## SEEMOUS 2013 + Solutions

Here are some of the problems of SEEMOUS 2013. **Update: the 4th problem has arrived; it is number 3 below.**

**1.** Let be a continuous mapping, such that

Find the form of the map .

*Solution:* Change the variable from to in the RHS integral and DO NOT calculate the last integral in the RHS. Get all the terms in the left and find that it is in fact the integral of a square equal to zero.

**2.** Let be nonzero matrices such that and . Prove that there is an invertible matrix such that and .

*Solution:* One solution can be given using the fact that can be written in that form, but for different matrices .

Another way to do it is to consider applications . We get at once and from these we deduce that and . First note that is not the zero application. Then there exists , i.e. there exists such that . We have . Consider .

Then are not collinear, . Consider now the basis formed by and take to be the change of base matrix from the canonical base to .

**3.** Find the maximum possible value of

over all continuously differentiable functions with and .

**4.** Let such that there is , with . Prove that either or .

*Solution:* Consider the minimal polynomial of , which has degree at most . The eigenvalues of satisfy . We have two cases: either the eigenvalues are real and therefore they are both equal to either they are complex and conjugate of modulus one. In both cases the determinant of is equal to . Therefore, by Cayley Hamiltoh theorem satisfies an equation of the type .

By hypothesis the minimal polynomial divides . If has degree one then and so .

If not, then the minimal polynomial is and we must have

It can be proved that is in fact an integer. (using the fact that the product of two primitive polynomials is primitive)

Since it follows that . The rest is just casework.

SEEMOUS is comprised of four problem or not?

Thank you by posting these problems.

Pedro.

Yes, SEEMOUS contest subject usually has four problems, but I didn’t receive the 4th problem yet. I’ve searched the internet to see if I can find the problems on other forums, but so far no success. I will post the 4th problem as soon as I find it.

Ok very nice. Beni Bogoşel I didn’t understand when you used to solve the problem 3, q = 2cos (x)…

Pedro.

The trace of the matrix is . I forgot a minus sign in the proof.

Pedro, the 4th problem has arrived. It was number 3 who was missing.

The solution of the problem 3 is perfect, but the original statement says:

\int_{1}^{0}|f'(x)|^{2}dx\leq1.

I thought that something was too easy there.

http://www.seemous.eu/assets/files/2013/problems_2013.pdf