Posted by: Dave Richeson | April 14, 2017

I have encountered the number 17 several times in the last few weeks—enough times that it caught my attention. So I challenged myself to write a list of seventeen interesting things about the number 17. I tried to be as mathematical as possible. I wasn’t able to get seventeen facts on my own, so I turned to the internet. As it turns out, there are quite a few web pages about the number 17 (shocker, I know!). In particular, it turns out that a mathematics professor at Hampshire College, David Kelly, has been lecturing about 17 for a while. When he retired, the college changed all of the 15 MPH speed limit signs to 17 MPH.

1. $17=2^{2^{2}}+1$ is a Fermat prime.
2. The teenage Gauss proved that the regular 17-gon is constructible by compass and straightedge (which is related to the previous bullet).
3. There are 17 wallpaper groups.
4. A haiku has 17 syllabes.
5. A Sudoku needs at least 17 clues to have a unique solution.
6. Theodorus proved that each of $\sqrt{3},$ $\sqrt{4},$ $\sqrt{5},$…, $\sqrt{17}$ is an integer or is irrational. (Actually, the wording in Plato’s Theaetetus is ambiguous; it could have been that $\sqrt{17}$ is the first one that Theodorus was unable to show was irrational.)
7. According to hacker lore, 17 is the “least (or most) random number.”
8. To the nearest order of magnitude, the universe is $10^{17}$ seconds old (approximately $4.32\times 10^{17}$ seconds).
9. It is the smallest number that is the sum of two distinct positive integers raised to the fourth power: $17=1^4+2^4.$
10. It is the smallest number that can be written as the sum of a square and a cube in two different ways $17=3^2+2^3=4^2+1^3.$
11. Some cicadas have a 17-year life cycle.
12. There are 17 ways to write 17 as the sum of primes.
13. The Italians think 17 is unlucky (apparently because XVII can be rearranged to be VIXI, which means “my life is over”).
14. Plutarch wrote “The Pythagoreans also have a horror for the number 17, for 17 lies exactly halfway between 16, which is a square, and the number 18, which is the double of a square, these two, 16 and 18, being the only two numbers representing areas for which the perimeter equals the area.”
15. There are 17 nonabelian groups of order at most 17.
16. 17 is the smallest whole number whose reciprocal contains all ten digits: $\frac{1}{17}=0.\overline{0588235294117647}.$
17. In Ramsey theory $R(3,3,3)=17.$ In other words, when $n\le 16$ it is possible to color the edges of any graph with vertices using three colors so that there are no monochromatic triangles. But this is impossible for $n=17$ (the complete graph with 17 vertices can’t be colored in this way). I don’t think this was what Stevie Nicks was signing about in her song “Edge[s] of [K_]Seventeen.”

Posted by: Dave Richeson | March 30, 2017

## A Geometric Proof of Brooks’s Trisection?

[UPDATE: we have a proof! I included it at the end of the blog post.]

Yesterday I was looking at a few methods of angle trisection.

For instance, I made this applet showing how to use the “cycloid of Ceva” to trisect an angle. (It is based on Archimedes’s neusis [marked straightedge] construction.)

I also found David Alan Brook’s College Mathematics Journal article “A new method of trisection.” He shows how you can use the squared-off end of a straightedge (or equivalently, a carpenter’s square) to trisect an angle. (This is a different carpenter’s square construction than the one I wrote about recently.)

To perform the trisection of $\angle ABC$ (see figure below), bisect the segment $BC$ at $D.$ Then draw the segment $DE$ perpendicular to $AB$. Draw a circle with center $C$ and radius $CD.$ Next, arrange the carpenter’s square so that one edge goes through $B$, one edge is tangent to the circle, and the vertex, $F$, sits on $DE$. Then $\angle ABF=\frac13\angle ABC$.

I made an applet to illustrate this trisection.

Brooks’s proof used trigonometry. My question is: Is there a geometric proof that $\angle ABF=\frac13\angle ABC?$ I spent a little while working on it yesterday and couldn’t find one. If you can, let me know!

UPDATE: We have a proof! Thank you Marius Buliga!

In the proof we are referring to the figure below. Let $\alpha=\angle ABF$ and $\beta=\angle CBF.$ Then it suffices to show that $\beta=2\alpha.$ Let $G$ be the point of tangency. Draw segment $CG$ and extend $FD$ to $J$ on $CG.$ Then draw segments $DG,$ $BJ,$ and $CF.$ Because $\angle CGF=90^\circ,$ lines $CG$ and $BF$ are parallel, and hence $BFCJ$ is a parallelogram. This implies that $D$ is the midpoint of the diagonal $FJ;$ so $DF=DJ.$ Moreover, Because $FJ$ is the hypotenuse of the right triangle $FJG$ and $DG$ is the median, $\angle DGF=\angle DFG.$ We see that $\angle DFG=\angle ABF$ so $\angle DGF=\alpha.$ Finally, because $DG$ is a chord of the circle, $\angle DCG=2\angle DGF;$ that is, $\beta=2\alpha.$

Update 2: Andrew Stacey just sent me an another proof:

In the proof we are referring to the figure below. Let $\alpha=\angle ABF$ and $\beta=\angle CBF.$ Then it suffices to show that $\beta=2\alpha.$ Let $G$ be the point of tangency of the carpenter’s square and the circle. Because $\angle BEF=\angle BFG=90^\circ,$ $\angle DFG=\angle EBF=\alpha.$ Construct $CG.$ Because $FG$ is a tangent line and $CG$ is a radius of the circle, $\angle CGF=90^\circ.$ Because $\angle CGF=\angle BFG= 90^\circ,$ $BF$ and $CG$ are parallel. Thus $\angle BCG=\beta.$ Construct a point $H$ so that $BCGH$ is a parallelogram. Notice that $F,$ $B,$ and $H$ are collinear. Let $J$ be the midpoint of $GH$. Construct the line segment $DJ$ and the circle with center $J$ and radius $GJ$. Note that the circle passes through $G,$ $H,$ $D,$ and $F$—the first three because $HJ=GJ=DJ,$ and the fourth because $\angle GFH=90^\circ$ and $GH$ is a diameter of the circle. Because $DJ$ is parallel to $FH,$ $\angle DJG=\beta.$ Finally, $\angle DJG$ is a central angle and $\angle DFG$ is an inscribed angle and both cut off the same arc of the circle; thus $\beta=\angle DJG=2\angle DFG=\alpha.$

Posted by: Dave Richeson | October 2, 2016

## Two More Impossible Cylinders

Earlier this year I wrote a couple blog posts about reverse engineering Sugihara’s impossible cylinder illusion. I then wrote it up more formally, and it has appeared in Math Horizons (pdf).

The example I gave on my blog and in the article was a cylinder that looked like a circular cylinder but like a square cylinder in the mirror. You can download a printable pdf template to make your own, and you can watch this video that I made.

Today I designed two more impossible cylinders. The first looks like a square cylinder and like a triangular cylinder in the mirror. Download a printable pdf template to make your own.

I also designed one that looks like a heart-shaped cylinder and like a diamond-shaped cylinder in the mirror. (The spade/club one is an exercise for the reader.) Here’s a template to make your own.

Update: I just created another cylinder. I’m not as pleased with this one. The shape of the cylinder is complicated enough that it is difficult to get it to look right. Here’s my best attempt at a star/moon cylinder. (Pdf template.)

Posted by: Dave Richeson | July 6, 2016

## Make a Sugihara Circle/Square Optical Illusion Out of Paper

Yesterday I explained the mathematics behind Sugihara’s Circle/Square Optical Illusion, which appears in this video.

Today I created a printable template from which you can make your own version of Sugihara’s object.

Making the shape and seeing the illusion is easy.

1. Cut out the figure at the top of the page.
2. Fold a sharp crease along the dotted line.
3. Tape the left and right edges together.
4. Fold a sharp crease along the taped seam.
5. Lightly squeeze together the creased sides so that the shape opens. Looking down on the shape it should have the shape at the bottom of the printout.
6. Close one eye. Look down on the shape at a 45 degree angle so that the two creases line up with each other.
7. Then turn it around 180 degrees and look again.
Posted by: Dave Richeson | July 5, 2016

## Sugihara’s Circle/Square Optical Illusion

[Update: Check out my second post in which I provide a template so you can make your own Sugihara circle/square object out of paper.]

Kokichi Sugihara created a video called Ambiguous Optical Illusion: Rectangles and Circles. In it he shows a variety of 3-dimensional objects that look like one shape when viewed from the front but look like a different shape in the mirror behind it.

In this blog post we show how he achieved the effect. For simplicity, we will show how he made a shape that looks like a circular cylinder from the front and a square cylinder in the mirror.

The following applet shows our final product (clicking the image links to the GeoGebra applet). It is a closed curve that represents the top rim of Sugihara’s shape. You can rotate the axes with your mouse. If you view the coordinate system with the positive green and blue axes lined up (1 with 1, 2 with 2, and so on), the curve will look like the unit circle in the green-red plane. If you drag the image so that the positive blue axis lines up with the negative green axis (1 with -1, 2 with -2, and so on), it will look like you are viewing a square (oriented as a diamond) in the green-red plane.

Here are screenshots showing the two views.

How does it work? It is all about perspective.

To set this up mathematically, we imagine two viewers in 3-dimensional space. One viewer is at ${P(0,a,a)}$ and the other is at ${Q(0,-a,a)}$ (in the video this second viewer is you, in the mirror). They are looking down on a curve ${\mathbf{r}(t).}$ However, from their vantage points it looks like they are seeing two different curves in the ${xy}$-plane: ${\langle u, f(u),0\rangle}$ and ${\langle u, g(u),0\rangle,}$ respectively.

In our example the two observed curves are the unit circle and the square passing through the points ${(\pm 1, \pm 1, 0)}$, as shown in the ${xy}$-plane below. We will have to break each of these shapes into two different curves, so we’ll have ${f_{1}(u)=1-|u|,}$ ${f_{2}(u)=|u|-1,}$ ${g_{1}(u)=\sqrt{1-u^{2}},}$ and ${g_{2}(u)=-\sqrt{1-u^{2}}.}$ Also, we could choose ${a}$ to be some suitably large number greater than 1, but in fact, as we will see, taking the limit as ${a}$ tends to infinity produces a lovely final expression. For now we will continue to work in generalities and will wait to insert these specifics later.

Our aim is now to define ${\mathbf{r}(t).}$ Let’s fix ${t}$, and let ${A(t,f(t),0)}$ and ${B(t,g(t),0)}$ be two points on the curves in the ${xy}$-plane. In order for the person at ${P}$ to view her shape, ${\mathbf{r}(t)}$ must lie on the line ${AP}$ (see figure below). Likewise, for the person at ${Q}$ to see his shape, ${\mathbf{r}(t)}$ must lie on the line ${BQ.}$ Thus, ${\mathbf{r}(t)}$ must be the point of intersection of lines ${AP}$ and ${BQ.}$ (We know that the lines are not skew because they lie in the plane containing the points ${P,}$ ${Q,}$ and ${(t,0,0),}$ and for appropriate choices of ${f,}$ ${g,}$ and ${a}$ the lines intersect and the point of intersection is below ${z=a.)}$

It is straightforward to show that ${\mathbf{r}_{AP}(s)=\langle ts, (f(t)-a)s+a, -as+a\rangle}$ is a parametrization of the line ${AP}$ and ${\mathbf{r}_{BQ}(s)=\langle ts, (g(t)+a)s-a, -as+a\rangle}$ is a parametrization of ${BQ.}$ A little algebra shows shows that their point of intersection is

$\displaystyle \left(\frac{2at}{g(t)-f(t)+2a},\frac{af(t)+ag(t)}{g(t)-f(t)+2a},\frac{ag(t)-af(t)}{g(t)-f(t)+2a}\right),$

and thus our desired curve is

$\displaystyle \mathbf{r}(t)=\tfrac{a}{g(t)-f(t)+2a}\cdot\langle2t,f(t)+g(t),g(t)-f(t)\rangle.$

Because ${a}$ is a large value, we can take the limit as ${a}$ goes to infinity. This yields the elegant expression

$\displaystyle \mathbf{r}(t)=\langle t,\tfrac12(f(t)+g(t)),\tfrac12(g(t)-f(t))\rangle.$

We may now plug in our functions. The portion of our curve with nonnegative ${y}$-coordinates is given by

$\displaystyle \mathbf{r}_{1}(t)=\langle t,\tfrac12(1-|t|+\sqrt{1-t^{2}}),\tfrac12(\sqrt{1-t^{2}}+|t|-1)\rangle$

for ${-1\le t\le 1,}$ and the other half by

$\displaystyle \mathbf{r}_{2}(t)=\langle t,\tfrac12(|t|-1-\sqrt{1-t^{2}}),(1-\sqrt{1-t^{2}}-|t|)\rangle$

for ${-1\le t\le 1.}$ This is the curve shown in the applet.

[Update: See the comment by Joshua and my reply for a simpler way of obtaining the parametrization.]

Posted by: Dave Richeson | April 21, 2016

## Measuring Tapes for Circles and Spheres

I’d like to thank Matt Parker for introducing me to diameter tapes (or D-tapes). These are measuring tapes used by foresters to measure the diameters of trees. The forester wraps the measuring tape around a tree as if measuring the circumference, but the scale on the tape is adjusted so that the measurement gives the diameter (assuming the tree has a circular cross section). It uses the fact that C=πD.

Parker mentioned the existence of these measuring tapes in his talk at Gathering 4 Gardner 12 and he gave every attendee a D-tape that he made (see below).

That got me thinking. If you know the circumference, you can compute the diameter. But you can also compute the area. Also, if you wrap the tape around the equator of a sphere, you can compute the the volume and the surface area of the sphere.

Thus, inspired by this, I made measuring tapes to measure the diameter and area of a circle and the volume and surface area of a sphere. The top of each tape is marked off in centimeters, so it measures the circumference. The bottoms can be used to measure the diameter of the circle, the cross-sectional area, the volume of the sphere, and the circumference of the sphere.

I’ve also created a printable pdf of these measuring tapes, which should, I hope, print to the right scale so that 1 cm on the measuring tape is actually 1 cm. (Note that although a 25 cm ruler looks long, when you wrap it around a circle, you find that the diameter is only 8 cm. Not very big.)

It was a fun exercise to figure out how to mark and make the rulers. Except for the diameter ruler, the relationships aren’t linear. (I set everything up in Excel, then I moved the info over to a Geogebra spreadsheet. I used GeoGebra to draw everything. I exported the images as pdfs. Then I cleaned them up and added the text in Adobe Illustrator.)

Posted by: Dave Richeson | April 11, 2016

## Zip-Apart Möbius Bands

I’ve taught topology many times. One of the highlights for the students (and for me) is the investigation of the Möbius band—the one sided, one edged, non-orientable surface with boundary. On the day we introduce the Möbius band I bring many strips of paper, clear tape, and scissors and have the students make conjectures about what would happen if we taped and cut apart various topological shapes. Here are some activities that are fun to do:

1. Twist the paper zero times, and tape the ends (making a cylinder). Cut down the midline.
2. Give the paper one half-twist, and tape the ends (making a Möbius band). Cut down the midline.
3. Give the paper two half-twists ,and tape the ends. Cut down the midline.
4. Give the paper three half-twists, and tape the ends. Cut down the midline.
5. Twist the paper zero times, and tape the ends. Cut into thirds.
6. Give the paper one half-twist, and tape the ends. Cut into thirds.
7. Give the paper two half-twists, and tape the ends. Cut into thirds.
8. Give the paper three half-twists, and tape the ends. Cut into thirds.

In fact, these activities are fun for people of any age. My senior math majors enjoy it, and my kids’ kindergarten classes have too.

Last week I attended the 12th biennial Gathering 4 Gardner conference—a wonderful meeting of people interested in mathematics, puzzles, games, magic, and skepticism. One of the speakers (Iwahiro Hirokazu Iwasawa) suggested making zip-apart Möbius bands. Genius! And perfect timing (since I’m teaching topology this semester).

When I got home I bought zippers and Velcro and used them to make reusable strips. Half of them were just one zipper with Velcro on the ends. These can be used to do activities 1-4 above. Then I took identical pairs of zippers and joined them side-by side to make strips for activities 5-8.

Some tips:

1. The zippers I used were 12″ or 14″ long. They seemed to work well.
2. The Velcro has adhesive on the back, so I could just stick them to the ends of the strips. I did it so that the velcro folded over the ends and so was on both sides. That gave me flexibility if I gave an even or an odd number of half twists.
3. For the doubled zippers, align them so that both zippers start at the same end of the strip and so that one is on the top side and one is on the bottom side. Then, when you join with an odd number of half-twists, one begins where the other ends and they’re both on the same “side” (locally) of the Möbius band.
4. Ideally I would have sewn the two zippers side-by-side. But we don’t own a sewing machine. So I stapled them.
5. I learned the hard way that there are separating zippers (think of the zipper on a parka that comes apart at the bottom) and closed-end zippers (think the zipper on your pants that stops at the bottom). You want the former type for this activity.
6. The zippers were about \$3 apiece, but my wife told me about a local fabric store that was going out of business, so I got them for 70% off!

### Video of me unzipping a Möbius band in thirds

Posted by: Dave Richeson | April 7, 2016

## A Trig-free Proof of Crockett Johnson’s Theorem

I recently wrote a post about Crockett Johnson’s neusis construction of a regular heptagon. Johnson’s proof that the construction was correct required heavy trigonometry. I asked if there was a geometric proof that didn’t use trigonometry.

My friend Dan Lawson came to the rescue—he posted the following lovely proof on Twitter. Thanks Dan!

Posted by: Dave Richeson | March 23, 2016

## A Geometry Theorem Looking for a Geometric Proof

[Update: Dan Lawson has proved the theorem without trigonometry. Thanks, Dan!]

I spent a good chunk of last week reading about David Johnson Leisk (1906–1975), who is better known by his nom-de-plum Crockett Johnson. Johnson is most well known as the author of Harold and the Purple Crayon, a children’s book from 1955, and its sequels. Johnson was also the author of the 1940s comic Barnaby.

Later in his life Johnson became interested in mathematics. He was particularly interested in geometry, and most specifically in the problems of antiquity (squaring the circle, trisecting the angle, doubling the cube, and constructing regular polygons). He turned many geometric theorems into works of art. Eighty of his paintings are now at the Smithsonian.

Johnson even created new mathematics. I would like to discuss one of his contributions here. (See his article “A Construction for a Regular Heptagon,” The Mathematical Gazette Vol. 59, No. 407 (Mar., 1975), pp. 17-21.)

The heptagon is notewothy because it is the regular polygon with the fewest number of sides that cannot be constructed with compass and straightedge alone. In his article, Johnson gives a way to construct the heptagon using a marked straightedge (this is called a neusis construction). Johnson did not give the first neusis construction of a heptagon—François Viète gave the first such construction in 1593. (Also, Archimedes gave an unorthodox neusis-like construction).

However, Johnson’s proof used trigonometry (including the law of cosines and several trigonometric identities). My question to you is: Is there a purely geometric proof of his result? I played around with it for a little while and couldn’t find one, and I couldn’t find a geometric proof in the literature.

## The construction

The key to Johnson’s construction is producing 3:3:1 triangle; that is, a triangle in which the angles are in a 3:3:1 ratio (they would be $3\pi/7,$ $3\pi/7,$ and $\pi/7).$ The three vertices of the triangle are three vertices of a regular heptagon. If we construct the circumcircle, then it is easy to construct the four remaining vertices with a compass and straightedge.

Here’s his construction of a 3:3:1 triangle using a marked straightedge—that is, an otherwise ordinary straightedge, but possessing a mark one unit from the end (or equivalently, two marks one unit apart). We begin by drawing a line segment AB of length one (see below, left). Construct a unit line segment AC perpendicular to AB. Also, construct the perpendicular bisector to AB; call it l. Then construct a circle with center B and radius BC. Now we perform the neusis construction with the marked straightedge: Construct a line AD so that D is on l and D is one unit from the circle. (That is, the end of the straightedge is at D, the mark is on the circle, and the edge passes through A.) Then $\triangle ADB$ is the 3:3:1 triangle.

## A geometric proof?

Johnson’s proof that $\triangle ABD$ is a 3:3:1 triangle used trigonometry (see his article for details). Is there a geometric proof?

Boiled down to its essence, here’s the question: Suppose $\triangle ABD$ is isosceles and E is on AD. Moreover, suppose $ED=1,$  $AB=1,$ and $BE=\sqrt{2}$, prove that $3\angle ADB= \angle DAB=\angle ABD.$

Here is a fact that may help. Johnson discovered this fact—ironically—when he was sitting in a café in Syracuse on Sicily (the birth place of Archimedes) playing with the wine and food menus and some tooth picks. If we have an isosceles triangle ABD with the point on BD and on AD  such that $DE=EF=BF=AB,$ then triangle ABD is a 3:3:1 triangle. (This is easy to prove. Suppose $\angle ADB=\theta.$ Then, because $\triangle DEF$ is isosceles, $\angle DFE=\theta$. So the exterior angle of  $\triangle DEF,$ $\angle BEF,$ is $2\theta.$ Because  $\triangle BEF$ is isosceles, $\angle EBF=2\theta.$ Lastly, observe that  $\triangle ABD$ and  $\triangle ABF$ are similar, so $\angle ABF=\theta.$ It follows that $\angle ABD=\angle BAD=3\theta.$)

Thus, a possible route to proving the theorem is to find a point in the original diagram so that $DE=EF=BF=AB=1.$

If you find a proof, I’d love to hear about it!

Here are two of the paintings Johnson made from his work with the heptagon.

Johnson’s “Heptagon from Its Seven Sides” (1973)

Johnson’s “Heptagon from Its Seven Sides” (1973)

Posted by: Dave Richeson | March 11, 2016

## A Trisectrix from a Carpenter’s Square

UPDATE: The article is now published. Read it in Mathematics Magazine.

Yesterday I posted an article to the arXiv, “A Trisectrix from a Carpenter’s Square.

Abstract: In 1928 Henry Scudder described how to use a carpenter’s square to trisect an angle. We use the ideas behind Scudder’s technique to define a trisectrix—a curve that can be used to trisect an angle. We also describe a compass that could be used to draw the curve.

I also made a GeoGebra applet to accompany the article. Give it a try.