By the way, last year I made my 200th post on Tau Day. After all, I posted on each of the 180 days of the school year plus a few extra posts during vacation periods, so it's not surprising for me to make my 400th post exactly one year after my 200th post. But don't expect me to make my 600th post next Tau Day -- this year my plan is to post three times a week during the school year rather than every single day of the school year.

This is what I wrote last year on Tau Day:

But what exactly is this constant "tau," anyway? It is not defined in the U of Chicago text -- if it were, it would have appeared in Section 8-8, as follows:

Definition:

tau =

*C*/

*r*, where

*C*is the circumference and

*r*the radius of a circle.

Now the text tells us that

*C*= 2pi *

*r*, so that

*C*/

*r*= 2pi. Therefore, we conclude that tau = 2pi. The decimal value of this constant is approximately 6.283185307..., or 6.28 to two decimal places. Thus the day 6/28, or June 28th, is Tau Day.

It was about ten or fifteen years ago when I first read about Pi Day, and one old site (which I believe no longer exists) suggested a few alternate Pi Days, including Pi Approximation Day, July 22nd. One day that this page mentioned was Two Pi Day, or June 28th. But this page did not suggest giving the number 2pi the name tau or any another special name.

Now about five years ago -- on another Pi Day, March 14th -- I decided to search Google for some Pi Day webpages. But one of the search results was strange -- this webpage referred to a "half tau day":

http://halftauday.com/

"The true circle constant is the ratio of a circle’s circumference to its

*radius*, not to its diameter. This number, called

**Half Tau Day**. (Of course, since

**Tau Day**itself.) Although it is of great

*historical*importance, the mathematical significance of

This linked to the webpage of a physicist, Michael Hartl, who lives right here in California. He has come up with "The Tau Manifesto," where he explains the rationale for considering 2pi as a new constant, to be named after the Greek letter "tau":

http://tauday.com/tau-manifesto

In “

*The Tau Manifesto*is dedicated to the proposition that the proper response to “

*really*.” And the true circle constant deserves a proper name. As you may have guessed by now,

*The Tau Manifesto*proposes that this name should be the Greek letter

Hartl then devotes the rest of his Manifesto to explaining what would happen if mathematicians were to write their functions in terms of tau rather than pi.

In my last post from about a week ago, I mentioned how Legendre sometimes took the measure of a central angle of a circle to be the length of its associated arc. In the case of a circle of radius 1, this gives us radian measure. Now Hartl discusses radian measure in Section 2.1 of his Manifesto:

There is an intimate relationship between circles and angles.Since the concentric circles have different radii, the lines in the figure cut off different lengths of arc (or

*arc lengths*), but the angle

Hartl then shows what happens when we write radian measures in terms of tau rather than pi:

We also see the genius of Bob Palais’ identification of the circle constant as “one turn”:

*turn*of a circle. Moreover, note that with

*nothing to memorize*: a twelfth of a turn is

In the following section, Hartl moves on to the trig functions, sine and cosine:

Although radian angle measure provides some of the most compelling arguments for the true circle constant, it’s worth comparing the virtues of

*unit circle*(i.e., a circle with radius

Of course, since sine and cosine both go through one full cycle during one turn of the circle, we have [the period] T=τ ; i.e., the circle functions have periods equal to the circle constant.As a result, the “special” values of θ are utterly natural: a quarter-period is τ/4 , a half-period is τ/2 , etc. In fact, at one point I found myself wondering about the numerical value of θ for the zero of the sine function. Since the zero occurs after half a period, and since τ≈6.28 , a quick mental calculation led to the following result:

That’s right: I was astonished to discover that

Later on, in Section 5.1, Hartl discusses what effect his new constant has in higher dimensions. Here is not merely referring to the sphere and its surface area and volume. Indeed, here he considers dimensions *I had already forgotten that*τ/2 is sometimes called “π ”.Perhaps this even happened to you just now. Welcome to my world.*higher*than 3D. The idea that there can exist a fourth dimension -- an idea anathema to Euclid and the ancient Greeks -- is a separate topic of its own (and I've discussed this earlier this year, back when we were reading Rudy Rucker's book).

We start our investigations with the generalization of a circle to arbitrary dimensions. This object, called a 0 -sphere is the empty set, and we define its “interior” to be a point. A 1 -sphere is the set of all points satisfying

*hypersphere*or an*, can be defined as follows. (For convenience, we assume that these spheres are centered on the origin.)A*n -sphere
which consists of the two points ±r . Its interior, which satisfies

is the line segment from −r to r .

Then a 2-sphere is a circle, and a 3-sphere is an ordinary sphere, and a 4-sphere would be a sphere in a 4D space (often called a "glome"). Hartl then uses calculus to derive formulas for the surface areas and volume of all of these hyperspheres, and he obtains formulas in terms of pi. I won't go into the details here -- you can read it from Hartl's Manifesto or from Wolfram MathWorld, which is where Hartl himself links:http://mathworld.wolfram.com/Hypersphere.html

The Wolfram formula involves things like the factorials of half-integers (which the TI-83 and higher calculators can handle) and double factorials (which aren't built in to the TI-83).

Hartl then shows that these formulas are simpler when written in terms of tau -- and they become even simpler when written in terms of "lambda" -- the measure of a right angle, which can also be written as either pi/2 or tau/4. (Some authors have proposed the name "eta" instead of "lambda" for the measure of a right angle. I prefer "lambda" because I wish to reserve "eta" for the constant e^(1/e), which appears in tetration. the next operation after exponentiation.)

Hartl concludes his Tau Manifesto as follows:

*The Tau Manifesto*first launched on Tau Day: June 28 (6/28), 2010. Tau Day is a time to celebrate and rejoice in all things mathematical.

^{19}If you would like to receive updates about

*Tau Manifesto*mailing list below. And if you think that the circular baked goods on Pi Day are tasty, just wait—Tau Day has twice as much pi(e)!

Notice that Hartl first wrote his manifesto on Tau Day 2010. It was the following Pi Day 2011 when his Half Tau Day page appeared in my search results, and so I have been following his Tau Day page for five years now.

So we see that according to Hartl, tau is superior to pi. In his Tau Manifesto, he suggests that mathematicians should use tau instead of pi in calculations. Each year, he writes a "State of the Tau" address in which he discusses the progress he has made in converting mathematicians to tau. Here is the beginning of this year's "State of the Tau" address:

Happy Tau Day, everyone! It’s hard to believe it’s already been six years since

*The Tau Manifesto*launched and the real battle against pi’s supremacy began.
I don’t keep careful statistics, but I can confidently state that awareness of tau is enormously widespread. As an active member of the Caltech alumni community, for example, I frequently meet Caltech undergraduates, and they’ve virtually always heard of Tau Day.

Even though the following link is five years old, it discusses an interview with Hartl in which the tauist explains exactly how one can celebrate the holiday:

http://www.mathgoespop.com/2011/06/second-annual-tau-day-interview-and-ideas.html

Q: A big contributing factor to Pi Day’s success has undoubtedly been the food. Besides eating twice as much pie, do you have any ideas on how to build Tau Day into a distinct mathematical holiday?

A: Tau Day happens during the summer, so perhaps we could add a distinctive outdoor component. Tau Day at the beach? I’m certainly open to suggestions!

Notice that one of the commenters on this page is none other than Fawn Nguyen -- a teacher with whom we are certainly familiar. Just like calendar reformers and tetraters, tauists are an esoteric bunch. Since around Pi Day I posted several videos related to pi, today I post several videos related to tau -- one of which Nguyen linked to in her five-year-old comment as well:

1. Numberphile:

Numberphile is a British mathematician who posts videos about many numbers. He has several videos about pi, as well as these two videos about tau. In the first, he explains the number tau itself, and in the other, two mathematicians debate whether pi or tau is the superior circle constant. D News has also posted a pi vs. tau debate:

2. Emmex Plusbee:

In this video, a group of high school students walk out of class to protest Pi Day. They begin to chant, "What Do We Want? Tau! When Do We Want It? Now!"

This group also posted another video -- Blowing Up a Pi(e) on Tau Day -- but unfortunately, I haven't been able to locate it on YouTube. I just remember this part, paraphrased:

Girl: Why are we blowing up a pie again?

Boy: Because you don't define a circle by the diameter!

Girl: Huh?

Other Boy: Because it's fun!

After the students blow up the pie, the resulting fire display lead them to call out, "Happy Early Fourth of July!" Incidentally, the first day on which fireworks may legally be sold in many cities here in California is Tau Day.

By the way, most schools, even Labor Day Start Schools, are out by Tau Day. In other posts, I've mentioned that Pi Approximation Day, July 22nd, is a great alternative to Pi Day for summer school Geometry classes. But at some Early Start Schools, even summer school has ended by July 22nd. I point out that the best alternative day for a Pi Day in an Early Start Summer School class would in fact be Tau Day.

3. Michael Blake

As I mentioned back on Pi Day, Michael Blake is one of several musicians who created pi music simply by taking the digits of pi and applying them to the C major scale -- so 3 denotes the third note of the scale or E, 1 is the first note or C, 4 is the fourth note or F, and so on. In this video, Blake does the same with tau.

4. Vi Hart

Vi Hart -- a self-described "mathemusician" -- also has a song about the circle constant tau, based on the notes of the major scale. There is also a video -- first posted on that very same Pi Day when I first discovered the number tau -- where Hart explains why tau is superior to pi. (This is the video Fawn Nguyen linked to earlier.)

Hart isn't as prolific a video poster as she was in the past, but every year on Pi Day, Hart posts her annual "Anti-Pi Rant." Some commenters have joked that Pi Day should be renamed "Vi Day" because it's the only day that Vi is guaranteed to post a new video.

Vi Hart posted a special 360-degree video for Tau Day:

Numberphile -- inspired by Vi Hart -- also has a video called "The Tau of Phi," which features a song that combines the digits of tau with those of the golden ratio phi.

5. 3Blue1Brown:

Mentioned in Michael Hartl's State of the Tau address, this is a brief video about the number tau. In the background, there appears a special equation known as Euler's identity: e ^ (i*pi) + 1 = 0. (Recall that Euler was the same 18th-century Swiss mathematician who solved the Bridges of Konigsberg problem.) This formula is often considered elegant as it includes the five important constants 0, 1, e, i, and pi, along with the (first) three operations addition, multiplication, and exponentiation (sorry, no tetration). Tauists note that if we replace pi with tau, we obtain e ^ (i*tau) = 1 + 0, so it still contains five important constants and the (first) three operations.

6. Sal Khan:

Sal Khan is perhaps the best known mathematician on YouTube. Even though most of Khan's video is about the number pi, at the end he shows how some of these formulas can be simplified by replacing pi with tau.

6.28. Michael Hartl himself posts a video where he recites pi to 40 digits, beating

*Futurama*creator Matt Groening, who gave only two digits. I don't link to this video as it is not on YouTube, but you can see the video on the Tau Manifesto itself. And besides, it's about reciting pi, not tau.

In his "State of the Tau" address, Hartl notes that he was able to take advantage of last year's Pi Day of the Century to raise awareness of tau -- but of course, he notes that the Tau Day of the Century will occur on 6/28/31, or June 28th, 2031, since tau begins 6.2831....

I agree that Hartl makes a strong case in favor of tau. By the way, I once tried to tell a Pre-calculus student I was tutoring about tau. I'd notice that he kept referring to 180 degrees as pi/2 and 90 degrees as pi/4, and told them that if we were to use tau, the radian measures tau/2 and tau/4 would be correct after all. I'm not sure whether this helped him or not -- after all, all my discussion about tau didn't change the fact that his teacher and tests will use pi, not tau.

Based on this, I'd argue that if I were to ask a student to draw (freehand) an angle measuring pi/4 radians, the student would first convert this to 45 degrees, then draw the angle. But if students were to learn about tau instead of pi, if I were to ask someone to draw an angle of tau/4 radians, one could tell that this is 1/4 of a circle and thus a right angle without converting to degrees at all. (Then again, I'm not sure whether tau/12 is that much easier than pi/6 -- both are likely to result in a student needing to convert to degrees before drawing.) Still, I bet that he would have fared better if he had learned tau all along and not have to use pi ever, since tau fit better with his intuition.

Last year, I wrote an analogy involving tau as part of my "How to Fix Common Core" series. This year, I'm not continuing that series, but nonetheless I will include an abridged version of what I wrote last year:

And so, let me propose the following attempt to fulfill the Tau Manifesto. The goal is for all mathematicians to use tau instead of pi as the circle constant. Math teachers would use tau when teaching the circle, so the high school students in the above posted video would get what they want, and the student I tutored wouldn't have been so confused about radian measure. And let's give a date by which we want the goal to be accomplished -- let's say Tau Day of the Century, or June 28th, 2031, exactly fifteen years from today.

If we really want to enforce the new constant tau, let's explicitly require the students to learn it as a proposed Common Core standard:

CCSS.MATH.CONTENT.6.G.B.4 (proposed)

Know that tau is defined as the ratio of the circumference of a circle to its radius; know the formulas for the area and circumference of a circle in terms of tau and use them to solve problems; give an informal derivation of the relationship between the circumference and area of a circle.

And of course, PARCC/SBAC questions would have to be written so that they use tau instead of pi -- otherwise there's no point to including it in the standards. On the surface, this proposal should work. Over the next 15 years, students would learn only about tau and not about pi -- and these would become our future mathematicians. With an entire generation of students now aware of tau, we should attain our goal and fulfill the Tau Manifesto by 2031.

For starters, there will be a backlash to including tau in the standards. There is already a Pi Manifesto written as a response to the Tau Manifesto:

http://www.thepimanifesto.com/

In the Pi Manifesto, the author, who only goes by the initials MSC, begins by giving the reason that pi was defined using the diameter in the first place:

So why did mathematicians define it using the diameter? Likely because it is easier to measure the diameter of a circular object than it is to measure its radius. In practice, the only way to measure the radius of a circle is to first measure the diameter and divide by 2.

Most calculators, including the TI-83 and higher graphing calculators, currently have pi built in, but of course they don't have tau built in. If tau were included in the Common Core, I could easily see TI coming up with a new calculator with tau built in. But then people will say that tau was created just so that Texas Instruments could make more money selling new calculators. (Interestingly enough, Hartl notes that the Google calculator now recognizes tau. Google simply treats "tau" as a word and merely gives search results, but if at least one operation is typed in, such as "tau/1," then Google returns the correct value 6.28318530718. And of course "tau/2" returns 3.14159265359.)

So we see the problem here. We wanted to include tau in the Common Core Standards, because we believe that tau makes learning math -- especially trig and radian measure -- simpler. But far from convincing the public that tau is superior to pi, our plan would end up alienating students and parents, who now believe that tau makes math much worse.

The problem is that innovation is difficult -- there may be a new way of teaching math or another subject that's much, much superior to current methods, but teachers, parents, and students will be skeptical about claims of superiority if the first place they see the new methods is in Common Core or similar national standards.

Some people oppose Common Core because they believe that the Tenth Amendment requires there to be fifty separate sets of state standards. We could, therefore, use

*federalism*to achieve our goal by having a single state include tau in its standards. If people in other states see that the tau state is educating its students more successfully than pi states, then the pi states will rush to adopt tau as well.

Here, we avoid politics by only considering those who can afford private school for the purposes of this thought experiment. Before accepting the tuition payment, a private school can inform the parents that they are teaching the students a new method of learning geometry, using tau instead of pi. They can show the prospective parents Hartl's Tau Manifesto and inform them that if they are successful, maybe someday pi will be obsolete and every mathematician will be using tau instead -- and their children will be the leaders of the new movement.

Also, notice that adoption of tau by some private school is unnecessary to avoid the "guinea pig" problem -- a private

*tutoring*company can teach its students tau (along with pi). We've already seen the Khan Academy video in which tau is mentioned, so that's a start.

So we see that trying to fix these problems with Common Core is not easy indeed. I used the debate of pi vs. tau as an example to show the problems with Common Core, but we see that the Common Core debate will be much more difficult to solve.

(Oh, and of course in my new middle school classes this fall, I'll definitely be teaching them

*pi*as I'm supposed to, not

*tau*.)

I wish to use today's post to relaunch my Spherical Geometry series, based on the 200-year-old Geometry text written by the French mathematician Adrien-Marie Legendre. I wish to start by repeating my final Spherical Geometry post from last year, so that you can remember where we left off. Since I'm posting this on Tau Day, I changed all values given in terms of pi so that they're given in terms of tau instead.

The three propositions that I'm covering today are, perhaps, the most important theorems in all of 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. For those who don't remember the link to the Legendre text, here it is again:

https://books.google.com/books?id=gs5JAAAAMAAJ&pg=PA157&source=gbs_toc_r&cad=4#v=onepage&q&f=false

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 tau/2 - 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 A*

*B*with the correct length

*AG*=

*DE*, and likewise

*H*on the ray

*opposite A*

*C*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.

*IHC*,

*CHK*supplementary, 17. Linear Pair Theorem

*HCK*,

*ICH*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 triangles 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 tau/2 - 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 tau/2 - alpha, tau/2 - beta, and tau/2 - gamma -- which adds up to 3tau/2 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 tau. So if we call the original angle sum

*sigma*, we obtain:
(tau/2 - alpha) + (tau/2 - beta) + (tau/2 - gamma) < tau

3tau/2 - sigma < tau

-sigma < -tau/2

sigma > tau/2

That is, the sum of the angle of a triangle must be greater than tau/2 radians, or 180 degrees. And since the perimeter must be greater than zero, we likewise have 3tau/2 - sigma > 0 or sigma < 3tau/2, which we have already proved. Therefore the sum can be anywhere between 180 and 540 degrees. QED

Note how easy it is to convert from Legendre's words to symbolic equations in terms of tau -- a "circumference" is tau, and a "

Note how easy it is to convert from Legendre's words to symbolic equations in terms of tau -- a "circumference" is tau, and a "

*semi*circumference" is tau/2 -- another advantage of using tau. An alternative to using "tau" to translate "circumference" is using "lambda" to translate Legendre's "right angle." So the theorem states that that the angle sum is between 2lambda and 6lambda. Using the right angle as a unit of measure (rather than degrees) is a tradition dating back to Euclid -- in no case is "pi" a natural unit of angle measure.
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.
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, and we should be going far enough this summer 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 tau/2 plus the

*area*of the triangle!)
But the question that many people ponder (including Rudy Rucker yet again) 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*.Happy Tau Day, everyone! My next post will be later week, when I will return to preparing for my new class.