## Saturday, June 17, 2017

### Van Brummelen Chapter 2: Exploring the Sphere

1. Pappas Page of the Day
2. Van Brummelen Chapter 2: Exploring the Sphere
3. Exercise 2.7: What Color Is the Bear?
4. Exercise 2.12: Spherical Polygon-Sum Theorem
5. Exercise 2.14: Polar Triangle Inequality
6. A Possible High School Geometry Course
7. Hilbert's Axioms for Spherical Geometry?

Pappas Page of the Day

This is what Theoni Pappas writes on page 168 of her Magic of Mathematics:

600 BC -- Thales introduces deductive geometry. It was developed over the years by such mathematicians and philosophers as Pythagoras, the Pythagoreans, Plato, Aristotle.

On this page, Pappas is providing us with a timeline of Euclidean and non-Euclidean geometry. Let me post some of the highlights from this timeline:

300 BC -- Euclid compiles, organizes and systematizes geometric ideas, which had been discovered and proven, into thirteen books, called The Elements.

140 BC -- Posedonius restates Euclid's 5th postulate.

1637 -- Rene Descartes formulates analytic geometry.

1795 -- Gaspard Monge (1746-1818) describes structures by plane projections.

1854 -- G.F. Bernhard Riemann (1826-1866) presents elliptical geometry.

1888 -- Giuseppe Peano (1858-1932) creates Peano space-filling curve (fractal).

1904 -- Helge von Koch (1870-1924) creates Koch snowflake curve (fractal).

Notice that Pappas credits Riemann with presenting spherical (elliptic) geometry. Even though Legendre had written about spherical geometry earlier, it was Riemann who recognized spherical geometry as a separate, non-Euclidean geometry.

Van Brummelen Chapter 2: Exploring the Sphere

Chapter 2 of Glen Van Brummelen's Heavenly Mathematics is "Exploring the Sphere." In Chapter 1 he focuses on the "trig" part of spherical trig, and here he writes about the "spherical" part, before he puts it all together in the next chapter. He begins:

"At first glance there's not much to see on a sphere; every point on its surface looks the same as any other. But give it some physical meaning -- call it the celestial sphere, or the Earth's surface -- place some identifying marks on it, and set it in motion, and visualizing what's happening can become rather complicated."

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.

Afterwards, Van Brummelen defines three coordinate systems, similar to longitude and latitude for earth's surface. One is called right ascension and declination, symbolized (alpha, delta). The second is called ecliptic longitude and latitude, symbolized (lambda, beta). The third is called azimuth and altitude -- this determines where a star is at a given time at night. Zero longitude for both the equatorial and ecliptic coordinates is set to the spring equinox. The symbol for this equinox can't be rendered in ASCII, but it's actually the symbol for "Aries."

Now Van Brummelen writes some basic theorems in spherical geometry. We've already proved some of these theorems last year with Legendre, so today's listing of them can serve as a review.

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

Given: Figure 2.6. O is the center of the sphere.

Proof:
Consider the cross-section in figure 2.6. Let D be any point on the cross-section, and let OC be the perpendicular line dropped from O onto the intersecting plane. Then Angle OCD is right, and the Pythagorean Theorem applies: OD^2 = OC^2 + CD^2. But OD is constant regardless of D's position on the cross-section, since it is the radius of the sphere; and clearly OC doesn't depend on where D is either. Therefore CD^2 cannot change as we move D around the cross-section, and so neither does CD. QED

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

Note: Van Brummelen defines the length of a segment AB on the sphere to be the measure, in degrees, of the central Angle AOB. Last year, I wrote that if we were to use radians instead of degrees on the unit sphere, then the measure of the central angle really is the length of the segment.

Given: Figure 2.8. O is the center of the sphere, and ABC is a spherical triangle.

Proof:
Examine the angles at O corresponding to the sides in figure 2.8. Imagine allowing segment OA to fall onto the plane OBC, leaving O in place but bringing A downward. Then two of the angles would fit perfectly within the third. But if we lift A back into its original position, the two angles at O that rise with it become larger. So their sum must be greater than BOC on the plane. QED

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

Proof:
In figure 2.8, join A, B, and C with straight lines, forming a tetrahedron with O. The nine angles in the tetrahedron, excluding the angles in face ABC, must add up to 3 * 180 = 540 degrees, since they form three triangles. Now, the sum of the two of those nine angles that are located at A exceeds Angle A in the plane triangle ABC (by the same argument that led to the lemma a few moments ago), and likewise for the pairs of angles at B and C. Therefore,

Sum of sides = Angles at O
= 540 - (angles at A + angles at B + angles at C)
< 540 - (Angle A + Angle B + Angle C)
= 540 - 180 = 360 degrees

QED

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

Given: Figure 2.10. The triangle ABC is shown with its polar triangle A'B'C'.

Proof: In figure 2.10, extend the arcs of the original triangle to intersect the sides of the polar triangle. Since C' is a pole of AB and A' is a pole of BC, both C' and A' are 90 degrees removed from B. So B must be a pole of A'C'. Likewise for the other arcs. QED

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.

Proof:
In figure 2.10 both D and E (extensions of the sides of the original triangle to the sides of the polar triangle) are 90 degrees removed from A; therefore Angle A = DE. Now since C' is a pole of ABD and B' is a pole of ACE, both C'D and B'E are 90 degrees. Therefore

B'C' = B'E + C'D - DE = 180 - DE = 180 - Angle A.

Similarly for the other sides of the polar triangle; we have now dispatched the first half of the theorem. The second half follows immediately from the duality relation; simply apply the result we have just established to the polar triangle and its polar (i.e., the original triangle), rather than the original and the polar triangle. QED

Theorem:
The angle sum of a triangle must exceed 180 degrees.

Proof:
We know that the sum of the sides of a polar triangle must be < 360 degrees. Since the sides of the polar triangle are the supplements of the angles of the original,

(180 - A) + (180 - B) + (180 - C) < 360,

so A + B + C > 180 degrees. QED

Van Brummelen writes, "And now, finally, we have enough spherical geometry under our belts to tackle some spherical trigonometry."

Exercise 2.7: What Color Is the Bear?

7. (a) A bear hunter walks one km south, then one km east, then one km north, and ends up back where he started. What color is the bear?

This question is a classic! I wrote about it here on the blog in November 2015. Instead of searching for that old post, let me just post the link I gave that day:

http://www.murderousmaths.co.uk/books/bearpuz.htm

(b) The puzzle in (a) has a particular location in mind, but there are actually many locations where this journey is possible. Identify the others and say what animal must replace the bear in the story.

http://www.murderousmaths.co.uk/books/bearpuz2.htm

Exercise 2.12: Spherical Polygon-Sum Theorem

12. Show that a spherical polygon with n sides (each less than 180 degrees) has a sum of interior angles greater than 180n - 360 degrees. [paraphrased from Cresswell 1816, p.54]

As it turns out, we prove this theorem exactly like the Euclidean Polygon-Sum Theorem from Lesson 5-7 of the U of Chicago text.

Choose a vertex on the polygon. Call it A. Draw the diagonals from A. For the n-gon, the diagonals form n - 2 triangles. Since the angle sum of each triangle is greater than 180 degrees, the total for the entire n-gon is more than (n - 2)180 = 180n - 360 degrees. QED

Exercise 2.14: Polar Triangle Inequality

14. Show that in any spherical triangle, the difference between any angle and the sum of the other two is less than 180 degrees. [paraphrased from Moritz 1913, p. 12]. (Hint: use the polar triangle.)

Well, let's follow Van Brummelen's hint and consider the polar triangle. Since the theorem asks us to prove something about the angles of a triangle, we look at the sides of the polar triangle. In particular, to find a relationship between one angle and the sum of the other two angles, we look in the polar triangle for a relationship between one side and the sum of the other two sides. The obvious relationship to consider is the Triangle Inequality.

Polar Triangle: A'B' < A'C' + B'C'
Original Triangle: 180 - C < (180 - A) + (180 - B)
180 - C < 360 - (A + B)
(A + B) + 180 - C < 360
(A + B) - C < 180

QED

A Possible High School Geometry Course

Van Brummelen suggests that his spherical trig book can be used as a course text. I wouldn't teach spherical trig to high school students, but I can see a situation where we might introduce spherical geometry as the last unit of a high school Geometry class. Yesterday, I mentioned that the Honors Geometry course at Barron Trump's new school does exactly that.

I like the idea of such a course. Here is a unit plan for my ideal Geometry course, which revolves around the Common Core transformations and ends up with spherical geometry:

1. Reflections
2. Rotations
3. Translations
4. Glide Reflections
(first semester finals)
5. Dilations
6. Two Dimensional Measurement
7. Three Dimensional Measurement
(state testing)
8. Spherical Geometry
(second semester finals)

This was actually what I had in mind during the first two years of this blog. The first unit, on Reflections, actually introduces the basics of Geometry (points, lines, angles, and so on). Congruence of triangles is lumped in with Rotations, parallel lines are part of Translations, and Glide Reflections includes other miscellaneous topics (most notably parallelograms). Then similarity is obviously included in the Dilations unit.

I actually like the idea of all eight units being named after a transformation. The 2D measurement unit can be named Transvections, since these preserve area, and a transvection maps a parallelogram to a rectangle of the same area (hence the parallelogram area formula). The 3D measurement unit can be named Screws, as these are 3D isometries that don't exist in 2D. (Transvections also exist in 3D and are closely related to Cavalieri's Theorem.) This doesn't imply that students actually focus on transvections or screws, but they are just made aware of their existence.

The final section on spherical geometry can't really be named for a transformation, though. Had this been hyperbolic geometry, a good name would be Horolations. But unlike hyperbolic geometry with its extra isometry, spherical geometry has one fewer isometry. This is because a translation is the composite of two reflections in parallel lines -- but in spherical geometry there are no parallel lines and thus no translations.

My idea is that parallel lines and their properties are introduced in the translations unit. Thus every theorem mentioned in the first two units is valid in both Euclidean and spherical geometry -- and this fact can be brought up in the final unit on spherical geometry. The eight units of the year readily correspond to the eight quavers of the year.

During the first two years of this blog, I wanted to organize my course with this in mind. But there were two problems. The first is that the U of Chicago text wasn't organized this way, so I had to jump around the text to fit this scheme. Eventually I decided that it's better to just to follow the text in order rather than jump around.

The other reason has to do with a difficulty in spherical geometry that doesn't exist in either Euclidean or hyperbolic geometry. To see what this is, we go back to Van Brummelen and look at his commentary following the proof of the Spherical Triangle Inequality:

"If you are not happy with the informality of this argument, I can bring none other than Euclid to my defense. The planar equivalent of this statement, that the third side of any plane triangle cannot exceed the sum of the other two sides, is Proposition 20 in the first book of the Elements. Curiously, Euclid's proof works for spherical triangles just as well...

"Euclid avoided [5th postulate] as long as he possibly could, until finally he was forced to use it in Proposition 29. Now, it turns out that spherical geometry is one of the non-Euclidean geometries that is consistent with Euclid's other axioms, but not with the parallel postulate. Since Proposition 20 comes before 29, Euclid's proof works on the sphere as well as it does on the plane."

The problem is that according to David Joyce, Euclid's Proposition 16 fails in spherical geometry, despite coming before 29:

http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI16.html

This theorem is known as the Triangle Exterior Angle Inequality, or TEAI. Proposition 20 is proved using 19, which is proved used 18, which in turn is proved using 16 or TEAI. Therefore, the proof of Proposition 20 is not valid in spherical geometry (even though the theorem itself is)!

One misconception about spherical geometry is that it, like hyperbolic geometry, is one of the three geometries that satisfies Euclid's first four postulates. It is elegant to assume that all we have to do is consider the parallel postulate:

-- Through a point not on a line, there is exactly one line parallel to the given line.

and replace "exactly one line" with "more than one line" for hyperbolic geometry and "no line" for spherical geometry, and be done. But as it turns out, this works for hyperbolic geometry, but not for spherical geometry. This is because spherical geometry does not satisfy the first four postulates!

The first postulate of Euclid is:

-- Through any two points, there is exactly one line.

And this fails in spherical geometry, since through the poles there are infinitely many lines. In fact, spherical geometry arguably satisfies the fifth postulate, since the postulate and many equivalent statements are written in the form "if bla bla bla, then the lines intersect" (which is trivially true as all lines intersect) or "if two lines are parallel, then bla bla bla" (which is vacuously true). And Playfair's axiom, the simplest equivalent of Euclid's fifth postulate, is correctly written as:

-- Through a point not on a line, there is at most one line parallel to the given line.

"At most one line" clearly includes the possibility that there are zero parallel lines -- which is exactly the case for spherical geometry.

Some people (like David Joyce) point out that spherical geometry fails the first postulate. Almost no one points out that spherical geometry actually satisfies the fifth postulate. And so technically, if we want to see how spherical geometry is different from Euclidean geometry, we shouldn't be looking at the fifth postulate at all, but rather the first postulate.

Hilbert's Axioms for Spherical Geometry?

Another way to look at this is to consider not Euclid's axioms, but Hilbert's axioms. Recall that Hilbert's axioms are more rigorous than Euclid's. Just as with Euclid's axioms, only changing the parallel postulate makes Euclidean geometry into hyperbolic geometry, but we must do more to make spherical geometry. Here is a link to a webpage that considers which of Hilbert's axioms are valid in spherical geometry:

http://math.iit.edu/~mccomic/420/notes/Bolin_spherical.pdf

According to this link, only three of Hilbert's axioms fail in spherical geometry:

(I1) Any two distinct points lie on a unique line. (Euclid's 1st postulate)
(B2) For any two distinct points A and B, there exists a point C such that A * B * C (that is, B is between A and C).
(B3) Given three distinct points on a line, exactly one lies between the other two.

Notice that a parallel postulate isn't listed among these axioms.

Arguably, Euclid's second postulate may fail in spherical geometry. This postulate states that any line segment can be extended, and (B2) involves extending a segment AB to a new segment AC. For simplicity, I interpret Euclid's second postulate as stating that a segment can be extended so that it divides the plane (or sphere) in half. (Joyce alludes to this at the link above.) This is so that denying a single postulate of Euclid's, the first, can give us spherical geometry (just as denying a single postulate of Euclid's, the fifth, can give us hyperbolic geometry). But in truth, we know that we must modify three of Hilbert's postulates to produce spherical geometry.

I often wonder how to rewrite these three postulates for spherical geometry. We might rewrite (I1):

(I1-sphere?) Any two distinct points lie on at least one line.

that is, we assert existence but not uniqueness of a line. Notice that we can even include uniqueness by adding a third point:

(I1-sphere?) Any two distinct points line on at least one line, and any three distinct points lie on at most one line.

This version of (I1-sphere?) is just a weaker version of (I1), and as such it still holds in Euclidean as well as spherical geometry.

Of course, the exceptions to both (I1) and (B2) for spherical geometry are antipodal points. Through antipodal points there are infinitely many lines, and indeed every other point lies between a pair of antipodal points. Somehow, a spherical version of either (I1) or (B2) should mention antipodal points.

As for (B3), this may be the hardest to rewrite spherically. Not all counterexamples to (B3) are antipodal -- for example, consider three points spaced 120 degrees apart on the Equator.

None of this matters for a high school Geometry course, where we don't use Hilbert's axioms. So instead, the idea is that in the first two units (Reflections and Rotations), we only prove theorems that are provable in both Euclidean and spherical geometry. There's already a term, neutral geometry, for geometry in which the first four of Euclid's postulates hold (Euclidean and hyperbolic). I propose referring to Euclidean and spherical geometry as natural geometry. (In previous posts, I used the term "normal geometry," but I like "natural" better.) In natural geometry we don't have Euclid's first postulate, but we do have his second though fifth postulates (under our interpretation of them).

The most obvious results of natural geometry are SSS, SAS, and ASA, since these hold in both Euclidean and spherical geometry. And so our first two units can focus on proofs that require only these three congruence theorems. On the other hand, AAS and HL aren't natural, since they hold in Euclidean and not spherical geometry. Also, AAA isn't natural, since it holds in spherical but not Euclidean geometry.

But there are a few problems here. We've seen that the Triangle Inequality holds in both Euclidean and spherical geometry, but we've yet to see a single rigorous proof that works for both. So the Euclidean proof that Van Brummelen mentions fails in spherical geometry due to the dependence of the proof on TEAI.

There are a number of inequalities that we prove in Geometry -- Triangle Inequality, SAS Inequality, SSS Inequality, Unequal Sides, and Unequal Angles. Euclid proves all of these using TEAI. But we wish to avoid TEAI in our proofs as it isn't part of natural geometry. The U of Chicago text assumes the Triangle Inequality as a postulate. But it seems as if we should be able to do better -- there ought to be a proof of one of the Inequality Theorems that doesn't ultimately depend on TEAI.

The concurrence theorems are also tricky. The angle bisectors of a triangle are concurrent in both Euclidean and spherical geometry, but no proof I can find is valid in both. The usual Euclidean proof uses AAS ot HL, while the following link gives a spherical proof using AAA!

Even the Converse Isosceles Triangle Theorem causes problems. It is true in both Euclidean and spherical geometry, but the usual Euclidean proof uses AAS, which isn't spherically valid. One way to unify the proofs (that is, give a single proof that works for both geometries) is the "quick and dirty" proof that a triangle with equal base angles is congruent to itself via ASA.

Notice that I'm trying to keep the theorems as natural as possible by avoiding theorems like AAS and HL as long as possible, and instead using only results valid in both geometries. This doesn't mean that we actually introduce spherical geometry (until the final unit). The following are to be avoided in a high school Geometry course:

-- terms like "Lambert quadrilateral" (or "Saccheri quadrilateral" for that matter). In both Euclidean and spherical geometries, a Lambert quadrilateral has three right angles. The difference is that in Euclidean geometry, the fourth angle is also right, but in hyperbolic geometry, it is obtuse.
-- theorems like "the sum of the angles of a triangle is at least 180 degrees." This is true for both Euclidean and spherical geometry (=180 for Euclidean, >180 for spherical), but the proofs are completely different. Even if we could unify the proofs, we should still avoid it in high school.

Again, during the first two years of this blog, I tried to rearrange the U of Chicago text so that it follows this pattern. It's tricky because SSS, SAS, and ASA appear so late (Chapter 7), while the first postulate that isn't natural appears in Lesson 3-4:

Corresponding Angles Postulate (Euclid's Proposition 28):
If corresponding angles have the same measure, then the lines are parallel.

Parallel Lines Postulate (Euclid's Proposition 29):
If two lines are parallel, corresponding angles have the same measure.

The first postulate holds in Euclidean and hyperbolic geometry, but fails in spherical geometry. The second postulate holds in Euclidean geometry and vacuously holds in spherical geometry, but fails in hyperbolic geometry. The second is equivalent to Euclid's fifth postulate, which we know fails in hyperbolic geometry. The first doesn't use Euclid's fifth at all, so it holds in hyperbolic geometry. The idea is that spherical geometry is the opposite of hyperbolic geometry -- the first somehow uses a postulate like Euclid's first which fails in spherical geometry, while the other avoids Euclid's first and so is valid in spherical geometry (even if vacuously so). In Lesson 3-5, the Two Perpendiculars Theorem follows from the Corresponding Angles Postulate and hence is spherically invalid.

Believe it or not, the next spherically invalid result isn't until Lesson 5-6. Notice along the way, in Lesson 5-5, we have several theorems involving isosceles trapezoids and rectangles, but these hold vacuously in spherical geometry. In Lesson 5-4, the theorems involve kites and rhombi. Kites aren't problematic -- they exist in spherical geometry, and all the theorems about them are true. The definition of "rhombus" is a quadrilateral with four sides equal in length -- and such equilateral quadrilaterals exist in spherical geometry. Nowhere in the definition of "rhombus" does it state that a rhombus is a parallelogram. Rhombi exist in spherical geometry, but not parallelograms.

In Lesson 5-6, we reach our next main non-natural (unnatural?) result -- the Alternate Interior Angles Test (abbreviated as AIA => || Lines Theorem). It's used to prove that a rhombus is a parallelogram, which we already know is spherically invalid since rhombi exist, but not parallelograms. (Also, notice that equiangular quadrilaterals exist as well. These quadrilaterals aren't rectangles -- instead all four angles are obtuse. Yet they have the same symmetry lines -- these symmetry lines divide the equiangular quadrilateral into two Saccheri or four Lambert quadrilaterals.)

Once we reach Lesson 5-6, we are fully Euclidean and no longer spherical. Still, it's bad style to claim that almost all theorems up to Lesson 5-6 are spherically valid by counting vacuous theorems as valid. I'd much rather go to non-vacuous results like SSS, SAS, and ASA before all of these vacuous quadrilateral theorems.

Before we leave my ideal geometry course, let's look at the Pearson Integrated Math I course, which was written with Common Core in mind. Geometry begins in Lesson 8-1 with translations on the coordinate plane, so we're already spherically invalid -- and this is followed in Chapter 9 with area formulas which are spherically invalid. But there are a few interesting things to note here.

In Lesson 11-2, there is an unusual postulate:

Same-Side Interior Angles Postulate:
If a transversal intersects two parallel lines, then same-side interior angles are supplementary.

This statement is equivalent to Euclid's fifth postulate, and so it fails in hyperbolic and holds in Euclidean and spherical geometry (vacuously). In fact, as every statement is equivalent to its contrapositive, we can rewrite this as:

If same-side interior angles are not supplementary, then the lines aren't parallel.

This is actually closer to Euclid's original fifth postulate than any parallel postulate that occurs in any modern text. The only difference is that Euclid's fifth tells us on which side of the transversal the two lines intersect. This statement holds in spherical geometry because lines are never parallel -- and they always intersect on both sides of the transversal.