m, the mirror. Our goal is to show that if P lies on l, then P', the reflection image of P, also lies on l.

Let M be a point where l and m intersect, and measure MP. Then choose P' to be the point on l at a distance MP from M, but on the other side of M. Then m is the perpendicular bisector of PP', which, by definition of reflection, means that P' is the image of P. Thus P' lies on P, and therefore l is its own reflection image. QED (Line Perpendicular to Mirror)

Proof (Line Parallel to Mirror):
This is a new proof, since I don't like my previously posted proofs. Instead, we use the U of Chicago definition of parallel -- two lines are parallel if they have no point in common or are identical. Thus the statement l is parallel to l' is equivalent to if l and l' have a point in common, then l and l' are the same line."

So let l | | m, and let l' be the reflection image of l across m. So our goal is to prove l | | l' -- that is, we show that if l and l' have a point P in common, then they are the same line:

Thus P lies on l, and P lies on l'. The question is, where does P', the image of P, lie? Well, we know that P' must lie on l', since l' contains the image of every point on l (definition of image of a line).

Case I:
P and P' are distinct points on l'. By definition of reflection, m is the perpendicular bisector of PP', so that m is perpendicular to l'. Therefore by Line Perpendicular to Mirror, l' is invariant. But to say that l' is invariant is to say that l itself is invariant. (You might ask, if T is an isometry, why does the fact that T(T(l)) = T(l) imply that T(l) = l? Well, according to the Reflection Theorem, we can show that if T(x) = T(y) then it follows that x = y. It follows from the idea that all isometries are invertible, since these are all one-to-one correspondences. The betweenness and collinearity clauses imply that they are invertible for lines as well as points -- no two distinct lines can ever have the same image.) So l and l' must be the same line.

Case II:
P and P' are the same point (known to be on both l and l'). But the only fixed points (that is, points such that P = P') of a reflection are points on the mirror. Thus l and m have a point in common -- but we are given that l | | m, so l and m are the same line -- and so l', the reflection image of l, must also be that same line. QED (Line Parallel to Mirror)

Notice that this proof reads more easily as an indirect proof -- we assumed that l is not parallel to l' and derived a contraction, therefore l is parallel to l'. The assumption I make here is that the students haven't learned about indirect proofs yet.

An important result based on these two lemma is the idea that if P is a fixed point of a certain isometry -- either a reflection of the composite of two reflections in distinct mirrors -- then P must lie on (both of) the relevant mirror(s). This is clear if there is only one mirror. If there are two mirrors, then suppose the first mirror maps P to some point P'. Then either P = P' and the point must lie on the mirrors, or else the second mirror would have to map P' back to P. But the two distinct mirrors can't both be the perpendicular bisector of PP'. Therefore P must be a fixed point of both reflections -- that is, it must lie on both mirrors.

Now let's try the main proof:

A line and its translation image are parallel.

Let l be the preimage and l' its image. Our goal is to show l | | l'. Let m and m' be the mirrors of the two reflections whose composite is the translation. By definition of translation, m | | m'.

Again, our goal is to show l | | l' -- that is, if some point P lies on both l and l', then l and l' must be the same line. We follow the same pattern as the Line Parallel to Mirror Theorem -- we know that P', the translation image of P, must lie on l' since l' contains the translation image of every point on l.

Now consider a line through P perpendicular to the first mirror m. By Line Perpendicular to Mirror, this line is invariant with respect to the first mirror. But any line perpendicular to the first mirror must be perpendicular to the second mirror (which is where this proof fails in hyperbolic geometry). So this same line is also invariant with respect to the second mirror. Thus the line is invariant with respect to both mirrors, hence it's invariant with respect to the composite translation. By the definition of invariant, if P lies on this line, then so does P'.

(Notice that P and P' must be distinct. As the composite of two reflection, a translation can have a fixed point only if it lies on both mirrors -- but the mirrors are parallel, so no point can lie on both of the mirrors. Thus a translation has no fixed points. If P lies on one of the mirrors, it can't lie on the other, and so PP' is still invariant with respect to that other mirror.)

That is, l', a line containing both P and P', is invariant. And so by the same reasoning as in the Line Parallel to Mirror proof, l is invariant -- that is, l and l' are the same line. QED

You might ask, how exactly would be prove the Perpendicular to Parallels Theorem -- the one step that fails in hyperbolic geometry? In some old posts, I actually assumed Perpendicular to Parallels as a postulate, but this is clumsy. We prefer to use the parallel postulate that appears in most Geometry texts, namely Playfair's Postulate. (Through a point not on a line, there exists at most one line parallel to the given point.)

As it turns out, we can prove Perpendicular to Parallels directly from Playfair:

Let a, b, t be lines such that a | | b and a perpendicular to t. We can say that t intersect a at A, and let t intersect b at B. (That t and b even intersect at all derives from Playfair -- if t | | b, then t and a are two distinct lines through A parallel to b, when Playfair tells us that there's at most one such line.)

Now let m be the perpendicular bisector of AB. Consider the reflection across m. This reflection maps A to B and t to itself by Line Perpendicular to Mirror. And since a | | m, by Line Parallel to Mirror, it maps a to a line parallel to a through B. By Playfair, at most one such line exists -- and that line must be exactly b. Thus our reflection maps t to itself and a to b.

Reflections preserve angle measure. Since the angle between t and a is 90, therefore the angle between their images t and b is also 90. QED

And from "a line is parallel to its translation image," we can finally prove the Corresponding Angles Consequence ("if two parallel lines are cut by a transversal,...") It turns out that this is proved exactly like Perpendicular to Parallels, except with a translation mapping a to b rather than a reflection (namely the translation that maps A to B, which leaves t invariant). This was our original goal -- to prove Corresponding Angles using transformations rather than assuming it as a postulate. It turns out that we need a reflection for the perpendicular case (one mirror), and then that case plus a translation is used for the general case (two mirrors).

My idea was that all of these proofs, while invalid in Euclidean geometry, would be valid in both Euclidean and spherical geometry. Some of these theorems (such as Perpendicular to Parallels) are vacuously true in spherical geometry, while others have a possible interpretation in spherical geometry (for example, the Case II of Line Parallel to Mirror shows us that if a line and its image are distinct and intersect, then the mirror must intersect there as well, in both geometries). But in no case did I use a postulate that holds in Euclidean but not spherical geometry.

Or did I? Consider the idea that if PP' is invariant and both P and P' lie on l', then l' is invariant. This appears to assume that there exists at most one line through P and P' (namely l'), but this fails in spherical geometry. But this is built into Line Perpendicular to Mirror -- it holds in spherical geometry that mirror m reflects P to P' and infinitely many lines pass through P and P', then every single such line is invariant with respect to m (because every such line is perpendicular to m).

We also note that Perpendicular to Parallels uses the Two Perpendiculars assumption (to show that a must be parallel to m). It seems silly to worry about this (since the theorem is vacuous unless parallel lines exist), but still, it's possible to avoid this:

Three Perpendiculars Theorem:
If ab, c are all perpendicular to t, then ab, c are either all parallel or all concurrent.

(Theorem is indeed "natural" in that it holds in both Euclidean and spherical geometry. Of course, "parallel" corresponds to Euclidean and "concurrent" to spherical geometry, but nowhere in the theorem or its proof is this stated directly, hence it's natural.)

Let A, B, C be the points where a, b, c (respectively) intersect t. Without loss of generality, our goal is to show that if a and b have a point P in common, then P must lie on c as well.

Let m be the perpendicular bisector of AB. This reflection leaves t invariant and maps A to B. Since reflections preserve angle measures, a (perp. to t at A) is mapped to b (perp. to t at B). And since a intersects its image b at P, the mirror m must pass through P as well (from that above argument about spherical Case II), and as reflections preserve distance, PA = PB.

Now let n and o be the perpendicular bisectors of AC and BC respectively. Both reflections leave t invariant, and map A to C and C to B respectively. Since reflections preserve angle measure, the reflections map a to c and c to b respectively. The composite isometry maps a to b -- and since isometries preserve distance and PA = PB, the composite isometry maps P to itself.

Since P is a fixed point of an isometry that's the composite of two reflections, we conclude (from earlier in this post) that P must lie on both mirrors n and o. Thus the reflection over n, which maps a to c, must map PA (on a) to PC (on c). Therefore P lies on c. QED

The sole purpose of Three Perpendiculars Theorem is to take a proof that would ordinarily use Two Perpendiculars and apply it to natural geometry, without using a single postulate that distinguishes between Euclidean and spherical geometry.

But if we need an awkward Three Perpendiculars Theorem just to prove that a line and its translation image are parallel, then we should just forget about it. I'll use this -- my 750th post -- to declare this for future reference. Unless I make a sudden insight, this is probably as far as I'll go with this idea.

There are more interesting problems that I'd like to solve with transformations anyway. For example, I've mentioned how graphing linear equations is always a dark spot in eighth grade math and Algebra I classes -- students tend to struggle with it. Can transformations be used to make this unit easier? We know the Core's suggestion that similarity/dilations be used to prove the existence of slope -- but does this insight make graphing easier for students to learn? And does this make related lessons in Math 8 or Algebra I easier, such as solving systems of equations?


Notice how much of our algebraic manipulation in today's post involves solving simple systems of equations (where x + y and x - y are known). I'm continuing to monitor how the students are doing in the B sessions of the summer school classes (which I was never scheduled to teach). The first unit test of Algebra 1B is indeed on systems of equations (which are sometimes covered in first semester, but is covered in second semester for us).

The top student from A session (with his grade of 87%) earned 90 on the first unit test. His overall grade so far in B session is 93%.

The bottom student from A session (with his grade of 59%) earned 65 on the first unit test. His overall grade so far in B session is 70.

It's notable that so far, both students are finding systems of equations easier than graphing. As it turns out, today the famous math teacher Sarah Carter writes one of her final Algebra I blog posts -- on systems of equations:

(Oh how I'd looked up her blog earlier in anticipation of summer school, only to wind up not teaching summer school!) Carter writes:

And, now the same for elimination. By far, students definitely preferred elimination over substitution.

Next, we addressed solving systems of equations by both substitution and elimination. Last year, I had these as a single skill of solving systems of equations algebraically. As a result, most students chose the method that they preferred and used that method every single time. This meant that I had a lot of students that became really good at solving by elimination but were rubbish at solving by substitution.

This year, I decided to make separate skills for solving by substitution and solving by elimination.

Hmm, well, the top student scored 90 on both quizzes, while the bottom student actually scored slightly higher on substitution (70) than on "linear combinations" -- which is the name that Edgenuity gives to elimination (68). But overall, Carter's observations seem to hold, with higher scores on elimination than substitution. This includes the only student who failed substitution with 50 and went on to earn 68 on elimination. Indeed, no one failed elimination, while four had perfect scores -- twice as many as substitution.

In today's post, we used elimination since our questions gave values to x + y and x - y.

Meanwhile, in the sports world, many Southern Californians have been talking about the signing of the NBA free agent LeBron James with the Los Angeles Lakers. However, the defending champions from Northern California, the Golden State Warriors, have also gotten stronger.

As a general rule, fans from Northern California and Southern California don't cheer for each other's teams, even though we're the same state -- for now, that is. If the long-shot ballot proposition passes, the Warriors would play in Northern California while the Lakers would play in California. (The third state, Southern California, would have no NBA teams.)

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