**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.)

Theorem:

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.

Theorem:

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

Theorem:

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.

Theorem:

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*-

*B*,

*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 AriesSun

*GZM*, from which we find sin delta/sin lambda = sin epsilon/1, so sin delta = sin lambda sin epsilon.

(2) Use figure

*E*Sun

*HQM*, from which we have sin delta/sin eta = sin(90 - phi)/1, or sin eta = sin delta/cos phi.

(3) Use figure

*N*Sun

*MQH*, which gives sin(90 - eta)/sin(90 - delta) = sin

*MQ*, or sin

*MQ*= cos eta/cos delta.

(4) Finally, use figure

*NGZM*Sun 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, Aries*A*= alpha, Sun*A*= 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*O*Aries is*E*, and the foot of perpendicular from*A*to*O*Aries is*F.*And finally, the foot of perpendicular from*C*to*OB*is*K.*

Van Brummelen proceeds as follows:

The two right triangles Sun

*ED*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
Sun

*D*/Sun*E*=*CK*/*CO.*

But Sun

*D*= sin delta (to see why, consider the vertical circular segment*ODA*Sun) and similarly Sun*E*= 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 AriesSun

*GZM*, 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

*E*Sun

*HQM*, 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

*N*Sun

*MQH*, 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

*NGZM*Sun 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:*

s

*CF*/s

*CE*= s

*BF*/s

*BD** s

*AD*/s

*AE*

*Now since*

*F*is the pole of

*BC*, both

*BF*and

*CF*are 90 degrees, and s90 (= sin 90) = 1. So we have:

1/s

*CE*= 1/s

*BD** s

*AD*/s

*AE*

*We multiply both sides by s*

*BD*to obtain the desired result:

s

*BD*/s

*CE*= s

*AD*/s

*AE*

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

s

*CE*/s

*EF*= s

*AC*/s

*AB** s

*BD*/s

*DF*

*Since the desired theorem has s*

*DF*/s

*EF*on the left side, we multiply both sides by s

*DF*and then divide both sides by s

*CE*:

s

*DF*/s

*EF*= s

*AC*/s

*AB** s

*BD*/s

*CE*

*And we have s*

*AB*on the right side, so all we need to make appear is s

*AD*. If we could somehow show that s

*AD*= s

*AC** s

*BD*/s

*CE*, then we're done. Let's return to the Four Quantities and solve for s

*AD*:

s

*BD*/s

*CE*= s

*AD*/s

*AE*

s

*AD*= s

*AE** s

*BD*/s

*CE*

*Oops -- we needed s*

*AC*there, not s

*AE.*In fact, we can multiply the previous formula by s

*AE*/s

*AE*:

s

*DF*/s

*EF*= s

*AC*/s

*AB** s

*BD*/s

*CE** s

*AE*/s

*AE*

s

*DF*/s

*EF*= (s

*AE** s

*BD*/s

*CE*) * 1/s

*AB** s

*AC*/s

*AE*

s

*DF*/s

*EF*= s

*AD*/s

*AB** s

*AC*/s

*AE*

*And so we have an extra factor of s*

*AC*/s

*AE*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: s

*AC*/s

*BC*= s

*AE*/s

*DE** s

*DF*/s

*BF*

Disjunction: s

*BC*/s

*AB*= s

*CF*/s

*EF** s

*DE*/s

*AD*

*Let's take the disjunction formula. We know that s*

*CF*= 1, so we have:

s

*BC*/s

*AB*= 1/s

*EF** s

*DE*/s

*AD*

s

*AD*/s

*AB*= 1/s

*EF** s

*DE*/s

*BC*

We now start working on the conjunction formula, aware that s

*BF*= 1:

s

*AC*/s

*BC*= s

*AE*/s

*DE** s

*DF*/1

s

*DE*/s

*BC*= s

*AE*/s

*AC** s

*DF*

*Let's plug this into the earlier equation:*

s

*AD*/s

*AB*= 1/s

*EF** s

*DE*/s

*BC*

s

*AD*/s*AB*= 1/s*EF** s*AE*/s*AC** s*DF*
s

s*AD*/s*AB*= s*DF*/s*EF** s*AE*/s*AC**DF*/s

*EF*= s

*AD*/s

*AB** s

*AC*/s

*AE*

*And so once again, we have this extra factor of s*

*AC*/s

*AE*that we can't eliminate.

By the way, you might wonder under what situation would we have s

*AC*/s

*AE*= 1. Well, if s

*AC*= s

*AE*, 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, s

*AC*can't equal s

*AE*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 (s

*BD*/s

*CE*= s

*AD*, as s

*AE*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:

s

*AD*/s

*ABD*= s

*AB*/s

*ADB*

*Now Angle*

*ABD*= 90, and so s

*ABD*= 1:

s

*AD*/1 = s

*AB*/s

*ADB*

s

*AD*/s

*AB*= 1/s

*ADB*

*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*:

s

*EF*/s

*EDF*= s

*DF*/s

*DEF*

1/s

*EDF*= s

*DF*/s

*EF** 1/s

*DEF*

*If we set 1/s*

*ADB*= 1/s

*EDF*, then we obtain:

s

*AD*/s

*AB*= s

*DF*/s

*EF** 1/s

*DEF*

s

*DF*/s

*EF*= s

*AD*/s

*AB** s

*DEF*

*So this time, the extra factor we can't eliminate is s*

*DEF.*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 s*

*BF*and s

*CF.*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 s

*AC*/s

*AE*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:

s

*DF*/s

*EF*= s

*AD*/s

*AB** s

*AC*/s

*AE*

**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:

Theorem:

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:

Lemma:

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!