# A Perfect Problem: Difference of Squares

This is a long post. The first half was written by Benjamin Dickman, who shared the problem with me. The second half was written by me, Michael Pershan. Enjoy these two different maps of the same mathematical terrain.

Part One: Benjamin’s Writeup

Part Two: Michael’s Writeup

[BD]

Let us use the word ‘problem’ to refer to a question for which the method of solution is unknown at the outset. Problems can be viewed on a continuum that ranges from “trivial” to “intractable,” but locating a problem is a function of the individual or group trying to solve it and the resources to which they have access. There is, of course, no single perfect problem for all; but, here is a proposal for one of many perfect problems for some.

There is a wonderful problem for secondary school students looking to be stretched that asks which numbers can be written as the difference of two squares; here, we are roughly in the area of mathematics referred to as “number theory,” and are speaking of numbers that are non-negative integers. It is deeply unfortunate that number theory is not an area of mathematics taught more in our secondary schools across the world; as a result, there are many examples of number theoretical problems that can be effectively posed for students, but for which the techniques or strategies involved are not yet familiar.

Here is a proposed method – in other words, a spoiler – for characterizing the numbers that can be written as a difference of squares: Observe a2 – b2 = (a-b)(a+b); if a and b have the same parity, then the factors a-b and a+b are both even, which means their product is a multiple of 4. If a and b are of different parity, then the factors a-b and a+b are both odd, which means their product is odd. So, a necessary criterion for a number to be expressed as the difference of squares is that it is either a multiple of 4 or odd; it turns out that this is a sufficient criterion, too, as established by the following two identities:

4n = (n+1)2 – (n-1)2 and 2n+1 = (n+1)2 – n2

Uncovering this result is already an opportunity for exciting mathematical exploration with students; their thinking is sure to proceed in a manner much less linear than the description above.

Whenever a problem is solved, there are several options around how to move forward. Here are three such possibilities:

1. Abandon the problem and move on to an unrelated one;
2. Try to derive the solution in a new way; or
3. Try to solve a related problem that is a bit more difficult.

Rich mathematics may (or may not) be uncovered through any of the aforementioned choices, but we will focus here on the third option. (There are slightly different ways of deriving the solution above that will be almost certainly unfamiliar to students; for example, we can observe that a number squared is always 0 or 1 modulo 4; so, the difference of two squares is necessarily odd or a multiple of 4, and this sort of phrasing allows one to bring in topics that are otherwise unseen in K-12 school mathematics: modular arithmetic, quadratic residues, and so forth.)

Our related problem is as follows: Given a nonnegative integer n, in how many ways can it be written as a difference of squares?

This can be viewed as a direct generalization of the earlier problem: if we know how to tell when the answer is “zero ways,” then we know which numbers cannot be written as a difference of squares.

One of the classical strategies for mathematical problem posing is to start with small cases; however, it is often presumed that “small” refers to magnitude (absolute value) and that the ordering from smaller to bigger proceeds additively. That is, one might try looking at 0, then 1, then 2, then 3, etc. But, for a problem whose underlying structure is multiplicative – for example, a problem that might be more easily expressed in the language of factors or factorizations – this additive procession can obfuscate important patterns.

For our problem, we know that the only numbers that can be expressed as a difference of squares are odd numbers and multiples of 4; so, let us begin by investigating the former and see where it leads.

If we have an odd number expressed as a2 – b2, then we also have a factor pair for that number: a-b and a+b. Indeed, any factor pair for an odd number can be written in this manner, for two factors of an odd number must both be odd, and this will mean that their average is a whole number, from which we can adjust up and down by the same amount to recover a representation in the a-b and a+b form. Specifically, we use a as the factor pair’s average and adjust by b. This all becomes more clear by way of example.

Consider the odd number 15, which has factor pairs (1, 15) and (3, 5). For the first factor pair, we find the average of 1 and 15 to be 8, and note that 1 = 8-7 and 15 = 8+7. As a result, we can express 15 as 1(15) = (8-7)(8+7) = 82 – 72. Similarly, we can look to the second factor pair and find the average of 3 and 5 to be 4, and note that 3 = 4-1 and 5 = 4+1. As a result, we can express 15 as 3(5) = (4-1)(4+1) = 42 – 12.

We have now established a matching between two representations of 15: the first representation is as a specific difference of squares, and the second representation is as a specific factor pair. The number of factor pairs is usually equal to half the number of factors; the only exception is if the number of factors is odd, which occurs precisely when we are dealing with a perfect square. As we deal with nonnegative (rather than positive) integers, let us establish in that setting that we will continue to use the number of factor pairs as the number of ways to express our number as a difference of squares; so, we will take the total number of factors, add 1, then divide by 2 for our result. Again, let us clarify by way of example.

Consider the odd square 9, which has factor pairs (1, 9) and (3, 3). As in the example with 15, we use these factor pairs to produce the following representations of 9 as a difference of squares: 9 = (5-4)(5+4) = 52 – 42 and 9 = (3-0)(3+0) = 32 – 02. Note that, if we were to adhere to positive integers only, we would not be able to use zero in our latter representation. The result of this is still that the number of representations, 2, is equal to our number of factor pairs; but, the number of factors is 3, so it is not quite right to say we have 3/2 representations. Instead, we add 1 to the number of factors, thereby double-counting the factor 3, which gives us 4 total factors (counted with multiplicity among factor pairs) and halving this gives us the desired result: 2 ways to represent the odd square 9 as a difference of squares.

As we segue to multiples of 4, we find that matters are slightly more complicated. For an illustrative example, consider that of 8: its factor pairs are (1, 8) and (2, 4). The latter factor pair generates a difference of squares: 2(4) = (3-1)(3+1) = 32 – 12. Unfortunately, matters go somewhat awry with the former factor pair: the average of 1 and 8 is 4.5; it is true, numerically, that 1(8) = (4.5-3.5)(4.5+3.5) = 4.52 – 3.52; however, we have decided only to use nonnegative integers, which means that this difference of squares is inadmissible for our present purpose.

The issue at hand for the above-described example is that 1 and 8 have different parity; as a result, their average is a non-integer. To resolve this, we need to ensure that every factor pair for the multiples of 4 has two factors with the same parity. As the product is even, this means, in particular, that each of the factors needs to be even; so, we propose the following resolution: Given a number n = 4m, factor out the 4, which is equal to 22, and consider all of the factor pairs for m. Next, we modify every factor pair by multiplying each factor by 2; as we double each of the factors, we end up with 4m as the product, which is equal to our starting number of n. Let us illustrate matters again by way of example.

Consider 60, which is an even multiple of 4. Let us now factor out a 4, which leaves us with the number 15 to consider. We saw earlier that 15 has factor pairs (1, 15) and (3, 5). We can now modify these pairs by doubling the factors in each to yield (2, 30) and (6, 10); these now give us all of the factor pairs with the same parity for 60, which means we can express 60 as a difference of squares using them: 2 and 30 have an average of 16, which leads to the representation 2(30) = (16-14)(16+14) = 162 – 142; similarly, 6 and 10 have an average of 8, which leads to the representation 6(10) = (8-2)(8+2) = 82 – 22.

The result of this line of thinking is that when n is a multiple of 4, the number of representations of n as a difference of squares is the number of factor pairs for n/4; as was the case for the odds, the number of factor pairs is usually half the number of factors, but in the case of a perfect square we would need to add 1 to the number of factors to count them with multiplicity among the various factor pairs. One more example should do the trick in clarifying this matter.

Consider 36, which is an even multiple of 4 and a perfect square. We can divide it by 4 to get 9, which we saw earlier yields the factor pairs (1, 9) and (3, 3). Multiplying each factor by 2, we arrive at (2, 18) and (6, 6). Respectively, these yield 102 – 82 and 62 – 02 as the two ways in which 36 can be represented as a difference of squares. Just as occurred with our odd square case examined above, we are using the number of factor pairs (here, for 36/4), but this is slightly different from the number of factors: there are only three factors across the relevant factor pairs, but we count one of them (the 6) with multiplicity as it appears twice in the pair (6, 6). As a result, we end up with 36/4 = 9, which has 3 factors; adding 1, we get 4 factors; dividing 4 by 2, we get our answer: there are two ways to represent 36 as a difference of squares.

If we decide to summarize the above thinking succinctly, then we can use the ceiling function (rounding, if necessary, to the nearest integer greater than or equal to its input) for our final result. Defining d(n) to be the number of divisors, or factors, of the natural number n, and S(n) to be the number of ways in which n can be represented as a difference of nonnegative squares, we have the following:

If n is odd, then S(n) = ceil(d(n)/2);

If n is even but not a multiple of 4, then S(n) = 0;

If n is even and a multiple of 4, then S(n) = ceil(d(n/4)/2)

Finally, we recall that the number of factors can be computed if we know a natural number’s prime factorization. In particular, if we write n as a product of the primes pk raised to the respective powers of ak, then the number of factors is the product of (ak + 1) across all k. We close out with one more example.

The number 180 is an even multiple of 4; so, S(180) = ceil(d(180/4)/2). But, what is d(180/4)? Since 180/4 = 45, and 45 has prime factorization 3251, we have that its number of factors is equal to (2+1)(1+1) = 3(2) = 6; so, we find d(180/4)/2 = 6/2 = 3, and ceil(3) = 3. This tells us that the number of representations of 180 as a difference of squares is 3. Indeed, we can verify this by listing them out exhaustively:

180 = 2(90) = (46-44)(46+44) = 462 – 442;

180 = 6(30) = (18-12)(18+12) = 182 – 122; and

180 = 10(18) = (14-4)(14+4) = 142 – 42

Q. E. D.

[MP]

Here is a table that is worth spending some time mulling over.

Here’s what it’s all about: differences of squares.

Given a number, can you tell whether it’s possible to write that number as a difference of squares? Is it possible to characterize all the numbers that are possible to write as a difference of squares? And is there a systematic way to tell how many ways a given number can be written as a difference of squares? An algorithm? A formula?

Let’s tackle these questions in two parts:

• Given a number, can you tell whether it’s possible to write it as a difference of squares, at all?
• If a number can be written as a difference of squares, how many different ways are there to do it?

I.

To start the first question, let’s note that every odd number can be written as a difference of squares. This is due to a wonderful property of squares — they can be decomposed into a sum of odd numbers. Every odd number can be seen as the difference between a large square and some inner, removed square.

That’s not really an explanation as much as restating the statement…ah, well. Here’s a picture:

So, let’s start checking out the small even numbers. Can they be written as a difference of squares?

Starting with the smallest, 2 can definitely not be written as a difference of squares. The smallest difference of squares is 22 – 12 = 3, so 2 is a no-go.

How about 4? That’s a no. (32 – 22 = 5.) How about 6? Also a no-go.

But, wait! 8 works: 32 – 12 = 8.

So…why is that? Why can some even numbers be written as a difference of squares, while others cannot?

Any difference of squares can be written as the product of two numbers: a2 – b2 = (a-b)(a+b). This factoring move can help explain what’s going wrong with so many of these even numbers.

If 6 were to have a representation, then 6 = a2 – b2 = (a-b)(a + b). But there are only so many ways to write 6 as the product of two factors. To make matters worse, the only ways to factor 6 involve one even and one odd factor. To see why this is a problem, note that while 6 = 3 x 2, this couldn’t produce a difference of squares:

a + b = 3

a – b = 2

2a = 5

a = 2.5

b = 0.5

And while it is true that 2.52 – 0.52 = 6, we were only looking for whole numbers.

The issue, then, is that some even numbers can be factored only into pairs of numbers where one is even and the other odd, i.e. of different parity. This explains why 8 works: 2 x 4 = 8, and setting a + b = 4 and a – b = 2 results in (3 + 1)(3 – 1) = 32 – 12 = 8.

As long as the prime factorization of an even number N has just one factor of 2 (as in e.g. 6, 14, 42, 30) then it can only ever be factored into an even and odd factor. That will never work.

As long as your even number is at least divisible by 4, it will always be possible to find at least one solution:

2n(2m + 1)

2(2nm + 2n-1)

a + b = 2nm + 2n-1

a – b = 2

2a = 2nm +  2n-1+ 2

a = 2n-1m +  2n-2+ 1

b = 2n-1m +  2n-2– 1

Both a and b are integers.

There is one other issue to worry about, and that’s 4 itself. 4 = 2 x 2 = (a + b)(a – b) demands that a = 2 and b = 0. Should we count that? It is true that 22 – 02 = 4. It’s not so interesting, which argues in favor of tossing it out of consideration. But sometimes these sort of uninteresting cases can help simplify formulas and generalizations.

Let’s keep an open mind, for now, as to whether we’d rather deal with differences of positive squares or might expand our focus (slightly) to include non-negative squares.

II.

Let’s start with odd numbers. Every odd number can be represented as a difference of squares. But how many representations are there for each odd number?

Consider numbers that are the products of primes, like 15 and 21. They can be all represented in two different ways as differences of squares.

E.g. 15 = (8 + 7)(8 – 7) = (4 + 1)(4 – 1)

E.g. 21 = (11 + 10)(11 – 10) = (5 + 2)(5 – 2)

Then again, this comes as no surprise. If N = pq for p and q both prime, the only factor pairs are 1 x pq and p x q. All the factors are odd, so there are no parity problems — they all produce OK differences of squares.

Not much different for numbers like 147 or 75, which are a product of a prime and a square of a prime:

E.g. 147 = (74 + 73)(74 – 73) = (26 + 23)(26 – 23) = (14 + 7)(14 – 7)

E.g. 75 = (38 + 37)(38 – 37) = (15 + 10)(15 – 10) = (10 + 5)(10 – 5)

All of this still makes sense — 147 and 75 have 6 factors, all odd. That leads to 3 factor pairs, all which work for differences of squares.

In other words, all we’re doing is counting factor pairs, i.e. counting factors and dividing by 2.

E.g. 225 = 32 * 52= (113 + 112)(113 – 112) = (39 +36)(39 – 36) = (25 + 20)(25 – 20)

= (17 + 8)(17 – 8) = (15 + 0)(15 – 0)

This makes sense for 225, which has 9 factors but (including its square root) 5 factor pairs; 9/2 = 4.5, round that up and you get 5.

There is a much better-known number theory function that counts divisors, and it’s multiplicative:

If m and n are relatively prime, divisors(mn) = divisors(m)divisors(n)

So, if you have the prime factorization of a number and its odd, no big deal, you can find out how many ways it can be represented as a difference of squares, no trouble:

E.g. p2 * q8 * r3 has 3 * 9 * 4 = 108 divisors, and can therefore be represented in 54 different ways as a difference of squares.

As a formula, for odd N, N can be described as a difference of squares in ceil[divisors(N)/2] ways.

III.

Even numbers

Now, how do we deal with even numbers? Meaning, those that are divisible by 4. (If they’re not divisible by 4, then they can never be expressed as differences of squares.)

When you have any even number, there are always parity problems you have when it comes to making a difference of squares:

E.g. for 24 = 23 * 3

1 x 24

2 x 12

3 x 8

4 x 6

E.g. for 80 = 24 * 5

1 x 80

2 x 40

4 x 20

5 x 16

8 x 10

E.g. for 84 = 22 * 3 * 7

1 x 84

2 x 42

3 x 28

4 x 21

6 x 14

7 x 12

Compare, in particular, the results for 21 and 84:

There are three times as many factor pairs you get from multiplying 21 by 22. But, of course, 4 of them result in mismatched parity.

Consider one other case, before we head towards a formula: 22 * 32 * 5 * 7. Let’s reason:

• 32 * 5 * 7 should have 3*2*2 = 12 factors and 6 factor pairs, all of which work for differences of squares
• 22 times 32 * 5 * 7 should therefore have 3*12 factors and 18 factor pairs
• Some of these will have mismatched parity, though. We’ll have to toss out all the pairs with an odd factor; as we’ve said, there are 12 odd factors.
• That means 18 – 12 = 6 factor pairs, and we are left with what we started with.

Generalizing, this means that for 2a * N, where N is odd:

• N will have divisors(N) factors (all odd) determined by its prime factorization, and ceil[divisors(N)/2] factor pairs
• 2a N should therefore have ceil[(a+1)divisors(N)/2] factor pairs
• But you have to throw out divisors(N) of those factor pairs, since they contain an odd factor.

That leaves as the number of ways to represent this even number as a difference of squares:

ds(2aN) =  ceil[(a+1)divisors(N)/2] – divisors(N)

This formula is certainly ugly, but it works for the cases above, plus a few more:

IV.

Can we extend this formula in a meaningful way to even numbers that aren’t divisible by 4?

42 = 2 x 3 x 7, ceil[2 x 2] = 4 = 0.

30 = 2 x 3 x 5, ceil[2 x 2] – 4 = 0

18 = 2 x 32, ceil[2 x 1.5] – 3 = 0

Yes, I think so!

For 21N, ceil[2 x divisors(N)/2] – divisors(N) = ceil[divisors(N)] – divisors(N)  = 0

Can we extend this formula to odd numbers, that aren’t even divisible by 2? Well, not really:

For 20N, ceil[1 x divisors(N)/2] – divisors(N)

But we can patch it up. Rather than subtracting divisors(N), let’s subtract badDivisors(N), which are the divisors of N that wouldn’t work for difference of squares tally. Of course, for odd numbers there are no bad divisors, so badDivisors(N) = 0 for ever odd.

Here is our ur-formula, then:

For odd N badDivisors(N) = 0 so the formula simplifies to:

ceil[(0+1)divisors(N)/2] – 0 = ceil[divisors(N)/2]

For 21N, this simplifies to:

ceil[2  x divisors(N)/2] – divisors(N) = 0

Note that for the ceiling function, ceil[x + n] = ceil[x] + n where n is an integer. We can use this to prove something for another special case.

For 22N, our formula simplifies to:

ceil[3 x divisors(N)/2] – divisors(N)

= ceil[divisors(N)/2 + divisors(N)] – divisors(N)

=ceil[divisors(N)/2] + divisors(N) – divisors(N)

= ceil[divisors(N)/2]

So multiplying an odd number by 4 does not alter the number of ways it can be represented as a difference of squares.

More generally, for 22mN:

ceil[(2m + 1) x divisors(N)/2] – divisors(N) =

= ceil[(divisors(N)/2 + 2m x divisors(N)/2] – divisors(N)

=ceil(divisors(N)/2) + (m – 1)divisors(N)

Though maybe it makes more sense to consider the patterns of growth in two separate cases — the case of square and non-square odd Ns.

# Geometry Journal #4

Back at it again, #16:

At first, I just drew the diagram and stared at it for a while. I tried to mark congruent sides and angles. I didn’t really get anywhere.

Then, I thought that it would be a good idea to play with parallels. This was largely because of two things:

• I knew those midpoints were in there, and midpoints sometimes create parallel lines
• Because of the previous problem I worked on, I was thinking about how lines that are perpendicular to a line are also perpendicular to all of its parallels. I thought that would be good to play with again.

You can see my first sketch, where I was mostly staring at the diagram and playing with angles. Then I added another line, thinking that it would probably create congruent triangles.

It did!

This happens to be the solution the book names as well:

This really is connected to the previous problem I worked on. It pulled off the exact same “add a parallel and get yourself a free right angle” move. It was immensely satisfying to be able to see the connection between these two very different problems.

These connections are so important for learning, and so difficult to notice on your own. This is one reason why it is important to look back on solutions and compare them when self-studying — while working on the problem for the first time, even if you get the solution, you won’t necessarily see what links the solutions of different problems together.

I like using my blog as a journal, but it seems like it would be so much fun to find a way to dynamically link these solutions as I see them. A wiki? A static website that I edit over time? I don’t know, but it’s still fun to try problems and see the connections.

I’m also feeling motivated to review some of the solutions from problems I have already studied. Maybe I’ll make flashcards in Anki for them? It would be cool to be able to really think about the connections, and it’s hard to do that without a bunch of solutions in mind. If I were taking a class, at some point the teacher might ask us to study for an exam — that’s the sort of thing I’m thinking would be a useful addition to my learning mix.

# The most interesting parts of “Respect: The Life of Aretha Franklin”

There is an endless number of great soul singers who marched through the Franklin household. Smokey Robinson was a friend of Aretha’s brother, Cecil:

“Cecil and I were kids when we met,” he told me. “We grew up on the same love of music — not just gospel, but jazz. The first great voice that influenced me was Sarah Vaughan. I don’t think Cecil and I were ten when we started digging progressive jazz.”

Aretha’s father, C.L. Franklin, was a famous preacher at the New Bethel Baptist Church at 4210 Hastings Street, in Detroit. His most famous sermon was “As The Eagle Stirreth,” a bestselling sermon recording:

“Hastings Street is ground zero for the Aretha Franklin story. Her father’s New Bethel Baptist Chruch was at 4210 Hastings, steps from the heart of the black entertainment district. It was the point where Saturday night merged into Sunday morning and sin met salvation at the crossroads of African American musical culture … Was it the grinding grooves of the club that got into the church, or was it the sensuous beat of the church that got into the club? Did C.L. Franklin get his blues cry from Muddy Waters the same way Bobby Bland borrowed his blues cry from C.L.?”

When I first visited Detroit from Chicago,” said Buddy Guy, “it was later in the fifties. I had to see two people. The first was Reverend C.L. Franklin, ’cause B.B. had told me he could preach better than Howlin’ Wolf could sing. B. was right. … Gospel music made folks happy. Blues made folks lose their blues. I didn’t see that much difference between the two, even if preachers did claim it was the difference between Jesus and the devil. B.B. King loved C.L. Franklin because he didn’t say that. He didn’t pit one against the other. He said all good music came from God.”

Reverend James Cleveland was one of my favorite coming into this book. His piano playing is so heavy. The author asks him “which came first — the spirituals or the blues?”

“Aretha’s father would laugh at that question,” said Reverand Cleveland, “because he knew there was no answer. It’s a riddle that can’t be solved. You could say that the spirituals came first, but if you broke it down further you could also say that the field shouts came before the spirituals. How do we know whether someone out there was picking cotton didn’t first start moaning about how tired he was, or about how much he wanted a woman? Then maybe a God-fearing woman heard that song and switched it up to where she was praying for God to save her. The fleshy needs and the godly needs are very close. We’re likely to use music to call out both those needs because they’re both so basic. Which comes first? You tell me.

Aretha’s first husband took control of her career and beat her violently. Aretha’s peers call him a “gentleman pimp.”

“You can’t understand the music culture of Detroit in the early sixties,” said R&B singer Bettye LaVette, who emerged from that culture, “without understanding the role of the pimp. Pimps and producers were often the same people. The sensibility was the same — get women working for you; get women to make you money. We demonize pimps now, but back then they were looked up to by men and sought out by women. They had power. They knew how to survive the ghetto and go beyond the ghetto. Some of my best men friends were pimps. Some of the women I admired most were working for them — classy, sophisticated, beautifully dressed women. I didn’t have what it took to be a high-class prostitute of the kind that the best pimps like to parade, but as a singer, I was certainly pimped by certain producers — and glad to be.”

“Back then, women were powerless. If we wanted to get ahead in show business, we had to operate in the system. The greatest example of that system was probably Motown, where Berry Gordy’s first wife, Raynoma Singleton, claimed that Berry himself had pimped women. He wasn’t good with whores, but he was great with singers. The parallel is strong.”

Aretha got her first recording contract with Columbia, where she released a series of eclectic, well-reviewed albums that didn’t yield any hits. She showed an acute desire to record showtunes and various standards.

“Al Jolson had sung “Rock-a-Bye with a Dixie Melody” in blackface in the twenties. Later it was recorded by Sammy Davis Jr., Judy Garland and Jerry Lewis. Though an essential American song, it seemed a strange choice for Aretha, especially at the start of the civil rights movement. Cecil [her brother and longtime manager] explained to me C. L. Franklin’s love of Al Jolson and his reason for urging Aretha to include the tune.”

“My father told me how Jolson harbored great affection for black people,” said Cecil. “His entire blackface act was a way of paying tribute to our musical genius. Dad knew the history of American entertainment and had read how Jolson had hired black writers and helped bolster the career of Cab Calloway. We forget now, but back in the day, Al Jolson and black people had a mutual-admiration society.”

“In spite of Cecil’s spirited defense of Aretha’s inclusion of the song, it’s difficult for me to listen to her version without cringing. Although her vocal is enthusiastic, the strings feel anemic, the horn chart cheesy, and the rock-’em-sock’em finale forced and false.”

Aretha had difficulty reaching widespread fame while recording jazz and standards for Columbia.

“While working with Etta [James] on her book, I asked her why she thought her string-heavy jazzy standard turned into a smash while, in that same year, Aretha couldn’t hit with a bluesy standard like “That Lucky Old Sun”

“The answer’s easy,” said Etta. “Aretha sang the shit outta those standards — just as good if not better than me. But Columbia didn’t know how to reach black listeners, and my company, Chess, did. Leonard Chess had a genius for feeling out the black community. Jerry Wexler was the same. They were white Jews who would never use the word [n-word], but they us [] better than we knew ourselves. Columbia didn’t have no one like that. They had John Hammon, but he was like a college professor up there in the ivory tower. He wasn’t street like Chess or Wexler.”

Not really sure what this was about. Apparently Jerry Wexler was almost … killed?

“I’d just come back from a music-industry convention in Miami, where I was to accept an award on Aretha’s behalf,” Wexler remembered. “In Floriday, what I thought would be a pleasure turned into a nightmare. Gangster elements had taken over the industry’s black-power movement. During the banquet, King Curtis came up to me and said, ‘We’re getting you outta here. You’ve been marked,’ King escorted me out to safety. Later I was hung in effigy. Phil Walden, Otis’s white manager, received death threats. Marshall Sehorn, a white promo man, was pistol-whipped. It was some scary shit.”

By the way, King Curtis seems like a great guy. He’s certainly a great musician.

Apparently Aretha’s live performances were hit or miss. But there are some really special performances. Here she is with her childhood friend Smokey:

By the way, is there anything more devastating in popular culture than the list of soul and R&B singers who died far, far too young under tragic circumstances? Donny Hathaway, Sam Cooke, Sam Cooke’s brother, Marvin Gaye, Otis Redding, Billy Holliday, Nat King Cole. Aretha was personal friends with nearly all these people, childhood friends with half of them. Her mother died when she was a child.

Two of my favorite piano players, Richard Tee and Billy Preston, also died too young. Here are Aretha and Billy playing together:

I haven’t quoted at length any of the emotional drama from the book, which I found sad and also tedious. The short story is that Aretha was a difficult person, prone to depression and delusion. She was a teen mother twice over, and largely left her children back in Detroit while pursuing her career in New York City. She canceled gigs on whim, had extravagant tastes, and created relationships for the press out of whole cloth. She comes across as a brilliant artist who was committed to staying relevant and having hits, even into her sixties. Those around her frequently express regret that she didn’t age more gracefully as an artist — all except Ray Charles, who completely identifies with her desire to stay on the charts.

What I’ll remember from this book is the picture of Detroit as a place that was just overflowing with musical innovation in every direction. It was the epicenter of something amazing happening in popular music, and it all seemed to be connected to Aretha’s father’s church. Everyone seemed to know Aretha.

There was another passage that struck me, though I can’t find it now. There was a bit where someone mentioned that a lot of the folk and rock music on the radio featured more cryptic lyrics, whereas soul and R&B tended to have more straightforward story songs. Country is like that too, and so was early Rock and Roll. I don’t know what to make of that, but it’s interesting.

# Geometry Journal #3

Problem 1-14 from Challenging Problems in Geometry:

This took me a while to understand, but once I had a sketch it didn’t take me very long to use isosceles triangles to show that there were a bunch of congruent triangles in the diagram. I did this on paper, but thought it would be nice to put this into Geogebra.

The most satisfying diagram came when I started with a square, ala Challenge 1:

From attempting this problem and then studying the solution I didn’t feel as if I learned any new techniques. But this result is really awesome, and it’s one that I want to remember. It reminds me of midpoint quadrilaterals — the quadrilaterals formed by connecting the midpoints of a 4-gon. It’s pretty.

Ooh! Just looking at this now, it occurs to me that I should be asking “what is the ratio of areas between the original square and the trisector-square?”

I just tried to calculate it … I got $\frac{2 - \sqrt{3}}{3}$, or about 8%. Seems to check out.

After the last few problems took a lot of energy, it was nice to feel this one come a bit easily.

***

The next problem wasn’t an immense struggle either, mostly because I didn’t have paper around so I just skipped to the hint.

Like I said, I didn’t have paper and I wasn’t sure what to do, so I skipped to the hint.

It’s a nice hint:

These theorems say:

• If a line is perpendicular to another line m, it’s also perpendicular to any lines m is parallel to.
• Side splitter: connect two midpoints of a triangle and you get a line parallel to the other base

I very much enjoyed taking those hints and figuring out how to use them, even though it didn’t take too long.

The result isn’t so interesting to me, but I think there is a problem-solving takeaway:

Takeaway(s): If a diagram has a lot of midpoints, there’s also a lot of parallel lines.

If you’re trying to prove that an angle is 90 degrees, try to start with all the other 90 degree angles in the diagram. See if they lead to other right angles.

***

I’ve been reading a lot about self-explanation lately. This is a notion emerging from cognitive science research that suggests an important part of learning is explaining things for yourself — making inferences beyond what is explicitly there. This is often seen as crucial for learning from worked examples, i.e. solutions. This is something I am doing as part of this geometry project.

In one of his papers (let me know if you want the reference) Renkl distinguishes between different types of self-explanation. One that he thinks is particularly relevant for learning math is principled explanations. In other words, connecting specific examples to general principles. It makes sense that this would be useful — it means you can use this specific example to help with other problems!

The thing that I’m realizing while studying geometry is that without a teacher’s help, I’m basically guessing as to what the general principles are. I think my “takeaways” above could be useful for tackling other problems, but I don’t really know for sure. This is relatively new to me! I won’t know for sure until I find a chance to use the techniques in another problem.

I guess this is just another way that learning from a teacher is easier than learning on one’s own.

# Geometry Journal, #2

The problem:

My process:

I put a rectangle into Geogebra. It actually took me a few tries to really nail down what the problem was saying, but once I did I started dragging the points around. I noticed that the angle made by the two perpendiculars was constant also. Cool! After noodling around with the diagram I understand why:

I started drawing a lot of lines. This is another one of those diagrams that has a bajillion similar triangles. I was playing with this while my kids were going to sleep, so it was about 30 minutes of lazy drawing and redrawing. I found a lot of congruent triangles, some I was pretty sure would help me, but I couldn’t make them work. Here was my favorite congruent triangle pair:

After skipping to the solution of the previous problem (and seeing Benjamin Leis’ solution) I was feeling motivated to find a solution to this one before checking the solution.

I grew frustrated with the computer and (following my takeaway from the previous post, actually!) I took out some paper and started writing equations. I was thinking about how I would get an expression for EF + EG (the two perpendiculars) and I had the idea that you could do that using similar triangles.

I took the two triangles above and set them similar to each other. Awful quality I know, sorry:

Here is that picture, cleaned up a bit:

The point is that if those triangles are similar, then those perpendiculars are in ratio also. Some fiddling tells us that if that’s so, the sum of the perpendiculars is equal to:

$\frac{EF*l}{x}$

Is EF a function of x, so that the x’s cancel out? If so, we would be done. Call that yellow angle in the top right corner theta.

$\Large \sin{\theta} = \frac{EF}{x}$

$\Large x \dot \sin{\theta} = EF$

And that works. The sum of perpendiculars is equal to $l \dot \sin{\theta}$.

***

Here is the book’s solution:

Here’s the point:

It’s a nice solution. When I was looking for a constant I was trying to do something like this — slide a segment over using a rectangle. The problem was I couldn’t get the right structure when I was trying this, I always ended up stuck with a trapezoid.

The big strategy here is “when you’re looking to prove something is constant, you can try to construct the thing out of both pieces.” I did try that first, but ended up stumped so I moved to a different approach.

Otherwise, I’m not sure what else to take away from this solution. The move of using an isosceles triangle to prove both sides are congruent is sweet.

# Geometry Journal, #1

The problem:

I spent about 15 minutes noodling around. I tried adding different lines. I tried writing ratios, because there are so many similar triangles here. I had trouble using the digital annotation tools I was using, which made it harder to mark congruent angles. Eventually I started thinking about how I would get EF + DG as an expression at all, so I started thinking about making rectangles. Maybe their perimeter would somehow be useful?

I was feeling ready for a hint, so I turned to the back of the book for this hint:

“Draw CD, CE, and the altitude from C to AB; then prove triangles congruent.”

I quickly saw how this would be useful. I had already tried drawing these lines, but I hadn’t seen how it would be useful. Now I saw that if I could prove these triangles congruent, I would immediately have DE = EF + DG. I don’t know how I missed it earlier:

I didn’t have the angles yet. I was confident I would need to use the isosceles triangles that BD = BC and AE = AC created. Still, I was having trouble, and was feeling impatient. I decided to look at the solution.

The solution:

Commentary:

That’s pretty much what I was expecting, though it’s a bit annoying that I didn’t get it easily. Was I just lazy? I was definitely a bit lazy. But I think there’s something else here worth remembering, which is that when the angles are all bunched up like this it can be useful to write the actual equations using the angles out. I think it would have been easier to see the relationships that way, as I had a hard time seeing all the relationships in the image. (I struggled to annotate the diagram in a way that contained this info after reading the solution.)

A very brief note: I tried to use a tablet to draw on the image, which sped up the process in some ways. It also makes me more reluctant to write out equations in a way that would have been helpful here. I used Google Jamboard for this problem, but I’d like to find something better.

Takeaways: don’t forget to write equations for angles; maybe do that on paper or find a better way to do geometry on the tablet.