Interestingly, when the legs are folded, the underside of the table encodes a famous proof of the Pythagorean theorem.

As we see in the photo below, the edge of the table and two adjacent legs form a right triangle with legs *a *and *b *and hypotenuse *c*. The table is square, so it has area But, the table is divided up into four right triangles each having area and a smaller square of area Thus,

Ta da!

After I posted this on Twitter I found out that Debra Borkovitz made the same observation about her card table.

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- is a Fermat prime.
- The teenage Gauss proved that the regular 17-gon is constructible by compass and straightedge (which is related to the previous bullet).
- There are 17 wallpaper groups.
- A haiku has 17 syllabes.
- A Sudoku needs at least 17 clues to have a unique solution.
- Theodorus proved that each of …, is an integer or is irrational. (Actually, the wording in Plato’s
*Theaetetus*is ambiguous; it could have been that is the first one that Theodorus was unable to show was irrational.) - According to hacker lore, 17 is the “least (or most) random number.”
- To the nearest order of magnitude, the universe is seconds old (approximately seconds).
- It is the smallest number that is the sum of two distinct positive integers raised to the fourth power:
- It is the smallest number that can be written as the sum of a square and a cube in two different ways
- Some cicadas have a 17-year life cycle.
- There are 17 ways to write 17 as the sum of primes.
- The Italians think 17 is unlucky (apparently because XVII can be rearranged to be VIXI, which means “my life is over”).
- 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.”
- There are 17 nonabelian groups of order at most 17.
- 17 is the smallest whole number whose reciprocal contains all ten digits:
- In Ramsey theory In other words, when it is possible to color the edges of any graph with
*n*vertices using three colors so that there are no monochromatic triangles. But this is impossible for (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.”

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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 (see figure below), bisect the segment at Then draw the segment perpendicular to . Draw a circle with center and radius Next, arrange the carpenter’s square so that one edge goes through , one edge is tangent to the circle, and the vertex, , sits on . Then .

I made an applet to illustrate this trisection.

Brooks’s proof used trigonometry. My question is: Is there a *geometric proof* that 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 and Then it suffices to show that Let be the point of tangency. Draw segment and extend to on Then draw segments and Because lines and are parallel, and hence is a parallelogram. This implies that is the midpoint of the diagonal so Moreover, Because is the hypotenuse of the right triangle and is the median, We see that so Finally, because is a chord of the circle, that is,

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

In the proof we are referring to the figure below. Let and Then it suffices to show that Let be the point of tangency of the carpenter’s square and the circle. Because Construct Because is a tangent line and is a radius of the circle, Because and are parallel. Thus Construct a point so that is a parallelogram. Notice that and are collinear. Let be the midpoint of . Construct the line segment and the circle with center and radius . Note that the circle passes through and —the first three because and the fourth because and is a diameter of the circle. Because is parallel to Finally, is a central angle and is an inscribed angle and both cut off the same arc of the circle; thus

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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.)

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

Click the following image to download the pdf.

Making the shape and seeing the illusion is easy.

- Cut out the figure at the top of the page.
- Fold a sharp crease along the dotted line.
- Tape the left and right edges together.
- Fold a sharp crease along the taped seam.
- 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.
- Close one eye. Look down on the shape at a 45 degree angle so that the two creases line up with each other.
- Then turn it around 180 degrees and look again.

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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 and the other is at (in the video this second viewer is you, in the mirror). They are looking down on a curve However, from their vantage points it looks like they are seeing two different curves in the -plane: and respectively.

In our example the two observed curves are the unit circle and the square passing through the points , as shown in the -plane below. We will have to break each of these shapes into two different curves, so we’ll have and Also, we could choose to be some suitably large number greater than 1, but in fact, as we will see, taking the limit as 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 Let’s fix , and let and be two points on the curves in the -plane. In order for the person at to view her shape, must lie on the line (see figure below). Likewise, for the person at to see his shape, must lie on the line Thus, must be the point of intersection of lines and (We know that the lines are not skew because they lie in the plane containing the points and and for appropriate choices of and the lines intersect and the point of intersection is below

It is straightforward to show that is a parametrization of the line and is a parametrization of A little algebra shows shows that their point of intersection is

and thus our desired curve is

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

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

for and the other half by

for 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.]

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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.)

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- Twist the paper zero times, and tape the ends (making a cylinder). Cut down the midline.
- Give the paper one half-twist, and tape the ends (making a Möbius band). Cut down the midline.
- Give the paper two half-twists ,and tape the ends. Cut down the midline.
- Give the paper three half-twists, and tape the ends. Cut down the midline.
- Twist the paper zero times, and tape the ends. Cut into thirds.
- Give the paper one half-twist, and tape the ends. Cut into thirds.
- Give the paper two half-twists, and tape the ends. Cut into thirds.
- 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:

- The zippers I used were 12″ or 14″ long. They seemed to work well.
- 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.
- 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.
- Ideally I would have sewn the two zippers side-by-side. But we don’t own a sewing machine. So I stapled them.
- 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.
- 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!

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

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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 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 and 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 is the 3:3:1 triangle.

Johnson’s proof that 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 is isosceles and *E *is on *AD*. Moreover, suppose and , prove that

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 *E *on *BD *and *F *on *AD * such that then triangle *ABD *is a 3:3:1 triangle. (This is easy to prove. Suppose Then, because is isosceles, . So the exterior angle of is Because is isosceles, Lastly, observe that and are similar, so It follows that )

Thus, a possible route to proving the theorem is to find a point *F *in the original diagram so that

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.

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