## Monday, August 17, 2015

### Spherical Geometry (Legendre 487-489)

This will be my final spherical geometry post of the summer. I had originally envisioned posting more often during the summer -- perhaps twice a week. Instead, I got lazy and posted only about once a week, and so I didn't cover as many propositions in Legendre as I was hoping.

Still, the three propositions that I'm covering today are, perhaps, the most important theorems in all of spherical geometry. Unlike my last post, which covered statements provable in both Euclidean and spherical geometry, today's post will cover statements that are theorems of spherical geometry, but not Euclidean geometry. That is, today's theorems will set spherical geometry apart.

Let's begin with Legendre's Proposition 487, which sounds innocent enough:

487. If two triangles described upon the same sphere or upon equal spheres are equiangular with respect to each other, they will also be equilateral with respect to each other.

But Legendre uses the words equilateral and equiangular differently from a modern geometer. We recall that he uses the word equilateral to denote the SSS condition. So it follows that he would use the word equiangular to denote the AAA condition. But surely AAA can't possibly be a valid congruence criterion -- or can it?

Well, we already know that, while SAS, ASA, and SSS are valid in both Euclidean and spherical geometry, AAS and HL are valid in Euclidean, but not spherical, geometry. So we are not surprised that there exists a congruence condition, AAA, that is valid in spherical, but not Euclidean, geometry.

Legendre gives two possible proofs of the AAA Congruence Theorem in spherical geometry. The first proof is quick and dirty -- it uses the concept of polar triangles:

Demonstration. Let A, B, be the two given triangles, P, Q, their polar triangles. Since the angles are equal in the triangles A, B, the sides will be equal in the polar triangles P, Q (476); but, since the triangles P, Q, are equilateral with respect to each other, they are also equiangular with respect to each other (482); and the angles being equal in the triangles P, Q, it follows that the sides are equal in their polar triangles A, B. Therefore the triangles A, B, which are equiangular with respect to each other, are at the same time equilateral with respect to each other.

In Legendre's simpler proof of AAA, A and B are not points but entire triangles, and P and Q are their respective polar triangles. Now Proposition 476 tells us that the angles of a triangle are related to the sides of its polar triangle. Indeed, we found out that there is a simple formula relating the angle measures of a triangle to the sides of the polar triangle, provided that the angles are measured in radians and the sphere has radius 1. In that case, if the first triangle has an angle of measure theta, then the polar triangle will have a side of length pi - theta. Then this converts two triangles that satisfy the AAA condition into their polar triangles that satisfy the SSS condition, which we have already proved sufficient for congruence in Proposition 482.

So the proof (not a full two-column proof, but more like a flow proof) looks something like this:

A and B congruent angles (given) --> P and Q congruent sides (476)
--> P and Q congruent triangles (SSS)
--> P and Q congruent angles (CPCTC)
--> A and B congruent sides (476)
--> A and B congruent triangles (SSS)

since the polar triangle of the polar triangle is the original triangle. Therefore AAA is a valid criterion for congruence in spherical geometry. QED

Below is Legendre's other proof of AAA. As Legendre himself writes, "This proposition may be proved without making use of polar triangles in the following manner":

Let ABC, DEF, be two triangles equiangular with respect to each other, having A = D, B = E, C = F; we say that the sides will be equal, namely, AB = DE, AC = DF, BC = EF.

Produce AB, AC, making AG = DE, AH = DF; join GH, and produce the arcs BC, GH, till they meet in I and K.

The two sides AG, AH, are, by construction, equal to the two DF, DE, the included angle GAH = BAC = EDF, consequently, the triangles AGH, DEF, are equal in all their parts (480); therefore the angle AGH = DEF = ABC, and the angle AHG = DFE = ACB.

In the triangles IBG, KBG, the side BG is common, and the angle IGB = GBK; and, since IGB + BGK is equal to two right angles, as also GBK + IBG, it follows that BGK = IBG. Consequently the two triangles IBG, GBK, are equal; therefore IG = BK, and IB = GK.

In like manner, since the angle AHG = ACB, the triangles ICH, HCK, have a side and the two adjacent angles of the one respectively equal to a side and the two adjacent angles of the other; consequently they are equal; therefore IH = CK, and HK = IC.

Now, if from the equals BK, IG, we take the equal CK, IH, the remainders BC, GH, will be equal. Besides, the angle BCA = AHG, and the angle ABC = AGH. Whence the triangles ABC, AHG, have a side and the two adjacent angles of one respectively equal to a side and the two adjacent angles of the other; consequently they are equal. But the triangle DEF is equal in all its parts to the triangle AHG; therefore it is also equal to the triangle ABC, and we shall have AB = DE, AC = DF, BC = EF; hence, if two spherical triangles are equiangular with respect to each other, the sides opposite to the equal sides will be equal.

We can try to convert Legendre's proof to two columns. But before we do so, let's see whether we can figure out what's going on here. He begins with "Produce AB, AC, making AG = DE, AH = DF." It seems that he is letting G be the point on ray AB such that AG = DE, and H be the point on ray AC such that AH = DF. But notice that our goal is to prove that AB = DE and AC = DF -- that is, to prove ultimately that B and G are the same point, and that C and H are the same point. This will be a bit awkward to prove.

Instead, we think about how the congruence theorems SAS, ASA, SSS are proved back in the U of Chicago text. In many ways, Legendre's proof of AAA and the U of Chicago proof of SSS are similar in that in both proofs, to prove Triangle ABC congruent to DEF, we prove that each triangle is congruent to a third triangle. For Legendre, the third triangle is AGH, while for the U of Chicago, the third triangle is DEC'.

Now the differences between the two proofs should become more apparent. In the U of Chicago proof, we are given congruent sides, and so we're able to use the given congruent sides AB and DE to line up the copy of ABC with the triangle DEF. But in the Legendre proof, we are instead given congruent angles, and so Legendre wants to be able to use the congruent angles A and D to line up a copy of DEF with the triangle ABC. But that's the problem -- it's easy to draw two distinct triangles with a side in common, but difficult to draw them with an angle in common.

Here's my solution -- I place G on the ray opposite AB with the correct length AG = DE, and likewise H on the ray opposite AC with the correct length AH = DF. Then the angles GAH and BAC are still congruent because they are vertical angles.

In the U of Chicago proof DEF and the image of ABC are reflection images of each other, with line DE as the mirror. In our version of Legendre's proof ABC and the image of DEF are rotation images of each other, with A as the pole of the rotation and a magnitude of 180 degrees.

Now we can finish our conversion of Legendre into a two-column proof:

Given: Angle A = D, Angle B = E, Angle C = F
Prove: Triangle ABC congruent to DEF

Proof:
Statements                               Reasons
1. Angle A = D, B = E, C = F  1. Given
2. G on ray BA st. AG = DE,   2. Ruler Postulate
H on ray CA st. AH = DF
3. Angle GAH = BAC              3. Vertical Angles Theorem
4. Angle GAH = EDF              4. Transitive Property of Congruence (steps 3, 1a)
5. Triangle AGH = DEF          5. SAS Congruence (steps 2, 4, 2)
6. Angle AGH = DEF,             6. CPCTC
Angle AHG = DFE
7. Angle AGH = ABC,             7. Transitive Property of Congruence (steps 6, 1b, then 6, 1c)
Angle AHG = ACB
8. Lines BC, GH meet at I, K   8. All lines intersect in two antipodal points.
9. BG = BG                               9. Reflexive Property of Congruence
10. Angle IGB = GBK              10. Same as step 7 (AGH = ABC) with the angles renamed.
11. IGB, BGK supplementary,  11. Linear Pair Theorem
GBK, IBG supplementary
12. Angle BGK = IBG               12. Angles suppl. to congruent angles (step 10) are congruent.
13. Triangle IBG = GBK           13. ASA Congruence (steps 10, 9, 12)
14. IG = BK                               14. CPCTC
15. CH = CH                              15. Reflexive Property of Congruence
16. Angle IHC = HCK               16. Same as step 7 (AHG = ACB) with the angles renamed.
17. IHCCHK supplementary,  17. Linear Pair Theorem
HCKICH supplementary
18. Angle CHK = ICH               18. Angles suppl. to congruent angles (step 16) are congruent.
19. Triangle ICH = HCK           19. ASA Congruence (steps 16, 15, 18)
20. IH = CK                               20. CPCTC
21. BC + CK = BK,                    21. Segment Addition Postulate
IH + HG = IG
22. IH + HG = BC + CK            22. Substitution Property of Equality (21 into 14)
23. GH = BC                              23. Subtraction Property of Equality (22 minus 20)
24. Triangle ABC = AGH           24. ASA Congruence (steps 7, 23, 7)
25. Triangle ABC = DEF           25. Transitive Property of Congruence (steps 24, 5)

When we look at this admittedly lengthy proof, it may take a while to figure why it doesn't work in Euclidean geometry. The problematic step for Euclidean geometry is step 8, where we have the lines BC and GH intersect at points I and K. Surely in Euclidean geometry, we can't have two distinct lines intersect in two distinct points, as we do in spherical geometry. In fact, we can also prove that in Euclidean geometry, BC and GH can't even intersect in one point, much less two. This is because the angles proved congruent in step 7 turn out to be alternate interior angles, and as these are congruent, the lines BC and GH must be parallel. In spherical geometry, there are no parallel lines. This is why this proof is valid in spherical geometry, but not Euclidean geometry.

We see that these points I and K are antipodal points. This may be easier to visualize in the special case where B lies due north of C. Then G will lie due south of H (since GH is simply BC rotated 180 degrees about point A). The points I and K end up being the North and South Poles, and we have the two meridians IBCK and IHGK. The region bounded by these two meridians is called a lune.

We can now divide this lune into two congruent spherical triangles in two different ways. The first is to cut the lune along the line BG (steps 9-14), and the other is to cut it along CH (steps 15-20). This is similar to how we can divide a Euclidean parallelogram into two congruent triangles by cutting it along its diagonal, and if we cut the original parallelogram along the other diagonal, we end up with two different congruent triangles.

In all, we had to prove five pairs of triangles to be congruent. Two of these pairs are the division of the lune, and the other three pairs follow the Transitive Property pattern from the U of Chicago text -- proving the original pair congruent by showing that each is congruent to a third.

The fact that AAA doesn't work in Euclidean geometry is, indeed, the subject of the next proposition, which Legendre calls a "scholium":

488. Scholium. This proposition does not hold true with regard to plane triangles, in which, from the equality of the angle, we can only infer the proportionality of the sides. But it is easy to account for the difference in this respect between plane and spherical triangles. In the present proposition, as well as those of articles 480, 481, 482, 486, which relate to a comparison of triangles, it is said expressly that the triangles are described upon the same sphere or upon equal spheres. Now, similar arcs are proportional to their radii; consequently upon equal spheres two triangles cannot be similar without being equal. It is therefore not surprising that equality of angles should imply equality of sides.

It would be otherwise, if the triangles were described upon unequal spheres; then, the angles being equal, the triangles would be similar, and the homologous sides would be to each other as to the radii of their spheres.

In the scholium, Legendre tells us that in spherical geometry, two triangle cannot be similar without being congruent. In terms of Common Core transformations, we are saying that the only similarity transformations that exist are isometries -- there's no such thing as a dilation with any scale factor other than 1.

Why can't we have dilations in spherical geometry? Legendre reminds us that spherical triangles are so-called because they lie on a sphere. Now the three spherical isometries -- rotations, reflections, and glide reflections -- correspond to the symmetries of the underlying sphere (rotation about an axis, reflection over a plane, and roto-reflection). But whereas the dilation image of the entire plane is the plane itself (provided the center of the dilation is in that plane), the dilation image of the entire sphere can never be that sphere itself. This is due to the Dilation Theorem -- a dilation with r as its scale factor doesn't preserve distance -- instead it multiplies all lengths by r, including the radius of the underlying sphere. So the dilation image of a sphere of radius 1 is a sphere of radius r -- that is, a different sphere. Therefore the dilation of scale factor r is not a transformation in spherical geometry, and so there are no similar triangles unless they are already congruent.

And so we see that the AAA condition already implies full congruence in spherical geometry, whereas in Euclidean geometry it only implies similarity. As it turns out, AAA implies full congruence in hyperbolic geometry as well. So we have the following congruence theorems:

Spherical Geometry: SSS, SAS, ASA, AAA
Euclidean Geometry: SSS, SAS, ASA, AAS, HL
Hyperbolic Geometry: SSS. SAS, ASA, AAS, HL, AAA

So Euclidean geometry is distinguished from the others by the existence of dilations and similarity.

We have one more proposition left to cover -- the sum of the angles of a triangle. We have already seen spherical triangles with two and even three right angles, so we already know that the sum of the angles of a triangle can't be 180 degrees. Let's see what Legendre has to say:

489. The sum of the angles of every spherical triangle is less than six, and greater than two right angles.

Demonstration. 1. Each angle of a spherical triangle is less than two right angles (see the following scholium); therefore the sum of the three angles is less than six right angles.

Let me intervene here before I give the rest of Legendre's proof. We know that Legendre, following Euclid, doesn't use degrees -- instead, he uses the right angle as a unit. So what Legendre is telling us is that the sum of the angles is strictly between 180 and 540 degrees. The first part of this proof -- that the sum is less than 540 -- is trivial. Since each angle of the triangle is less than 180, the sum of all three of them must be less than 540.

So now we must get to the nontrivial part of the proof -- namely that the sum is more than 180:

2. The measure of each angle of a spherical triangle is equal to the semicircumference minus the corresponding side of the polar triangle (476); therefore the sum of the three angles has for its measure three semicircumferences minus the sum of the sides of the polar triangle. Now, this last sum is less than a circumference (461); consequently, by subtracting it from three semicircumferences, the remainder will be greater than a semicircumference, which is the measure of two right angles; therefore the sum of the three angles of a spherical triangle is greater than two right angles.

We observe that Legendre's proof really depends on two previous propositions -- 476 and 461. In fact, I actually mentioned after giving the proof of 476 that we could have jumped directly to the sum of the angles right then and there. But I decided to wait until we reached the proof in Legendre.

Legendre's proof goes back to polar triangles. As I mentioned earlier in this post, an angle of measure theta corresponds to a side of length pi - theta in the polar triangle -- once again, assuming that we are using radians and the unit sphere. So if the angles of our triangle have measure alpha, beta, and gamma, then the sides of the polar triangle have length pi - alpha, pi - beta, and pi - gamma -- which adds up to 3pi minus the original angle sum. But Proposition 461 gives us a strict upper bound on the perimeter of the polar triangle -- it must be less than the circumference of the sphere, or 2pi. So if we call the original angle sum sigma, we obtain:

(pi - alpha) + (pi - beta) + (pi - gamma) < 2pi
3pi - sigma < 2pi
-sigma < -pi
sigma > pi

That is, the sum of the angle of a triangle must be greater than pi radians, or 180 degrees. And since the perimeter must be greater than zero, we likewise have 3pi - sigma > 0 or sigma < 3pi, which we have already proved. Therefore the sum can be anywhere between 180 and 540 degrees. QED

We already know that the sum of the angles of a Euclidean triangle is exactly 180. As it turns out, the sum of the angles of a spherical triangle isn't a constant -- for any value between 180 and 540, there exists a triangle whose angle sum is that value. As it turns out, the sum of the angles of a triangle in hyperbolic triangle must be less than 180. So we have the following triangle-sum theorems:

Spherical: greater than 180 degrees
Euclidean: exactly 180 degrees
Hyperbolic: less than 180 degrees

As neutral geometry incorporates both Euclidean and hyperbolic, but not spherical, geometry, it is a theorem of neutral geometry that the sum of the angles of a triangle is at most 180 degrees. This can be proved ultimately from the Triangle Exterior Angle Inequality, which, as we've said before, holds in neutral geometry.

I've said that I will give a Parallel Postulate at the start of the second quarter of my geometry course, so that the first quarter will ultimately be neutral geometry. But the proof that the sum of the angles of a triangle is at most 180 degrees in neutral geometry -- often given in college courses -- is not appropriate for high school audiences. It would be highly confusing to students, for a teacher to say that the sum could be less than 180, then later on say that it must be exactly 180.

But now we can appreciate the start of Section 5-7 of the U of Chicago text, which describes an experiment by the 19th century German mathematician Karl Friedrich Gauss. He wanted to find the angle sum of a triangle to determine whether the real world matched Euclidean, hyperbolic, or spherical geometry.

Of course, we know that Gauss should have determined that geometry was spherical -- since after all, he was working on the spherical earth. The problem was that the sum of a spherical triangle is actually closer to 180 for small triangles and closer to 540 for large triangles. Since Gauss's triangle was small compared to the size of the earth, the sum of the angles was 180, to within the accuracy of his instruments.

(As it turns out, the relationship between the angle sum and size of a triangle can be made exact, though we won't have time to go far enough in Legendre to prove it. If we choose the correct units -- as usual, that means radian measure on the unit sphere -- the sum of the angles is exactly pi plus the area of the triangle!)

But the question that many people ponder is, can the universe have spherical geometry? That is, if we travel forever without changing direction, can we eventually reach the earth again?

Notice that in our spherical geometry proofs based on Legendre's text, we actually used the underlying sphere in some of our proofs (including solid angles with vertices at the center). But if the universe is spherical, we actually wouldn't be aware of the center, or even that it's a sphere. We'd only know that the universe is spherical indirectly -- by examining the angles of a triangle, for example.

And so another way to approach spherical geometry -- without referring to the sphere itself or any of its Euclidean properties -- is to start with some postulates that describe spherical geometry and use these to prove the theorems of spherical geometry. We aren't allowed to use phrases like "great circle" which betray knowledge of the underlying sphere. Instead, we must refer to great circles by the role that they serve in spherical geometry -- as lines.

The first postulate in the U of Chicago text is the Point-Line-Plane Postulate. We see that part (a) of this postulate reads:

(a) Unique line assumption: Through any two points, there is exactly one line.

As we know, this already fails in spherical geometry, for through the North and South Poles, there exist infinitely many lines. So we could try replacing it with a new postulate, such as "through any two points, there is at least one line." Some other possible postulates are:

-- Any two lines intersect.
-- Every line has exactly two poles. (Here "pole" is an undefined term.)
-- Every point is the pole of exactly one line.
-- The pole of a line never lies on the line.
-- If l and m are lines and a pole of l lies on m, then the other pole of l lies on m, and both poles of m lie on l. (Two lines l and m satisfying this property are defined to be perpendicular.)

There can be an Angle Measure Postulate, except that it may be more convenient to give the measure in radians rather than degrees. Then the distance between two points A and B can be defined as the measure of the angle APB, where P is the pole of the line AB.

It may be interesting to see whether this will be sufficient to prove any of the spherical geometry theorems in Legendre, or otherwise what I would need to add to these to prove the theorems. But as I said earlier, unfortunately I was too lazy, and so I ran out of time this summer.

There will be one more post this summer. It will be a "How to Fix Common Core" post, which I will post in about a week.