**Problem 2019-1/A**

*Three gamblers select a non-negative probability distribution with mean 1. Say these distributions
are F,G,H. Then x is sampled from F, y is sampled from G, z is sampled from H. Biggest number wins.
What distributions should the gamblers choose?*

__Solution:__

At first sight it doesn't matter which distributions with mean 1 the gamblers choose, because the
expectation value is always 1.

However, if X chooses the distribution F that yields the sample x = 1 with certainty, and Y and Z both use the
distribution that yields each of the samples 0.1 and 1.9 with chance 1/2, then the chance that X wins is only 1/4.

To guarantee that the chance of winning solo is at least 1/3, the gamblers have to choose distributions that yield
big samples with a relatively big chance and equal samples with zero chance.

So each of them might best choose a continuous distribution E whose density function e(x) is slowly decreasing.

I propose they choose e(x) = a/(x+b)^{2+c} for some c > 0.

From ∫_{0}^{∞} e(x) dx = 1 = ∫_{0}^{∞} x*e(x) dx we deduce
b = c and a = (c+1)c^{c+1}.
Remark e(0) = (c+1)/c.

For c > 1 the variance is finite and equal to (c+1)/(c-1), which descends to the limit 1
if c tends to ∞.

The 50-percentile is c*2^{1/(c+1)}-c, which ascends to the limit ln(2) = 0.69315.. if c tends to ∞.

So each gambler may choose some c for which the variance is not too big nor too small and the 50-percentile is not
too small and the calculating facilities not too troublesome.

If they should agree to choose the same c, for instance c = 10, then each of them only has to take a random sample r
from (0,1), because the sample s is found from the equation ∫_{0}^{s} e(x) dx = r, so the largest s
comes from the largest r.

**Problem 2019-1/B**

*For given m ≥ 3, consider the regular m-gon inscribed in the unit circle. We denote the area of this m-gon by
A _{m}. Suppose m is odd. Prove that 2A_{m} and A_{2m} have the same minimal polynomial.*

__Solution:__

We find A_{m} = (m/2)*sin(2π/m).

Let y_{1} = A_{2m} = m*sin(π/m) and y_{2} = 2A_{m} = m*sin(2π/m).
Let x_{1} = m*cos(π/m) and x_{2} = m*cos(2π/m).

According to de Moivre, (cos(φ) + i sin(φ))^{k} = cos(kφ) + i sin(kφ),
so Im(x_{1} + i y_{1})^{m} = 0 and Im(x_{2} + i y_{2})^{m} = 0.

Moreover, since m is odd, none of these last two equations holds for an exponent less than m.

But both for x + iy = x_{1} + i y_{1} and for x + iy = x_{2} + i y_{2}
we have also

Im(x + i y)^{m} =
Im (Σ_{k=0}^{m} (m over k) x^{m-k} (iy)^{k}) =
m yx^{m-1} - (m over 3) y^{3}x^{m-3} + ... ± y^{m}, where the exponents of x are all even.

Using cos^{2} = 1 - sin^{2}, we can turn this into y times the minimal polynomial p(y) over the
rationals, which is the same for y = y_{1} and y = y_{2}.

For instance, for m=3 we get p(y) = 27 - 4y^{2} and for m=5 we get 3125 - 500y^{2} + 16y^{4}.

**Problem 2019-1/C**

*Let n be a prime number. Show that there are no groups with exactly n elements of order n. What happens
with this statement if n is not a prime number?*

__Solution:__

We deal with the general case at once, so let n have prime factorization
p_{1}^{a1}...p_{k}^{ak} with p_{1} ≤ ... ≤ p_{k}.
Let φ denote Euler's totient function.

Suppose a group G contains an element x of order n. Then x generates a cyclic subgroup C of G
that contains exactly φ(n) elements of order n, which generate the same subgroup C.

If G contains another element y of
order n, outside C, then y generates a cyclic subgroup D of G that has only elements of order less than n in common
with C. Etc.

So the union of all these cyclic subgroups of order n, if finite, contains mφ(n) elements of order n for some
natural m.

This number mφ(n) is equal to n iff m(p_{1}-1)..(p_{k}-1) = p_{1}..p_{k}.

Comparing the number of factors 2 at the left and right hand sides, we establish p_{1} = 2 and k ≤ 2. If
k=2 then by comparing the number of small prime factors at the left and right hand sides, we establish that
p_{2} must be 3.

We now distinguish between the cases k=1 and k=2.

1) Let k=1, so n = 2^{a} and m=2. So there are exactly two cyclic subgroups of order 2^{a}.

First suppose a=1. If the two cyclic subgroups of order 2 are generated by x and y, respectively, then there must be a
third cyclic subgroup
of order 2, generated by xy=yx. Contradiction. Here we finish the first part of the problem.

Next suppose a>1. There are several groups with 4 elements of order 4, for instance the direct product of C_{4}
and C_{2}, where C_{s} is cyclic of order s.

2) Let k=2, so n = 2^{a}3^{b} and m=3. There are several groups with 6 elements of order 6, for instance
the direct product C_{3}*C_{2}*C_{2}.

So the statement in the problem holds if n is prime, but not in general.

**Problem 2019-2/A** (edited)

*Let A = (a _{i,j} ) be the matrix on the natural base e in threedimensional euclidean space of a rotation R
(not equal to identity).
Suppose none of a_{2,3}+a_{3,2}, a_{1,3}+a_{3,1}, a_{1,2}+a_{2,1} is 0.
*

Prove that (1/(a_{2,3}+a_{3,2}), 1/(a_{1,3}+a_{3,1}), 1/(a_{1,2}+a_{2,1})) is a direction vector of the axis of R.

__Solution:__

There is an orthonormal base b, whose first vector is eigenvector of R with eigenvalue 1 and direction vector of its
axis, so that the matrix of R on base b
is [R]_{b,b} = M with matrix elements m_{1,1} = 1, m_{1,2} = m_{1,3} = m_{2,1} =
m_{3,1} = 0, m_{2,2} = m_{3,3} = cos(u), m_{3,2} = - m_{2,3} = sin(u) for some u
∈ (0,2π).

The last two vectors of b may be rotated around the axis over an arbitrary angle without changing this matrix M.
This will come in handy.

The relation between A and M is given by the formula A = [R]_{e,e} = [I]_{e,b} [R]_{b,b}
[I]_{b,e} = S M S^{-1}, where I is identity and the columns of S are the coordinates of the vectors of b
on the natural base e and S^{-1} is the inverse of S. Since b is orthonormal, S^{-1} is the transposed of S.

So we can express each matrix element a_{i,j} of A in cos(u), sin(u) and the matrix elements s_{m,n}
of S.

Now we have to prove that s_{1,1}(a_{2,3}+a_{3,2}) = s_{2,1}(a_{1,3}+a_{3,1}) =
s_{3,1}(a_{1,2}+a_{2,1}).

We may check these identities by taking the last two vectors of basis b in such a way that for some
v,w ∈ [0,2π) we have

s_{1,1} = cos(v)cos(w), s_{1,2} = - sin(v)cos(w), s_{1,3} = - sin(w),

s_{2,1} = cos(v)sin(w), s_{2,2} = - sin(v)sin(w), s_{2,3} = cos(w),

s_{3,1} = sin(v), s_{3,2} = cos(v), s_{3,3} = 0.

**Problem 2019-2/B** (adapted)

*1) Let k be a nonnegative integer and X a collection of infinite sets of positive integers such that no two sets of X
share more than k positive integers. Prove that X is countable.
*

*2) Does there exist an uncountable collection X of infinite sets of positive integers such that for all distinct sets
A and B of X the intersection of A and B is finite?*

__Solution:__

1) For every A in X, let A' be the set containing the smallest k+1 elements of A.

Let X' be the collection of all sets of k+1 positive integers.

Then A → A' is 1-1, so X is countable if X' is countable.

And we can count the sets A' of X' according to the principle: the greater the sum of the elements of A', the greater
its number in the count.

2) Yes: take the collection X of all infinite sets of the form {a_{1}, a_{1}a_{2},
a_{1}a_{2}a_{3}, a_{1}a_{2}a_{3}a_{4}, ... }, where a_{i}
∈ {0,1,2,...,9} (a_{1} > 0) and we use the decimal representation, so for instance
1000 ≤ a_{1}a_{2}a_{3}a_{4} ≤ 9999, etc.

We can prove that X is uncountable with the well-known Cantor diagonal argument. In fact
{a_{1}, a_{1}a_{2},
a_{1}a_{2}a_{3}, a_{1}a_{2}a_{3}a_{4}, ... } →
Σ a_{i} 10^{-i} is (almost) a bijection from X to [0.1,1].

The intersection of two distinct sets of X, if not empty, has always the form
{a_{1}, a_{1}a_{2}, ... ,
a_{1}a_{2}..a_{k}} for some positive integer k, so is always finite.