The key to the trick is that there is a hole at the bottom of the post that leads to a pipe that runs up through the post and down through the base of the cup. As the cup fills, so does the pipe. When the liquid gets to the top, it flows down the pipe and gravity does the rest. The pipe acts as a siphon much as the plumbing does in a modern toilet.

Here’s a video of me using the cup and me using a Pythagorean cup that I made.

To make the cup, I pushed a bendy straw through the bottom of a styrofoam cup. It was bent over on the inside so that the opening of the straw was at the base of the cup. (I also put a bead of clue around the hole in the cup so that the liquid wouldn’t leak out around the straw.) It worked beautifully!

If I were a talented potter or had any skills in designing objects for a 3D printer, I’d make a mug with the tube running through the handle. Here’s what I have in mind. If any of you make one, send me a photo—or better, a video—or better still, the real thing!

As for the name “Pythagorean cup”: Do any math/physics/Greek historians have information about the history of this cup? Based on nothing but a gut feeling, I highly doubt this was created by Pythagoras. But I’d definitely be interested in learning the source of this attribution.

Update: On January 1, I received a printable STL file from Pablo Untroib from Argentina. I haven’t had a chance to print it yet. But Pablo did, and it is fantastic. See his video below. He said that when it is sitting on a flat table, it may not leak, but when you pick up the cup, it starts pouring out. A great prank cup!

Further update: On January 4, I received an email from blogger James Stanley who also created a Pythagorean cup—one with the slightly modified design shown below. He wrote about it on his blog. It looks fantastic—I’ll have to print that one too!]

]]>In the logic section of my Discrete Mathematics class (our “intro-to-proofs” class), the students learned about the converse of a conditional statement: the converse of “if A, then B” is “if B, then A.” Most notably, a conditional statement is not logically equivalent to its converse. “If I am over six feet tall, then I am over five feet tall” is a true statement (in my case, the hypothesis is false and the conclusion is true). But the converse, “If I am over five feet tall, then I am over six feet tall” is false.

On my exam, I had a page with the following instructions and the following problem:

*For each of the following problems, determine if such an example exists. If not, state IMPOSSIBLE and give a brief explanation. If so, give an explicit example that satisfies the conditions.*

*A statement that begins “If x≥1, then . . .” and its converse such that the statement is***true**and its converse is**false**.

The answer that I was looking for was something like:

(True) statement: If *x*≥1, then *x*≥0.

(False) converse: If *x*≥0, then *x*≥1.

I class we had discussed that typically when you encounter a conditional like “if *x*≥1, then *x*≥0″ in mathematics, there is an implied universal quantifier. So, mentally, we read it as “for all real numbers *x*, if *x*≥1, then *x*≥0.” My first statement (if *x*≥1, then *x*≥0) is definitely true for all *x*, but the second statement (if *x*≥0, then *x*≥1) is not true for all *x* (for instance, it is not true when *x=*0).

Not everyone answered in this way. Interestingly, I discovered that it was not the case that the more prepared the student, the more likely she or he would get the problem correct (my intended answer). Here’s what I mean:

**Least prepared student:**This student might get it totally wrong, leave the problem blank, not know what the converse of an if-then statement is, and so on.**More prepared:**If a student knew what a converse is, then she or he might get the problem correct without realizing that there were any subtleties.**Yet more prepared:**In class, we talked about the fact that we can only assign truth values to statements. If there is a variable in a sentence, then it is not a statement (it is a predicate). For instance, “the integer*n*is even” is not a statement. It is not true or false unless we know the value of*n*. So, some students wrote on my exam that it was IMPOSSIBLE to solve because we don’t know the value of*x—*it is a predicate, not a statement*.*This was not the answer I was looking for, but I gave them full credit because they were, technically speaking, correct. (In a sense, their answer was more correct than my intended answer.)**Most prepared:**These students would have understood the previous argument, but would also have recalled our discussion of the implied universal quantifier and would have remembered that we had homework problems on this, and thus they would have given the answer I was looking for.

Interestingly, students (2) and (4) may have given the same answer on the paper. But because I know these students and because I saw how they did with the rest of their exam, I honestly suspect (2) and (4) are two different groups. In other words, many of the strongest students in the class got my desired answer and many of the students who struggled elsewhere on the exam got this problem correct. On the other hand, some very strong students fell into category (3).

All-in-all, I wish I’d not asked this question. I ended up giving almost all of the students full credit on the problem, whether they got they got an answer that I had intended to be “the right answer” (i.e., students (2) or (4)) or the more technically correct answer that I did not have in mind (3). But I’m upset that I gave this problem that could be equally interpreted in two different ways.

]]>I spent a while thinking about an example I could show them where this is the case. I found this Math Overflow link on this same question. They gave a few examples. Here’s one of them—the multi-function product rule for derivatives. The base case requires using the definitions of continuity and differentiability and the theorem that every function differentiable at a point is continuous at the point. It is not a simple “1=1” base case. The inductive step is much easier, although it is still not totally trivial because it requires using both the base case and the inductive hypothesis.

**Theorem.** *Let be functions that are differentiable at some point . Then *

**Proof. **This is a proof by induction on Let be the statement that given differentiable functions it follows that

Base case: Let and be differentiable at Recall that because is differentiable at it is continuous at Then

The penultimate equality follows from the continuity of at and the final inequality follows from the differentiability of and at Thus holds.

Inductive step. Suppose holds for some integer Let be functions that are differentiable at some point . Then By the base case, this equals By the inductive assumption, this equals

Thus, holds.

It follows that holds for all #

If you have another example—especially one that would be suitable for into-to-proof students (who don’t have an extensive math background)—post them in the comments.

]]>In 1667, James Gregory did the same, but he used areas: He discovered the following beautiful double-recurrence relation that can be used to compute the areas of inscribed and circumscribed *n*-gons:

**Gregory’s Theorem. ***Let I _{k} and C_{k} denote the areas of regular k-gons inscribed in and circumscribed around a given circle. Then for all n, I_{2}_{n} is the geometric mean of I_{n} and C_{n}, and C_{2n} is the harmonic mean of I_{2n} and C_{n}; that is,*

and

We can use these formulas to approximate π. For instance, a square inscribed in a unit circle has area *I*4=2 and a square circumscribed about the unit circle has area *C*_{4}=4. Applying the recurrence relations, we obtain the following sequence of bounds:

n |
In |
Cn |

4 | 2 | 4 |

8 | 2.828427125 | 3.313708499 |

16 | 3.061467459 | 3.182597878 |

32 | 3.121445152 | 3.151724907 |

64 | 3.136548491 | 3.144118385 |

128 | 3.140331157 | 3.14222363 |

256 | 3.141277251 | 3.141750369 |

512 | 3.141513801 | 3.141632081 |

1024 | 3.14157294 | 3.14160251 |

2048 | 3.141587725 | 3.141595118 |

This summer I tweeted this theorem:

Let I_n and C_n be the areas of the inscribed and circumscribed regular n-gons for the unit circle. In 1667James Gr… twitter.com/i/web/status/1…

—

Dave Richeson (@divbyzero) June 20, 2018

My friend Tom Edgar—a mathematician at Pacific Lutheran University and a master at finding “proofs without words”—emailed me to see if I wanted to try finding a proof without words of Gregory’s theorem. This is what we came up with.

The two parts of Gregory’s theorem follow from the two parts of the following lemma. We give the proof… without words.

**Lemma. ** and

*Proof.*

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Möbius bands have the surprising properties that they are one-sided and have only one edge. This inspired me to write the words MÖBIUS and BAND as ambigrams—in particular, as ambigrams that appear the same after a 180° rotation.

Then I created a Möbius band with the words running along the band.

If you want to make your own, you can download this printable pdf.

However, after I made the Möbius band, it occurred to me that we don’t actually need a Möbius band to achieve this effect with the ambigram. In particular, if you tape the band together as a cylinder, it still looks the same when it is flipped over.

Now, if you were really good, you could create a version of MÖBIUS and BAND that have the 180° rotation and look the same in a mirror. Then you could write the words on a transparent Möbius band. . . .

]]>It got a lot of interest, so I thought I’d try making my own. Here’s the final product.

You can download a printable pdf if you want to make your own.

]]>We do not know who discovered the cardioid. In 1637 Étienne Pascal—Blaise’s father—introduced the relative of the cardioid, the limacon, but not the cardioid itself. Seven decades later, in 1708, Philippe de la Hire computed the length of the cardioid—so perhaps he discovered it. In 1741, Johann Castillon gave the cardioid its name.

Got your coffee? Turn on the flashlight feature of your phone and shine the light into the cup from the side. The light reflects off the sides of the cup and forms a caustic on the surface of the coffee. This caustic is a cardioid.

The Mandelbrot set is one of the most beautiful images in all of mathematics. It is the set of complex numbers *c* such that the number 0 does not diverge to infinity under repeated iterations of the function *f _{c}*(

Cardioids even show up in audio engineering. Sometimes engineers need a uni-directional microphone—one that is very sensitive to sounds directly in front of the microphone and less sensitive to sounds next to or behind it. When they do, they reach for a *cardioid microphone.* The microphone is so-named because the graph of the sensitivity of the microphone in polar coordinates is a cardioid.

In this blog post, we present a few favorite places that cardioids appear. In particular, we will look how we can use lines to construct the curved cardioid. At the end of the blog post, we provide a template that you can use to make your own cardioid. And we provided printable pages that can be used to make a cardioid flip book.

A common kids math doodle is to draw a set of coordinate axes and then draw line segments from (0,10) to (1,0), from (0,9) to (2,0), and so on. This procedure magically produces a suite of lines that, when viewed together, has what appears to be a curved boundary. This curve is called the *envelope* of the family of lines.

Let *C _{t}* denote a family of curves parametrized by

Let us look at some features of this envelope. First, each line *C _{t}* is tangent to the curve. Second, if we take two nearby lines

In the following definition we let denote the partial derivative of *F* with respect to *t*.

**Definition.** Let be a differentiable function. The *envelope* of the set of curves *F*(*x,y,t*)=0 is the set of points (*x,y*) such that both *F*(*x,y,t*)=0 and *F _{t}*(

This is a mysterious definition. Why does it produce the envelope? For a fixed *t* and any the curves *F*(*x,y,t*)=0 and *F*(*x,y,t+h*)=0 (that is, *C _{t}* and

Returning to our “kids doodle” example, *F _{t}*(

It turns out that we can construct the cardioid as the envelope of curves, and we can do so in a number of different ways. For instance, pick a point *P* on a circle (the blue circle below, say). Draw circles with centers on the original circle that pass through *P.* The envelope of these circles is a cardioid.

But we will focus on a different example. Begin with a circle (the red circle below). Mark a certain number of evenly spaced points around the circle, *N,* say, and number them consecutively starting at some point *P*: 0, 1, 2,…, *N*-1. Then for each *n,* draw a line between points *n* and 2*n* (mod *N*). In our example, *N*=54, so we would join points 5 and 10, 19 and 38, and 31 and 8 (since 8 is 62 mod 54). The envelope of these lines is a cardioid.

Let’s see why this is the case. Suppose our circle has center (1,0) and radius 3 and that *P*=(4,0). Now, starting at *P,* find points *t* and 2*t* radians around the circle from *P,* and draw the line segment joining them. We will show show that the envelope of all such lines is the cardioid with polar equation

The two points on the circle—corresponding to *t* and 2*t*—have coordinates and The line joining them is After some some algebra and some applications of double angle formulas, we can express this line as In particular, the expression on the left is our function *F*(*x,y, t*). Taking the partial derivative of *F* with respect to *t* we obtain

Now, we want to show that the *x* and *y* coordinates at which *F*(*x,y,t*)=*F _{t}*(

It turns out that this analysis explains the cardioid in the coffee cup. We can view the caustic as an envelope of lines. As we see below, if we draw lines emanating from a single point *P* on the circle and allow them to reflect off the circle (the angle of incidence equalling the angle of reflection), then the cardioid is the envelope of these lines.

If the light source is located at point *P,* then a beam of light will reflect off a point *Q* on the circle and strike the circle again at *R* (see figure below). Since arc *PQ* equals arc *QR* arc *PR* is twice arc *PQ.* But then segment *QR* is a line that we would have drawn in the previous construction.

[Update: When I wrote this post I debated to myself whether to include the following info. Thanks to the nudge by Rick Wicklin in the comments, I decided to add it.] The coffee cup example requires one final comment. In practice, the light source is not at the edge of the coffee cup, but rather, far away. So the rays of light are roughly parallel when they reach the cup. In this case, the curve won’t be a cardioid, but its cousin—a *nephroid*. This is the envelope of lines one obtains by joining *n* and 3*n*. In particular, as we see below, arc *QR* is twice arc *PQ*. (So in our numbering, *n*=0 sits at the point *P*.)

This printable pdf has a circle with 60 numbered points. Connect each number *n* to the number 2*n* mod 60 to obtain a cardioid. For a little extra fun, try connecting *n* to 3*n* or 4*n* or 5*n* to see what shapes you obtain.

This 12-page pdf is a printable flip book. Print the pages double-sided. The pages are designed so that the mathematical figure is on one side and the flip book page number is on the reverse side. Cut out each page, put in numerical order, and secure with a binder clip. Flip through the pages and see the animation in action!

]]>My son is now in Algebra 2, and for the first time, he showed me something that I’ve never seen before—the relationship between polynomials and finite differences.

Take any polynomial, such as and any arithmetic sequence, such as 0, 2, 4, 6,… Plug these values into the polynomial. Take the neighboring pairwise differences. So, for instance, Then take the neighboring pairwise differences of those values, and so on. It turns out that the *n*th level will consist entirely of the same nonzero value if, and only if, the polynomial has degree *n*. Wow! That’s so cool!

Here’s my worked-out degree-3 example. Notice that after three levels, we yield the constant value 336:

My son’s textbook (*Algebra 2*, by Larson, Boswell, Kanold, and Stiff) shared this fact but provided no explanation for why it is true (which, as a mathematician, I find very disappointing). So, I had to work it out myself.

It turns out that **more** is true—if the polynomial has degree *n *with leading coefficient *c, *and *a *is the difference between terms in the arithmetic sequence, then the final value is In particular, if *a*=1 and *c*=1, then the pairwise difference process will terminate with *n*!. Notice that in my example, the final value is

Why is this true?

Here’s a proof by induction on the degree of the polynomial. As the base case, consider a degree-1 polynomial: *p*(*x*)=*cx*+*b*. Then,

so the base case holds.

Now, assume that the result is true for any polynomial of degree *n*-1, for some *n*≥2. We will prove that it is true for a polynomial of degree *n.* Let where c≠0 and “l.o.t.” means “lower order terms.” We see that

Let us call this polynomial *q*(x).

Notice that because the leading coefficient *acn** *is nonzero, *q*(*x*) has degree *n*-1. By our inductive hypothesis, after *n*-1 pairwise differences, the polynomial *q*(*x*) will yield a constant value Thus, for *p, *the process terminates after *n *steps with the constant value This proves the theorem.

After playing around with this, I googled it, and—no surprise—the mathematics of finite differences has a long history. Also, it is not difficult to see the resemblance of these calculations to the calculation of the derivative (using the definition of the derivative).

]]>I am interested in the so-called “problems of antiquity”—squaring the circle, trisecting the angle, doubling the cube, and constructing regular polygons. If you look in reference books, we *now* know that three of the four problems (all but squaring the circle) were proved impossible in 1837 by a French mathematician named Pierre Wantzel (1814–1848). I emphasized the word “now” because his contribution was largely ignored for over 100 years!

There has not been much written about Pierre Wantzel, especially in English. For instance, he does not appear in the massive 27-volume *Complete Dictionary of Scientific Biography, *and his Wikipedia entry is very minimal. The MacTutor site has the most thorough treatment I’ve seen. About 100 years ago Florian Cajori wrote an 8-page bio in the *Bulletin of the AMS*. In it, he cites three 19th century French sources. So far I’ve tracked down two of them.

I have a passable knowledge of French. It has been many years since I’ve studied French, so I can slowly go through and puzzle out a French math document. But it takes a while and I’m not always confident in the outcome. Fortunately, online translators like Google Translate and Microsoft Translator have gotten much better, so they provide a quick way to get the gist of a document in another language.

So, I took the two French articles, copied-and-pasted them into a Word file, cleaned up any errors, used the online translators to get a first approximation of the translation, and then started cleaning them up. Then it occurred to me: Maybe this would make a good crowdsourcing project. This is where you come in.

I made Latex files of these two documents, split them into two columns (French on the left, English on the right) and posted them to ShareLatex. I made them publicly editable. If you are interested in working on this translation project—go for it. I have no idea what is going to happen. It may be a huge success, it may be a disaster, or it may be somewhere in between. We may have one person do almost all the work or we may have an end-product that is the work of many people. If you work on this, leave your name as a comment at the start of the document. We’ll see how this goes! You can post comments, questions, and suggestions in the comment section at the end of this blog post.

Here’s the first document: the original 11-page article and the ~~editable latex~~. It was written by one of Wantzel’s collaborators, Jean-Claude Saint-Venant in 1848 shortly after Wantzel died. [UPDATE! I found out that there’s already a translation of this one online! So, I copied and pasted that version into my Latex file and gave the proper citation; here is a pdf version.]

Here’s the second document: the original 3-page article and the editable Latex. It was written by A. de Lapparent in 1895.

Thank you! I look forward to watching this take shape.

]]>I came across this neat pdf by Troy Jones about using salt to do geometry. So, over Thanksgiving break I got my kids and their cousins together to do a little mathematics. We cut various shapes out of paper, propped them up on glasses, and poured salt over them. The salt is a natural bisector. The ridges can be used to bisect angles and to find the locus of points equidistant from two curves. We had fun making triangle centers, Voronoi diagrams, and conic sections. I had a good time thinking about why this worked (it is a fun exercise to see why these ridges form the various conic sections).

In a recent paper by John Sharp I learned about tying a strip of paper into a regular pentagon. It goes back to ‘Tom Tit,’ which was the pen-name of Arthur Good (1853–1928).

The American Institute of Physics gave templates to make physics-related snowflakes. I used their template to make this Isaac Newton snowflake. (As cool as it is, it doesn’t have six-fold symmetry like a true snowflake.) They also have a crystallography and a Nikola Tesla snowflake.

I have been wanting to make a mathematical flip book for a long time. Yesterday was the day that I made it happen. My 13-year old son and I figured out how to do it using the Adobe suite (my son is becoming a self-taught Adobe wiz). I started by creating the first four stages of the Koch snowflake using Illustrator. My son imported them into After Effects and had them morph from one to the other. He used Media Encoder to export them as an animated gif:

Then we exported the frames of the gif as 90 separate images. I printed them on card stock with nine per page (here’s a pdf). I cut them out and used a binder clip to hold them together. It seemed like 90 frames was perhaps too many, so I took every other frame and made two books of 45 frames. Those worked well. Lastly, I put the two books back to back and took following video of the triangle turning into a snowflake turning back into a triangle. I’m excited to make more flip books.

I found this blog post, which has instructions on how to make a fractal Christmas tree. I made one and thought it looked great. In fact, it folds up nicely into a card. So my daughter made one to give to her teacher.

After that, I tried making one of my own design. Rather than cutting the paper in half at each stage, I cut it into thirds. The result is below. It looks cool, but when it is folded as a card, part of the inside sticks out.

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