Friday, June 15, 2018

Van Brummelen Chapter 4: The Medieval Approach

Table of Contents

1. Summer School Announcement
2. Review: What Is Spherical Geometry?
3. Review: What Is Spherical Trig?
4. Van Brummelen Chapter 4: The Medieval Approach
5. Exercise 4.2
6. Exercise 4.3
7. Exercise 4.9
8. A Resolution of the Cliffhanger?

Summer School Announcement

Well, this is exactly what I feared would happen -- not enough students signed up for first semester Algebra I in summer school, and so I no longer have a class to teach this summer. And so let the Great Post Purge 2018 begin...

...just kidding! I already wrote that I won't delete any posts this year. And so all those posts I wrote over the past month about both Algebra I that I'd teach and the music that I'd play will remain posted, even though I don't have a class.

I'd actually run one of the Mocha programs to create a song just minutes before I received the email message that summer school was cancelled. The song would have been a redo of "Solve It," one of the last songs I played at the charter school -- since after all, one of the earliest lessons in the Algebra I class is solving equations.

Meanwhile, I was originally planning on writing about one last scale this week, 22EDL. But since I'm not teaching a class, there's no reason to write about 22EDL any more. And my original original plan (that is, from back in April before the possibility of summer school even arose) was to write about EDL scales throughout the summer. I decided that I've already written enough on EDL scales over the last month that I don't need to spend the summer writing about music some more.

And so instead, today I'll begin my other other plan for this summer -- returning to spherical trig. Last summer, if you recall, we began Glen Van Brummelen's Heavenly Mathematics: The Forgotten Art of Spherical Trigonometry. Last year we covered the first three chapters, and so this year my hope is to finish the book with Chapters 4 through 9.

Spherical trig might not be compatible with Euclidean geometry -- but hey, at least it is Geometry, as opposed to the Algebra I and music topics that have dominated this blog for a month now.

Today's chapter is on the medieval approach to spherical trig. During this time, Europe was stuck in the Dark Ages, and so many of the mathematicians who were active this time were Muslim. It's only fitting because depending on the moon sighting, today is Eid al-Fitr, one of the most important holidays in the Islamic calendar. In fact, some school districts (such as New York, observing the holiday for the second time in 2018) are closed today for Eid al-Fitr, just as many districts are closed on Jewish holidays.

Review: What Is Spherical Geometry?

This is what I wrote last summer about spherical geometry and spherical trig. Today I begin by quoting Chapter 2 of Van Brummelen, which is the chapter in which he introduces a little bit of spherical geometry. I wrote:

Van Brummelen assumes that we're already familiar with earth's surface, and so he devotes part of this chapter to describing the celestial sphere. He writes:

"We've already seen the most obvious feature of the celestial sphere, namely its daily rotation around us. Given the sphere's unfathomably large size, rendering the Earth as an infinitesimal pin prick at its center, one can only imagine how quickly it is actually moving."

The author defines several key terms here:

-- The celestial equator rises from the east point of the horizon and sets in the west.
-- The ecliptic is the path the sun takes as it makes a complete circuit around the celestial sphere.
-- The obliquity of the ecliptic, symbolized by the Greek letter "epsilon," is the tilt between the celestial equator and the ecliptic. Its current value is 23.44 degrees.
-- The equinoxes are the two points where the celestial equator and the ecliptic intersect.
-- The summer solstice is the most northerly point on the ecliptic, halfway between the equinoxes.

I'll quote -- but not prove -- the some key theorems of spherical geometry. Again, you can go back to last year's posts to relearn these theorems. Better yet, you can go back to posts from previous summers, when I first started writing about spherical geometry. (Oh, and by the way, those old posts from previous years were originally based on the writings of 19th-century French mathematician Adrien Legendre.)

Every cross-section of the sphere by a plane is a circle.

Lemma (Triangle Inequality):
The third side of any spherical triangle cannot exceed the sum of the other two.

The sum of sides in a spherical triangle cannot exceed 360 degrees.

The polar triangle of a polar triangle is the original triangle.

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

The angle sum of a triangle must exceed 180 degrees.

Review: What Is Spherical Trig?

And in the following chapter, Van Brummelen introduces spherical trig:

Van Brummelen begins by writing about Hipparchus of Rhodes, the founder of trigonometry. But unfortunately, not much of this Greek scholar's work survives today. Van Brummelen proceeds:

"We must therefore move more than two centuries ahead, to a figure almost as elusive as Hipparchus. We are aware that Menelaus of Alexandria lived in Rome in the late first century AD because Ptolemy tells us he made some observations there, but that is all we know."

At this point Van Brummelen states and proves the key theorems of the book -- the theorems of ancient Greek mathematician Menelaus, which have both plane and spherical versions. Again, let me only state the main theorem and all lemmas used to prove it -- I won't prove the theorem again:

Menelaus's Plane Theorem: In figure 3.2, AK/KB = AT/TD * DL/LB.
Given (I insert a given section here, since you can't see figure 3.2):
A-D-T (that is, D is between A and T), A-K-BDB and TK intersect at L

Lemma A: In figure 3.5, AB/BC = sin alpha/sin beta.
Given: A-B-C, with A and C on the surface of the sphere (so B is interior to the sphere). A vertical diameter is drawn through B. The arcs drawn from A to the diameter and C to the diameter are labeled alpha and beta, respectively.

Lemma B: In figure 3.6, AC/AB = sin alpha/sin beta.
Given: A-B-C, with B and C on the surface of the sphere (so A is interior to the sphere). A horizontal diameter is drawn through A. The arcs drawn from C to the diameter and B to the diameter are labeled alpha and beta, respectively.

Menelaus's Theorem A: sin AZ/sin BZ = sin AG/sin GD * sin DE/sin EB.
By the way, figure 3.4 is the same as 3.2, except there are additional arcs on the sphere. Point Z is now on Arc AB, and G is chosen so that D is on Arc AG, Arcs DB and GZ intersect at point E. (Once again, this is much easier seen than described.)

Menelaus's Theorem B: sin AB/sin AZ = sin BD/sin DE * sin GE/sin GZ.

Van Brummelen then leaves Menelaus and describes al-Kumi, an Iraqi mathematician. (Yes, here we begin with our Islamic mathematicians.) I don't wish to repeat everything I wrote last year, so again I'll state only the givens and the goal as he applies the Menelaus result to the celestial sphere:

  • The equator and the ecliptic intersect, as always, at Aries.
  • The equator intersects the horizon (another great circle) at E. The arc from Aries to E is measured as theta.
  • The Sun is always along the ecliptic, and it's rising on the horizon.
  • We drop a perpendicular from Sun to horizon at point M. As always, the arc from Sun to M is given as delta (declination).
  • The arc from E to Sun is given as eta.
  • The arc from E to M is given as n. As always, the arc from Aries to M is given as alpha (right ascension), therefore theta + n = alpha.
  • The arc from Sun to M can also be extended upward to N, the North Pole.
  • Arc NGZ is the "equator" to the "pole" at Aries. (Recall that on the sphere, all great circles have "poles," and all points have "equators.") Point G lies on the ecliptic (so the arc from Aries to G must be 90) and Z lies on the equator (so the arc from Aries to Z must be 90), and as before, the arc from G to Z must be epsilon (obliquity of the ecliptic).
  • Arc NHCQ is the "equator" to the "pole" at E. Point H lies on the horizon (so the arc from E to H must be 90), C lies on the ecliptic, and Q lies on the equator (so the arc from E to Q is 90). The arc from N to H is labeled as phi (which equals our latitude).

And here is the final result:

sin 90/sin MZ = sin 90/sin(90 - lambda) * sin(90 - delta)/sin 90

or sin MZ = cos lambda/cos delta. Van Brummelen explains that theta = Arc ZQ = Arc MQ - MZ.

Again, you can return to last year's posts for further review. For now, let's start the new chapter.

Van Brummelen Chapter 4: The Medieval Approach

This is what Van Brummelen writes in Chapter 4 of his book, "The Medieval Approach":

"Reading al-Kuhi's statement defending the advantages of Menelaus's Theorem in the previous chapter is a bit like eavesdropping on someone holding a telephone conversation. We have a rough idea of what was said, but important parts of the debate are a blank to us. We are never told the name of the advocate of the new theorem, nor even what the new theorem was."

So in this chapter, the author writes about some of the theorems of al-Kuhi's contemporaries. We begin with Abu Nasr, the discoverer of the polar triangle. He proposes two theorems in his Book of the Azimuth (which unfortunately no longer exists):

Rule of Four Quantities: sin BD/sin CE = sin AD/sin AE.
Abu Nasr's Second Theorem: sin DF/sin EF = sin AD/sin AB.

Here are the givens (in Figure 4.1): both BD and CE are perpendiculars to ABC that intersect at F. It's been a while since we've thought about spherical geometry, so now's a good time to remember that perpendiculars to the same line aren't parallel on the sphere -- indeed, no lines are parallel in spherical geometry. So this is how both BD and CE can be perpendicular to the same line yet intersect at F. In fact, we call point F the pole of line ABC.

Van Brummelen states that the proof of this theorem is easy:

"At first it appears that these theorems are nothing more than corollaries to Menelaus, and in a mathematical sense they are."

Proof of the Rule of Four Quantities: Apply Menelaus's conjunction theorem to figure 4.1: we get 1/sin CE = 1/sin BD * sin AD/sin AE. QED

At this point the author gives us a trick to help remember some of these spherical theorems:

"The Rule of Four Quantities is also our first example of the principle of locality. Imagine a spherical triangle shrinking in size until it almost vanishes. As it gets smaller it begins to resemble a plane triangle; and when it is very small, it almost becomes one. Therefore any statement about a spherical triangle, applied to a triangle shrinking to nothingness, becomes a statement about a plane triangle. In our case, imagine the configuration of figure 4.1 shrinking until it is so small that the sides are almost straight. In radian measure, as x -> 0, the value of sin x essentially becomes x itself. So replacing the sines of the arcs in the Rule of Four Quantities with the arcs themselves, we find that for two nested right triangles, the ratios of the altitudes to the hypotenuses are equal. It's similar triangles."

This explains why, as we proceed through this chapter and the book, many of the spherical theorems look just like Euclidean theorems with lengths replaced with sines. Again, very small triangles on the sphere look like Euclidean triangles -- which is why if we draw a small triangle on the spherical earth, the triangle looks Euclidean, with the sum of its angles being essentially 180.

At this point, Van Brummelen shows how the Rule of Four Quantities can be used to prove al-Kumi's rising times problem -- the problem I quoted again in the last section of this blog. Indeed, figure 4.2 is a repeat of the figure I mentioned earlier (the equator and ecliptic intersecting at Aries, etc.), and so I don't need to repeat the givens. Let's just jump into the proof:

Solution of al-Kuhi's rising times problem, using the Rule of Four Quantities:
(1) We begin with figure AriesSunGZM, from which we find sin delta/sin lambda = sin epsilon/1, so sin delta = sin lambda sin epsilon.
(2) Use figure ESunHQM, from which we have sin delta/sin eta = sin(90 - phi)/1, or sin eta = sin delta/cos phi.
(3) Use figure NSunMQH, which gives sin(90 - eta)/sin(90 - delta) = sin MQ, or sin MQ = cos eta/cos delta.
(4) Finally, use figure NGZMSun to get sin(90 - lambda)/sin(90 - delta) = sin MZ/1, or sin MZ = cos lambda/cos delta.

"Just as before, theta = MQ - MZ, and we are done. There is no doubt about it: the Rule of Four Quantities is much easier to apply, and we get results much more quickly. Menelaus and al-Kuhi didn't stand a chance."

Van Brummelen now proceeds to derive a better-known theorem -- the Spherical Law of Sines. The following proof is due to Abu 'l-Wafa, the author of the Almagest (not to be confused with Ptolemy):

Derivation of Spherical Law of Sines, given Triangle ABC (figure 4.3):
Choose C to be one of the vertices, so that its perpendicular projection onto the opposite side AB lands between A and B, at D. [In other words, let C be the largest angle -- dw]. Let EZ be the equator corresponding to pole A, and let HT be the equator for pole B; extend the sides of the original triangle as shown. [Recall that just as every line has two poles, every point is the pole of an equator. I point out that by definition of "equator" of a point, several right angles are formed -- AEZ, AZE, BHT, as well as BTH. Also, sides AE, AZ, BH, BT are quadrants.] Then apply the Rule of Four Quantities to two configurations, both involving CD. Firstly, on ACZED we get

sin CD/sin B = sin EZ/sin AZ, or sin CD = sin A * sin b.

Secondly, on BCTHD we get

sin CD/sin a = sin TH/sin TB, or sin CD = sin B * sin a.

Combine the two equations and eliminate the shared term sin CD. A little juggling results in

sin a/sin A = sin b/sin B.

But we could have started the argument equally well with any of the three vertices, not just C. If we had applied it to A, for instance, we would have ended up with sin b/sin B = sin c/sin C. Combining these two results, we are left with the breathtakingly simple

Spherical Law of Sines: sin a/sin A = sin b /sin B = sin c/sin C.

It goes without saying that Van Brummelen compares this to the Euclidean Law of Sines:

Planar Law of Sines: a/sin A = b/sin B = c/sin C.

Again, the author uses the principle of locality to explain why the spherical law looks like the planar law with sines of the sides rather than the lengths of the sides.

At this point, Van Brummelen takes an aside. He informs us that even though most mathematicians working during this time were Islamic, there were a few working on spherical trig in India as well.

The author gives one example of Indian trig here -- finding declinations of arcs of the ecliptic. Most of the problem is in stating all the givens from the diagram (figure 4.4):
  • The equator and the ecliptic intersect, as always, at Aries.
  • The Sun lies on the ecliptic. The foot of perpendicular dropped from Sun to equator is A. Arc lengths are as follows: AriesSun = lambda, AriesA = alpha, SunA = delta.
  • B lies on equator, and C lies on ecliptic, each a quadrant away from Aries. Thus BC is epsilon, the angle of the ecliptic (about 23.4 degrees).
  • O is the center of the sphere.
  • Other points are in the interior of the sphere, the plane through the center and equator. The foot of perpendicular from Sun to this plane is D. The foot of perpendicular from D to OAries is E, and the foot of perpendicular from A to OAries is F. And finally, the foot of perpendicular from C to OB is K.
Van Brummelen proceeds as follows:

The two right triangles SunED and COK, called the "kranki-setras" or "declination triangles," are similar since they share the angle epsilon between the planes of the equator and the ecliptic. Therefore

SunD/SunE = CK/CO.

But SunD = sin delta (to see why, consider the vertical circular segment ODASun) and similarly SunE = sin lambda and CK = sin epsilon. CO is the radius, so it is equal to 1, and the standard formula sin delta = sin lambda sin epsilon follows. (He omits the derivation of alpha here.)

Now the author returns to Arabic culture. He points out that two of the Five Pillars of Islam require knowledge of astronomy -- the Ramadan fast, and the five daily prayers. Since today, after all, is Eid al-Fitr (that is, the end of fasting), let's look more closely at Ramadan. Van Brummelen writes:

"Consider the monthly fast. The Arabic calendar is lunar, so each month begins when the lunar crescent reappears from behind the Sun after New Moon. Miss the crescent on a particular day, and you may end up violating the fasting requirement unawares. Muslim scientists worked hard attempting to predict the first appearance of the lunar crescent, with varying degrees of success."

Indeed, missing today's Shawwal new moon is just as bad as missing the Ramadan new moon, since just as fasting is required on 1st Ramadan, fasting is forbidden on 1st Shawwal. So this shows how important it was for medieval Islamic mathematicians to predict the crescent dates.

But Van Brummelen now turns to the final pillar -- the daily prayers. As he explains:

"When the moment occurs, worshipers are enjoined to face the Ka'ba, the most sacred site of Islam. The Ka'ba, a cubical building (figure 4.5) that houses the Black Stone, is the destination of the pilgrimage that Muslims are asked to embark upon once in their lives. The direction of the Ka'ba -- the qibla -- serves several purposes besides the daily prayers, including determining the direction in which Muslims should face when they are buried."

So let's begin the calculations:

"On the face of it the qibla does not seem difficult to calculate. Since the positions of both Mecca and the worshiper are given, we know the local latitude phi_L, phi_Mecca = 21.67 degrees, and the difference in longitude. So we would seem to have a right triangle on the Earth's surface with values of the two sides adjacent to the right angle (figure 4.6)."

But, as the author points out, the side of this "triangle" isn't a line (that is, a great circle), since parallels of latitude aren't great circles. By the way, notice what effect this fact has on the qibla -- it means that Muslims on the same latitude as Mecca (that is, 21.67N) do not face due east or west during prayer, even though the holy city lies due east or west of them. (The shortest path between two points directly east or west of each other is in general not due east or west.) And it's even possible that a worshiper north of 21.67N might even have to face slightly north of east (or west) to Mecca.

Anyway, Van Brummelen performs the calculations for Ghazna, Afghanistan. Let's start with his description of the givens in the diagram:

"To give a reader a taste of ancient and medieval diagrams, we have reproduced al-Biruni's diagram in figure 4.7. Although it looks two-dimensional, appearances are deceiving. Imagine that you are looking directly down on Ghazna from above the celestial sphere. All the curves on the figure (even the two straight lines) are great circle arcs on the celestial sphere seen from above, so G is the zenith directly above Ghazna. The line connecting north and south through G, actually a great circle called the meridian of Ghazna, passes through the north pole P; the outer circle is Ghazna's horizon. M is the point on the celestial sphere that an observer at Mecca would perceive as the zenith. WM connects the west point on the horizon to M, and extends to A on the meridian. PMB is the meridian of Mecca."

And now let's start the calculation of the qibla:

Al-Biruni's geographical coordinates for Ghazna and Mecca were phi_L = 33.58 degrees, phi_M = 21.67 degrees, and a longitude difference of lambda = 27.37 degrees. Now phi_L is the altitude NP of the North Pole, the northernmost segment of Ghazna's meridian; but both NG and PC are 90, so GC = phi_L = 33.58. [Van Brummelen doesn't say this explicitly, but C is where Ghazna's meridian meets the equator -- dw] So the arc from the worshiper's zenith perpendicularly down to the equator is equal to the local latitude. This fact must also apply to the zenith of Mecca, so MB = phi_M = 21.67. Finally, the difference in longitude is equal to the angle at the North Pole between the two zeniths, so Angle MPG = BC = 27.37. Now that we have transferred all the data onto arcs in the diagram, we are ready to begin the actual mathematics.

We shall use nothing but the Rule of Four Quantities. Starting with configuration CAPMB we have

sin PM/sin MA = sin PB/sin BC, or sin(90 - phi_M)/sin MA = 1/sin lambda,

so sin MA = cos phi_M sin lambda, which gives the "modified longitude" MA = 25.29. Our second configuration is WMACB, from which we get

sin WM/sin MB = sin WA/sin AC or sin(90 - MA)/sin phi_M = 1/sin AC,

so sin AC = sin phi_M /cos MA, and we have the "modified latitude" AC = 24.11. Then GA = GC - AC = phi_L - 24.11 = 9.47.

With the modified longitude and latitude in hand, we turn out attention to the outer horizon circle for Ghazna, which is where the qibla resides. It will take two steps. Firstly, from WMASD [where S is the southern horizon, a quadrant south of Ghazna -- dw]:

sin WM /sin MD = sin WA/sin AS or sin(90 - MA)/sin MD = 1/sin(90 - GA),

so sin MD = cos MA cos GA, which gives MD = 63.10. Our final step applies the Rule of Four Quantities to figure GMDSA:

sin GM /sin MA = sin GD/sin DS or sin(90 - MD)/sin MA = 1/sin DS,

so sin DS = sin MA/cos MD. This gives us the qibla, because DS = 70.79 is the number of degrees west of south that we must turn to face Mecca.

Van Brummelen wraps up the chapter by describing the medieval qibla tables:

"The best of these tables was a set composed by Shams al-Din al-Khalili, an astronomical timekeeper employed by the Umayyad mosque in Damascus. Its sixteen pages contain almost 3000 entries of the qibla for every degree of latitude and difference in longitude for all Earthly locations that mattered. The effort involved must have been Herculean."

Exercise 4.2

Let's try some of the exercises from this chapter. I choose Exercise 4.2 -- and you'll find out why I selected it in a moment:

Repeat question 2 of chapter 3, but use only the Rule of Four Quantities.

Oh, now my choice makes sense. Last last we did Exercise 3.2, so this year we do Exercise 4.2. This is what I wrote last year about Exercise 3.2 (in preparation for today's 4.2):

Choose a particular date (say May 20), and a particular latitude (say 49.3 degrees N). Use Menelaus's Theorem to calculate the following quantities:

(a) the Sun's declination delta
(b) the ortive amplitude eta
(c) the equation of daylight n
(d) the rising time theta

First of all, I don't choose a latitude of 49.3N. Instead, I choose my home latitude, which is 34N.

For today's exercise, of course I'm changing the date to June 15th, today's date. (Notice that today is just a few days before the summer solstice.) And now we may begin our solution using the Rule of Four Quantities. Again, we refer to the description of figure 4.2 from earlier in today's post for the givens from the relevant diagram.

(a) We begin with figure AriesSunGZM, from which we find sin delta/sin lambda = sin epsilon/1. For this solution, I'll plug the values directly into the equation -- lambda = 82.9 degrees from the chart (it becomes 90 degrees around June 22nd-23rd, the summer solstice) and epsilon is always 23.4 degrees:

sin delta/sin lambda = sin epsilon/1
sin delta/sin 82.9 = sin 23.4
sin delta/0.99233 = 0.39714
sin delta = 0.3941
delta = 23.2 degrees

(b) Use figure ESunHQM, from which we have sin delta/sin eta = sin(90 - phi)/1.

sin delta/sin eta = sin(90 - phi)/1
0.3941/sin eta = sin (90 - 34)
0.3941/sin eta = 0.82904
sin eta = 0.47537
eta = 28.4 degrees

(c) Use figure NSunMQH, which gives sin(90 - eta)/sin(90 - delta) = sin MQ.

sin(90 - eta)/sin(90 - delta) = sin MQ
sin (90 - 28.4)/sin(90 - 23.2) = sin MQ
0.87978/0.91907 = sin MQ
0.95726 = sin MQ
MQ = 73.2 degrees
n = 90 - MQ (from last year)
n = 16.8 degrees

(d) Finally, use figure NGZMSun to get sin(90 - lambda)/sin(90 - delta) = sin MZ/1.

sin(90 - lambda)/sin(90 - delta) = sin MZ
sin (90 - 34)/sin(90 - 23.2) = sin MZ
0.82903/0.91907 = sin MZ
0.90204 = sin MZ
MZ = 64.4 degrees

theta = MQ - MZ
theta = 73.2 - 64.4
theta = 8.8 degrees
theta = about 35 minutes

Exercise 4.3

Prove Abu Nasr's second theorem using Menelaus.

To do this proof, let's repeat the givens and the goal from earlier:

Abu Nasr's Second Theorem: sin DF/sin EF = sin AD/sin AB.

Here are the givens (in Figure 4.1): both BD and CE are perpendiculars to ABC that intersect at F. It's been a while since we've thought about spherical geometry, so now's a good time to remember that perpendiculars to the same line aren't parallel on the sphere -- indeed, no lines are parallel in spherical geometry. So this is how both BD and CE can be perpendicular to the same line yet intersect at F. In fact, we call point F the pole of line ABC.

Van Brummelen makes this problem sound so simple. We take the theorem of Menelaus, plug in the given values, simplify the equation, and voila! The whole proof should take less than five minutes.

But I struggled with this problem for more than five hours. And no matter how many times I try it, I can never obtain sin DF/sin EF = sin AD/sin AB as a result.

Let's review how to get from Menelaus to the Rule of Four Quantities, since Van Brummelen implies that the proof here is similar. The Menelaus Conjunction Theorem for this problem is:

sin CF/sin CE = sin BF/sin BD * sin AD/sin AE

Since we're going to be finding the sine of everything, let's agree to abbreviate sine as "s," so we don't have to keep writing "sin" over and over:

sCF/sCE = sBF/sBD * sAD/sAE

Now since F is the pole of BC, both BF and CF are 90 degrees, and s90 (= sin 90) = 1. So we have:

1/sCE = 1/sBD * sAD/sAE

We multiply both sides by sBD to obtain the desired result:

sBD/sCE = sAD/sAE

Now we attempt to prove Abu Nasr 2. We recall that there's a second Menelaus theorem -- often known as the Disjunction Theorem:

sCE/sEF = sAC/sAB * sBD/sDF

Since the desired theorem has sDF/sEF on the left side, we multiply both sides by sDF and then divide both sides by sCE:

sDF/sEF = sAC/sAB * sBD/sCE

And we have sAB on the right side, so all we need to make appear is sAD. If we could somehow show that sAD = sAC * sBD/sCE, then we're done. Let's return to the Four Quantities and solve for sAD:

sBD/sCE = sAD/sAE
sAD = sAE * sBD/sCE

Oops -- we needed sAC there, not sAE. In fact, we can multiply the previous formula by sAE/sAE:

sDF/sEF = sAC/sAB * sBD/sCE * sAE/sAE
sDF/sEF = (sAE * sBD/sCE) * 1/sAB * sAC/sAE
sDF/sEF = sAD/sAB * sAC/sAE

And so we have an extra factor of sAC/sAE that we can't get rid of.

Our second attempt is to relabel the Menelaus diagram. We can interchange A with F and B with E to obtain a new Menelaus diagram. The two Menelaus formulas become:

Conjunction: sAC/sBC = sAE/sDE * sDF/sBF
Disjunction: sBC/sAB = sCF/sEF * sDE/sAD

Let's take the disjunction formula. We know that sCF = 1, so we have:

sBC/sAB = 1/sEF * sDE/sAD
sAD/sAB = 1/sEF * sDE/sBC

We now start working on the conjunction formula, aware that sBF = 1:

sAC/sBC = sAE/sDE * sDF/1
sDE/sBC = sAE/sAC * sDF

Let's plug this into the earlier equation:

sAD/sAB = 1/sEF * sDE/sBC
sAD/sAB = 1/sEF * sAE/sAC * sDF
sAD/sAB = sDF/sEF * sAE/sAC
sDF/sEF = sAD/sAB * sAC/sAE

And so once again, we have this extra factor of sAC/sAE that we can't eliminate.

By the way, you might wonder under what situation would we have sAC/sAE = 1. Well, if sAC = sAE, and AC = AE, then Triangle ACE would be isosceles. The Isosceles Triangle Theorem is valid in spherical geometry, and so Angle AEC = ACE, which is given to be 90. This would then force A to be the pole of CEF, so both AC and AE would be quadrants. In other words, sAC can't equal sAE unless both equal 1. And we know for a fact that no, angle AEC isn't intended to be a right angle, otherwise the Rule of Four Quantities would transform into the Rule of Three Quantities (sBD/sCE = sAD, as sAE would equal 1).

My next attempt involves the Spherical Law of Sines (even though we're not asked to use this to prove the theorem). In Triangle ABD, we have:


Now Angle ABD = 90, and so sABD = 1:

sAD/1 = sAB/sADB
sAD/sAB = 1/sADB

OK, so there's the right side of Abu Nasr 2. We notice that Angle ADB = EDF as these are vertical angles, and so let's try to apply Spherical Law of Sines to Triangle DEF:

1/sEDF = sDF/sEF * 1/sDEF

If we set 1/sADB = 1/sEDF, then we obtain:

sAD/sAB = sDF/sEF * 1/sDEF
sDF/sEF = sAD/sAB * sDEF

So this time, the extra factor we can't eliminate is sDEF. Notice that the only way to make this extra factor equal 1 is for Angle DEF = 90. Once again, it all works out only when A is the pole of CEF.

If we think about, we notice that our goal is to take an equation with six values (Menelaus) and wind up with an equation with four quantities (Abu Nasr 2). The only two values that are known to disappear to 1 are sBF and sCF. This implies that we need a Menelaus equation that connects the four values in Abu Nasr 2 (DF, EF, AD, AB) to BF and CF. Yet there's no way to rearrange either form of Menelaus so that it uses only these six values.

At this point, I give up. Instead, I attempted to find a counterexample. To do so, I wanted to choose values of each arc so that both Menelaus and Four Quantities are satisfied, yet Abu Nasr 2 isn't. Of course, I tried to use as many measures of 30, 45, 60, and 90 degrees as possible, since we know the sines of all of these angles.

I actually found values that satisfy both Menelaus Theorems and the Rule of Four Quantities and not Abu Nasr 2. This implies that Abu Nasr 2 can't be algebraically derived from Menelaus. But recall that we found two more Menelaus equations (the ones we found after we switched A-F and B-E), and the values I chose do not satisfy those equations. This means that, while I've found an algebraic counterexample, I don't have a geometric counterexample.

Of course, even if I stumble upon a proof of Abu Bakr 2, we still have a valid proof of the same equation with the sAC/sAE factor thrown in -- a factor which is never 1 except in a special case. And since Van Brummelen doesn't use Abu Nasr 2 at all in the rest of the chapter, it almost makes me believe that the author made a mistake -- that the correct Abu Nasr equation is:

sDF/sEF = sAD/sAB * sAC/sAE

Exercise 4.9

In this question we shall work through another of al-Biruni's methods for finding the qibla of Ghazna in The Determination of the Coordinates of Cities. The diagram is identical to figure 4.7, except that AMW is omitted, and GFH is drawn from G perpendicular to PM. BC = lambda = 27.37 degrees,
GC = phi_L = 33.58 degrees, and MB = phi_M = 21.67 degrees.

[By the way, I was considering changing this from Ghazna to my hometown to find my own qibla, but I don't know whether these methods work if Mecca is more than a quadrant away.]

(a) Use the Rule of Four Quantities on figure PGCBF to find FG.
sin FG/sin BC = sin PG/sin PC
sin FG/sin 27.37 = sin 56.42/1 (as PG is 90 - 33.58)
sin FG = sin 56.42 sin 27.37
sin FG = 0.38301
FG = 22.52 degrees

(b) Use the Rule of Four Quantities on figure EFPNH to find PE, and from it find ME.
sin FH/sin PN = sin EF/sin PE
sin 67.48/sin 33.58 = 1/sin PE (as E is the pole of HFG, implied but not stated)
sin PE = 0.59876
PE = 143.22 degrees (also implied by the diagram is PE > 90)
ME = PE - PM
ME = 143.22 - 68.33 (as PM is 90 - 21.67)
ME = 74.89 degrees

(c) Use the Rule of Four Quantities on figure EMFHD to find MD.
sin MD/sin FH = sin ME/sin EF
sin MD/sin 67.48 = sin 74.89/1
sin MD = sin 67.48 sin 74.89
sin MD = 0.8918
MD = 63.1 degrees

(d) Finally, use the spherical Law of Sines on Triangle PGM to find PGM, the direction of the qibla.
sin p/sin P = sin g/sin G
sin (90 - 63.1)/sin lambda = sin(90 - 21.67)/sin PGM
sin 26.9/sin 27.37 = sin 68.33 /sin PGM
sin PGM = sin 27.37 sin 68.33/sin 26.9
sin PGM = 0.94434
Angle PGM = 70.79 or 109.21 degrees

Technically speaking, PGM is 109.21 degrees. From the graph, this is the number of degrees west of north (since P is the North Pole). The other value of the arcsine, 70.79 degrees, is the number of degrees west of south. This matches the qibla value we found earlier in this post.

A Resolution of the Cliffhanger?

Last summer, at the end of my last Van Brummelen post, I wrote the following:

In fact, I believe I now have a proof of the following:

A line and its translation image are parallel.

Last year on the blog, I believed that I had a proof of this statement. But that proof is invalid -- and I know it because nowhere in the proof did I use the Fifth Postulate. So the proof would work in hyperbolic geometry -- yet the result is false in hyperbolic geometry. Once again, I use the third geometry to check the validity of a proof.

This post is getting long enough, and I definitely want to double- and triple-check this proof before I attempt to post it. But the following appears to be a preliminary lemma needed for the proof:

Suppose line l is the image of a reflection and l' is its image.
(a) If l and l' are parallel (and not identical), then there is exactly one possible position for the mirror.
(b) If l and l' intersect, then there are exactly two possible positions for the mirror.
(c) If l and l' are identical, then there are infinitely many possible positions for the mirror.

This lemma holds in both Euclidean and spherical geometry. Of course, case (a) doesn't apply in spherical geometry, so only (b) and (c) matter here.

Well, I'm now wondering whether I ever had a valid proof. But let me at least say a little about what I was thinking at the time.

The theorem to prove is:

A line and its translation image are parallel.

Now a translation, if you recall, is the composite of reflections in parallel mirrors. So we can rewrite the theorem as:

Given a composite of two reflections, if the mirrors are parallel, then so are the line and its image.

And this is equivalent, via the contrapositive, to:

Given a composite of two reflections, if the line and its image are nonparallel, then so are the mirrors.

But now let's recall what "nonparallel" is. Here I'm using the U of Chicago definition of "parallel," which allows a line to be parallel to itself. (This is significant because it's easy to map a line to itself via a translation -- just slide the line in the direction of that line.) Thus, we must show that if the line and its image are distinct and intersecting, then the mirrors are also distinct and intersecting.

Here's the idea behind the lemma -- we let l be the original line and l" the image of the composite. We can now let l' be any line. Now we must find mirrors m and m' so that a reflection over m maps l to l' , and a reflection over m' maps l' to l". The lemma tells us how many possibilities for m and m' exist.

But this is where I'm stuck. I don't remember exactly what I was thinking last year -- and even if I did, that proof might never have been valid (which is why I didn't just post it right then last year).

Last year, I wrote that I like thinking about spherical trig because it reminds me how it feels to be a student in our math classes. I worked on Question 4.3, struggled, and gave up -- just as many students do when searching for proofs in our Euclidean geometry classes.

Of course, this last problem applies to Euclidean geometry, not (just) spherical geometry. I don't expect students to figure it out on their own -- the idea was for me to find the proof just once and use it as a new way to teach transformations and parallel lines in Geometry. But as of now, I don't have a valid proof. I will definitely continue to think about this problem.

I don't know when my next post will be yet -- it won't be Monday, since I'm no longer teaching a summer school class that day. So I return to the randomness of summer posting.

To any and all Muslim readers, have a blessed Eid al-Fitr!