1. Introduction
2. Definitions in Natural Geometry
3. Placement of U of Chicago Lessons in Natural Geometry
4. Coordinate Plane in Natural Geometry?
5. Three Perpendiculars Revisited
6. Actual High School Proofs
7. Number Bases
8. Feynman, pages 161-250
9. Conclusion
Introduction
This is the fifth post of my summer "Natural Geometry" series. The fourth post of this series was my Pi Approximation Day post. There I discussed translations as well as a special "Three Perpendiculars Theorem" that's supposed to hold in both Euclidean and spherical (hence natural) geometry.
I decided to give glide reflections their own post, especially since the translations have been absorbed into the Pi Approximation Day post. Of course, there really isn't much more to say about glide reflections except that they are the composite of a reflection and a translation (in Euclid) or a rotation (on the sphere).
Instead, this post will tie up some loose ends regarding the first five units of natural geometry and what they would look like in an actual high school course.
Definitions in Natural Geometry
To review, our first five units are the following:
- Point-Line-Plane
- Reflections
- Rotations
- Translations
- Glide Reflections
These five units are supposed to correspond to the first semester of a high school Geometry course -- that is, the first seven chapters of the U of Chicago text.
The most important idea is to be certain of our definitions. We've seen that just as there are inclusive and exclusive definitions of words such as "trapezoid" within Euclidean geometry, there are inclusive and exclusive definitions of words across geometries. If an object exists in Euclidean geometry but not spherical geometry, such as a "rectangle," then we must decide whether we want an inclusive definition of a "rectangle" (such as "equiangular quadrilateral") or an exclusive definition (a rectangle must have four right angles). Then under the inclusive definition, rectangles exist on the sphere, but under the exclusive definition, spherical rectangles don't exist.
I've been thinking about this in the last few days since my most recent posts. And anyway, here is what I decided:
First of all, the word "parallel" needs to keep its definition -- coplanar lines that don't intersect. After all, it's the existence or nonexistence of parallel lines that allows us to distinguish between Euclidean and non-Euclidean geometry.
Of course, I was playing around with "ultraparallel" in recent posts. The sole purpose of this was to define "translation" as "composite of reflections in ultraparallel mirrors" and then consider all rotations on the sphere to be translations (as some authors do). We're not going to use "ultraparallel" moving forward, so this means that we won't consider rotations to be translations anymore.
This means that the translations of Unit 4 exist only in Euclid. Thus this unit signifies the end of the theorems that are valid in both Euclidean and spherical geometry -- and the start of the theorems that are valid in Euclid only.
Since the word "parallelogram" contains the word "parallel" and parallel lines don't exist on the sphere, neither do parallelograms. As for "rectangles," even though this doesn't contain the word "parallel," the root "rect" means "right," so rectangles should have something to do with right angles.
Actually, there's a good reason to say that rectangles don't exist on the sphere, rather than try to consider equiangular, Saccheri, or Lambert quads to be rectangles. Here's the reason -- I've written that besides Point-Line-Plane, I don't want to change any other postulate in the text. This includes the Area Postulate of Lesson 8-3. Included in this postulate is the rectangle area formula A = lw. There is no spherical quad for which this area applies. Therefore, we conclude that there must be no rectangle on the sphere, for otherwise we would contradict this postulate.
Of course, since there is no spherical rectangle, there is no spherical square either. But there is one quad on the hierarchy which we might wish to keep on the sphere -- a "rhombus." As defined in the U of Chicago text:
- A quadrilateral is a rhombus if and only if its four sides are equal in length.
Nowhere in this definition are parallel lines mentioned. Indeed, such equilateral quads definitely exist on the sphere. Of course, in Euclid we prove that all rhombi are parallelograms -- and of course, this proof, given in Lesson 5-6, is spherically invalid. Notice that most of the quad hierarchy appears in Lesson 5-2 and follows almost immediately from the definition, but only "every rhombus is a pgram" requires additional thought -- and Euclidean geometry.
Thus we can allow rhombi to exist on the sphere. Actually, kites exist on the sphere as well -- and it follows trivially from the definition that every rhombus is a kite.
So far, we seem to be keeping all of the U of Chicago definitions, which force the definitions of "parallel" and of the special quads to be exclusive except for "rhombus" and "kite." Actually, let's look at the U of Chicago definition of parallel once more:
- Two coplanar lines m and n are parallel lines, written m | | n, if and only if they have no points in common, or they are identical.
Those last four words, "or they are identical," are the key here. According to this definition, every line is parallel to itself. Extending this to spherical geometry, we conclude that two lines are parallel iff they are identical. Thus the only translation that exists is the composite of reflections in identical mirrors -- in other words, the identity transformation.
So on the sphere, "parallel" means "identical" and "translation" means "identity." This degeneracy reminds us of similarity and dilations. On the sphere, "similar" means "congruent," and so "dilation" also means "identity" as well.
There might eventually be a situation where this definition is useful. We might be able to prove in natural geometry that two lines m and n are parallel. On the sphere, we then conclude that m and n are the same line, whereas in Euclid they need not be identical.
Placement of U of Chicago Lessons in Natural Geometry
Once again, the five units of natural geometry are supposed to correspond to the first seven chapters of the U of Chicago text. So let's start placing U of Chicago lessons into the five units.
Unit 1, "Point-Line-Plane," needs to include not just points, lines, and planes, but angles as well. So it needs to incorporate Lessons 1-1 through 3-3 of the text.
Unit 2, "Reflections," starts with Chapter 4, of course. It's definitely a good place to include Lesson 5-1 on isosceles triangles, since these have reflection symmetry. In the past, I also suggested that the triangle congruence theorems can be placed here as well, Lessons 7-1 to 7-4.
But Unit 3, "Rotations," is a problem. If we want to include figures with rotational symmetry in the rotations chapter, then parallelograms would be placed here. Yet we wish Unit 3 to be the last natural (spherically valid) unit, and so we don't want there to be "parallelograms" in this unit.
Let's think about this from another perspective -- the school calendar. Again, we want these five units to span the first semester of the year -- August through December. It's easy to think of the five units as corresponding to months -- Unit 1 in August, Unit 2 in September, up to Unit 5 in December.
If you think about it, we probably want Unit 1 in August to be light. Depending on the school, the first day of school might be midway through the month, plus it might take several days for the students to settle. So perhaps a good Unit 1 might start with something fun (like Lessons 1-4 or 1-5, as I did the first three years of this blog) and then go through the rest of Chapter 1, plus Lessons 3-1 and 3-2 to introduce angles.
Similarly, Unit 5 in December can be light as well. Once again -- depending on the calendar -- prep for finals, the exams themselves, and winter break can easily take up half the month. Thus "Glide Reflections" is an excellent December unit, as Lesson 6-6 is the only lesson that must be here.
Then Units 2-4 are the "meaty" units. It actually might make more sense to include all of Chapter 2 in the "Reflections" unit before moving on to Chapter 4.
The congruence theorems of Chapter 7 are placed in Unit 3, "Rotations" -- after Lesson 6-3. We do this because besides the congruence theorems, not much left in the U of Chicago text is natural.
This leaves Unit 4, "Translations," to contain Lesson 6-2 and all of the lessons that depend on the existence of parallel lines. This includes Lesson 3-4 (parallel line consequences) and parts of Chapter 5 (quadrilaterals) and 7 (parallelograms). Actually, notice that by doing this, all of our triangle theorems land in Unit 3 and quad theorems in Unit 4 anyway.
There are a few lessons that remain difficult to place. There might be a way to sneak parallelogram symmetry into Unit 3 (since there exist quads on the sphere with 180-degree rotational symmetry) without actually using the words "parallel" or "parallelogram." Once again, I recently pointed out that the following is easy to prove:
- If the opposite sides of a quadrilateral are congruent, then so are the opposite angles.
without any mention of pgrams. To do so, we simply divide the quad into two triangles and then prove those triangles congruent by SSS. We thus include this proof as an exercise shortly after teaching the students SSS and proofs with auxiliary lines. Later on, we then state the properties of parallel lines and pgrams, and then redo the proof.
We also need to place lessons required by the Common Core Standards but not included anywhere in the U of Chicago text. This includes the concurrency proofs.
Coordinate Plane in Natural Geometry?
We shouldn't do slopes of parallel and perpendicular lines until after similarity in the second semester, but I still like the idea of seeing the coordinate plane in October, just before the PSAT, even if the Geometry students aren't actually reviewing linear equations.
It's possible, even in Unit 3 in October (where geometry should still be natural), to define some sort of coordinate plane. To do this, we let any point be the origin, and draw two perpendicular lines through this point to be the x- and y-axes. Provided that some sort of distance has been defined, then the points on these axes have coordinates. Then x = h is defined to be the line perpendicular to x-axis through point with coordinate h, and y = k is likewise defined to be perpendicular to the y-axis.
It's interesting to see what happens with this coordinate plane on the sphere. First of all, notice that the coordinates can't be made to correspond to latitude (and longitude), since the curves of constant latitude (the "parallels") aren't great circles, and hence aren't lines. If the origin is placed on the Equator, then the lines x = h do become meridians (of constant latitude). But if the origin is placed at one of the Poles, then both axes are meridians.
There is a problem with antipodal points -- if the two axes intersect at the origin, then they must also intersect at the point antipodal to the origin. Also, if the origin is on the Equator, then lines x = h (all meridians), intersect at the Poles. But the coordinate plane works provided that we stick to a single hemisphere (excluding the boundary).
Notice that if we consider the quad whose sides are the two axes, x = h, and y = k, then we notice that by definition, the angle between x -axis and x = h is right, as is that between y-axis and y = k. Since the angle between the axes is always right, we now have a Lambert quadrilateral. We can't prove that the angle between x = h and y = k is right (and on the sphere it can't possibly be right, since otherwise it would be a rectangle).
In fact, there might be enough here to prove some of the coordinate properties of reflections here, such as the image of (h, k) reflected over the x-axis is (h, -k). Notice that the quad whose sides are the y-axis, x = h, y = k, and y = -k is a Saccheri quadrilateral. In fact, we might be able to sneak in some properties of Lambert, Saccheri, and equiangular quads here just by trying to prove the coordinate properties of reflections and rotations. (At no point do we actually use the words "Lambert," "Saccheri," or "equiangular" here.)
In the special case that h = k, the Lambert quad becomes a kite. (On the sphere, this kite is not a rhombus, since a Lambert quad can't be a rhombus unless it's a square.) It might be possible to use the symmetry properties of a kite to prove that the image of (h, k) reflected over y = x is (k, h). (Note that the kite is used to prove that y = x is even a line, since we don't have any slope properties.) Of course, this requires that we teach kite symmetry in Unit 3 rather than wait for Unit 4.
Once again, I like the idea of placing these lessons in October just before the PSAT, even if only to remind the students of how the coordinate plane works. Of course, there's no harm in teaching the students linear equations, but only as a review of Algebra I rather than something that needs to be proved rigorously. I remind the readers that we need to wait until the second semester before we can fully prove the existence and properties of slope.
Three Perpendiculars Revisited
In my Pi Approximation Day post, I alluded to a natural "Three Perpendiculars Theorem":
Three Perpendiculars Theorem:
If three lines are perpendicular to the same line, then the three lines are either parallel or concurrent.
When I first posted this theorem last year, I wrote that, while it's valid in both Euclidean and spherical geometry, it's more of a curiosity than something we might want to teach students. Unless there's a compelling reason why we'd want to teach Three Perpendiculars, we probably shouldn't.
But let's think about it -- there's another theorem that students need to know, and it definitely has something to do with three perpendiculars and concurrency. You guessed it -- it's the concurrency of the three perpendicular bisectors of a triangle.
Let's think again about that incomplete proof of the concurrency of perpendicular bisectors as given in Lesson 4-5 of the U of Chicago text:
Construct the circle through the three noncollinear points A, B, and C.
Solution:
Step 1. Subroutine: m, the perpendicular bisector ofAB
Step 2. Subroutine: n, the perpendicular bisector ofBC
Step 3. m and n intersect at O. (Point Rule)
Step 4. Circle O containing A (Compass Rule)
If m and n intersect, it can be proven that this construction works. Because of the Perpendicular Bisector Theorem with line m, OA = OB. With line n, this theorem also justifies the conclusion OB = OC. By the Transitive Property of Equality, the three distances OA, OB, and OC are all equal. Thus Circle O with radius OA contains points B and C also. QEF
Earlier, we discovered that this proof is incomplete, because it takes it for granted that the lines m and n actually intersect. Also, we notice that the noncollinearity of A, B, and C isn't used in the proof. We found out that these are related depending on which geometry we're in:
Euclidean: If A, B, C are collinear then m | | n. If A, B, C are noncollinear then m, n intersect.
Spherical: All lines intersect, therefore m, n intersect.
Hyperbolic: If A, B, C are collinear than m | | n. If A, B, C are noncollinear then there's no conclusion.
Because of this, we can write a single theorem that incorporates both cases:
Theorem:
Let A, B, C be any three distinct points. Then the three perpendicular bisectors ofAB, BC, AC are either concurrent or parallel.
Proof:
It's essentially the same as the U of Chicago proof. All we really need to change is drop Circle O and instead, use Converse Perpendicular Bisector Theorem to show that since O is equidistant from A, C, we conclude that O lies on the perpendicular bisector ofAC. QED
Notice that without loss of generality, we can start by knowing that any two of the three perpendicular bisectors must intersect (AB & BC, AB & AC, or BC & AC) and conclude that the third bisector must pass through O as well. Thus the only way to avoid the existence of O is for all three perpendicular bisectors to be parallel. This is why we have validly shown that the three bisectors must be either concurrent or parallel.
This proof is valid in all three geometries -- Euclidean, spherical, and hyperbolic.
Actual High School Proofs
But even with this version of the Three Perpendiculars Theorem, we still haven't actually reached the proofs that we really want to see. Once again, Three Perpendiculars should ultimately lead to "a line and its translation image are parallel."
Indeed, as written, the theorem doesn't yet distinguish between the parallel and concurrent cases. We know that the lines are always concurrent in spherical geometry, and if A, B, C are noncollinear (as in the vertices of a triangle), then they are concurrent in Euclid as well. Thus, we should somehow be able to prove in neutral geometry that if A, B, C are noncollinear then the lines are concurrent. Such a proof can be placed near the end of Unit 3 -- but I can't figure out yet what the proof would look like.
There might also be a path to some interesting proofs if we were to begin with the concurrence of angle bisectors, rather than perpendicular bisectors. In this case, instead of points A, B, C, we begin with three lines a, b, c, and show that if these lines intersect then so do the bisectors of the angles that they form. Even though this proof would be mainly about intersecting lines, an indirect proof could be set up so that it refers to the properties of parallel lines instead.
Don't forget that we do have Playfair's Parallel Postulate available. Officially, Playfair is neutral (due to the "at most" one line clause). We know that Playfair must somehow be used in the concurrence of perpendicular bisectors proof, since the proof fails in hyperbolic geometry. This means that Playfair would be introduced before Unit 4 as well, even as other Euclidean parallel line properties aren't taught until Unit 4.
So there are still several proofs that I still need to figure out here. Still, it would be nice if we could could adapt the concurrency proofs so that they could help us out with the parallel line proofs.
Once we reach Unit 4, the geometry becomes fully Euclidean. So the original Euclidean version of our Point-Line-Plane postulate is used in proofs. Once again, we don't tell the students this -- we just quietly start using the "through any two points, there is exactly one line" clause.
Number Bases
In my Pi Approximation Day post, I briefly mentioned number bases -- mainly because I decided to include the Eleven Calendar (and hence its implied undecimal base) in my annual FAQ post.
Number bases is still one of my favorite topics, though I rarely have an opportunity to write about this topic on the blog, since it has nothing to do with Geometry. But since I'm thinking about it right now, let me squeeze some number bases into today's post.
And indeed, it helps that Pappas has a number base problem on her calendar today:
113 (base x) = 805 (base 9)
Determine x.
We might notice that just as 805 (base 9) means 8 + 9^2 + 0 * 9 + 5, we can expand 113 (base x) so that it becomes 1 * x^2 + 1 * x + 3:
113 (base x) = 805 (base 9)
x^2 + x + 3 = 805 (base 9)
This is just a quadratic equation in x, and so it can be solved like any other quadratic. We could do all the arithmetic in base 9 since the right side is already nonary. But to us native decimalists, it's easier just to convert the right side to decimal:
x^2 + x + 3 = 653
x^2 + x - 650 = 0
(x + 26)(x - 25) = 0
x = -26 or x = 25
There actually do exist such a thing as negative bases, but these are a bit awkward -- it's generally assumed in a number base problem that bases are natural numbers unless directly stated. Therefore the desired base is x = 25 -- and of course, today's date is the twenty-fifth.
The two bases in this problem are base 9 and base 25. Both of these are square bases. It turns out that square bases have a special property that non-square bases lack. Recall that a full repetend prime is a prime whose decimal expansion is of maximum length -- one less than the prime itself:
1/7 = 0.142857... (6 digits repeat)
1/17 = 0.0588235294117647... (16 digits)
1/19 = 0.052631578947368421... (18 digits)
Let's try to find the full repetend primes in bases 9 and 25. The calculations have already been completed on the Dozens Online website:
Base 9:
https://www.tapatalk.com/groups/dozensonline/everything-done-up-to-the-nines-t723.html
Base 25:
https://www.tapatalk.com/groups/dozensonline/25-bingo-pentavigesimal-t787.html
And we see that unless we count 2, none of the primes are full repetend primes. It turns out that in odd square bases, 2 is the only full repetend prime -- and in even square bases (such as hexadecimal) there are no full repetend primes.
There's a simple reason for this -- each digit in a perfect square base corresponds to two digits in another base. For example, each digit of nonary (base 9) is two digits of ternary (base 3), and likewise with each digit of base 25 being two digits of base 5. Thus the maximum length of a repetend in a square base must be half of that same repetend in the square root base:
1/7 = 0.125... (base 9) = 0.010212... (base 3)
1/7 = 0.3e7... (base 25) = 0.032412... (base 5)
and so on. Thus 7 is a full repetend prime in bases 3 and 5, but not bases 9 and 25.
Before we leave number bases, here's a link to the Dozens Online for base 11:
https://www.tapatalk.com/groups/dozensonline/undecimal-the-unbelievable-t514.html
According to this link, undecimal has many full repetend primes: 2, 3, 12 (thirteen), 16 (seventeen), 21 (twenty-three), and 27 (twenty-nine).
It's also mentioned that one French Revolution mathematician, Joseph-Louis Lagrange, had (perhaps jokingly) considered an undecimal metric system. I'm not sure whether time (or angle measure) was included in Lagrange's proposal -- otherwise he might have stumbled upon the Eleven Calendar (or something similar).
Feynman, pages 161-250
It's time for us to continue our summer side-along graphic novel, Jim Ottaviani's Feynman. We left off with our hero offering $1,000 prizes for nanotechnology -- a miniature motor and a tiny book.
Unfortunately, our summary of this section of the book must be relatively brief. The book is due back at the library soon, plus there's a second book for us to read.
"I didn't expect it to take many years for someone to claim the money," Feynman narrates. "I also didn't expect all the kooks, and people who didn't get it."
Feynman (to various people holding large objects): No. Um, no. No, I said small.
"This went on for a few months, and I saw a lot of big boxes containing, well, big motors."
Feynman: Look, thanks for coming, Mr. McClellan. But anything that needs a crate of that size...
McClellan: No no, the motor's in here.
Feynman (seeing the small envelope): Uh-oh.
As it turns out, McClellan's envelope contained the proper motor, and so Feynman had to pay up -- if only he had set aside the prize money.
Unfortunately, in 1978, Feynman catches cancer. He loses his appetite, and so his doctor suggests that someone feed him his favorite foods so that he can eat again:
"So she [my friend Alix] came running from Brentwood to Pasadena with her vichyssoise, chocolate mousse, cakes, etc.," he narrates. "It worked very well. I loved to talk to Alix too. She was so interested in art and archaeology. She'd go to sites, and could stretch her imagination so fully so visualize how things had been."
Shortly thereafter, Feynman lectures on QED (quantum electrodynamics) in New Zealand. His goal is to explain QED so that an amateur like his friend Alix can understand it. And here is a small portion of this lecture:
Student: And naturally, the length of the arrow -- the amplitude -- is bigger for this one than it is for that one.
Feynman: No! They're almost exactly the same. It's the time that's different. Suppose you had to start at the light source and get to the eye, fast. You...running. And you had to touch the mirror along the way...
Picking A to touch isn't a good idea (too close to the light). Neither is touching point M (too close to the eye). The time it takes, if we plot it right underneath, looks like this (a parabola). The time for point A is much longer than for point B (between A and M near A). The stopwatch arrows turn with time, remember? So toward the center the different in direction of the arrows gets small, and then increases again as you go farther away...
But when we add them together, instead of two like I did before, it's millions of 'em, and we get a net result -- a total amplitude -- for the photon to arrive. It's this tremendous line. And it's length is mostly made up of arrows D through J.
Student: So, QED predicts that light reflects off the mirror.
Feynman: Very good! The contribution of these crazy bits near the edges is almost nothing -- all that staggering around cancels out.
Of course, this lecture would be easier to understand if you could see the pictures. The important thing here is that reflections here are much more complex than reflections in Geometry. This is because a mirror isn't perfectly flat -- Feynman draws various "arrows" (or vectors) to show which direction the photons are going. Only on average, he explains, does the angle of incidence actually equal the angle of reflection.
In fact, some Geometry texts provide the solution for a perfectly flat mirror. We consider the image of the eye in the mirror, and then draw a straight line from the light to the image of the eye. Then the light will bounce off the mirror in just the right place to reach the eye.
There are many other events in this chapter. Feynman ultimately receives his own prize -- the Nobel Prize -- for his work on QED. And the book ends shortly after the Challenger disaster, when the physicist -- after speaking to Houston NASA officials -- is invited to the Soviet Union to lecture on the causes of the disaster. He muses about the high school history teacher who died that day, namely Christa McAuliffe:
Feynman: She trained with astronauts. She had to know the true situation better than NASA officials.
"NASA has to deal in a world of reality to understand technological weaknesses well enough to actively try to eliminate them," he narrates. "They should propose only realistic flight schedules they have a reasonable chance of meeting. If they means we wouldn't support them, well...so be it. For a successful technology, reality must take precedence over public relations, for nature cannot be fooled. That was it. I was done. I only gave two interviews -- one to a local paper, one to the National Enquirer."
Feynman: ...Hey, if I said something dopey and they quote it, nobody will blame me! It's good to be home, but I'm going to leave this bag packed...we're going to Kyzy1!
Richard Feynman didn't make it to Tuva. The official, formal invitation from Moscow permitting the visit was dated February 19th, 1988 -- four days after he died. His last words, upon forcing himself awake from a coma: "I'd hate to die twice. It's so boring."
(I'm sorry, but I can't help but think about Niels Henrik Abel, the 19th century Norwegian who was invited to teach math in Berlin -- just days after he died. But at least Feynman lived a full life -- Abel lived only to 26.)
Thus concludes our reading of Ottaviani's Feynman. We'll move on to his latest book, Hawking -- as in Stephen Hawking -- in my next post.
Conclusion
At this point, this is what our natural geometry course looks like:
Unit 1: Point-Line-Plane
covers at least Lessons 1-4 through 1-9, 3-1, 3-2 of the U of Chicago text
Unit 2: Reflections
covers Chapters 2 and 4 of the U of Chicago text
Unit 3: Rotations
covers Lesson 6-3, 7-1 though 7-4, coordinate reflections/rotations, Playfair, concurrency proofs
Unit 4: Translations
covers Lesson 6-2, Chapter 5, Lessons 7-5 through 7-8
Unit 5: Glide Reflections
covers Lesson 6-6, review for first semester final
There are still several gaps to be filled in. These missing proofs won't be easy to find, but hopefully they'll be worth it. They'll shed some insight on parallel lines and what exactly distinguishes Euclidean from spherical geometry.
Three Perpendiculars Revisited
In my Pi Approximation Day post, I alluded to a natural "Three Perpendiculars Theorem":
Three Perpendiculars Theorem:
If three lines are perpendicular to the same line, then the three lines are either parallel or concurrent.
When I first posted this theorem last year, I wrote that, while it's valid in both Euclidean and spherical geometry, it's more of a curiosity than something we might want to teach students. Unless there's a compelling reason why we'd want to teach Three Perpendiculars, we probably shouldn't.
But let's think about it -- there's another theorem that students need to know, and it definitely has something to do with three perpendiculars and concurrency. You guessed it -- it's the concurrency of the three perpendicular bisectors of a triangle.
Let's think again about that incomplete proof of the concurrency of perpendicular bisectors as given in Lesson 4-5 of the U of Chicago text:
Construct the circle through the three noncollinear points A, B, and C.
Solution:
Step 1. Subroutine: m, the perpendicular bisector of
Step 2. Subroutine: n, the perpendicular bisector of
Step 3. m and n intersect at O. (Point Rule)
Step 4. Circle O containing A (Compass Rule)
If m and n intersect, it can be proven that this construction works. Because of the Perpendicular Bisector Theorem with line m, OA = OB. With line n, this theorem also justifies the conclusion OB = OC. By the Transitive Property of Equality, the three distances OA, OB, and OC are all equal. Thus Circle O with radius OA contains points B and C also. QEF
Earlier, we discovered that this proof is incomplete, because it takes it for granted that the lines m and n actually intersect. Also, we notice that the noncollinearity of A, B, and C isn't used in the proof. We found out that these are related depending on which geometry we're in:
Euclidean: If A, B, C are collinear then m | | n. If A, B, C are noncollinear then m, n intersect.
Spherical: All lines intersect, therefore m, n intersect.
Hyperbolic: If A, B, C are collinear than m | | n. If A, B, C are noncollinear then there's no conclusion.
Because of this, we can write a single theorem that incorporates both cases:
Theorem:
Let A, B, C be any three distinct points. Then the three perpendicular bisectors of
Proof:
It's essentially the same as the U of Chicago proof. All we really need to change is drop Circle O and instead, use Converse Perpendicular Bisector Theorem to show that since O is equidistant from A, C, we conclude that O lies on the perpendicular bisector of
Notice that without loss of generality, we can start by knowing that any two of the three perpendicular bisectors must intersect (
This proof is valid in all three geometries -- Euclidean, spherical, and hyperbolic.
Actual High School Proofs
But even with this version of the Three Perpendiculars Theorem, we still haven't actually reached the proofs that we really want to see. Once again, Three Perpendiculars should ultimately lead to "a line and its translation image are parallel."
Indeed, as written, the theorem doesn't yet distinguish between the parallel and concurrent cases. We know that the lines are always concurrent in spherical geometry, and if A, B, C are noncollinear (as in the vertices of a triangle), then they are concurrent in Euclid as well. Thus, we should somehow be able to prove in neutral geometry that if A, B, C are noncollinear then the lines are concurrent. Such a proof can be placed near the end of Unit 3 -- but I can't figure out yet what the proof would look like.
There might also be a path to some interesting proofs if we were to begin with the concurrence of angle bisectors, rather than perpendicular bisectors. In this case, instead of points A, B, C, we begin with three lines a, b, c, and show that if these lines intersect then so do the bisectors of the angles that they form. Even though this proof would be mainly about intersecting lines, an indirect proof could be set up so that it refers to the properties of parallel lines instead.
Don't forget that we do have Playfair's Parallel Postulate available. Officially, Playfair is neutral (due to the "at most" one line clause). We know that Playfair must somehow be used in the concurrence of perpendicular bisectors proof, since the proof fails in hyperbolic geometry. This means that Playfair would be introduced before Unit 4 as well, even as other Euclidean parallel line properties aren't taught until Unit 4.
So there are still several proofs that I still need to figure out here. Still, it would be nice if we could could adapt the concurrency proofs so that they could help us out with the parallel line proofs.
Once we reach Unit 4, the geometry becomes fully Euclidean. So the original Euclidean version of our Point-Line-Plane postulate is used in proofs. Once again, we don't tell the students this -- we just quietly start using the "through any two points, there is exactly one line" clause.
Number Bases
In my Pi Approximation Day post, I briefly mentioned number bases -- mainly because I decided to include the Eleven Calendar (and hence its implied undecimal base) in my annual FAQ post.
Number bases is still one of my favorite topics, though I rarely have an opportunity to write about this topic on the blog, since it has nothing to do with Geometry. But since I'm thinking about it right now, let me squeeze some number bases into today's post.
And indeed, it helps that Pappas has a number base problem on her calendar today:
113 (base x) = 805 (base 9)
Determine x.
We might notice that just as 805 (base 9) means 8 + 9^2 + 0 * 9 + 5, we can expand 113 (base x) so that it becomes 1 * x^2 + 1 * x + 3:
113 (base x) = 805 (base 9)
x^2 + x + 3 = 805 (base 9)
This is just a quadratic equation in x, and so it can be solved like any other quadratic. We could do all the arithmetic in base 9 since the right side is already nonary. But to us native decimalists, it's easier just to convert the right side to decimal:
x^2 + x + 3 = 653
x^2 + x - 650 = 0
(x + 26)(x - 25) = 0
x = -26 or x = 25
There actually do exist such a thing as negative bases, but these are a bit awkward -- it's generally assumed in a number base problem that bases are natural numbers unless directly stated. Therefore the desired base is x = 25 -- and of course, today's date is the twenty-fifth.
The two bases in this problem are base 9 and base 25. Both of these are square bases. It turns out that square bases have a special property that non-square bases lack. Recall that a full repetend prime is a prime whose decimal expansion is of maximum length -- one less than the prime itself:
1/7 = 0.142857... (6 digits repeat)
1/17 = 0.0588235294117647... (16 digits)
1/19 = 0.052631578947368421... (18 digits)
Let's try to find the full repetend primes in bases 9 and 25. The calculations have already been completed on the Dozens Online website:
Base 9:
https://www.tapatalk.com/groups/dozensonline/everything-done-up-to-the-nines-t723.html
p
|
Len(M)/Maximum
|
M
|
2
| 1/1 | 4 |
3
| –REGULAR– | |
5
| 2/4 | 17 |
7
| 3/6 | 125 |
12
| 5/11 | 07324 |
14
| 3/13 | 062 |
18
| 8/17 | 04678421 |
21
| 10/20 | 042327518 |
25
| 12/24 | 03462311507 |
32
| 15/31 | 02712148617674 |
Base 25:
https://www.tapatalk.com/groups/dozensonline/25-bingo-pentavigesimal-t787.html
| n | dec. | 1/n | |
| 2 |
0.ccc…
| c | |
| 3 |
0.888…
| 8 | |
| 5 |
0.5
| 5 | |
| 7 |
0.3e73e7…
| 3e7 | |
| b | 11 |
0.26kb926kb9…
| 26kb9 |
| d | 13 |
0.1n1n1n…
| 1n |
| h | 17 |
0.1bj2nd5m1bj2nd5m…
| 1bj2nd5m |
| j | 19 |
0.17m956ebl17m956ebl…
| 17m956ebl |
| n | 23 |
0.1248h9je36d1248h9je36d…
| 1248h9je36d |
And we see that unless we count 2, none of the primes are full repetend primes. It turns out that in odd square bases, 2 is the only full repetend prime -- and in even square bases (such as hexadecimal) there are no full repetend primes.
There's a simple reason for this -- each digit in a perfect square base corresponds to two digits in another base. For example, each digit of nonary (base 9) is two digits of ternary (base 3), and likewise with each digit of base 25 being two digits of base 5. Thus the maximum length of a repetend in a square base must be half of that same repetend in the square root base:
1/7 = 0.125... (base 9) = 0.010212... (base 3)
1/7 = 0.3e7... (base 25) = 0.032412... (base 5)
and so on. Thus 7 is a full repetend prime in bases 3 and 5, but not bases 9 and 25.
Before we leave number bases, here's a link to the Dozens Online for base 11:
https://www.tapatalk.com/groups/dozensonline/undecimal-the-unbelievable-t514.html
According to this link, undecimal has many full repetend primes: 2, 3, 12 (thirteen), 16 (seventeen), 21 (twenty-three), and 27 (twenty-nine).
It's also mentioned that one French Revolution mathematician, Joseph-Louis Lagrange, had (perhaps jokingly) considered an undecimal metric system. I'm not sure whether time (or angle measure) was included in Lagrange's proposal -- otherwise he might have stumbled upon the Eleven Calendar (or something similar).
Feynman, pages 161-250
It's time for us to continue our summer side-along graphic novel, Jim Ottaviani's Feynman. We left off with our hero offering $1,000 prizes for nanotechnology -- a miniature motor and a tiny book.
Unfortunately, our summary of this section of the book must be relatively brief. The book is due back at the library soon, plus there's a second book for us to read.
"I didn't expect it to take many years for someone to claim the money," Feynman narrates. "I also didn't expect all the kooks, and people who didn't get it."
Feynman (to various people holding large objects): No. Um, no. No, I said small.
"This went on for a few months, and I saw a lot of big boxes containing, well, big motors."
Feynman: Look, thanks for coming, Mr. McClellan. But anything that needs a crate of that size...
McClellan: No no, the motor's in here.
Feynman (seeing the small envelope): Uh-oh.
As it turns out, McClellan's envelope contained the proper motor, and so Feynman had to pay up -- if only he had set aside the prize money.
Unfortunately, in 1978, Feynman catches cancer. He loses his appetite, and so his doctor suggests that someone feed him his favorite foods so that he can eat again:
"So she [my friend Alix] came running from Brentwood to Pasadena with her vichyssoise, chocolate mousse, cakes, etc.," he narrates. "It worked very well. I loved to talk to Alix too. She was so interested in art and archaeology. She'd go to sites, and could stretch her imagination so fully so visualize how things had been."
Shortly thereafter, Feynman lectures on QED (quantum electrodynamics) in New Zealand. His goal is to explain QED so that an amateur like his friend Alix can understand it. And here is a small portion of this lecture:
Student: And naturally, the length of the arrow -- the amplitude -- is bigger for this one than it is for that one.
Feynman: No! They're almost exactly the same. It's the time that's different. Suppose you had to start at the light source and get to the eye, fast. You...running. And you had to touch the mirror along the way...
Picking A to touch isn't a good idea (too close to the light). Neither is touching point M (too close to the eye). The time it takes, if we plot it right underneath, looks like this (a parabola). The time for point A is much longer than for point B (between A and M near A). The stopwatch arrows turn with time, remember? So toward the center the different in direction of the arrows gets small, and then increases again as you go farther away...
But when we add them together, instead of two like I did before, it's millions of 'em, and we get a net result -- a total amplitude -- for the photon to arrive. It's this tremendous line. And it's length is mostly made up of arrows D through J.
Student: So, QED predicts that light reflects off the mirror.
Feynman: Very good! The contribution of these crazy bits near the edges is almost nothing -- all that staggering around cancels out.
Of course, this lecture would be easier to understand if you could see the pictures. The important thing here is that reflections here are much more complex than reflections in Geometry. This is because a mirror isn't perfectly flat -- Feynman draws various "arrows" (or vectors) to show which direction the photons are going. Only on average, he explains, does the angle of incidence actually equal the angle of reflection.
In fact, some Geometry texts provide the solution for a perfectly flat mirror. We consider the image of the eye in the mirror, and then draw a straight line from the light to the image of the eye. Then the light will bounce off the mirror in just the right place to reach the eye.
There are many other events in this chapter. Feynman ultimately receives his own prize -- the Nobel Prize -- for his work on QED. And the book ends shortly after the Challenger disaster, when the physicist -- after speaking to Houston NASA officials -- is invited to the Soviet Union to lecture on the causes of the disaster. He muses about the high school history teacher who died that day, namely Christa McAuliffe:
Feynman: She trained with astronauts. She had to know the true situation better than NASA officials.
"NASA has to deal in a world of reality to understand technological weaknesses well enough to actively try to eliminate them," he narrates. "They should propose only realistic flight schedules they have a reasonable chance of meeting. If they means we wouldn't support them, well...so be it. For a successful technology, reality must take precedence over public relations, for nature cannot be fooled. That was it. I was done. I only gave two interviews -- one to a local paper, one to the National Enquirer."
Feynman: ...Hey, if I said something dopey and they quote it, nobody will blame me! It's good to be home, but I'm going to leave this bag packed...we're going to Kyzy1!
Richard Feynman didn't make it to Tuva. The official, formal invitation from Moscow permitting the visit was dated February 19th, 1988 -- four days after he died. His last words, upon forcing himself awake from a coma: "I'd hate to die twice. It's so boring."
(I'm sorry, but I can't help but think about Niels Henrik Abel, the 19th century Norwegian who was invited to teach math in Berlin -- just days after he died. But at least Feynman lived a full life -- Abel lived only to 26.)
Thus concludes our reading of Ottaviani's Feynman. We'll move on to his latest book, Hawking -- as in Stephen Hawking -- in my next post.
Conclusion
At this point, this is what our natural geometry course looks like:
Unit 1: Point-Line-Plane
covers at least Lessons 1-4 through 1-9, 3-1, 3-2 of the U of Chicago text
Unit 2: Reflections
covers Chapters 2 and 4 of the U of Chicago text
Unit 3: Rotations
covers Lesson 6-3, 7-1 though 7-4, coordinate reflections/rotations, Playfair, concurrency proofs
Unit 4: Translations
covers Lesson 6-2, Chapter 5, Lessons 7-5 through 7-8
Unit 5: Glide Reflections
covers Lesson 6-6, review for first semester final
There are still several gaps to be filled in. These missing proofs won't be easy to find, but hopefully they'll be worth it. They'll shed some insight on parallel lines and what exactly distinguishes Euclidean from spherical geometry.
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