Lesson 7-2: Triangle Congruence Theorems (Day 72)

This is what Theoni Pappas writes on page 35 of her Magic of Mathematics:

"Why, sometimes I've believed as many things as six impossible things before breakfast."
-- Lewis Carroll.

This is the first page of the first section of Chapter 2, "How Mathematical Worlds Are Formed." And some of the material mentioned on this page should look familiar to Geometry students.

Here are excerpts from this page:

"Little did Euclid know in 300 B.C. when he began to organize geometric ideas into a mathematical system that he was developing the first mathematical world. All form the universe of mathematics.

"Every mathematical world exists in a mathematical system. It explains how its objects are formed, how they generate new objects, and how they are governed. Undefined terms can be described, so one has a feeling of what they mean, but technically they cannot be defined. It takes terms to form definitions, and you have to begin with some terms.

"The best way to understand a system is to look at one. Assume this mini world's undefined terms are points and lines. In addition to undefined terms, a mathematical system also has axioms, theorems, and definitions."

Yes, this definitely sounds like the world formed Chapter 1 of the U of Chicago text. But actually, as we'll find out tomorrow, this world will be much simpler than the Euclidean world. Indeed, we already see that planes is an undefined term for Euclid and the U of Chicago, but there are no planes in the world Pappas is forming here.

Before we continue, let me make an important announcement. Tomorrow I will be substitute teaching at a high school in the district where I work. But there is something special happening tomorrow:

Announcement: High School Putnam Exam

The [name redacted] district has been selected to pilot the administration of the William Lowell Putnam Mathematical Exam on Wednesday, December 6, 2017. All seniors are expected to take the math test on Wednesday. The special bell schedule for Wednesday will be:

8:00-11:00 Period 1/Putnam A Session
11:00-11:45 Lunch
11:50-2:50 Period 2/Putnam B Session

Students in grades 9-11 are expected to attend periods 1 and 2 during the indicated blocks. All math teachers are highly encouraged to give the Pre-Putnam practice test to students in grades 9-11 in order to prepare them for the Putnam test next year.

Some people applaud the arrival of the new test. "The highest level of CCSS success leads only to no remediation for college algebra," says Steve H., an advocate of traditional math. "CCSS institutionalizes low expectation, no-STEM math." while the Putnam is a high-expectations test. The old CCSS tests only assess up to Algebra II, but the new Putnam test will assess students on Calculus and other college-level topics. Success on the new Putnam test will demonstrate a high school student to be prepared for success in a STEM major.

The State of California is planning to replace the high school math portion of the SBAC with the Putnam for all students by the 2019-2020 school year. A more rigorous replacement for the ELA section of the SBAC has yet to be determined.

OK, the above announcement is satire. It's actually based on a dream I had last night -- I was in a high school classroom one day, and suddenly I had to administer the Putnam to all students.

Of course, this is a horrible idea. Recall that the Putnam is a test that even most math majors don't take -- and most math majors who do take the test earn a score of zero! But based on recent trends, the Putnam being administered to all high school seniors isn't as far-fetched as it ought to be.

We know that districts, states, and the federal government all embrace standardized testing and the idea that all students should be prepared for college. And we know that traditionalists criticize the SBAC and PARCC for their lack of rigor and failure to prepare students for college math. So surely the Putnam qualifies a sufficiently rigorous test. And it's true that any high school student with a decent Putnam score is more than prepared for success in a college-level course. (Oh, and before you say that the Putnam will never be given in high school because it's always on a Saturday, remember that the same used to be true of the PSAT.)

Yesterday, I wrote that I've told my students as young as eighth grade about the Putnam exam, and even give them one of the questions as an example. But under no circumstances would I ever advocate forcing all high school students to take even AMC (American Mathematics Competition for high school students), much less Putnam.

Anyway, here is the Putnam question that I want to discuss, Problem B1:

https://artofproblemsolving.com/community/c7h1554577_putnam_2017_b1

As usual, Kent Merryfield is the one who posts the problem at the Art of Problem Solving forum:

Let  and  be distinct lines in the plane. Prove that  and  intersect if and only if, for every real number  and every point  not on  or  there exist points  on  and  on  such that 

Notice that I cut-and-paste the LATEX from the forum, but everything else in my post is ASCII. It's mostly safe to click the above link -- Merryfield gives a proof using vectors, but it's the Geometry proof given by CantonMathGuy at the bottom of the page that I want to discuss here. So as long as you don't scroll down at the link, you can avoid spoilers until we reason this question out. But I will tell you this -- CantonMathGuy's proof will use one of the Common Core transformations.

First of all, we notice the words "if and only if" (abbreviated "iff" last week). As we learned in Lesson 2-5, "if and only if" means that there are two statements to prove:

• If L_1 and L_2 intersect, then for every real number lambda =/= 0 and every point P not on....
• If for every real number lambda =/= 0 and every point P not on..., then L_1 and L_2 intersect.

And we must prove both of these statements in order for the Putnam graders to award us any points.

The next thing we notice here are quantifiers -- "for every" and "there exist." High school Geometry students are usually shielded from having to deal with quantifiers, but they appear in higher-level math all the time.

One analogy often used to help understand quantifiers is "opponent" and "proponent." The proponent is the person trying to prove the theorem, and the opponent is trying to stump the proponent. The idea is that "for every" number or object that the opponent throws out there, "there exists" a response that the proponent can give to complete the proof. For example, we can write Goldbach's Conjecture as:

For every even number greater than 2, there exist two prime numbers whose sum is the even number.

So the opponent's job is to name any even number, and the proponent's job is to name the two primes that add up to it. If the opponent can stump the proponent, then the opponent wins and the conjecture is proved false. If the proponent can always avoid being stumped, then the proponent wins and the conjecture is proved.

And so let's rewrite the forward direction of the statement to be proved in terms of the proponent and the opponent:

• The opponent names intersecting lines L_1 and L_2, a nonzero real number lambda, and a point P not on either line.
• The proponent now names points A_1 on L_1 and A_2 on L_2 such that PA_2 = lambda PA_1.

There are a few more things we must clean up before we begin the proof. First of all, notice that PA_1 and PA_2 are actually vectors. The U of Chicago text uses this notation in Lesson 14-5 -- P is the initial point of each vector. Point A_1 is the terminal point of the first vector, and A_2 is the terminal point of the second vector.

Then lambda is a Greek letter to represent the scalar by which we are multiplying the first vector to obtain the second vector. Scalar multiplication appears in Lesson 14-6 of the U of Chicago text. You might not like using Greek letters, but the most common variable to represent a scalar factor in mathematics is lambda. How about this -- let's use the variable k instead, since Lesson 14-6 actually uses k when defining scalar multiplication.

OK, so let's begin the forward direction of the proof. You, by the way, are the proponent trying to prove the theorem. The opponent has given you the intersecting lines L_1 and L_2, the nonzero real number k, and the point P not on either line. What will you, as the proponent, do next to find the points A_1 and A_2? Keep in mind that I already mentioned a hint earlier -- a Common Core transformation will be used in the proof.

Think about it for a moment -- you've been given a number k and a point P. So what transformation has something to do with a number and a point? That's right -- it's a dilation. So let's consider the dilation with center P and magnitude k. (By the way, how do we know it's not a rotation, since rotations also have centers and magnitudes? Well, our k seems to have nothing to do with degrees, as you would expect of magnitudes of rotations. The Greek letters also give it away -- if an angle were intended, the writers would have named it theta instead of lambda.)

OK, so we have the dilation with center P and magnitude k. What happens to the lines L_1 and L_2 under this dilation? Well, Common Core expects us to know that a line and its dilation image are parallel, and so L_1 | | L_1' and L_2 | | L_2'.

But how are, say, L_1' and L_2 related? You are given that L_1 intersects L_2 and L_1 | | L_1', and so we conclude that L_1' and L_2 also intersect. The proof of this statement is that it follows from either Transitivity of Parallelism (Lesson 3-4 -- if L_1 | | L_1' and L_1' | | L_2 then L_1 | | L_2, which is false) or Playfair (Lesson 13-6 -- if we name the intersection point of L_1 and L_2 as Q, then there are two lines parallel through Q parallel to L_1' and Q -- namely L_1 and L_2 -- where there's only supposed to be one such line).

Anyway, since you proved that L_1' and L_2 intersect, you want to give this intersection point a name, so you give it the name A_2. After all, A_2 is indeed a point on L_2, and your job as proponent is to name a point on L_2.

Now you need to find a point A_1 on L_1. Well, you could repeat the process above and find the point where L_2' and L_1 intersect, but you're going to do something else. The point you named earlier, A_2, is on L_1' as well as L_2. But what is L_1'? It's the dilation image of L_1. In other words, every point on L_1' is the dilation image of a point on L_1. So A_2 must also be the dilation image of a point on L_1 -- and that's the point you name A_1.

So A_1 and A_2 aren't just any points on L_1 and L_2 -- indeed, A_2 is the dilation image of A_1. But how does that help you, as the proponent, win the proof? Let's look at the definition of dilation (or "size change") as given in Lesson 12-2 of the U of Chicago text:

Definition:
Let O be a point and k be a positive real number. For any point P, let S(P) = P' be the point on ray OP with OP' = k * OP. Then S is the size change with center O and magnitude k.

Pay close attention to the location of P' -- it is on ray OP with OP' = k * OP. Now think about what it would mean to take vector OP and multiply it by the scalar k. That's right -- the result is exactly the vector called OP'! So the text could have written:

Definition:
Let O be a point and k be a positive real number. For any point P, choose S(P) = P' such that vector k * OP = vector OP'. Then S is the size change with center O and magnitude k.

Most Geometry texts don't use vectors to define dilations. If vectors are used to define any transformation at all, they are used to define translations, not dilations. But we can clearly see that multiplying a vector by a scalar k is equivalent to dilating a point by magnitude k. Indeed, the fact that the U of Chicago text uses the variable k to refer to both is not a coincidence!

Thus, returning to the original problem, the statement that vector PA_2 = k PA_1 is exactly equivalent to the statement that A_2 is the image of A_1 under a dilation with center P and magnitude k. And guess what -- that's exactly how you selected points A_1 and A_2. You, the proponent, wins the proof.

Well, there's just one problem -- notice that k (or lambda) in the original problem is nonzero -- that is, it could be negative. But dilations in the U of Chicago text are defined mainly for positive k. You don't win until you specify what happens for negative k.

It actually is possible to define dilations for negative magnitudes. These do exactly what they need to do for the proof to work -- map vectors to new vectors pointing in the opposite direction. Indeed, the dilation of magnitude -1 is equivalent to a rotation of 180 degrees with the same center, and a dilation of magnitude -k is the composite of this rotation and a dilation with magnitude k, each with the same center of course.

In the modern Third Edition of the U of Chicago text, vectors appear in Chapter 4, well before dilations in Chapter 12. Thus the new text could use vectors to define dilations, yet it doesn't. On the other hand, at least the new text allows k to be negative, so that the proof works. That text gives a longer definition to accommodate negative magnitudes, but it's easier just to change "positive" to "nonzero" in the above definition, since negative scalar multiplication is already defined.

Definition:
Let O be a point and k be a nonzero real number. For any point P, choose S(P) = P' such that vector k * OP = vector OP'. Then S is the size change with center O and magnitude k.

OK, this wraps up the forward direction of the proof. Don't forget that this is an "if and only if" statement, so we need to prove the converse before we're done with the proof:

If for every real number lambda =/= 0 and every point P not on..., then L_1 and L_2 intersect.

This direction is a bit awkward to think of in terms of proponent and opponent. You might think that we can just reverse the proponent and opponent (so that the opponent finds A_1 and A_2 instead of the proponent). This is sort of, but not quite, the case.

Most mathematicians would use an indirect proof for this direction. To prove that the two lines intersect, we assume the opposite is true -- that the lines are parallel. This is actually what the opponent does -- give two parallel lines L_1 and L_2. The proponent's job in this case is to provide the contradiction. The statement to contradict is that every lambda (or k) and every point P works -- so all the proponent needs is a counterexample, a single point P and value of k that don't work.

OK, so your opponent has already given you two parallel lines L_1 and L_2. How are you going to find the point P and value of k?

Well, let's say you choose a point P in between the two parallel lines. Now you need to find a k that doesn't work -- that is, so that PA_2 can never equal k * PA_1. You can think of this in terms of the opponent again -- as soon as you name the k, the opponent will try as hard as he can to find the A_1 and A_2 that do make PA_2 = k PA_1, so you must find a k that will stump the opponent. (So in a way, the positions have reversed, since you're indeed finding P and k while the opponent is finding the points A_1 and A_2.)

In order to find a k that doesn't work, let's rule out values of k that do work. Since vector PA_2 is a scalar multiple of PA_1, the points P, A_1, and A_2 must be collinear (as scalar multiples point in the same or opposite directions). So we draw any line through P that intersects both L_1 and L_2 and label the points of intersection A_1 and A_2. We can find k such that PA_2 = k PA_1 -- we can't divide vectors, but we divide their lengths, so PA_2 / PA_1 (as lengths) is our k, as long as we choose the correct sign. Since P is between the lines, the vectors point in opposite directions, so k is negative. Of course, you just found a k that works, so you don't give that k to the opponent.

Let's try to find a second value of k to rule out. We draw another line through P that intersects both L_1 and L_2, and this time you label the points of intersection B_1 and B_2. This time, you found another value to rule out, namely PB_2 / PB_1.

But hold on a minute. Notice that we have drawn two triangles here -- PA_1B_1 and PA_2B_2. As soon as we have two triangles in a proof, the first thing you should ask is, how are these two triangles possibly related?

Well, notice that we have two parallel lines L_1 and L_2 cut by a transversal A_1A_2. Thus the alternate interior angles PA_1B_1 and PA_2B_2 are congruent. We also have the two parallel lines cut by a transversal B_1B_2. Thus the alternate interior angles PB_1A_1 and PB_2A_2 are congruent. So by AA Similarity, triangles PA_1B_1 and PA_2B_2.

Since corresponding sides of similar triangles are proportional, PB_2 / PB_1 = PA_2 / PA_1 -- and recall that PA_2 / PA_1 is exactly the k that we found earlier. And so even though we choose two new points, we ended up with the same k.

In fact, the points we choose are arbitrary. This means that no matter what transversal we draw through point P, it will lead to similar triangles and the exact same value of k. And this is great news, because you, the proponent, are hoping to prove that not all values of k work, and we just showed that only one specific value of k works! So all you have to give your opponent is any value of k other than the one that works, and this forces the contradiction that leads to a proponent victory.

Notice that if P is chosen outside the parallel lines instead of inside, the same proof works. The alternate interior angles become corresponding angles instead. In this case, we could also use the Side-Splitting Theorem of Lesson 12-10 to arrive at the same conclusion -- only a single value of k works, and in this case k will be positive.

We actually went beyond what is required -- we only needed one point P and one value of k to win the proof, and we found that for each P, almost any value of k wins the proof. In fact, we don't even need similar triangles -- any P inside the lines requires k to be negative and any P outside the lines requires k to be positive, so just give the opponent a value of k with the wrong sign to win.

It's also possible to make the backward direction of the proof look like the forward direction -- we look at dilations centered at P with magnitude k. In the case where L_1 and L_2 are parallel, with the correct k, the dilation image of L_1 is exactly L_2. All other k's map L_1 to a line parallel to L_2, which is why there's no point of intersection A_2 as in the forward direction. It's because the image of L_1 always intersects L_2 in the forward direction that makes that direction of the proof work.

Let's scroll down to see how CantonMathGuy worded his response (cut-and-paste):

Here's a geometric solution!

If , then selecting  to be on the line midway between them always forces , so the problem holds.

Suppose  and  are not parallel. Let  be the dilation with center  and scale . Since , lines  and intersect at a point . Let . It is clear that , , and .

Here we see that in the reverse direction, CantonMathGuy chooses a point between the lines where lambda (our k) is known to be -1, so we can give the opponent any value of k other than -1. In the forward direction, he comes with the dilation, which he calls h, to make the proof work.

Let's take a brief look at some of the other problems. Here's a link to A1:

https://artofproblemsolving.com/community/c7t310571f7h1554571_putnam_2017_a1

Let  be the smallest set of positive integers such that

a)  is in 
b)  is in  whenever  is in  and
c)  is in  whenever  is in 

Which positive integers are not in 

(The set  is ``smallest" in the sense that  is contained in any other such set.)

This question is worded a bit awkwardly -- indeed, in the thread, even some test takers where confused by the word "whenever." Here, "q whenever p" means the same as "if p then q." Also, some high school students may be confused by the mention of a set S. So let's rewrite this with some adjective (perhaps starting with S) to represent the members of set S. We use the word "sexy" in order to grab the students' attention:

Some positive integers are sexy and some of them aren't:

a) 2 is sexy
b) If n^2 is sexy, then n is sexy.
c) If n is sexy, then (n + 5)^2 is sexy.
d) No other numbers are sexy unless a)-c) force them to be sexy.

In fact, we can write b) even better:

b) If n is sexy, then sqrt(n) is sexy.

Technically, we should write +/- sqrt(n), but this is unnecessary because the problem states that these are positive integers. (And of course if sqrt(n) isn't an integer, rule b) doesn't apply.)

Notice that we just created mathematical world as described by Pappas! We have an undefined term, "sexy," with axioms to generate more sexy numbers. At this point we can have the students prove some theorems in this mathematical world:

Theorem: 7 is sexy.

Proof:
Statements          Reasons
1. 2 is sexy.         1. Rule a)
2. 49 is sexy.       2. Rule c) [1]
3. 7 is sexy.         3. Rule b) [2]

If we add more steps, we can prove that 144 and hence 12 are sexy. In fact, we have the following theorem, which Merryfield calls a "lemma":

Theorem: If n is sexy, then n + 5 is sexy.

Given: n is sexy.
Prove: n + 5 is sexy.

Proof:
Statements               Reasons
1. n is sexy              1. Given
2. (n + 5)^2 is sexy. 2. Rule c) [1]
3. n + 5 is sexy.        3. Rule b) [2]

And so as soon as we have 2 sexy, we also have 7, 12, 17, 22, 27, and so on all sexy. Try having the students prove that 69 is sexy:

Proof:
Statements          Reasons
1. 2 is sexy.         1. Rule a)
2. 49 is sexy.       2. Rule c) [1]
3. 54 is sexy.       3. Lemma [2]
4. 59 is sexy.       4. Lemma [3]
5. 64 is sexy       5. Lemma [4]
6. 69 is sexy.       6. Lemma [5]

Notice that after step 5, we could have applied Rule b) to prove that 8 is sexy as well.

Teachers might want to write the numbers in rows of 5, and then as soon as a number is proved sexy, all numbers below it are automatically sexy. You can read the final proof at the link above, where it's proved that the only numbers that aren't sexy are 1 and the multiples of 5.

Another questions that could be worth mentioning are Problem A4:

A class with  students took a quiz, on which the possible scores were  Each of these scores occurred at least once, and the average score was exactly  Show that the class can be divided into two groups of  students in such a way that the average score for each group was exactly 

There's nothing in this problem that a high school student shouldn't understand. (Last year, the problem I gave my eighth graders was also numbered A4 -- it's a bit eerie that a simple-sounding problem lands in the A4 spot.)

There's one more problem that involves Geometry, Problem B5:

A line in the plane of a triangle  is called an equalizer if it divides  into two regions having equal area and equal perimeter. Find positive integers  with  as small as possible, such that there exists a triangle with side lengths  that has exactly two distinct equalizers.

But this problem leads to some very messy Algebra, and so it's not recommended for high school.

Meanwhile, nearly every year, at least one Putnam problem mentions the date. This year, there are two problems involving the number 2017 -- B2 and B6. I won't post the problems here, but I do point out that both of them depend on the fact that 2017 is prime.

In fact, perhaps the first thing a student should do the night before the Putnam is to look at number theoretic properties of the year number, especially its prime factorization. Two years ago, someone got a problem wrong because they factored 2015 = 5 * 403 and thought that 403 was prime, instead of factoring it as 13 * 31. Well, let me save you some work for next year -- 2018 = 2 * 1009, and 1009 is indeed prime.

By the way, what happened in my dream when I tried to administer the Putnam in high school? Well, it turned into a classroom management nightmare. With twenty minutes left to go in the A Session, one student starts asking, "Why do I have to take the Putnam?" (And of course, there's no good reason -- I expect that most students don't even know where to begin on Problem A1.) It turns into a big argument, and when this is happening, all other students leave the room early with no intention of returning for the B Session.

Of course, for college students, "Why should I take the Putnam?" has a definite answer -- in order to qualify for a monetary prize, including $12K and a Harvard scholarship for the top student. (Indeed, three Putnam fellows have gone on to earn Fields medals.) There's another monetary prize that I want to discuss now -- the prize for solving the Poincare conjecture.

Chapter 14, the final chapter of George Szpiro's Poincare's Prize is called "The Prize." It begins:

"Did I say that the International Congress of Mathematicians in Madrid was the endpoint of the saga? Well, it wasn't quite."

In this chapter, Szpiro writes about the use of money as a reward for solving math problems. Indeed, mathematicians have been winning prizes for centuries:

"Leonhard Euler won no less than twelve prizes from various academies. Joseph-Louis Lagrange won the French prizes in 1764, 1766, 1772 (jointly with Euler), 1774, and 1778."

On the other hand, the French academy offers prizes for solving certain problems, then withdraws the award because no one can solve the problem:

"At this stage the academy again decided it was time to lie low for a while. When the prize was offered in 1894 with a question on differential equations, guess what? No entries."

The idea of offering the Millennium Prize Problems goes back to Harvard professor Arthur Jaffe. So who exactly is this Jaffe?

"He has written more than 160 scientific articles, mainly in quantum field theory; authored or coauthored four books; edited seven more; served as chair of the mathematics department at Harvard, as president of the International Association of Mathematical Physics, as president of the American Mathematical Society; received the Dannie Heinemann Prize in mathematical physics, a medal from the College de France; has been named a member of the National Academy of Sciences."

Szpiro also tells us about the financial backer of the award, Landon T. Clay. (He actually died just four months before this post.) He also is the donor of several archaeology grants, but one such grant may be controversial:

"Nobody accuses Clay himself of having done anything untoward, but on that occasion the savvy investor may have been taken in by grave robbers."

Clay and Jaffe decide to create a mathematical institute, even though it would have to compete with large software companies:

"It would be much more effective, financially and in impact, to create a foundation devoted to mathematics."

Nonetheless, the Clay Mathematics Institute (CMI) is born. Szpiro describes this organization, first formed in 1998:

"Since then, CMI has fulfilled its mission by supporting scholars, funding research projects, organizing summer programs, and publishing research results. But the program for which CMI is best known among the public is the creation of the Millennium Prizes."

A scientific advisory board is formed to choose seven problems which would make their solvers rich:

"The scientific advisory board asked specialists to write up an account for each problem. The recorder for the Poincare Conjecture was John Milnor from SUNY at Stony Btook, who had won a Fields Medal in 1962 for his work on seven-dimensional spheres."

The next issue is to set the amount for each prize. For example, the British publishing house Faber & Faber announced a million-dollar (or should that be million-pound?) reward for the solution of Goldbach's Conjecture (mentioned earlier in this post). The catch is that the reward two years after it is published -- a conjecture that has lasted over 250 years without a proof.

"Jaffe's initial idea was to arrange for a prize fund the increased, albeit slowly, each year. So a problem solved after sixty or more years would yield a substantial reward."

In the end, the prize amount is set at $1 million for each problem. Of course, there's also the matter of a proof that is eventually revealed to be incorrect:

"The assumption in the 'test of time' is that if a proof contains a hole, eventually someone, somewhere, will spot it."

Soon after the prizes are announced in the year 2000 (hence the name "Millennium Prizes"):

"The prizes hit the front page of Le Monde, France's leading newspaper, an Associated Press report was carried out by several hundred U.S. newspapers, and Nature published an editorial."

But Anatoly Vershik, a colleage of Perelman's, thinks that the prizes lead to too much hype:

"As a case in point, he remarks that the Clay Institute had played no role whatsoever in furthering or hastening the solution of any of the seven Millennium Problems."

The worry, of course, is that the other six problems might take centuries to prove:

"After all, the Poincare Conjecture was a hundred years old before Grigori Perelman came along, the Four-Color Problem two hundred when it was proven, Fermat's conjecture three hundred, and Kepler's conjecture four hundred. In December 1999, I asked Robert MacPherson from the Institute for Advanced Study what question he would ask his colleagues if he went to sleep and woke up at the start of the fourth millennium. His answer: 'Has the Riemann conjecture finally been proven?'"

This is likely a reference to the character Fry from Futurama, who likewise is frozen in December 1999 and wakes up at the turn of the fourth millennium.

And so the question is, should Perelman be awarded the million dollars? One requirement is that the proof actually be published, followed by a two-year waiting period:

"If the members of CMI's scientific advisory board decide that Perelman's arXiv submissions of 2002 and 2003 constitute a valid form of publication, then the two-year waiting period has long since elapsed without any errors or gaps having been found."

Furthermore, should the million be awarded solely to Perelman?

"During the past century, hundreds of mathematicians contributed in one way or another to the proof of the Poincare Conjecture, starting with Poincare himself and the early topologists, to those who have reduced the question from topology to a manifold question, to those who adapted it to geometry."

The author adds that "Hamilton even uses a technique first applied by John Nash that actually appeared on a blackboard in the background of the movie A Beautiful Mind." (And you thought that the equations written in movies were fake -- at least until Hidden Figures came around.)

And we're not even sure if Perelman even wants the money. After all, he is mugged in Berkeley when carrying only a few dollars:

"How much more dangerous it would be to carry thousand of rubles around in Russia/ Whatever the reasons, Perelman is again keeping up the suspense."

Szpiro concludes the book as follows:

"It is easy to remain humble in the face of mathematics; so many problems have not been solved. Let's allow the magnificent enterprise to continue."

And we allow math to continue every time a Millennium Problem is solved, a Putnam test-taker receives full marks, or even a Geometry student writes the last step in a proof.

But that doesn't answer the question about the Poincare million-dollar prize. The status of the prize was still unknown at the time Szpiro published his book. Well, let's find out the answer:

http://www.claymath.org/sites/default/files/millenniumprizefull.pdf

The Clay Mathematics Institute (CMI) announces today [March 18th, 2010] that Dr. Grigoriy Perelman of St. Petersburg, Russia, is the recipient of the Millennium Prize for resolution of the Poincaré conjecture.  The citation for the award reads: The Clay Mathematics Institute hereby awards the Millennium Prize for resolution of the Poincaré conjecture to Grigoriy Perelman.

And of course, Perelman never accepts the prize money.

In the end, Szpiro's book is an enjoyable read. If you don't have access to the book, you can read the CMI link above, which summarizes everything mentioned in the book. I undoubtedly left parts out as I tried to describe it here on the blog.

Szpiro takes pains to explain what some of the mathematical terms mean via analogies -- and I posted some, but not all, of these analogies on the blog. One of these words is "tensor" -- I've heard that word mentioned several times before, but I never knew what it meant -- only that it is something similar to a vector. But I knew it wasn't the same as "vector," otherwise why wouldn't mathematicians just say "vector"? Szpiro writes:

"The tensor is a generalization of the notion of scalars, vectors, and matrices. It is an array of numbers that describe physical or geometric quantities."

Ahh, so tensors can have various "ranks," or dimensions. On the other hand, all scalars are dimension zero, vectors are dimension one, and matrices are dimension two. Now I'm finally starting to understand what a tensor is, thanks to Szpiro.

Every time I finish a side-along reading book, I compare it back to our own students. Just as I was confused with what a "tensor" is, our students might be just as confused with "vector." This is likely why Geometry books define dilations without confusing students with vectors -- and the U of Chicago text even defines translations without mentioning vectors. (On the other hand, other texts figure that vectors make translations easier to comprehend.)

It has been said that proof is the currency of mathematics. As we've seen with Poincare and Putnam, this could be literal currency as in thousands or millions of dollars. But actually, nothing is worth anything in math unless it has been proved. Mathematicians from Poincare to Perelman spent 100 years trying to find an elusive proof, Putnam test-takers spend six hours trying to find proofs, and our Geometry students spend minutes trying to find proofs. And for Geometry students, it all starts with the four big triangle theorems -- SSS, SAS, ASA, and AAS.

Lesson 7-2 of the U of Chicago text is called "Triangle Congruence Theorems." As you already know, this is one of the most important lessons in the entire text.

This is what I wrote two years ago about today's lesson:

And so we finally reach Lesson 7-2 of the U of Chicago text, the Triangle Congruence Theorems. I will be able to demonstrate how SSS, SAS, and ASA follow from the definition of congruence in terms of isometries.

Let's start with ASA, since as I said earlier this week, we can use the same proof of ASA directly out of the U of Chicago text. Here is the proof as given in the text:

ASA Congruence Theorem:
If, in two triangles, two angles and the included side of one are congruent to two angles and the included side of the other, then the triangles are congruent.

Proof:
Given AB = DE, Angle A = FDE, and Angle B = FED. Consider the image Triangle A'B'C' of Triangle ABC under an isometry mapping AB onto DE. Triangle A'B'C' and DEF form a figure with two pairs of congruent angles.

Think of reflecting Triangle A'B'C' over line DE. Applying the Side-Switching Theorem to Angle C'DF, the image of Ray A'C' is Ray DF. Applying the Side-Switching Theorem to Angle C'EF, the image of Ray B'C' is Ray EF. This forces the image of C' to be on both Ray DF and Ray EF, and so the image of C' is F. Therefore the image of Triangle A'B'C' is Triangle DEF.

So if originally two angles and the included side are congruent (AB = DE, Angle A = D, Angle B = E) then Triangle ABC can be mapped onto Triangle DEF by an isometry. (First map AB onto DE, then reflect the image of Triangle ABC over the line DE.) Thus, by the definition of congruence, Triangle ABC is congruent to Triangle DEF. QED

Now as we said earlier, the U of Chicago text uses the Isosceles Triangle Theorem to prove SAS, when we instead want to use SAS to prove the Isosceles Triangle Theorem. But as it turns out, we can write a proof of SAS that's not much different from the ASA proof given above. In this proof, we will discuss more in detail how we perform the first isometry -- the one that maps Triangle ABC to the position A'B'C', from which we can perform the final reflection.

SAS Congruence Theorem:
If, in two triangles, two sides and the included angle of one are congruent to two sides and the included angle of the other, then the triangles are congruent.

Proof:
Given AB = DE, AC = DF, and Angle A = FDE. Our first isometry will be to map A onto D. Now we simply reflect A onto D -- that is, the mirror is the perpendicular bisector of AD. This may be a bit trickier to visualize that the translation, but it works. The image A' is D, while the images B' and C' can be anywhere on the plane.

Next, we must map the whole segment A'B' onto our destination, DE. Since A' is already at D, we can perform (just as Euclid did) a rotation centered at D. How many degrees should this rotation be? The answer is that it's exactly the measure of Angle B'DE. Then this rotation maps Ray DB' to Ray DE, and therefore segment DB' to DE as they both have the same length as the original segment AB (that is, we already have a point on Ray DE that's the correct distance from D, and it has the name E). At this point, after the reflection and rotation, the image A" is still D and the image B" is now E, so all we have to do is figure out where C" is.

Now it could be the case that C" is already F -- in which case, we'd already be done. (Euclid erroneously made the assumption that C" is always F.) This is why the U of Chicago text always draws the case where C" is not F, in hopes that one final reflection, over line DE, will map C" to F.

So far, this part of the proof isn't particular to SAS. All of the congruence theorems will begin with this same isometry -- reflect A to D, then rotate DB' to DE. (By the way, it's possible to dispense with the rotation and use only reflections in the proof. Instead of rotating, use the angle bisector of B'DE as the mirror. Then by the Side-Switching Theorem, Ray DB' maps to DE.)

Notice that I have written C" double-primed. This is because C mapped to C' under the first reflection and then C' maps to C" under the rotation (or second reflection). We would have to write a third prime for the final reflection, to show that C'" is F. (The U of Chicago text shows only a single isometry mapping AB to DE, so it uses one fewer prime symbol.) To avoid having to write multiple prime symbols, we will abuse notation and simply refer to the image of Triangle ABC under the isometry mapping AB to DE as Triangle ABC, without any prime symbols. I'm hoping that this will actually be less confusing to the students -- calling the image ABC drives home that the fact that all six parts of both triangles are already known to be congruent, and that it's only the congruence of the parts of ABC and DEF that remains to be proved.

Now here's the part of the proof that's particular to SAS. Instead of using isosceles triangles as in the U of Chicago proof, we notice that just as in the ASA proof, since Angles CAB (which has been moved to CDE) and FDE are given to be congruent, the Side-Switching Theorem once again tells us that the reflection over line DE maps Ray DC to Ray DF, and thus segment DC to DF as it's given that they both have the same length as the original segment AC (that is, we already have a point on Ray DC that's the correct distance from D, and it has the name F), so the image of C is F. Therefore the image of Triangle ABC is Triangle DEF.

So if originally two sides and the included angle are congruent (AB = DE,  AC = DF, Angle A = D) then Triangle ABC can be mapped onto Triangle DEF by an isometry. (First map AB onto DE, then reflect the image of Triangle ABC over the line DE.) Thus, by the definition of congruence, Triangle ABC is congruent to Triangle DEF. QED

When presenting this proof in class, we can start with this SAS proof so that students can see how to perform the opening reflection and rotation. Then when we get to ASA and the other proofs, we can just say "there exists an isometry" mapping ABC to the reflection of DEF, so that the proof only needs to discuss the final reflection.

Now let's go for SSS. I mentioned earlier that the proof in the U of Chicago text creates a kite -- and the properties of kites ultimately go back to isosceles triangles. I wrote that we can avoid the Isosceles Triangle Theorem by using the Converse to the Perpendicular Bisector Theorem instead. Here is the new proof:

SSS Congruence Theorem:
If, in two triangles, three sides of one are congruent to three sides of the other, then the triangles are congruent.

Proof:
Given AB = DE, AC = DF, and BC = EF. We begin, as in the other proofs, by performing the isometry that maps AB to DE. So A is mapped to D, and B is mapped to E, and we are now ready to reflect C over line DE.

We are given that AC (mapped to DC) = DF -- that is, D is equidistant from C and F. Therefore by Converse Perpendicular Bisector, D lies on the perpendicular bisector of CF. We are given that BC (mapped to EC) = EF -- that is, E is equidistant from C and F. Therefore by Converse Perpendicular Bisector, E lies on the perpendicular bisector of CF. So we already know two points on the perpendicular bisector of CF, namely D and E. Since two points determine a line, this tells us that the perpendicular bisector of CF is exactly line DE -- and that's convenient, since DE is exactly the mirror over which we wish to reflect!

So line DE is the perpendicular bisector of CF. Therefore, by the definition of reflection (meaning), C reflected over line DE must be F -- which is exactly what we want to prove. QED

But Lesson 7-2 contains another congruence theorem -- AAS. The U of Chicago, like most texts, prove AAS using ASA plus the Triangle Sum Theorem. I like the following proof of AAS:

AAS Congruence Theorem:
If, in two triangles, two angles and a non-included side of one are congruent respectively to two angles and the corresponding non-included side of the other, then the triangles are congruent.

Proof:
Given AB = DE, Angle A = FDE, Angle C = DFE. We begin, as in the other proofs, by performing the isometry that maps AB to DE. And as in the proofs of ASA and SAS, the Side-Switching Theorem implies that using line DE as a mirror, Ray DC is mapped to Ray DF. So we know that the image of C is somewhere on Ray DF, but we don't know yet that C' is exactly F.

We know that Angle ACB (same as DCE) is mapped to Angle DC'E, and as reflections preserve angle measure, these angles are congruent. And we are given that Angles ACB and DFE are congruent, so this tells us that DC'E and DFE are congruent. But this isn't sufficient to identify the images of any rays, since we don't know whether the vertices of the angles, C' and F, are the same point yet.

So let's try an indirect proof -- assume that C' and F are not the same point. This may be tough to visualize, so try drawing a picture. We already know that C' lies on Ray DF, so we can draw C' to be any point on Ray DF other than F. It doesn't matter whether C' is between D and F or on the opposite side of F from D -- both will lead to the same contradiction.

After we label the two angles known to be congruent, DC'E and DFE, we notice something about the diagram we've drawn. We see that lines C'E and FE are in fact two lines cut by the transversal DF, and the two angles DC'E and DFE turn out to be corresponding angles that are congruent. Thus, by the Corresponding Angles Test, lines C'E and FE are parallel! And so we have two parallel lines that intersect at E, a blatant contradiction. So the assumption that C' is not F must be false, and so C' is exactly F. QED

Like previous indirect proofs involving parallel lines, this can be converted into a direct proof if we use the U of Chicago definition of parallel. Then C'E and FE are parallel lines with E in common, so they are identical line -- that is, C' lies on FE. Then just as in the ASA proof, C' lies on both DF and FE, so C' is exactly F.

I like this proof as it has the same flavor as the SAS, ASA, and SSS proofs. Now I have an alternate proof of HL that avoids AAS and uses only theorems that have been proved previously on the blog so far.

HL Congruence Theorem:
If, in two right triangles, the hypotenuse and a leg of one are congruent to the hypotenuse and a leg of the other, then the two triangles are congruent.

Proof:
Given AB = DE, BC = EF, A and D are right angles. We begin, as in the other proofs, by performing the isometry that maps AB to DE. Now since BC (same as EC) = EF, just as in the proof of SSS, we see that E is equidistant from C and F, so that E lies on the perpendicular bisector of CF.

Now since CAB (same as CDE) and FDE are right angles, line DE is perpendicular to CF. By the Uniqueness of Perpendiculars Theorem, DE is the only line through E perpendicular to CF, so line DE must be that perpendicular bisector that we were discussing earlier. So just as in the proof of SSS, we have by the definition of reflection (meaning), C' is exactly F. QED

[2017 update: Two years ago, I wrote some information about an alternate proof of AAS below. I retain this information for technical interest, but none of it is relevant any longer. We should just stick to the proof of AAS given in the text --  the usual proof with ASA and the Third Angle Theorem. The proofs are the same in the Second and Third Editions of the text, except that SAS gets a new proof in the Third Edition.]

I was wondering whether there's a proof of AAS that avoids TEAI (or the Corresponding Angles Test, which can also be proved as a result of TEAI). As the U of Chicago uses AAS to prove HL and we've already been reversing many of the proofs in the text, I'm wondering whether we might possibly use HL to prove AAS (drawing in altitudes in order to generate right triangles) -- but I wasn't able to find such a proof.

Dr. Randall Holmes, a math professor at Boise State, also tried to find a proof of AAS that avoids the TEAI, but he could not find such a proof. Two years ago, he wrote:

http://math.boisestate.edu/~holmes/math311/M311S13announcements.html

"AAS proof note: I'm convinced that there is no way to prove AAS without using the exterior angle theorem, which makes it less attractive as a test proof (because of the need for cases – but see that I actually handle the cases quite compactly below). By the way, the ASA proof does not need cases, because the application of the Angle Construction Postulate in it does not depend on the position of the new point in the same way the application of the Exterior Angle theorem in the AAS proof does."

We can easily why we need two cases in our proof of AAS using TEAI. Recall that in our above proof of AAS, the image C' could either be on the segment DF or on the other side of F from D. Well in the former case, Angle DC'E is exterior to Triangle C'FE, so by TEAI, Angle DC'E > DFE. And in the latter case, Angle DFE is exterior to that same triangle, so by TEAI, Angle DFE > DC'E. In either case, we have a contradiction since angles DC'E and DFE are known to be congruent.

Now Dr. Holmes isn't merely a geometer -- he's also a set theorist. Set theory ultimately goes back to the mathematician Georg Cantor. (Yes, the same Cantor for whom the Cantor dust is named. But no, he is not one of the mathematicians whose biography is given in Mandelbrot's book.) Now as it turned out, Cantor's original theory led to contradictions (for example, a set containing all sets was problematic for Cantor). Ever since then, set theorists have been trying to find new theories that avoid the contradiction of the Cantor's theory.

Now here's where Holmes comes in -- two years ago, he completed a proof that an alternate set theory, called New Foundations, allows for a set of all sets without contradictions. This theory is complicated -- recall that I once called the Axiom of Choice the set theorists' Parallel Postulate. Well, the Axiom of Choice is not even compatible with New Foundations.

Thus Holmes is undoubtedly an expert at proofs and determining which theorems can be proved using which axioms or postulates. So if someone like Holmes is unable to come up with a proof of AAS without using TEAI (or a theorem derived from TEAI), then who am I even to try?

And so today I post worksheets for SAS, ASA, and SSS, but not AAS yet. We'll get to HL next week since it's not until Lesson 7-5 of the U of Chicago text. I'm still holding out hope that I can find a proof of AAS from HL by the time we get to Lesson 7-5 (but considering what Holmes wrote above, don't count on it).

Here are the worksheets for today. I've actually written a worksheet for ASA last year, but I couldn't post it due to the computer problems I had last Thanksgiving. I'd written it using the prime-notation from the U of Chicago text, where we begin with some isometry mapping Triangle ABC to A'B'C' as we prepare for the final reflection. The new worksheets for SAS and SSS that I created this year do not refer to Triangle A'B'C' -- instead we abuse the name ABC again.