Opening and closing treasure chests.

Suppose there are 500 treasure chests, each labeled 1 to 500. There are also 500 pirates. Initially all 500 treasure chests are closed. Then the first pirate opens all the treasure chests. Next, the second pirate opens every other treasure chest, labeled 2, 4, 6, and so on. Then the third pirate changes the state of the treasure chest from closed to open, or open to close on chests labeled 3, 6, 9, and so on. In general, the `n`th pirate changes the state of the chests that has label a multiple of `n`. 

After the 500th pirate, how many chests are open?

For which rational degree is its tangent rational?

For which rational number `r in QQ` is its tangent `tan r^o` rational? What about `cos r^o` and `sin r^o`?

We know the obvious ones, `tan 45^o = 1` is rational, is there another rational angle with its tangent rational?

Rationality of `tan 1^o`.

 Is `tan 1^o` a rational number?

Constructibility of trigonometric values.

If we can construct the lengths `cos alpha` or `sin alpha`, can we then construct `tan alpha`? How about the converse, if `tan alpha` is constructible, is `cos alpha` or `sin alpha` constructible? 

Probability of three segments forming a triangle.

Consider the unit interval `I = [0,1]` and pick two points `x,y` uniformly from `I` independently at random. This creates three segments. What is the probability that these three segments can form the sides of a triangle?

Would you guess this is "likely", `> 50%`, or "less likely", `< 50%`?

Probability of a self-intersecting brokenline.

On the plane starting at some point `O`, pick a random angle uniformly and draw a unit segment `OA`. Again pick a random angle uniformly and draw a unit segment `AB`. Repeat two more times to get `BC` and `CD`. What is the probability of this broken line `OABC` self-intersects?

Test post: jxgbox

Testing jxgbox to draw with javascript library jxgbox

Self-intersections: 0

Non-constructible angle of `20^o`.

 In the popular lore of straightedge and compass construction, we have

One cannot trisect an arbitrary angle.

It is often used as an example that `60^o` cannot be trisected into `20^o`. And this is demonstrated by the fact that the length `cos 20^o` is not a constructible length. By the way, the quoted lore above about "arbitrary angles cannot be trisected" itself has nuances, but I will explain it in another post.

As for `cos 20^o` is not constructible, we shall use the lore that 

Given the length `1` on the plane, then the length `r` is constructible by straightedge and compass, if and only if `r` lies in a finite tower of quadratic extensions from `QQ`.

As a corollary,

If `r` is a constructible length, then `r` is a root of some irreducible polynomial over `QQ` of degree `2^m`, for some `m`. 

(Note the converse is not necessarily true.)

So it suffices to show that `r = cos 20^o` is a root of some irreducible polynomial of an inappropriate degree. Indeed, recall the triple angle formula, we have 
$$
\cos 3\theta = 4\cos^3\theta - 3\cos \theta, 
$$ (cf. Chebyshev polynomials) we see that `r` is a root of the polynomial equation `1/2 = 4 r^3 - 3r` or `8r^3 -6r-1 =0`. Since this is a cubic polynomial, if it is reducible over `QQ` then it would have a rational root. But by rational root test one sees that `8r^3-6r-1` has no rational root. Hence it is irreducible of degree 3. Since `3 ne 2^m` for any non-negative integer `m`, we conclude that `cos 20^o` is not constructible!

Construction of a regular pentagon.

It is well-known that a regular pentagon is constructible. Here is the simplest way that I know of.

Note that in a regular pentagon inscribed in a unit circle shown on the right, the angle `angle BAC = 36^o`. This makes the length `AB = (1+sqrt(5))/2`, as `triangle ABC` is a right triangle.

So all we need is to be able to construct the length `(1+sqrt(5))/2`. To do so, observe the diagram on the right, where two smaller circle of radius `1/2` is inscribed in a unit circle. If we extend the segment of `AP` to `AQ`, then we will find that `AQ = (1+sqrt(5))/2`, precisely what we need!

Constructible angles of integral degrees.

For which positive integer `n` is `n^o` constructible (classically with straightedge and compass)? There is a satisfactory answer:

Let `n` be a positive integer. Then the angle `n^o` is constructible if and only if `n` is a multiple of `3`.

The proof is straightforward as well, provided we believe some other facts. First note the classic fact that

A regular pentagon is constructible, hence `72^o` is constructible.

And that

If `A` and `B` are any two constructible angles, then so is any integer combination `xA + yB`, where `x, y in ZZ`.

So, since we can also construct an equilateral triangle, we can construct `60^o`. Hence we can construct `72^o - 60^o = 12^o`. Now, using the fact that

If `A` is a constructible angle, then so is its bisection `A/2`.

This shows we can construct `6^o` and `3^o`. Now using the fact of integer combinations of constructible angles, we have the first direction:

For any integer `x`, the angle `3x^o` is constructible.

Now we show if integer `n` is not divisible by `3`, then `n^o` is not constructible.

 

We show this by contradiction. Suppose to the contrary that some integer `n` is not divisible by `3` but is constructible. Then by repeated subtraction of `3^o`, we can construct the angles `1^o` or `2^o`, as `n mod 3` is `1` or `2`. And since we can bisect, constructing `2^o` means we can construct `1^o`. So far, this says if such integer `n` that is not a multiple of `3` but `n^o` is constructible, then we can construct `1^o`. But wait, this is crazy! Because of the venerated fact that

The angle `20^o` is not constructible.

We should not be able to construct `1^o`, otherwise by integer combination of constructible angles, we could construct any integral angles!