- Many applications in the sciences use hyperbolic trigonometry; for example, the use of Laplace transforms in solving differential equations, various applications in physics, modeling population growth (the logistic model is a hyperbolic tangent curve);
- Hyperbolic trigonometric substitutions are, in many instances, easier than circular trigonometric substitutions, especially when a substitution involving or is involved;
- Students get to see another
*form*of trigonometry, and compare the new form with the old; - Hyperbolic trigonometry is
*fun.*

OK, maybe that last reason is a bit of hyperbole (though not for me).

Not everyone thinks this way. I once had a colleague who told me she did not teach hyperbolic trigonometry because it wasn’t on the AP exam. What do you say to someone who says that? I dunno….

In any case, I want to introduce the subject here for you, and show you some interesting aspects of hyperbolic trigonometry. I’m going to stray from my habit of not discussing things you can find anywhere online, since in order to get to the better stuff, you need to know the basics. I’ll move fairly quickly through the introductory concepts, though.

The hyperbolic cosine and sine are defined by

I will admit that when I introduce this definition, I don’t have an accessible, simple motivation for doing so. I usually say we’ll learn a lot more as we work with these definitions, so if anyone has a good idea in this regard, I’d be interested to hear it.

The graphs of these curves are shown below.

The graph of is shown in blue, and the graph of is shown in red. The dashed orange graph is which is easily seen to be asymptotic to both graphs.

Parallels to the circular trigonometric functions are already apparent: is an even function, just like Similarly, is odd, just like

Another parallel which is only slight less apparent is the fundamental relationship

Thus, lies on a unit hyperbola, much like lies on a unit circle.

While there isn’t a simple parallel with circular trigonometry, there is an interesting way to characterize and Recall that given any function we may define

to be the *even* and *odd parts* of respectively. So we might simply say that and are the even and odd parts of respectively.

There are also many properties of the hyperbolic trigonometric functions which are reminiscent of their circular counterparts. For example, we have

and

None of these are especially difficult to prove using the definitions. It turns out that while there are many similarities, there are subtle differences. For example,

That is, while some circular trigonometric formulas become hyperbolic just by changing to and to sometimes changes of sign are necessary.

These changes of sign from circular formulas are typical when working with hyperbolic trigonometry. One particularly interesting place the change of sign arises is when considering differential equations, although given that I’m bringing hyperbolic trigonometry into a calculus class, I don’t emphasize this relationship. But recall that is the unique solution to the differential equation

Similarly, we see that is the unique solution to the differential equation

Again, the parallel is striking, and the difference subtle.

Of course it is straightforward to see from the definitions that and Gone are the days of remembering signs when differentiating and integrating trigonometric functions! This is one feature of hyperbolic trigonometric functions which students always appreciate….

Another nice feature is how well-behaved the hyperbolic tangent is (as opposed to needing to consider vertical asymptotes in the case of ). Below is the graph of

The horizontal asymptotes are easily calculated from the definitions. This looks suspiciously like the curves obtained when modeling logistic growth in populations; that is, finding solutions to

In fact, these logistic curves are hyperbolic tangents, which we will address in more detail in a later post.

One of the most interesting things about hyperbolic trigonometric functions is that their inverses have closed formulas — in striking contrast to their circular counterparts. I usually have students work this out, either in class or as homework; the derivation is quite nice, so I’ll outline it here.

So let’s consider solving the equation for Begin with the definition:

The critical observation is that this is actually a quadratic in

All that is necessary is to solve this quadratic equation to yield

and note that is always negative, so that we must choose the positive sign. Thus,

And this is just the beginning! At this stage, I also offer more thought-provoking questions like, “Which is larger, or These get students working with the definitions and thinking about asymptotic behavior.

Next week, I’ll go into more depth about the calculus of hyperbolic trigonometric functions. Stay tuned!

]]>and whose graph looks like:

I replied that yes, I had success, and what I found out would make a nice follow-up post rather than just a reply to his comment. So let’s go!

Just a brief refresher: if, for example, we wanted to describe the behavior of where it crosses the *x*-axis at we simply retain the term and substitute the root into the other terms, getting

as the best-fitting parabola at

Now consider another way to think about this:

For examples like the polynomial above, this limit is always trivial, and is essentially a simple substitution.

What happens when we try to evaluate a similar limit with the Folium of Descartes? It seems that a good approximation to this curve at (the U-shaped piece, since the sideways U-shaped piece involves writing as a function of ) is as shown below.

To see this, we need to find

After a little trial and error, I found it was simplest to use the substitution and so rewrite the equation for the Folium of Descartes by using the substitution which results in

Now it is easy to see that as we have giving us a good quadratic approximation at the origin.

Success! So I thought I’d try some more examples, and see how they worked out. I first just changed the exponent of looking at the curve

shown below when

What would be a best approximation near the origin? You can almost eyeball a fifth-degree approximation here, but let’s assume we don’t know the appropriate power and make the substitution with yet to be determined. This results in

Now observe that when we have

so that Thus, in our case with we see that is a good approximation to the curve near the origin. The graph below shows just how good an approximation it is.

OK, I thought to myself, maybe I just got lucky. Maybe introduce a change which will really alter the nature of the curve, such as

whose graph is shown below.

Here, the curve passes through the *x*-axis at with what appears to be a linear pass-through. This suggests, given our previous work, the substitution which results in

We don’t have much luck with here. But if we move the to the other side and factor, we get

Nice! Just divide through by to obtain

Now a simple calculation reveals that And sure enough, the line does the trick:

Then I decided to change the exponent again by considering

Here is the graph of the curve when

It seems we have two roots this time, with linear pass-throughs. Let’s try the same idea again, making the substitution moving the over, factoring, and dividing through by This results in

It is not difficult to calculate that

Now things become a bit more interesting when is even, since there is always a root at in this case. Here, we make the substitution move the over, and divide by resulting in

But since is even, then is a factor of so we have

Substituting in this equation gives

which immediately gives as well! This is a curious coincidence, for which I have no nice geometrical explanation. The case when is illustrated below.

This is where I stopped — but I was truly surprised that everything I tried actually *worked.* I did a cursory online search for Taylor series of implicitly defined functions, but this seems to be much less popular than series for

Anyone more familiar with this topic care to chime in? I really enjoyed this brief exploration, and I’m grateful that William Meisel asked his question about the Folium of Descartes. These are certainly instances of a larger phenomenon, but I feel the statement and proof of any theorem will be somewhat more complicated than the analogous results for explicitly defined functions.

And if you find some neat examples, post a comment! I’d enjoy writing another follow-up post if there is continued interested in this topic.

]]>Our first presenter was Stan Isaacs. There was no real title to his presentation, but he brought another set of puzzles from his vast collection to share. He was highlighting puzzles created by Wayne Daniel.

Below you’ll see one of the puzzles disassembled. The craftsmanship is simply remarkable.

If you look carefully, you’ll see what’s going on. The outer pieces make an icosahedron, and when you take those off, a dodecahedron, then a cube…a wooden puzzle of nested Platonic solids! The pieces all fit together *so* perfectly. Stan is looking forward to an exhibition of Wayne’s work at the International Puzzle Party in San Diego later on this year. For more information, contact Stan at stan@isaacs.com.

Our second speaker was Scott Kim (www.scottkim.com), who’s presentation was entitled *Motley Dissections.* What is a motley dissection? The most famous example is the problem of the squared square — that is, dissecting a square with an integer side length into smaller squares with integer side lengths, but with *all the squares different sizes.*

One property of such a dissection is that no two edges of squares meet *exactly* corner to corner. In other words, edges *always* overlap in some way.

But there are of course many other motley dissections. For example, below you see a motley dissection of one rectangle into five, one pentagon into eleven, and finally, one hexagon into a triangle, square, pentagon and hexagon.

Look carefully, and you’ll see that no single edge in any of these dissections exactly matches any other. For these decompositions, Scott has proved they are minimal — so, for example, there is no motley dissection of one pentagon to ten or fewer. The proofs are not exactly elegant, but they serve their purpose. He also mentioned that he credits Donald Knuth with the term *motley dissection,* who used the term in a phone conversation not all that long ago.

Can you *cube the cube?* That is, can you take a cube and subdivide it into cubes which are all different? Scott showed us a simple proof that you can’t. But, it turns out, you can *box the box.* In other words, if the length, width, and height of the larger box and all the smaller boxes may be different, then it *is* possible to box the box.

Next week, Scott is off to the Gathering 4 Gardner in Atlanta, and will be giving his talk on Motley Dissections there. He planned an activity where participants actually build a boxed box — and we were his test audience!

He created some very elaborate transparencies with detailed instructions for cutting out and assembling. There were a very few suggestions for improvement, and Scott was happy to know about them — after all, it is rare that something works out perfectly the first time. So now, his success at G4G in Atlanta is assured….

We were so into creating these boxed boxes, that we happily stayed until 5:30 until we had two boxes completed.

I should mention that Scott also discussed something he terms *pseudo-duals* in two, three, and even four dimensions! There isn’t room to go into those now, but you can contact him through his website for more information.

As usual, we went out to dinner afterwards — and we gravitated towards our favorite Thai place again. The dinner conversation was truly exceptional this evening, revolving around an animated conversation between Scott Kim and magician Mark Mitton (www.markmitton.com).

The conversation was concerned with the way we perceive mathematics here in the U.S., and how that influences the educational system. Simply put, there is a *lot* to be desired.

One example Scott and Mark mentioned was the National Mathematics Festival (http://www.nationalmathfestival.org). Tens of thousands of kids and parents have *fun* doing mathematics. Then the next week, they go back to their schools and keep learning math the same — usually, unfortunately, boring — way it’s always been learned.

So why does the National Mathematics Festival have to be a one-off event? It doesn’t! Scott is actively engaged in a program he’s created where he goes into an elementary school at lunchtime one day a week and let’s kids play with math games and puzzles.

Why this model? Teachers need no extra prep time, kids don’t need to stay after school, and so everyone can participate with very little needed as far as additional resources are concerned. He’s hoping to create a package that he can export to any school anywhere where with minimal effort, so that children can be exposed to the joy of mathematics on a regular basis.

Mark was interested in Scott’s model: consider your Needs (improving the perception of mathematics), be aware of the Forces at play (unenlightened administrators, for example, and *many* other subtle forces at work, as Mark explained), and then decide upon Actions to take to move the Work (applied, pure, and recreational mathematics) forward.

The bottom line: we all know about this problem of attitudes toward mathematics and mathematics education, but no one really knows what to do about it. For Scott, it’s just another puzzle to solve. There are solutions. And he is going to find one.

We talked for over two hours about these ideas, and everyone chimed in at one time or another. Yes, my summary is *very* brief, I know, but I hope you get the idea of the type of conversation we had.

Stay tuned, since we are planning an upcoming meeting where we focus on Scott’s model and work towards a solution. Another theme throughout the conversation was that mathematics is *not* an activity done in isolation — it is a communal activity. So the Needs will not be addressed by a single individual, rather a group, and likely involving many members of many diverse communities.

A solution *is* out there. It will take a lot of grit to find it. But mathematicians have got grit in spades.

]]>

Differentiating is particularly nice using this method. We first approximate

Then we factor out a from the denominator, giving

As we did at the end of last week’s post, we can make as small as we like, and so approximate by considering as the sum of an infinite series:

Finally, we have

which gives the derivative of as

We’ll look at one more example involving approximating with geometric series before moving on to the product, quotient, and chain rules. Consider differentiating We first factor the denominator:

Now approximate

so that, to first order,

This finally results in

giving us the correct derivative.

Now let’s move on to the product rule:

Here, and for the rest of this discussion, we assume that all functions have the necessary differentiability.

We want to approximate so we replace each factor with its linear approximation:

Now expand and keep only the first-order terms:

And there’s the product rule — just read off the coefficient of

There is a compelling reason to use this method. The traditional proof begins by evaluating

The next step? Just add and subtract (or perhaps ). I have found that there is just no way to convincingly motivate this step. Yes, those of us who have seen it crop up in various forms know to try such tricks, but the typical first-time student of calculus is mystified by that mysterious step. Using linear approximations, there is absolutely *no* mystery at all.

The quotient rule is next:

First approximate

Now since is small, we approximate

so that

Multiplying out and keeping just the first-order terms results in

Voila! The quotient rule. Now usual proofs involve (1) using the product rule with and but note that this involves using the chain rule to differentiate or (2) the mysterious “adding and subtracting the same expression” in the numerator. Using linear approximations avoids both.

The chain rule is almost ridiculously easy to prove using linear approximations. Begin by approximating

Note that we’re replacing the *argument* to a function with its linear approximation, but since we assume that is differentiable, it is also continuous, so this poses no real problem. Yes, perhaps there is a little hand-waving here, but in my opinion, no rigor is really lost.

Since is differentiable, then exists, and so we can make as small as we like, so the “” term acts like the “” term in our linear approximation. Additionally, the “” term acts like the “” term, resulting in

Reading off the coefficient of gives the chain rule:

So I’ve said my piece. By this time, you’re either convinced that using linear approximations is a good idea, or you’re not. But I think these methods reflect more accurately the intuition behind the calculations — and reflect what mathematicians do *in practice.*

In addition, using linear approximations involves *more* than just mechanically applying formulas. If all you ever do is apply the product, quotient, and chain rules, it’s *just* mechanics. Using linear approximations requires a bit more understanding of what’s really going on underneath the hood, as it were.

If you find more neat examples of differentiation using this method, please comment! I know I’d be interested, and I’m sure others would as well.

In my next installment (or two or three) in this calculus series, I’ll talk about one of my favorite topics — hyperbolic trigonometry.

]]>This idea is certainly not new. But I think it is sorely underexploited in the calculus classroom. I like it because it reinforces the idea of derivative as *linear approximation.*

The main idea is to rewrite

as

with the note that this approximation is valid when Writing the limit in this way, we see that *as a function of* is linear in in the sense of the limit in the definition actually existing — meaning there is a good linear approximation to at

Moreover, in this sense, if

then it must be the case that This is not difficult to prove.

Let’s look at a simple example, like finding the derivative of It’s easy to see that

So it’s easy to read off the derivative: ignore higher-order terms in and then look at the coefficient of as a function of

Note that this is perfectly rigorous. It should be clear that ignoring higher-order terms in is fine since when taking the limit as in the definition, only one divides out, meaning those terms contribute to the limit. So the coefficient of will be the only term to survive the limit process.

Also note that this is nothing more than a rearrangement of the algebra necessary to compute the derivative using the usual definition. I just find it is more intuitive, and less cumbersome notationally. But every step taken can be justified rigorously.

Moreover, this method is the one commonly used in more advanced mathematics, where functions take vectors as input. So if

we compute

and read off

I don’t want to go into more details here, since such calculations don’t occur in beginning calculus courses. I just want to point out that this way of computing derivatives is in fact a natural one, but one which you don’t usually encounter until graduate-level courses.

Let’s take a look at another example: the derivative of and see how it looks using this rewrite. We first write

Now replace all functions of with their *linear* approximations. Since and near we have

This immediately gives that is the derivative of

Now the approximation is easy to justify geometrically by looking at the graph of But how do we justify the approximation ?

Of course there is no getting around this. The limit

is the one difficult calculation in computing the derivative of So then you’ve got to provide your favorite proof of this limit, and then move on. But this approximation helps to illustrate the essential point: the differentiability of at does, in a real sense, imply the differentiability of everywhere else.

So computing derivatives in this way doesn’t save any of the hard work, but I think it makes the work a bit more transparent. And as we continually replace functions of with their linear approximations, this aspect of the derivative is regularly being emphasized.

How would we use this technique to differentiate ? We need

and so

Since the coefficient of on the left is so must be the coefficient on the right, so that

As a last example for this week, consider taking the derivative of Then we have

Now since and we have and so we can replace to get

Now what do we do? Since we’re considering near then is small (as small as we like), and so we can consider

as the sum of the infinite geometric series

Replacing, with the linear approximation to this sum, we get

and so

This give the derivative of to be

Neat!

Now this method takes a bit more work than just using the quotient rule (as usually done). But using the quotient rule is a purely mechanical process; this way, we are constantly thinking, “How do I replace this expression with a good linear approximation?” Perhaps more is learned this way?

There are more interesting examples using this geometric series idea. We’ll look at a few more next time, and then use this idea to prove the product, quotient, and chain rules. Until then!

]]>The idea is to give a geometrical meaning to an algebraic procedure: factoring polynomials. What is the *geometry* of the different factors of a polynomial?

Let’s look at an example in some detail:

Now let’s start looking at the behavior near the roots of this polynomial.

Near the graph of the cubic looks like a parabola — and that may not be so surprising given that the factor occurs quadratically.

And near the graph passes through the *x*-axis like a line — and we see a linear factor of in our polynomial.

But which parabola, and which line? It’s actually pretty easy to figure out. Here is an annotated slide which illustrates the idea.

All you need to do is set aside the quadratic factor of and substitute the root, in the remaining terms of the polynomial, then simplify. In this example, we see that the cubic behaves like the parabola near the root Note the scales on the axes; if they were the same, the parabola would have appeared much narrower.

We perform a similar calculation at the root

Just isolate the linear factor substitute in the remaining terms of the polynomial, and then simplify. Thus, the line best describes the behavior of the graph of the polynomial as it passes through the *x*-axis. Again, note the scale on the axes.

We can actually use this idea to help us sketch graphs of polynomials when they’re in factored form. Consider the polynomial Begin by sketching the three approximations near the roots of the polynomial. This slide also shows the calculation for the cubic approximation.

Now you can begin sketching the graph, starting from the left, being careful to closely follow the parabola as you bounce off the *x*-axis at

Continue, following the red line as you pass through the origin, and then the cubic as you pass through Of course you’d need to plot a few points to know just where to start and end; this just shows how you would use the approximations near the roots to help you sketch a graph of a polynomial.

Why does this work? It is not difficult to see, but here we need a little calculus. Let’s look, in general, at the behavior of near the root Given what we’ve just been observing, we’d guess that the best approximation near would just be

Just what does “best approximation” mean? One way to think about approximating, calculuswise, is matching derivatives — just think of Maclaurin or Taylor series. My claim is that the first derivatives of and match at

First, observe that the first derivatives of both of these functions at *must* be 0. This is because will always be a factor — since at most derivatives are taken, there is no way for the term to completely “disappear.”

But what happens when the th derivative is taken? Clearly, the th derivative of at is just What about the th derivative of ?

Thinking about the product rule in general, we see that the form of the th derivative must be When a derivative of is taken, that means one factor of survives.

So when we take we *also* get This makes the th derivatives match as well. And since the first derivatives of and match, we see that is the best th degree approximation near the root

I might call this observation *the geometry of polynomials**. *Well, perhaps not the *entire* geometry of polynomials…. But I find that any time algebra can be illustrated graphically, students’ understanding gets just a little deeper.

Those who have been reading my blog for a while will be unsurprised at my geometrical approach to algebra (or my geometrical approach to *anything*, for that matter). Of course a lot of algebra was invented *just* to describe geometry — take the Cartesian coordinate plane, for instance. So it’s time for algebra to reclaim its geometrical heritage. I shall continue to be part of this important endeavor, for however long it takes….

The first problem I’ll share is a “ball and urn” problem. These are a staple of mathematical contests everywhere.

You have 20 identical red balls and 14 identical green balls. You wish to put them into two baskets — one brown basket, and one yellow basket. In how many different ways can you do this if the number of green balls in either basket is less than the number of red balls?

Another popular puzzle idea is to write a problem or two which involve the year of the contest — in this case, 2014.

A positive integer is said to be

fortunateif it is either divisible by 14, or contains the two adjacent digits “14” (in that order). How many fortunate integersnare there between 1 and 2014, inclusive?

The other two problems from the contest I’ll share with you today are from other contests shared with me by my colleagues.

In the figure below, the perimeters of three rectangles are given. You also know that the shaded rectangle is in fact a square. What is the perimeter of the rectangle in the lower left-hand corner?

I very much like this last problem. It’s one of those problems that when you first look at it, it seems totally impossible — how could you consider *all* multiples of 23? Nonetheless, there is a way to look at it and find the correct solution. Can you find it?

Multiples of 23 have various digit sums. For example, 46 has digit sum 10, while 8 x 23 = 184 has digit sum 13. What is the smallest possible digit sum among all multiples of 23?

You can read more to see the solutions to these puzzles. Enjoy!

First we’ll look at the solution to the ball and urn problem. Since the number of green balls is less than the red in any one basket, we know that every red ball has a green partner, so we consider the 14 pairs of red/green balls along with the six remaining red balls.

There are 15 ways to put the 14 pairs into the two baskets; the brown basket may contain 0, 1,…,14 pairs. There are 5 ways to put the remaining six red balls in the brown basket — 1, 2, 3, 4, or 5 red balls. We cannot put either 0 or 6, or else that would leave one basket with the same number of red as green balls.

Thus, there are 15 x 5 = 75 ways to put the balls in the baskets to fulfill the conditions of the problem.

Now let’s see how many fortunate numbers there are between 1 and 2014. We first count the number of integers which contain the digits “14” in that order. There are 100 fortunate numbers of the form 14AB, 20 of the form A14B (A could be 1 or 0; in the latter case, the number has just three digits), and 21 of the form AB14 (A or B may be 1 or 0, and we have the lone case with A=2 being 2014). However, the number (and the *only* number) counted twice in this scheme is 1414. This gives 100 + 20 + 21 – 1 = 140 numbers so far.

There are 143 multiples of 14 between 0 and 2014. Now 140 + 143 = 283, but some numbers have been counted twice — those which *both* contain the digits “14” and are multiplies of 14. There are just 12 of these: 14, 140, 714, 1148, 1400, 1414, 1428, 1442, 1456, 1470, 1484, and 1498. Thus, the final count of fortunate numbers is 283 – 12 = 271.

The perimeter of the shaded rectangle is 68. To see this, we label various segments in the figure as follows:

The information about the perimeters of the smaller rectangles results in the following system of equations: *b* + *e* = 24, *a* + *b* + *c* = 68, and *a* + *c* + *d* = 78. We are looking for the quantity 2(*d *+ *e*). But adding the first and third equations results in *a* + *b* + *c* + *d* + *e* = 102, and then subtracting the second equation from this gives *d* + *e* = 34. So the perimeter of the rectangle in the lower left corner is 2(34) = 68.

As for the smallest possible digit sum — it’s just 2! It should be clear that no multiple of 10 can be a multiple of 23. However, it is possible to have a digit sum of 2: 100,000,000,001 is a multiple of 23. But how would you go about finding this multiple?

Just start a long division problem where a number beginning 10000…. is divided by 23. 23 goes into 100 four times, with a remainder of 8, bring down a 0. 23 goes into 80 three times with a remainder of 11, bring down a 0, and so forth.

Now bringing down a 0 will *never* result in an even division (because no power of 10 is a multiple of 23), but perhaps bringing down a 1 will help. Of the numbers 11, 21, 31,…, 211, 221, only 161 is a multiple of 23. So if at any time we have a remainder of 16, we may add a 1 (instead of a 0) to the divisor, with the result an even division. This in fact happens after 10 zeros, giving a minimum digit sum of 2. Amazing!

I hope you enjoyed this set of puzzles. Stay tuned for the next installment of The Puzzle Archives for more problem-solving fun!

]]>Our first speaker was Frank A. Farris, our host at Santa Clara University. (Recall that last month, he presented a brief preview of his talk.) His talk was about introducing a sound element into his wallpaper patterns.

In order to do this, he used frequencies based on the spectrum of hexagonal and square grids. It’s not important to know what this means — the main idea is that you get frequencies that are *not* found in western music.

Frank’s idea was to take his wallpaper patterns, and add music to them using these non-traditional frequencies. Here is a screenshot from one of his musical movies:

Frank was really excited to let us know that the San Jose Chamber Orchestra commissioned work by composer William Susman to accompany his moving wallpaper patterns. The concert will take place in a few weeks; here is the announcement, so you are welcome to go listen for yourself!

Frank has extensive information about his work on his website http://math.scu.edu/~ffarris/, and even software you can download to make your very own wallpaper patterns. Feel free to email him with any questions you might have at ffarris@scu.edu.

The second talk, *Salvador Dali — Old and New*, was given by Tom Banchoff, retired from Brown University. He fascinated us with the story of his long acquaintance with Salvador Dali. It all began with an interview in 1975 with the Washington Post about Tom’s work in visualizing the fourth dimension.

He was surprised to see that the day after the interview, the article *Visual Images And Shadows From The Fourth Dimension* in the next day’s Post, as well as a picture of Dali’s Corpus Hypercubus (1954).

But Tom was aware that Dali was very particular about giving permission to use his work in print, and knew that the Post didn’t have time to get this permission in such a short time frame.

The inevitable call came from New York — Dali wanted to meet Tom. He wondered whether Dali was simply perturbed that a photo of his work was used without permission — but luckily, that was not the reason for setting up the meeting at all. Dali was interested in creating stereoscopic oil paintings, and stereoscopic images were mentioned in the Post article.

Thus began Tom’s long affiliation with Dali. He mentioned meeting Dali eight or nine times in New York (Dali came to New York every Spring to work), three times in Spain, and once in France. Tom remarked that Dali was the most fascinating person he’d ever met — and that includes mathematicians!

Then Tom proceeded to discuss the genesis of *Corpus Hypercubus.* His own work included collaboration with Charles Strauss at Brown University, which included rendering graphics to help visualize the fourth dimension — but this was back in the 1960’s, when computer technology was at its infancy. It was a lot more challenging then than it would be today to create the same videos.

He also spent some time discussing a net for the hypercube, since a hypercube net is the geometrical basis for Dali’s *Corpus **Hypercubus.* What makes understanding the fourth dimension difficult is imagining how this net goes together.

It is not hard to imagine folding a flat net of six squares to make a cube — but in order to do that, we need to fold some of the squares *up* through the third dimension. But to fold the hypercube net to make a hypercube *without distorting the cubes* requires folding the cubes up into a fourth spatial dimension.

This is difficult to imagine! Needless to say, this was a very interesting discussion, and challenged participants to definitely think outside the box.

Tom remarked that Dali’s interest in the hypercube was inspired by the work of Juan de Herrera (1530-1597), who was in turn inspired by Ramon Lull (1236-1315).

Tom also mentioned an unusual project Dali was interested in near the end of his career. He wanted to design a horse that when looked at straight on, looks like a front view of a horse. But when looked from the side, it’s 300 meters long! For more information, feel free to email Tom at banchoff@math.brown.edu.

Suffice it to say that we all enjoyed Frank’s and Tom’s presentations. The change of venue was welcome, and we hope to be at Santa Clara again in the future.

Following the talks, Frank generously invited us to his home for a potluck dinner! He provided lasagna and eggplant parmigiana, while the rest of us provided appetizers, salads, side dishes, and desserts.

As usual, the conversation was quite lively! We talked for well over two hours, but many of us had a bit of a drive, so we eventually needed to make our collective ways home.

Next time, on April 7, we’ll be back at the University of San Francisco. At this meeting, we’ll go back to shorter talks in order to give several participants a chance to participate. Stay tuned for a summary of next month’s talks!

]]>This year, I thought I’d give you a virtual tour of the exhibit! So I asked contributing artists to submit their own personal statement about their work and/or mathematical art in general, as well as an image of one of their displayed artworks. I’ll let the artists speak for themselves…. (By the way, the order the artists are presented in is the order in which they sent me their information. There is no ranking implicit in the order.)

Shirley Yap (shirley.yap@csueastbay.edu)

I created this image out of golden spirals. While working on a math demonstration for my students, I unpacked a roll of netting. During the unraveling process, I had a vision of a two-dimensional golden spiral unravelling, which I tried to recreate with code. I wanted to viewer to not just witness the unraveling, but also be inside the web of the fabric. I often create code that has a lot of randomness in it, so that it captures a moment in time that can never be recreated.

Frank A. Farris (ffarris@scu.edu and math.scu.edu/~ffarris)

Mathematicians can feel lonely to find ourselves face to face with the most beautiful thoughts humans have ever known, only to realize that communicating our experience is unreasonably difficult. I have found comfort in visual art, digitally computed images that are the best I can do (short of giving an hour lecture) to say, ‘This is the beauty of mathematics.’

David Honda (snaporigami@gmail.com and snaporigami.weebly.com)

I’m primarily a middle school mathematics teacher with one of my hobbies being Origami and other paper crafts. Some years back I became interested in the work of Heinz Strobl which uses joined, folded strips of paper to create various structures. Much like unit origami, the structures are held together solely by the folds, no adhesives. My interest soon became an obsession and I’ve been neck-deep in little strips of paper ever since. Lately I’ve been exploring concepts in Topology. This particular work is my attempt at a Klein Bottle.

Dan Bach (www.dansmath.com,** **art@dansmath.com**, and **@dansmath)

“I’m a career college mathematics teacher, now an interactive book author and 3D math artist. I like to illustrate number theory and vector calculus principles with surprising and colorful images, using a software palette of Mathematica, Cheetah3D, and iBooks Author. My math art encourages viewers to think, notice, and wonder. And hopefully say, ‘That’s cool! That’s math?’ “

Roger Antonsen (http://rantonse.no/ and rantonse@ifi.uio.no)

This is a visualization of seven repeated perfect out-shuffles of a deck with 128 cards. The horizontal lines represent the particular orders of the cards throughout the shuffling, and the vertical curves represent the path each card takes from start to finish. The curves are colored from black to white in order to show the mechanics of the shuffling. The dots are colored from black to red to black in order to show that each perfect out-shuffle preserves the so-called “stay-stack principle”. Notice that the order of cards returns to the original order after seven shuffles.

Linda Beverly (lbeverly@horizon.csueastbay.edu and https://mathcsresearch.wordpress.com/)

This image is an embedding of a photograph of a set of watercolors. The embedding was performed using Locally Linear Embedding (LLE), a nonlinear dimensionality reduction technique, introduced by Saul & Roweis in 2000. LLE is an unsupervised machine learning algorithm.

I am pursuing a double major in Mathematics and Computer Science. I am currently working on research in geometric dimensionality reduction and unsupervised machine learning. This work will be extended into neural networks and deep learning. I enjoy seeking out interesting intersections between mathematics, computer science, and art.

Jason Herschel (jherschel@gmail.com)

Undulating organic light show with minimal code generated by a 16 MHz processor calculating color and brightness values through a Perlin noise algorithm. Blobs appear to grow, shrink, and drift relaxingly over the LED grid. 100 ping pong balls covering 100 individually addressable LEDs on poster board with Arduino Nano v.3 controller and battery pack.

Paul Gonzalez-Becerra (pgonzbecer@gmail.com and http://www.pgonzbecer.com/)

Programming is my art. I might not be a good “designer”, but I am a good developer where I am able to take a structured approach on art. I specialize in computer graphics, thus my understanding of the mathematics behind geometry, 3D models, and 2D sprites are better than my ability to free-form draw them.

Phil Webster (phil@philwebsterdesign.com, http://www.philwebsterdesign.com/ )

All of my work stems from one core impulse: to celebrate the inherent beauty of mathematical forms. Since traveling to India in 2012, I have been particularly focused on blending traditional Islamic motifs with polyhedra and fractals. The results are distinctly Islamic in flavor but with a modern twist.

This piece has global and local 8-fold rotational symmetry around each gold star. Star centers occupy the nodes of 8 fractal quaternary trees, which are pruned at the octant boundaries. The original central star is replaced by an inward progression of the same fractal diminution.

Vince Matsko (vince.matsko@gmail.com, www.vincematsko.com)

This piece is based on fractal binary trees. The usual way of creating a binary tree is to move forward, then branch to the left and right some fixed angle as well as shrink, and repeat recursively. Recent work involves specifying the branching by arbitrary affine transformations. In this piece, the affine transformations were chosen so that as the tree grows, nodes are repeatedly visited. The nodes are covered by disks which become smaller with each iteration, accounting for the overlapping circles. The research needed to produce these images was undertaken jointly with Nick Mendler.

I hope you enjoyed this virtual tour of the Art Exhibition at the recent MAA Regional Conference. The upcoming conference is next March; stay tuned for another virtual tour in about a year!

]]>So I and my colleagues created three levels of contests — Beginning, Intermediate, and Advanced — since it seemed that it would be difficult to create a single contest that everyone could enjoy. But I did include three problems that were the same at *every* level, so all participants could talk about some aspect of the contests with each other.

Participants had a few days to get as many answers as they could, and we even had books for prizes! Many remarked how much they enjoyed working out these puzzles.

Now this conference took place before I started writing my blog. I have written several similar contests over the years for various audiences, and so I thought it would be nice to share some of my favorite puzzles from the contests with you. And so *The Puzzle Archives* are born!

First, I’ll share the three puzzles common to all three contests. I needed to create some puzzles which were fun, and didn’t require any specialized mathematical knowledge. As I’m a fan of cryptarithms and the conference took place in Denver, I created the following puzzle. Here, no letter stands for the digit “0.”

For the next puzzle, all you need to do is complete the magic square using the even numbers from 2 to 32. Each row, column, and diagonal should add up to the same number. There are two solutions to this puzzle — and so you need to find them both!

And of course, I had to include one of my favorite types of puzzles, a CrossNumber puzzle. Remember, no entry in a CrossNumber puzzle can begin with “0.”

I also included a few geometry problems, staples of *any* math contest. For the first one, you need to find the area of the *smallest* circle you could fit the following figure into. Both triangles are equilateral; the smaller has side length 1 and the larger has side length 2.

And for the second one, you need to find the radius of the larger circle. You are given that the smaller circle has a diameter of 2 units, and the sides of the square are 2 units long. Moreover, the smaller circle is tangent to the square at the midpoint of its top edge, and is also tangent to the larger circle.

The last two problems I’ll share from this contest are number puzzles. The first is a word problem, which I’ll include verbatim from the contest itself.

Tom and Jerry each have a bag of marbles. Tom says, “Hey, Jerry. I have four different colors of marbles in my bag. And the number of each is a different perfect square!” Jerry says, “Wow, Tom! I have four different colors of marbles, too, but the number of each of mine is a different perfect

cube!”If Tom and Jerry have the

sametotal number of marbles, what is the least number of marbles they can have?

And finally, another cryptarithm, but with a twist. In the following multiplication problem, *F,* *I,* *N,* and *D* represent different digits, and the *x*‘s can represent *any* digit. Your job is to find the number *F I N D. *(And yes, you have enough information to solve the puzzle!)

Happy solving! You can read more to see the solutions; I didn’t want to just put them at the bottom in case you accidentally saw any answers. I hope you enjoy this new thread!

(Note: The *FIND* puzzle was from a collection of problems shared by a colleague. The first geometry problem may have come from elsewhere, but after four years, I can’t quite remember….)

The simplest entry point into solving the first cryptarithm is to look at the fifth column. D would have to be 0 or 9, but it can’t be 0 since we are given there are no zeroes in the puzzle. The final solution:

For the magic square puzzle, you need to first add the even numbers from 2 to 32, which is 272. Since the four rows use all 16 numbers, each row, column, and diagonal sum to 272/4, or 68. The two solutions (which differ only by interchanging the middle two columns):

The solution to the CrossNumber puzzle:

Here is a diagram illustrating the solution to the first geometry puzzle. The essential idea is to use the geometry of the equilateral triangles to show that *ABC* is a right triangle, meaning that *AC* is a diameter of the circle. Since *AC* is then the area of the circle is

The second geometry problem is a little trickier. The diagram below will help, where *q* is the center of the larger circle. If we say the length of *pq* is 2 + *x,* then so is the length of *qr*, since they are both radii of the circle. Therefore, the length of *rs* must be 2 – *x.* Since the length of *sq* is 1, we may apply the Pythagorean theorem to triangle *qsr,* which gives *x* to be 1/8, so that the radius of the larger circle is 17/8.

As for marbles, Tom has 9 + 16 + 36 + 100 = 161 marbles, and Jerry has 1 + 8 + 27 + 125 = 161 marbles. Here, begin by looking for the smallest possible sum of four cubes (which won’t work); the next smallest sum is the one you want.

And for the final cryptarithm, start by observing that *N* must be 0, and because each partial product is only four digits long, this means that *F* must be fairly small. The answer is *FIND* = 3201.