Are there any rational values of for which the line is tangent to the graph of

Clearly the answer is yes: But my gut feeling was that this was the only such After some head scratching, I obtained the following proof.

Suppose they are tangent at . Then the point of tangency is Moreover, the slope of the tangent line is . Thus, must satisfy the two equations:

and

.

Using a trig identity,

.

So,

Note that is an algebraic number—it is the root of a polynomial with integer coefficients.

In 1882 Ferdinand von Lindemann famously proved that is transcendental (that is, non-algebraic) for every non-zero algebraic number and a similar proof holds for the sine and cosine functions.

Thus, because and is algebraic, it must be the case that . In particular,

This concludes the proof.

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The bag of gifts that I received at the end of the conference was amazing. It contained exotic dice, clever puzzles, unique playing cards, mathematical earrings, 3d printed objects, cool pencils, puzzle books, artwork, etc. Nearly all of the gifts were handmade. We haven’t received the written gifts yet. What a fantastic idea!

My gift was a set of instructions and templates for making *Gabriel’s horn* out of paper cones. Gabriel’s horn is the surface obtained by revolving the curve about the -axis. Mathematics professors wow introductory calculus students by sharing its paradoxical properties: it has finite volume, but infinite surface area. As they say, “you can fill it with paint, but you can’t paint it.”

Photos of the paper horn (in white and in rainbow) and a photo of the cones are shown below. I’m linking to a pdf copy of the article here in case you want to make one.

Note: my inspiration was this blog post by Dan Walsh in which he makes a pseudosphere out of cones.

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I received quite a few links. I’m looking forward to reading them. It seemed like there was a lot of interest, so I’m posting the unedited list here.

- Visualizing Math
- Jimi Cullen
- James Campbell
- Complex Projective 4-Space
- Fix epsilon>0
- Not Only Truth But Supreme Beauty
- 1ucasvb’s lab
- Curiosa Mathematica
- Math is Fun
- Math Professor Quotes
- Math Major Problems
- Herp, Derp, Math Nerd
- Eat, Sleep, Math
- None Sequiturs
- Maxwell’s Equations
- Euler, Erdős
- Yellow Pigs
- Prof. John Golden has had his students blog. Here are links to past semesters

If you know of any others, post them in the comments.

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Pretty cool! Let’s see why

Two terms are easy to deal with:

and

But why is

One way to prove this is via the inverse trigonometric identity

In this case

Because and are between 0 and , .

I didn’t use this approach on my first attempt to solve the problem. (To be honest, I didn’t know this identity existed before finding it online.) I used geometry and trigonometry. We are interested in the angle in the diagram below.

The law of cosines tells us that

This implies that and hence

(I wonder if there is a year of tau coming up sooner than the year 112233.)

Update: I just saw that Cut-the-Knot has a page devoted to this topic too.

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When I got back, the GPS said that I had paddled for 2.74 miles (14470 feet). Let’s say that I stayed exactly 50 feet off the shore for the entire trip. What is the perimeter of the lake?

We should probably make some assumptions about the shoreline and about my path. For simplicity, let’s say that my path is differentiable, that it bounds a region , and that the lake is the set , where

is the *radius neighborhood of *. Intuitively, the boundary of is always units off the port side of the boat. (See Ravi Vakil’s article “The Mathematics of Doodling” for more about this set.)

You may not think you have enough information to solve the problem, but you do! The clue is found in a well-known puzzler. Suppose we have an electrical wire running round the equator of the earth. We want to get it up off the ground and put it at the top of 50-foot electrical poles. How much longer must the wire be? [If you haven't seen this puzzler before, stop now and think about it.] If the radius of the Earth is feet, then the original wire must be feet long. The new wire is a circle with a radius 50 feet longer than before, so its length must be feet. That is, it must be feet longer.

It turns out that this argument can be generalized in various ways (again, see Vakil’s article for more information). First of all, if is any convex set with a polygonal boundary, then, just like in the puzzle, the perimeter of is units longer than the perimeter of . Vakil illustrates this with the picture below.

It turns out that the exact same result holds if is nonconvex—such as the region enclosed by my kayak trip. As the kayak goes in coves and around peninsulas, the “extra” regions of convexity and nonconvexity cancel each other out. In the end it is the same as going around a circle one time. Thus the perimeter of the lake is feet longer than my kayak path: feet.

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Here is the source code for this figure. If you click on the link you can get an editable copy of the document in WriteLatex

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Fun Friday question—who best fits this criteria: great mathematician + great expositor + living today?

—

Dave Richeson (@divbyzero) April 12, 2013

I got a great repsonse. Here is the complete—unedited—list of names (in alphabetical order).

- Micheal Atiyah
- Art Benjamin
- Andrea Bertozzi
- Manjul Bhargava
- Joan Birman
- Bela Bollobas
- Stephen Boyd
- María Chudnovsky
- Fan Chung
- John H. Conway
- Ingrid Daubechies
- Keith Devlin
- Marcus du Sautoy
- Jordan Ellenberg
- Joe Gallian
- Rob Ghrist
- Tim Gowers
- Judith Grabiner
- Brian Greene
- Benedict Gross
- Dusa McDuff
- Danica McKeller
- Curt McMullen
- Maryam Mirzakhani
- John Allen Paulos
- Mary Rees
- Jean-Pierre Serre
- Michael Sipser
- Ian Stewart
- Steven Strogatz
- Daina Taimina
- Terence Tao
- Eva Tardos
- Ravi Vakil
- Robin Wilson

If you have other suggestions, add them in the comments.

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I’ve put a sample document on WriteLatex. If you click on this link you can get an editable copy of the document. You can edit it and it won’t change my copy—so go wild with it! (This cool feature of WriteLatex is described on their WriteLatex for Education page.)

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Abstract. We answer the question: who first proved that is a constant? We argue that Archimedes proved that the ratio of the circumference of a circle to its diameter is a constant independent of the circle and that the circumference constant equals the area constant (). He stated neither result explicitly, but both are implied by his work. His proof required the addition of two axioms beyond those in Euclid’s

Elements; this was the first step toward a rigorous theory of arc length. We also discuss how Archimedes’s work coexisted with the 2000-year belief—championed by scholars from Aristotle to Descartes—that it is impossible to find the ratio of a curved line to a straight line.

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