But first, I do want to say something about female mathematicians and scientists from my last post. In particular, yesterday the Google Doodle featured the American scientist Nettie Stevens. At the turn of the century, Stevens helped to discover that she, as a woman, had XX chromosomes, while males have XY chromosomes, but unfortunately, sexism stripped her of recognition. She performed her research right here in California (at Stanford). As a math teacher, I want to emphasize female mathematicians and physical scientists (as physics is more closely allied with math) rather than biologists like Stevens. But still, I wrote in my last post that I want to encourage girls to pursue STEM careers -- to create a world in which females are recognized for their contributions to STEM as much as males -- and biology definitely falls under the STEM umbrella. And so I celebrate the scientist Nettie Stevens here on the blog.
Now here's a female mathematician -- Theoni Pappas, author of the Mathematical Calendar 2016. As it turns out, her feature for July is "Geometric Worlds: From Euclidean to digital geometry." She writes about several different types of geometry -- including spherical geometry. And so let me post what Pappas has to say about this particular brand of non-Euclidean geometry:
Now here's a female mathematician -- Theoni Pappas, author of the Mathematical Calendar 2016. As it turns out, her feature for July is "Geometric Worlds: From Euclidean to digital geometry." She writes about several different types of geometry -- including spherical geometry. And so let me post what Pappas has to say about this particular brand of non-Euclidean geometry:
"In elliptic geometry, such as on the surface of a sphere, its lines are also not straight, but are defined as great circles. In this world two lines (i.e. two great circles) always intersect in two points -- no parallel lines exist here. Unlike Euclidean geometry, the sides of a triangle are curved and its three angles always total more than 180 degrees."
The question for May 2nd was, "In spherical geometry, all pairs of lines intersect in ____ points." As Pappas writes above, the correct answer is two -- and of course, this question appeared on the second day of a month.
Pappas tells us that another way to distinguish among Euclidean, hyperbolic, and elliptic (spherical) geometry is curvature:
"The world of a plane in Euclidean geometry is flat, and therefore is said to have zero curvature. A hyperbolic plane, on the other hand, curves inward and therefore has negative curvature through it. In an elliptic geometric world such as a sphere, it continually curves outward which means it has positive curvature."
There are a few more interesting geometries mentioned in the Pappas article. She continues:
"In 1736, Leonhard Euler solved the famous Konigsberg Bridge problem in an innovative way using networks, and his work launched the geometry of topology."
This was, of course, our first day of school activity posted here on the blog! Indeed, I'm considering using the Konigsberg bridge problem as a first day of school activity in my actual classroom this upcoming school year!
Notice that the title of this article mentions "digital geometry." Pappas explains:
"Today a new geometry -- digital geometry -- is evolving in the world of the computer monitor."
I've made a big deal about this in the past as well. In Lesson 1-1 of the U of Chicago text, we see the following profound statement:
A point is a dot.
Whenever students wonder why they have to learn Geometry, we can remind them that they most likely enjoy the products of digital geometry everyday! (But let's not get carried away here -- recall that I'll be teaching middle school math in the fall, not Geometry.)
With that, let's get back to Legendre. But as Pappas points out, one of the most important theorems of spherical geometry is Triangle-Sum -- whereas the sum of the angle measures of a triangle is always exactly 180 degrees in Euclidean geometry, it's always more than 180 in spherical geometry. Using radian measure, this angle sum always exceeds pi -- or should I say, tau/2. (Just like last summer, I plan on using the new circle constant tau as much as possible from Tau Day until Pi Approximation Day, when I'll return to using pi.)
Now Triangle-Sum was Legendre's Proposition 489. As it turns out, most of today's theorems are basically corollaries of Triangle-Sum. Here's Legendre's Proposition 490:
490. Corollary I. The sum of the angles of a spherical triangle is not constant like that of a plane triangle; it varies from two right angles to six, without the possibility of being equal to either limit. Thus, two angles being given, we cannot thence determine the third.
Proposition 490 follows trivially from 489. Legendre -- like Euclid two millennia earlier -- uses right angles, not degrees, as units. In my last post, I mentioned that we can use lambda as a constant to denote the right angle, so we conclude that the sum can vary from 2lambda to 6lambda. In terms of tau, the sum varies from tau/2 to 3tau/2.
Legendre tells us that we can't determine the third angle of a triangle in spherical geometry given two angles as we can in Euclidean geometry. Thus there is no equivalent of what some Geometry texts call the "Third Angle Theorem."
Let's move on to Legendre's Proposition 491:
491. Corollary II. A spherical triangle may have two or three right angles, also two or three obtuse angles.
This is in stark contrast to a Euclidean triangle in which at least two angles must be acute. It's easy to see why this is the case -- if a Euclidean triangle had two right angles, then the sum of these angles would be 2lambda, which is the triangle sum for all three angles. This would leave us a measure of zero for the third angle. A triangle with two obtuse angles would be even worse -- the third angle would have to have a negative measure! On the other hand, in spherical geometry the sum of the angles is always more than 2lambda, so we can comfortably fit two right, or even two obtuse, angles and still have some measure left for the third angle.
When thinking about a triangle with three right angles, I can't help but think about the following brainteaser, which I've posted several times here on the blog:
- A bear hunter sets out from camp and walks one mile south.
- He sees a bear and is about to shoot it.
- The bear grabs his gun and eats it.
- The hunter runs away one mile east.
- He then walks one mile north and gets back to his camp and changes his underwear.
- What colour was the bear?
The answer is that the "colour" (sorry -- this is obviously from a British website) of the bear is white, since the puzzle describes a polar bear at the North Pole. Technically, this is not a spherical triangle, since the "one mile east" is along a parallel of latitude, not a great circle. It's not even close to being a great circle -- if the hunter ran approximately six miles east he would have walked in a complete circle around the pole.
Indeed, Legendre tells us what a true spherical triangle with two or three right angles would look like:
"If the triangle ABC has two right angles B and C, the vertex A will be the pole of the base BC (467); and the sides AB, AC will be quadrants.
"If the angle A also is a right angle, the triangle ABC will have all of its angles right angles, and all of its sides quadrants. The triangle having three right angles is contained eight times in the surface of the sphere."
Here Legendre tells us that if ABC has right angles B and C, we can still let A be at the North Pole just as in the original "What color was the bear?" puzzle. But then points B and C would end up lying on the Equator! Proposition 467 cited by Legendre here is one that we discussed last year -- it refers to the relationship between a line (great circle) and its poles. Every line has two poles -- the name "pole" is justified by the fact that the poles of the Equator are the actual North and South Poles. (Last year I came up the habit of using lowercase "pole" to refer to the poles of any line, but the capitalized "Poles" are the North and South Poles only.) The two sides opposite the right angles (the two "hypotenuses"?) are each a quadrant in length -- a quadrant being the distance from the North Pole to the Equator.
Then Legendre's triangle with three right angles has all three sides equal to a quadrant (equiangular triangles are always equilateral in both Euclidean and spherical geometry). He writes that the entire globe can be divided into eight such triangles. It's easy to see how to divide the Northern Hemisphere into four such triangles -- the sides of these triangles can be the Prime Meridian, the 90th Meridians (both east and west), and the 180th Meridian (the International Date Line). We can divide the Southern Hemisphere into four triangles in exactly the same manner, so that there are eight total triangles with three right angles each. (This also tells us how to find the poles of any meridian -- the poles of a meridian lie on the Equator, 90 degrees east and west of the original meridian.)
It's trickier to find a triangle with two obtuse angles. If we again take the North Pole to be the vertex angle A of an isosceles triangle, where must B and C be placed so that both B and C are obtuse? The answer is that they must lie in the Southern Hemisphere. This is difficult to visualize -- B and C might be on the same parallel of latitude and any parallel forms a right angle with any meridian, but the problem is that parallels of latitude (other than the Equator) are not great circles. The symbol BC can refer to the actual great circle joining B and C, but then it's not obvious why BC must form obtuse angles with both AB and AC. Indeed, the only great circles that are easy to visualize are the Equator and all meridians, and BC is neither. A full proof would be quite long, and would only distract us from the theorems that actually appear in Legendre. This is probably why Legendre doesn't give us a triangle with two obtuse angles in the first place.
Here is Legendre's Proposition 492:
492. Scholium. We have supposed in all that precedes, conformably to the definition, art. 442, that spherical triangles always have their sides less each than a semicircumference; then it follows that the angles are always less than two right angles. For the side AB is less than a semicircumference, as also AC; these arcs must both be produced in order to meet in D. Now the two angles ABC, CBD taken together, are equal to two right angles; therefore angle ABC is by itself less than two right angles.
Here Legendre hints at the idea that a triangle may have an angle greater than 180 degrees. (Some texts give the name "reflex angle" to an angle that is larger than a straight angle.) In Euclidean geometry, the angles of a triangle add up to 180, so no angle can be greater than 180 without one of the other angles being negative. But the sum of the angles of a Euclidean quadrilateral is 360, and so there can -- and do -- exist quadrilaterals with an angle greater than 180. Polygons containing reflex angles are called "concave" (or "nonconvex").
So Legendre is telling us that in spherical geometry, there might exist concave triangles. But, as he writes, all triangles with sides less than a "semicircumference" are convex. Most of the time, we expect sides of a triangle to be less than a semicircumference indeed -- after all, to get from A to B, we usually want to go the short way (which is a minor arc less than a semicircumference) rather than the long way (a major arc greater than a circumference). Legendre proves that in this case, the angles must all be less than 180. This is because Legendre extends AB and AC to meet at D -- recall that all great circles passing through A meet at the antipodal point of A, which is D. Now D is exactly a semicircumference away, so we truly are extending AB and AC to meet at D. Then the two angles ABC and CBD form a linear pair which adds up to 180. So each angle must be less than 180.
Legendre continues:
"We will remark, however, that there are spherical triangles of which certain sides are greater than a semicircumference, and certain angles greater than two right angles. For if we produce the side AC till it becomes an entire circumference ACE, what remains, after taking from the surface of the hemisphere the triangle ABC, is a new triangle, which may also be designated by ABC, and the sides of which are AB, BC, AEDC. We see, then, that the side AEDC is greater than the semicircumference AED, but, at the same time, the opposite angle B exceeds two right angles by the quantity CBD."
The easiest way to visualize what Legendre is writing here is to place B at the North Pole (rather than A) and all the other named points on the Equator. Since he calls AED a "semicircumference," his intent is for D to be the antipodal point of A. He doesn't state where E is, but we might as well let E be the antipodal point of C. Then there are two triangles with vertices ABC -- one where we go the short way AC, and the other where we go the long way AEDC. Now it becomes obvious why the angle B opposite AEDC must be more than 180 -- the large angle ABC must equal ABD plus CBD, but ABD is exactly 180 as A and D are antipodal. That is, the great circle through AB (a meridian, as B is the North Pole) must pass through D since all great circles through A pass through D. So ABD is a straight line (great circle), and all straight angles measure 180. Then the excess CBD makes the full angle ABC a reflex angle.(Adding the smaller angle ABC to this gives 360 degrees or tau.)
Let's proceed with Legendre's Proposition 493, our only major theorem for today:
493. The lunary surface AMBNA is to the surface of the sphere as the angle MAN of this surface is to four right angles, or as the arc MN, which measures this angle, is to the circumference.
What is this theorem stating? First of all, let's recall what a "lunary surface" is -- often abbreviated to lune, a lunary surface is the region bounded by two intersecting line segments (great circle arcs). As all lines intersect at antipodal points, the two arcs must be semicircles. A lune is essentially a polygon with two sides -- which doesn't exist in Euclidean geometry.
The name lune or lunary surface reminds us of the word lunar, which means "of the moon." And indeed, every time we look up at the moon, we see a lune -- the portion of the moon illuminated at any time (regardless of the phase of the moon) is a lune. Another way to imagine a lune is to let points A and B be two antipodal points -- say the North and South Poles. The points M and N can be placed anywhere on the Equator -- this allows us to measure either the angle MAN or the arc MN, just as Legendre does in the statement of the theorem.
This is the first theorem in Legendre that hints at the concept of area, but he doesn't use "square units" when considering area. Instead, he uses ratios -- just as the use of a right angle as a unit of angle measure, the use of ratios to determine area is a tradition going back to Euclid. But notice that there's an obvious natural unit of area to consider -- the surface area of the sphere itself. Let's assume that our sphere is the unit sphere -- that is, it has radius 1 -- and we will use radians throughout.
Legendre mentions the ratio of the angle MAN to four right angles -- that is, the ratio of the angle MAN to tau. Notice that the ratio of the arc MN to the circumference is this exact same ratio, because we chose radians and the unit sphere.
Now the theorem tells us that this equals the ratio of the area of the lune AMBNA to the surface area of the entire sphere. The surface area of a sphere is 4pi r^2 and this is a unit sphere, so the area of the sphere is 4pi -- oops, make that 2tau. So the theorem is essentially telling us:
Area(AMBNA)/(2tau) = Angle MAN/tau
Let's see how Legendre proves this theorem:
Demonstration. Let us suppose, in the first place, that the arc MN is to the circumference MNPQ in the ratio of two entire numbers, as 5 to 48, for example. [That is, suppose that the measure of the arc MN is 5tau/48 -- dw] The circumference MNPQ can be divided into 48 equal parts, of which MN will contain 5; then joining the pole A and the points of division by as many quadrants, we shall have 48 triangles in the surface of the hemisphere AMNPQ, which will be equal among themselves, since they have all of their parts equal. The entire sphere will therefore contain 96 of these partial triangles, and the lunary surface AMBNA will contain 10 of them; therefore the lunary surface is to the sphere as 10 is to 96, or as 5 is to 48, that is, as the arc is to the circumference.
So we can easily see what Legendre is doing here -- he is essentially dividing both the lune and the entire sphere into many small triangles. In his example, the lunes has angle measure 5tau/48, so he divides each half of the sphere into 48 triangles, with the lune taking up 5 of these triangles. We must consider both halves of the lune or sphere, which is why the entire lune takes up 10 out of the 96 triangles into which the entire sphere is divided. Using algebra, we can generalize this by saying that if the measure of the angle is (p/q)tau, we can divide each hemisphere into q triangles of which the lune takes up p. So the entire lune contains 2p out of 2q triangles into which the entire sphere is divided, and 2p/2q reduces to p/q, so the lune takes up p/q of the surface area of the sphere -- which, if you remember, is 2tau. So the area of the lune is (p/q)2tau.
The proof seems to be complete, yet Legendre adds the following cryptic lines:
"If the arc MN is not commensurable with the circumference, it may be shown by a course of reasoning, of which we have already had many examples, that the lunary surface is always to that of the sphere as the arc MN is to the circumference. QED"
What is Legendre saying here? Here's the problem -- what if the angle measure equals k tau for some irrational value k? ("Commensurable" essentially means "rational" -- a tradition going all the way back to Pythagoras and his sqrt(2) irrationality proof.) Then we can't just blindly write k = p/q and divide the hemisphere into q triangles. Instead, Legendre resorts to a "course of reasoning" that allows us to conclude that the theorem holds for both rational and irrational k. Dr. Hung-Hsi Wu gives this "course of reasoning" a name -- the Fundamental Assumption of School Mathematics.
By the way, notice that the equation:
Area(AMBNA)/(2tau) = Angle MAN/tau
can be rewritten as:
Area(AMBNA) = 2Angle MAN
And notice that the lune contains another angle at the South Pole, MBN, that is congruent to MAN (as both angles have the same measure as arc MN). So we can also write this as:
Area(AMBNA) = Angle MAN + Angle MBN
This begins to hint at a relationship between area and angle measure that exists in spherical, but not Euclidean, geometry.
The last two propositions are corollaries of this theorem:
494. Corollary I. Two lunary surfaces are to each other as their respective angles.
This is too trivial to prove as it follows directly from that equation we just wrote:
Area(AMBNA) = 2Angle MAN
Let's look at the final proposition for today:
495. Corollary II. We have already seen the entire surface of the sphere is equal to eight triangles having each three right angles (491); consequently, if the area of one of these triangles be taken for unity, the surface of the sphere will be represented by eight. This being supposed, the lunary surface, of which the angle is A, will be expressed by 2A [Yes, this is the formula we just wrote again -- dw], the angle A being estimated by taking the right angle for unity; for we have 2A : 8 :: A : 4. Here there are two kinds of units; one for angles, this is the right angle; the other for surfaces, this is the spherical triangle, of which all the angles are right angles, and the sides quadrants.
We see that Legendre's equilateral right triangle has area one-eighth that of the entire sphere -- since the sphere has area 2tau, the triangle has area 2tau/8 = tau/4 = lambda. So his two units -- the right angle for angle measure and the equilateral right triangle for area measure -- are both equal to lambda (again, assuming the unit sphere and radian measure).
In fact, we see that this is true for any triangle with two right angles -- if the third angle has measure A, then the triangle is half of a lune with angle A and measure 2A, so its measure is also A:
Area(AMN) = Angle A
And notice that the sum of the angles of Triangle AMN is exactly two right angles (tau/2) plus the measure of this Angle A. Legendre mentions the special case where Angle A = lambda. Again we hint at a connection between angle measure and area.
This is a good place for us to end this post. My next post will be tomorrow, when I'll continue with the rules of my new classroom.
Let's proceed with Legendre's Proposition 493, our only major theorem for today:
493. The lunary surface AMBNA is to the surface of the sphere as the angle MAN of this surface is to four right angles, or as the arc MN, which measures this angle, is to the circumference.
What is this theorem stating? First of all, let's recall what a "lunary surface" is -- often abbreviated to lune, a lunary surface is the region bounded by two intersecting line segments (great circle arcs). As all lines intersect at antipodal points, the two arcs must be semicircles. A lune is essentially a polygon with two sides -- which doesn't exist in Euclidean geometry.
The name lune or lunary surface reminds us of the word lunar, which means "of the moon." And indeed, every time we look up at the moon, we see a lune -- the portion of the moon illuminated at any time (regardless of the phase of the moon) is a lune. Another way to imagine a lune is to let points A and B be two antipodal points -- say the North and South Poles. The points M and N can be placed anywhere on the Equator -- this allows us to measure either the angle MAN or the arc MN, just as Legendre does in the statement of the theorem.
This is the first theorem in Legendre that hints at the concept of area, but he doesn't use "square units" when considering area. Instead, he uses ratios -- just as the use of a right angle as a unit of angle measure, the use of ratios to determine area is a tradition going back to Euclid. But notice that there's an obvious natural unit of area to consider -- the surface area of the sphere itself. Let's assume that our sphere is the unit sphere -- that is, it has radius 1 -- and we will use radians throughout.
Legendre mentions the ratio of the angle MAN to four right angles -- that is, the ratio of the angle MAN to tau. Notice that the ratio of the arc MN to the circumference is this exact same ratio, because we chose radians and the unit sphere.
Now the theorem tells us that this equals the ratio of the area of the lune AMBNA to the surface area of the entire sphere. The surface area of a sphere is 4pi r^2 and this is a unit sphere, so the area of the sphere is 4pi -- oops, make that 2tau. So the theorem is essentially telling us:
Area(AMBNA)/(2tau) = Angle MAN/tau
Let's see how Legendre proves this theorem:
Demonstration. Let us suppose, in the first place, that the arc MN is to the circumference MNPQ in the ratio of two entire numbers, as 5 to 48, for example. [That is, suppose that the measure of the arc MN is 5tau/48 -- dw] The circumference MNPQ can be divided into 48 equal parts, of which MN will contain 5; then joining the pole A and the points of division by as many quadrants, we shall have 48 triangles in the surface of the hemisphere AMNPQ, which will be equal among themselves, since they have all of their parts equal. The entire sphere will therefore contain 96 of these partial triangles, and the lunary surface AMBNA will contain 10 of them; therefore the lunary surface is to the sphere as 10 is to 96, or as 5 is to 48, that is, as the arc is to the circumference.
So we can easily see what Legendre is doing here -- he is essentially dividing both the lune and the entire sphere into many small triangles. In his example, the lunes has angle measure 5tau/48, so he divides each half of the sphere into 48 triangles, with the lune taking up 5 of these triangles. We must consider both halves of the lune or sphere, which is why the entire lune takes up 10 out of the 96 triangles into which the entire sphere is divided. Using algebra, we can generalize this by saying that if the measure of the angle is (p/q)tau, we can divide each hemisphere into q triangles of which the lune takes up p. So the entire lune contains 2p out of 2q triangles into which the entire sphere is divided, and 2p/2q reduces to p/q, so the lune takes up p/q of the surface area of the sphere -- which, if you remember, is 2tau. So the area of the lune is (p/q)2tau.
The proof seems to be complete, yet Legendre adds the following cryptic lines:
"If the arc MN is not commensurable with the circumference, it may be shown by a course of reasoning, of which we have already had many examples, that the lunary surface is always to that of the sphere as the arc MN is to the circumference. QED"
What is Legendre saying here? Here's the problem -- what if the angle measure equals k tau for some irrational value k? ("Commensurable" essentially means "rational" -- a tradition going all the way back to Pythagoras and his sqrt(2) irrationality proof.) Then we can't just blindly write k = p/q and divide the hemisphere into q triangles. Instead, Legendre resorts to a "course of reasoning" that allows us to conclude that the theorem holds for both rational and irrational k. Dr. Hung-Hsi Wu gives this "course of reasoning" a name -- the Fundamental Assumption of School Mathematics.
By the way, notice that the equation:
Area(AMBNA)/(2tau) = Angle MAN/tau
can be rewritten as:
Area(AMBNA) = 2Angle MAN
And notice that the lune contains another angle at the South Pole, MBN, that is congruent to MAN (as both angles have the same measure as arc MN). So we can also write this as:
Area(AMBNA) = Angle MAN + Angle MBN
This begins to hint at a relationship between area and angle measure that exists in spherical, but not Euclidean, geometry.
The last two propositions are corollaries of this theorem:
494. Corollary I. Two lunary surfaces are to each other as their respective angles.
This is too trivial to prove as it follows directly from that equation we just wrote:
Area(AMBNA) = 2Angle MAN
Let's look at the final proposition for today:
495. Corollary II. We have already seen the entire surface of the sphere is equal to eight triangles having each three right angles (491); consequently, if the area of one of these triangles be taken for unity, the surface of the sphere will be represented by eight. This being supposed, the lunary surface, of which the angle is A, will be expressed by 2A [Yes, this is the formula we just wrote again -- dw], the angle A being estimated by taking the right angle for unity; for we have 2A : 8 :: A : 4. Here there are two kinds of units; one for angles, this is the right angle; the other for surfaces, this is the spherical triangle, of which all the angles are right angles, and the sides quadrants.
We see that Legendre's equilateral right triangle has area one-eighth that of the entire sphere -- since the sphere has area 2tau, the triangle has area 2tau/8 = tau/4 = lambda. So his two units -- the right angle for angle measure and the equilateral right triangle for area measure -- are both equal to lambda (again, assuming the unit sphere and radian measure).
In fact, we see that this is true for any triangle with two right angles -- if the third angle has measure A, then the triangle is half of a lune with angle A and measure 2A, so its measure is also A:
Area(AMN) = Angle A
And notice that the sum of the angles of Triangle AMN is exactly two right angles (tau/2) plus the measure of this Angle A. Legendre mentions the special case where Angle A = lambda. Again we hint at a connection between angle measure and area.
This is a good place for us to end this post. My next post will be tomorrow, when I'll continue with the rules of my new classroom.
No comments:
Post a Comment