Tuesday, August 8, 2017

Van Brummelen Chapter 3: The Ancient Approach

Table of Contents

1. Pappas Page of the Day
2. Blaugust?
3. Van Brummelen Chapter 3: The Ancient Approach
4. Exercise 3.2
5. Exercise 3.3
6. Exercise 3.4
7. Natural Geometry
8. A Natural Proof?

Pappas Page of the Day

This is what Theoni Pappas writes on page 220 of her Magic of Mathematics:

"Over the centuries paradoxes have created glitches in traditional logic. True or false logic could not be used to explain such paradoxes as Eublides' pile of sand paradox or Bertrand Russell's class membership paradox."

This page marks the final section of the computer chapter, "Fuzzy Logic & Computers." Here Pappas explains that traditional logic based on true or false -- also known as Boolean logic -- can often lead to paradoxes.

Back in January 2016, I wrote about the mathematician David Kung, who lectured about paradoxes such as the two mentioned by Pappas here. I don't think that Kung specifically mentions the pile of sand paradox (also known as the heap paradox or sorites paradox), though. Here is how Pappas describes this paradox:

"4th century B.C. Greek philosopher Eublides considered how many grains of sand make a pile."

The idea is that a single grain clearly isn't enough to constitute a pile, and simply adding a grain of sand to a non-pile doesn't make it into a pile. But combining these makes a proof by induction that no amount of sand can ever make up a pile:

Initial Step: 1 grain of sand isn't a pile.
Induction Step: If n grains of sand aren't a pile, then neither are n + 1 grains of sand.
Conclusion: No finite number of grains of sand is a pile.

We also think of this by starting with clearly enough sand to make a pile -- such a million grains -- and then subtracting one grain at a time:

Initial Step: 1,000,000 grains of sand are a pile.
Induction Step: If n grains of a sand are a pile, then so are n - 1 grains of sand.
Conclusion: Any finite number of grains of sand is a pile.

This is the paradox. The only way to avoid it is if there some discrete natural number N such that any fewer number of grains aren't a pile, and any greater number is a pile. But no such bound N exists.

The other paradox, Russell's Paradox, does appear in Kung's lectures. This is what I wrote about Russell's Paradox back in January 2016:

I've mentioned the British mathematician Bertrand Russell a few times on the blog before. Kung begins by describing a few details of Russell's life. In addition to a mathematician, Russell was also a peace activist and a Nobel Prize laureate. But in mathematics, he is especially known for a paradox, Russell's Paradox:

-- Does the set of all sets that don't contain themselves contain itself?

Kung explains that most sets don't contain themselves as elements -- for example, the set of all prime numbers is not itself a number, much less a prime number. But the set of abstract ideas is an abstract idea, so it does contain itself as an element.

The professor then points out that Russell's Paradox is the same as the Barber's Paradox, except replacing "shaves" with "contains (as an element)." The barber shaves himself if and only if it doesn't shave himself, and the Russell set contains itself if and only if it doesn't contain itself.


You might remember how last year, there was an MTBoS challenge called "Blaugust," where math teachers would write about the start of the upcoming school year.

As of now, I'm not a math teacher. Therefore I won't participate in Blaugust this year. There are some MTBoS challenges in which I participated even as a sub, such as the January Blogging Initiative. But other challenges make sense only for actual teachers. I can't really write about my new classroom for the first day of school if I'm not in a classroom. Of course, if I'm hired at the last minute as a teacher, then (as long as it's by August 31st) I will make an official Blaugust post.

Actually, it's uncertain whether there will be an official Blaugust challenge this year. Last year, there was a true Blaugust challenge, with one teacher hosting the challenge and providing a list of over 40 possible prompts. This year, no one appears to be hosting Blaugust or posting a list of prompts.

Yet some bloggers are still labeling their posts as "Blaugust." For example, here's a link to Jackie Stone, a Illinois high school math teacher:


Indeed, not only am I not sure whether there's a Blaugust this year, it's debatable whether there is even an MTBoS this year. In previous posts, I've referred to Dan Meyer as the King of the MTBoS, but now he wants to retire the MTBoS label:


Instead, Meyer proposes using "I teach math" as a replacement to MTBoS. But others disagree, including Fawn Nguyen:


Interestingly enough, Jackie Stone, in her first Blaugust post, writes about the debate. Apparently, as for now, she will remain neutral and use both labels:


I, of course, can't use the "I teach math" label since as of now, I don't teach math. I'll wait to choose either the MTBoS or "I teach math" label if and when I'm actually teaching math again.

Anyway, I can't really call Dan Meyer the King of the MTBoS anymore since he is currently rejecting the MTBoS label. In fact, if anyone should called a monarch of the MTBoS, it should be Fawn Nguyen, thanks to her defense of the label in the post "Long Live the MTBoS!"

So long live the MTBoS! And long live the Queen of the MTBoS, Fawn Nguyen!

Of course, Dan Meyer is now the King of the "I teach math" label, since he's the one proposing its widespread use.

Van Brummelen Chapter 3: The Ancient Approach

Oops -- well, as soon as I found out that I wasn't getting a new teaching job and had to return to the job hunt, my reading of Glen Van Brummelen's Heavenly Mathematics fell apart. (Yesterday would have been the first day of school had I been hired there.)

My plan had been to read all nine chapters of Van Brummelen before today. After the disappointment, let me instead just complete one more chapter finished this summer. We can wait until next year to finish the book, just as it took us two summers to finish Legendre's treatment of spherical geometry.

This is what Van Brummelen writes in Chapter 3 of his book, "The Ancient Approach":

"We tend to think of the growth of mathematical knowledge that that of a glacier. The boundaries spread outward gradually as new bits of knowledge are added to the existing structure."

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."

According to Van Brummelen, Menelaus wrote one major work -- the Sphaerica. As its title implies, his work focused on spherical geometry and trigonometry. But one of the first theorems in this work actually applies to Euclidean geometry:

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, DB and TK intersect at L

Proof: Draw DX parallel to TLK, then Triangle XAD ~ KAT and Triangle DBX ~ LBK. Therefore

As we can see here, we can prove plane Menelaus mainly using the Side-Splitting Theorem, found in Lesson 12-10 of the U of Chicago text. The auxilary segment DX is a side-splitter for triangle KAT, while a new triangle, DBX, is formed with side-splitter LK.

Now the goal is for Menelaus to prove the spherical analog of his theorem. We begin by proving a few lemmas first:

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.

Proof: Project A and C perpendicularly onto the vertical diameter. Since the circle has radius 1, the two dotted line segments have lengths sin alpha and sin beta. The two right triangles are similar, so the ratio holds. QED

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.

Proof: Project B and C perpendicularly onto the horizontal diameter. The result follows immediately from the fact that the two triangles are similar. QED

You might be wondering where the sines of alpha and beta come from in this proofs. We can think of alpha and beta in Lemma B not just as arcs, but as central angles. Then we know that the vertical coordinate of an angle on the unit circle is the sine (and the horizontal coordinate is the cosine). The same is actually true in Lemma A, even though the diameter is vertical. The initial sides of the angles are on this vertical diameter (so this counts as the "x-axis" in determining sine and cosine).

According to Van Brummelen, the spherical Menelaus theorem is now upon us. Since every ratio of line segments in his planar theorem may be replaced with the sines of the corresponding arcs in figure 3.4, we conclude:

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.)

Van Brummelen points out that if there is a Theorem A, then there must be a Theorem B. Here is his derivation of this theorem:

Figure 3.7 begins with the same configuration as before. Extend Arcs BA and BD to form two semicircles connecting B to its antipodal point X. This action leads us to a new Menelaus configuration XAZEGD, to which we can apply Theorem A:

sin AZ/sin AX = sin GZ/sin GE * sin DE/sin DX.

But since BZAX is a semicircle, sin AX = sin(180 - AB) = sin AB; likewise, sin DX = sin BD. Substituting and shuffling a bit gives us

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

(That's right -- points on the sphere have antipodal points.) The author admits that these formulas are awkward to remember. So he provides us with figure 3.8 and assigns the following labels to the arcs:

a = AD, b = GD, c = AZ, d = BZ, e = DE, f = EB, g = EZ, h = GE

With these substitutions, the theorems become:

Disjunction: sin a/sin b = sin(c + d).sin d * sin g/sin h.
Conjunction: sin(a + b)/sin a = sin (g + h)/sin g * sin f/sin (e + f)

Van Brummelen explains that "disjunction" implies that the two arcs on the left hand side are disjoint (a and b), while in the "conjunction" they overlap (a and a + b).

He then gives one more way to write the equation, which he attributes to John Holte, a retired professor from a liberal arts college in Minnesota:

sin AG/sin GD * sin DE/sin EB * sin BZ/sin ZA = 1
sin BA/sin AZ * sin ZG/sin GE * sin ED/sin DB = 1

The author adds that in each case, the order of the vertices mentioned in each formula (A, G, D, E, B, Z, A in the first) produces a Hamiltonian (or Eulerian) circuit of the Menelaus diagram. (Think back to the Euler bridges problem, from Lesson 1-4 of the U of Chicago text as well as the first day of school activity in my own class.)

Now Van Brummelen shows us how to use these formulas. We know that the sun travels through the sky during the year along the ecliptic, but now we wish to find the sun's equatorial coordinates. He labels coordinates in his figure 3.3 -- we knew lambda (ecliptic longitude) from today's date; and Arc BC equals epsilon (obliquity of ecliptic) = 23.44 degrees, since both B and C are 90 degrees away from the spring equinox. (He uses a non-ASCII character for the equinox, but here I'll just render it as "Aries," as in the sign of Aries.) Oh, and you can see where the 90-degree angles come from -- in a triangle with two right angles, the measure of the third angle equals the opposite arc.

If we apply the conjunction formula correctly, we can find the declination, delta, as:

sin 90/sin epsilon = sin 90/sin delta * sin lambda/sin 90

since Arc BN = AN = AriesC = 90, or, more simply, sin delta = sin lambda * sin epsilon. He explains that tables of the form sin z = sin x * sin y were used to solve spherical trig problems -- especially in the case where sin y = sin 23.44 (since epsilon, remember, is a constant).

If we apply the disjunction formula correctly, we can find the right ascension, alpha, as:

sin (90 - alpha)/sin alpha = sin 90/sin(90 - epsilon) * sin(90 - lambda)/sin lambda

which we can simplify by recalling that sin(90 - x) = cos x (as in "complementary sine"). A little bookkeeping leaves us with tan alpha = tan lambda cos epsilon.

Van Brummelen writes:

"Menelaus's Theorem became the standard tool of spherical astronomy for the next 900 years. Menelaus may have Claudius Ptolemy to thank for his fame."

The author explains that Menelaus and his Spherica, along with Ptolemy and his Almagest, were the cornerstones of spherical trig for a very long time. As in many mathematical and scientific fields, the next major advance in spherical trig occurred during the Islamic Enlightenment.

Abu Sahl al-Kuhi was an Iraqi astronomer who lived during the last years of the first millennium. He, according to Van Brummelen, calculated pi to be 3 1/9. This value, of course, is less accurate than the Babylonian 3 1/8, which in turn is not as accurate as the Archimedean 3 1/7. (Hey, we could use this as an excuse to celebrate three Pi Approximation Days -- July 22nd for Archimedes, August 25th for the Babylonians, and September 28th for al-Kuhi.)

Anyway, let's look at al-Kuhi's work. Van Brummelen gives us figure 3.9 -- and again, I have to describe everything that's in the figure:

-- 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).

Here are al-Kuhi's calculations:

Step 1: Apply the conjunction theorem to configuration NGZMAriesSun as we did earlier, to get     sin delta = sin lambda * sin epsilon.

Step 2: Next find what in medieval times was called the ortive or rising amplitude eta = Arc ESun, the distance along the horizon between Sun and the east point E. For this al-Kuhi uses another Menelaus configuration, NHQMESun:

sin 90/sin (90 - phi) = sin 90/sin delta * sin eta/sin 90

or sin eta = sin delta/cos phi.

Step 3: Return to the Menelaus configuration of the previous step and apply conjunction again, but this time assign the arcs the other way:

sin 90/sin MQ = sin 90/sin(90 - eta) * sin (90 - delta)/sin 90

or sin MQ = cos eta/cos delta. The significance of MQ is that it is the complement of n = Arc EM, known to Muslim astronomers as the ascensional difference or equation of daylight.

Step 4: Our final step is a return to configuration NGZMAriesSun from Step 1, again applying conjunction but assigning the arcs the other way. The result is

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.

Van Brummelen explains that all of the results that al-Bruni derived were already discovered by either Menelaus or Ptolemy. The difference is that al-Bruni used only conjunction, while the previous mathematicians used both conjunction and disjunction. As the author writes, "It is a marvel of compact mathematics." This is the same reason why we try to avoid the Fifth Postulate as long as possible in Euclidean geometry, as well as avoid the Axiom of Choice in set theory.

Exercise 3.2

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 May 20th. Instead, I choose today's date, August 8th. And I don't choose a latitude of 49.3N. Instead, I choose my home latitude, which is 34N.

(Speaking of which, last week I visited my home confluence of 34N, 118W again. I notice that there is now a "No Parking" sign near the confluence. I ended up walking about a mile round trip from the last legal parking spot -- in 90-degree weather.)

(a) Notice that these questions correspond to the four al-Bruni steps above. So we follow Step 1:

sin delta = sin lambda * sin epsilon

The chart to find lambda is in the appendix. On August 8th, lambda is about 134.6 degrees.

sin delta = sin 134.6 sin 23.4
sin delta = 0.28278
delta = 16.4 degrees

(b) We follow Step 2:

sin eta = sin delta/cos phi
sin eta = 0.28278/cos 34
sin eta = 0.34109
eta = 19.9 degrees

(c) We follow Step 3. Notice that Arc MQ is the complement of n, so sin MQ = cos n, even though Van Brummelen never writes this:

cos n = cos eta/cos delta
cos n = cos 19.9/cos 16.4
cos n = 0.98003
n = 11.4 degrees (that is, Arc MQ is 78.6 degrees)

And according to Van Brummelen, the equation of daylight represents the difference between the rising time of the arc for an observer at our location and the rising time if the observer were at the terrestrial equator. (This really is a time -- 15 degrees is one hour, so 11.4 is about 45 minutes.)

(d) We follow Step 4:

sin MZ = cos lambda/cos delta
sin MZ = cos 134.6/cos 16.4
sin MZ = -0.73203
MZ = -47.1 degrees

That's right -- MZ is negative. This is because Z is the celestial longitude of the summer solstice (that is, 90 degrees past Aries), and so M is on the opposite side of Z from E. We finally calculate theta:

theta = MQ - MZ
theta = 78.6 - (-47.1)
theta = 125.7 degrees
theta = about 8 hours, 22 minutes

You may notice that for today's date, lambda is almost exactly 135 degrees, which is convenient since sin 135 is exactly sqrt(2)/2. Indeed, today is almost exactly halfway between the summer solstice and the fall equinox. Last year, I wrote about Lammas, the cross-quarter date. So apparently, today is the astronomical date of Lammas. We can use Van Brummelen's chart to find the other cross-quarters:

Spring Equinox -- March 20th
Beltane -- May 6th
Summer Solstice -- June 22nd or 23rd (given as 89.5 and 90.5 degrees on the chart)
Lammas -- August 8th
Fall Equinox -- September 23rd
Samhain -- November 6th
Winter Solstice -- December 20th
Imbolc -- February 2nd or 3rd (given as 314.5 and 315.5 degrees on the chart)

The debate last year was whether Lammas or the equinox was the first day of fall. Based on the temperature, fall starts at the equinox, but the fall semester at increasingly many schools is closer and closer to Lammas.

Since there are 360 degrees and 365 days, each day is approximately one degree. Notice that if I had started my Pappas day count on the day I bought the book (which was right around the spring equinox) instead of January 1st retroactively, we'd be only near page 135 now instead of 220. Then the page number would correspond to the sun's angle along the ecliptic.

Then again, knowing that today is day 135 doesn't really help us with the calculation. After all, in part (a) we had to calculate:

sin delta = sin lambda * sin epsilon

And even if we know that sin lambda = sqrt(2)/2 (which we can find on a simple calculator without the sine key), we'd still have to use the scientific calculator to find sin epsilon = sin 23.4. It does help us that sin epsilon is very close to 0.4 (indeed, the arcsine of 0.4 is 23.6 degrees), but even then, we could only find that sin delta = 0.2828, so we can't avoid the scientific calculator completely in calculating delta. (Again, Menelaus himself would have used charts to find delta.)

Exercise 3.3

In this and the following question, we will demonstrate the conjunction version of Menelaus's Theorem directly, rather than piggyback on the disjunction theorem. For our first step, demonstrate the following result related to the plane Menelaus Theorem (figure 3.2):


(Hint: draw a line segment KY, parallel to BD.)

Let's write this proof in two-column format, keeping Van Brummelen's "hint" in mind:

Given (I repeat this 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
Prove: AB/AK = BD/LD * LT/TK

Statements                                  Reasons
1. A-D-T, A-K-B, D-L-B, T-L-K 1. Given
2. Draw KY | | BD (A-Y-D)         2. Existence of Parallels (Van Brummelen's hint)
3. Angle ADB = AYK,                3. Corresponding Angles Consequence
    Angle DLT = YKT
4. Angle A = A,                          4. Reflexive Property of Congruence
    Angle T = T
5. Triangle ADB ~ AYK,             5. AA Similarity
    Triangle DLT ~ YKT
6. AB/AK = BD/KY,                    6. Corresponding sides of similar triangles are proportional.
    TK/LT = KY/LD
7. AB/AK * TK/LT = BD/LD      7. Multiplication Property of Equality
8. AB/AK = BD/LD * LT/TK      8. Multiplication Property of Equality

Exercise 3.4

(Continued from question 3. Originally, I was only going to choose two questions to do on the blog, but questions 3 and 4 go together. And so we might as well try this question as well.)

To move from the plane to the sphere we will need a slightly different diagram than before. In figure E-3.4, begin with the original spherical configuration. Then extend BZ and HA until they meet at a point X outside of the sphere. Next extend ZE and HG until they meet at a point Y. Finally join BE and extend to W on the line connecting X and Y. From this diagram, prove the conjunction version of Menelaus's Theorem.

Notice that figure E-3.4 is related to figure 3.4 (hence the numbering), and so the points A, B, D, E, G, H, and Z are the same as in that first figure. (Scroll up earlier in this post for figure 3.4.)

Here is the statement to be proved:

Prove: sin AB/sin AZ = sin BD/sin DE * sin GE/sin GZ

Clearly, the trick is to take the formula proved in question 3 and then use Lemmas A and B in order to replace the lengths of line segments with the sines of spherical arcs:

Statements                                  Reasons
1. W, X, Y as described               1. Given
2. XB/XZ = BW/BE * YD/YZ      2. Proved in question 3
3. XB/XZ = sin AB/sin AZ           3. Lemma B
4. BW/BE = sin BD/sin BE         4. Lemma B
5. YD/YZ = sin GE/sin GZ          5. Lemma B
6. sin AB/sin AZ = sin BD/sin DE 6. Substitution Property of Equality
    * sin GE/sin GZ

We see that only Lemma B is needed, since the auxiliary points W, X, and Y are all outside the sphere.

Natural Geometry

We have finished our reading Van Brummelen for the year. We'll return to Chapter 4 of his book beginning next summer.

In previous posts, I've written of a vision as to how we could actually fit spherical geometry into a high school class. This is my idea:

-- Begin with results provable in both Euclidean and spherical geometry.
-- Focus on results provable in Euclidean geometry only.
-- Give the end-of-year state tests in Euclidean geometry.
-- Introduce spherical geometry during the extra time left after testing.

Recall that this idea comes from Presidential Consistency. We saw how the Honors Geometry class at Barron Trump's new school teaches spherical geometry at the end of the year.

We know that this task is much easier if we replace "spherical" with "hyperbolic." Indeed, the results provable in both Euclidean and hyperbolic geometry are called "neutral geometry." Neutral geometry is based on the first four postulates of Euclid. To implement hyperbolic geometry, we need only replace the Fifth Postulate with a new hyperbolic parallel postulate -- one which states that there are many lines parallel to a given line through a point not on the line.

But we don't want to mention hyperbolic geometry in a high school class. Spherical geometry is more relevant as we live on a spherical earth. (After all, this is what Van Brummelen's book is about.) Of course, it's possible that we live in a hyperbolic universe, based on Einstein's theories -- as of now, the shape of the universe is unknown. Still, even if the universe is hyperbolic, the spherical earth is much more familiar.

Unfortunately, spherical geometry has a huge problem not present in hyperbolic geometry. Spherical geometry is not neutral geometry. So it's not enough to keep the first four postulates of Euclid and replace the fifth. In fact, the Fifth Postulate is possibly true in spherical geometry (as there really is at most one line parallel to any line in spherical geometry). Instead, we see that it's the First Postulate that fails in spherical geometry -- through any two points, there is at least one line, and infinitely many lines if the two points are antipodal.

Just as all results of Euclidean and hyperbolic geometry are provable in neutral geometry, all results of Euclidean and spherical geometry are to be proved in natural geometry. Just as Postulates 1-4 incorporate neutral geometry, Postulates 2-5 would incorporate natural geometry.

But still, trying to prove theorems in natural geometry is problematic. Let's consider one of the best known theorems of neutral (Euclidean-hyperbolic) geometry. It is named after two mathematicians that we've associated with non-Euclidean geometry -- Giovanni Saccheri and Adrien Legendre:

Saccheri-Legendre Theorem of Neutral Geometry:
The sum of the measures of the angles of a triangle is at most 180 degrees.

Of course, the sum is exactly 180 degrees in Euclidean geometry, and strictly less than 180 degrees in hyperbolic geometry. It would seem as if there is a corresponding theorem in natural geometry:

Theorem of Natural Geometry???
The sum of the measures of the angles of a triangle is at least 180 degrees.

Returning to neutral geometry, Saccheri-Legendre is closely related to the following theorem. In fact, it appears as the first step to proving Saccheri-Legendre:

Theorem of Neutral Geometry:
The sum of the measures of two angles of a triangle is less than 180 degrees.

This makes sense -- if the sum is at most 180, then removing one angle makes it less than 180. The corresponding theorem in natural geometry would look like:

Theorem of Natural Geometry???
The sum of the measures of four angles of a triangle is more than 180 degrees.

But this makes no sense at all -- where does this mysterious fourth angle come from? And so we see that any attempt to follow this line of thinking in natural geometry is doomed.

In an earlier post, I mentioned other problems with natural geometry. For example, the angle bisectors of a triangle are concurrent in both Euclidean and spherical geometry. But the usual proofs of this theorem in Euclidean geometry require either AAS or HL, which are invalid in spherical geometry. In fact, it appears that the proof of the theorem in spherical geometry requires AAA, which of course is invalid in Euclidean geometry.

Then again, we certainly don't want our natural proofs to look like the following:

Concurrency of Angle Bisectors in Natural Geometry:
The angle bisectors of a triangle are concurrent.

Case 1: Euclidean geometry
Using AAS, ....

Case 2: Spherical geometry
Using AAA, ...

This violates Van Brummelen's notion of compactness. We wish to be able to prove the theorem simply by omitting the postulate that distinguishes between Euclidean and spherical geometry (the First Postulate), just as we prove neutral theorems by omitting the postulate that distinguishes between Euclidean and hyperbolic geometry (the fifth postulate).

We've discussed the concurrency of the perpendicular (rather than angle) bisectors before. An almost complete proof is provided in Lesson 4-5 of the U of Chicago text. Here we have Triangle ABC, where m and n are the perpendicular bisectors of AB and BC respectively:

"If m and n intersect [at point O], it can be proven that this construction works. Because of the Perpendicular Bisector Theorem with line m, OA = OB. With line n, the theorem also justifies the conclusion OB = OC. By the Transitive Property of  Equality, the three distances OA, OB, and OC are equal. Thus Circle O with radius OA contains points B and C also."

That is, since OA = OC, O lies on the perpendicular bisector of AC also. But this proof still has a gap because of those key words "if m and n intersect." As it turns out, this is non-trivial to prove -- and in fact, in hyperbolic geometry, m and n could be parallel. In fact, the concurrency theorem is actually false in hyperbolic geometry! (Notice that even though I don't wish to introduce hyperbolic geometry in class, looking at hyperbolic geometry helps me to find errors and omissions in proofs in the other two geometries.)

Fortunately, in Euclidean and spherical geometry, m and n do intersect and the theorem is true. But how exactly do we know this?

Well, in spherical geometry, all lines intersect, so of course m and n intersect! So we really only have something to prove in Euclidean geometry. The U of Chicago doesn't provide this proof -- but later in Lesson 6-2, the proof of the Two Reflections Theorem for Translations is similar:

Indirect Proof (of m intersect n):
Assume to the contrary that m | | n. By definition of perpendicular bisector, AB is perpendicular to m and BC is perpendicular to n. Thus by the Perpendicular to Parallels Theorem, BC must also be perpendicular to m. So, by the Two Perpendiculars Theorem, AB | | BC. But B is on each line. So Line AB = BC and the three points are collinear. But A, B, and C are the vertices of a triangle, so they can't be collinear, a contradiction. Therefore m and n intersect. QED

Let's use hyperbolic geometry to check the validity of this proof. Since we know that the result (that m and n intersect) is false in hyperbolic geometry, at least one step used in the proof must be invalid in hyperbolic geometry. That step is the Perpendicular to Parallels Theorem.

Our problem is that this proof still isn't compact in natural geometry. We must divide into two cases to show that m and n intersect:

Concurrency of Perpendicular Bisectors in Natural Geometry:
The perpendicular bisectors of a triangle are concurrent.

Case 1: Spherical geometry
All lines intersect, therefore m and n intersect...

Case 2: Euclidean geometry
Assume that m | | n. Then by the Perpendicular to Parallels and Two Perpendiculars Theorems...(see above argument)...a contradiction. Therefore m and n intersect...

Each case uses a step that is invalid in the other geometry. In the spherical case, we used the fact that all lines intersect, which is false in Euclidean geometry. In the Euclidean case, we used the Two Perpendiculars Theorem, which is false in spherical geometry. As of yet we have no single proof that works in both geometries, even though the statement holds in both geometries.

There are other results which hold in both Euclidean and spherical geometry. One of them is the Converse to the Isosceles Triangle Theorem. Even though this is valid in spherical geometry, its usual proofs use either AAS or HL, which are invalid in spherical geometry. Closely related are the Unequal Sides and Unequal Angles Theorems. These are valid in spherical geometry, yet the proofs given in the U of Chicago text ultimately go back to TEAI, which is invalid in spherical geometry.

Notice that in a true natural geometry, we can't distinguish between the Euclidean plane and the sphere, so there shouldn't be any steps that require us to make such a distinction. In particular, our proofs shouldn't refer to points "inside the sphere" or "outside the sphere." Yet last summer, we saw that Legendre's proof of Triangle Inequality requires forming solid angles at the center of the sphere, so we must refer to "inside the sphere."

Of course, the Menelaus Theorem that we gave today also requires applying the Euclidean version to points inside or outside the sphere. But technically, the Menelaus Theorem is not a natural result -- we need different formulas for the Euclidean and spherical cases. Even though it appears that we obtain the spherical version by taking the "sine of everything" in the Euclidean version, we've seen for ourselves that these statements require rigorous proof. (The Euclidean version is clearly not natural as both Existence of Parallels and AA Similarity are used.)

A Natural Proof?

There might be a way to fix the Concurrency of Perpendicular Bisectors proof so that it works in both Euclidean and spherical geometry.

Let's begin by cutting off the Euclidean proof at the step which fails in spherical geometry. This step turns out to be the Two Perpendiculars Theorem. Notice that Perpendicular to Parallels Theorem holds vacuously in spherical geometry. It asserts that if we have two parallel lines, then any line parallel to one is parallel to the other (and yet says nothing about what happens if no parallel lines exist at all). On the other hand, Perpendicular to Parallels fails in hyperbolic geometry -- and we know that at least one step must fail in hyperbolic geometry. Thus Perpendicular to Parallels is exactly the theorem we want to use, as it holds in two out of the three geometries.

So far we have:

Indirect Proof (of m intersect n):
Assume to the contrary that m | | n. By definition of perpendicular bisector, AB is perpendicular to m and BC is perpendicular to n. Thus by the Perpendicular to Parallels Theorem, BC must also be perpendicular to m.

So far, we have that AB is the perpendicular bisector to m -- that is, there exists point D on AB such that AD = BD and the angles at D are both right angles.

It also means that the reflection image of B over m is exactly A. Notice that we're allowed to use the Reflection Postulate since it's valid in both Euclidean and spherical geometry. Let's look at that postulate once more:

Reflection Postulate:
Under a reflection:
a. There is a 1-1 correspondence between points and their images This means that each preimage has exactly one image and each image comes from exactly one preimage.

It might not be obvious that this holds in spherical geometry in one particular case -- when the preimage is a pole of the mirror. There are infinitely many lines through this pole that are perpendicular to the mirror. Yet reflections are well-defined in this case -- the image of a pole of the mirror is in fact the other pole. This is because the poles are the only points that are exactly the same distance (a quadrant) away from every point on the mirror. Thus a pair of antipodal points have infinitely many "midpoints" yet only one perpendicular bisector (their equator), on which all of these "midpoints" lie.

The well-definition of reflection tells us that since B reflected over m is A, it means that every single segment that is perpendicular to m and has B as one of its endpoints must have A as the other endpoint -- that is, every line perpendicular to m and passing through B also passes through A.

Let's look at the proof again:

Indirect Proof (of m intersect n):
Assume to the contrary that m | | n. By definition of perpendicular bisector, AB is perpendicular to m and BC is perpendicular to n. Thus by the Perpendicular to Parallels Theorem, BC must also be perpendicular to m.

So line BC is a line perpendicular through m and obviously passing through B. Therefore A must lie on line BC as well -- that is, A, B, and C are collinear. This is exactly what we wanted to prove:

But A, B, and C are the vertices of a triangle, so they can't be collinear, a contradiction.
Therefore m and n intersect. QED

Let E be the point where m and line BC intersect. Notice that in Euclidean geometry, we actually have that D and E are the same point. In spherical geometry, D and E can be distinct points -- then we have that ADBE is a lune. This is still a contradiction because a lune is not a triangle -- lunes have only two sides while triangles have three. (This is also how we know that m and n aren't identical -- if they were, then point C would be the same point as A, since reflecting B over n gives both C and A.)

Now that we have a natural proof of Concurrency of Perpendicular Bisectors, we also have a natural proof of Two Reflection Theorem for Translations as well. In this case, we have no contradiction, since we want to prove that A, A', and A" are collinear.

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.

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