## Romanian Masters in Mathematics contest – 2018

**Problem 1.** Let be a cyclic quadrilateral an let be a point on the side The diagonals meets the segments at The line through parallel to mmets the extension of the side beyond at The line through parallel to meets the extension of the side beyond at Prove that the circumcircles of the triangles and are tangent .

**Problem 2.** Determine whether there exist non-constant polynomials and with real coefficients satisfying

**Problem 3.** Ann and Bob play a game on the edges of an infinite square grid, playing in turns. Ann plays the first move. A move consists of orienting any edge that has not yet been given an orientation. Bob wins if at any point a cycle has been created. Does Bob have a winning strategy?

**Problem 4.** Let be positive integers such that and . Let be the set of values attained by as runs through the positive integers. Show that is the set of all positive divisors of some positive integer.

**Problem 5.** Let be positive integer and fix distinct points on a circle. Determine the number of ways to connect the points with arrows (oriented line segments) such that all of the following conditions hold:

- each of the points is a startpoint or endpoint of an arrow;
- no two arrows intersect;
- there are no two arrows and such that , , and appear in clockwise order around the circle (not necessarily consecutively).

**Problem 6.** Fix a circle , a line to tangent , and another circle disjoint from such that and lie on opposite sides of . The tangents to from a variable point on meet at and . Prove that, as varies over , the circumcircle of is tangent to two fixed circles.

Source: Art of Problem Solving forums

**Some quick ideas:** For Problem 1 just consider the intersection of the circle with the circle . You’ll notice immediately that this point belongs to the circle . Furthermore, there is a common tangent to the two circles at this point.

For Problem 2 we have . Eliminate the highest order term from both sides and look at the next one to get a contradiction.

Problem 4 becomes easy after noticing that if divides and then divides .

In Problem 5 try to prove that the choice of start points determines that of the endpoints. Then you have a simple combinatorial proof.

Problem 6 is interesting and official solutions use inversions. Those are quite nice, but it may be worthwhile to understand what happens in the non-inverted configuration.

I will come back to some of these problems in some future posts.

## SEEMOUS 2018 – Problems

**Problem 1.** Let be a Riemann integrable function. Show that

**Problem 2.** Let and let the matrices , , , be such that

Prove that .

**Problem 3.** Let such that and , where is the identity matrix. Prove that if then .

**Problem 4.** (a) Let be a polynomial function. Prove that

(b) Let be a function which has a Taylor series expansion at with radius of convergence . Prove that if converges absolutely then converges and

Source: official site of SEEMOUS 2018

**Hints: **1. Just use and . The strict inequality comes from the fact that the Riemann integral of strictly positive function cannot be equal to zero. This problem was too simple…

2. Use the fact that , therefore is symmetric and positive definite. Next, notice that . Notice that is diagonalizable and has eigenvalues among . Since it is also positive definite, cannot be an eigenvalue. This allows to conclude.

3. First note that the commutativity allows us to diagonalize using the same basis. Next, note that both have eigenvalues of modulus one. Then the trace of is simply the sum where are eigenvalues of and , respectively. The fact that the trace equals and the triangle inequality shows that eigenvalues of are a multiple of eigenvalues of . Finish by observing that they have the same eigenvalues.

4. (a) Integrate by parts and use a recurrence. (b) Use (a) and the approximation of a continuous function by polynomials on compacts to conclude.

I’m not sure about what others think, but the problems of this year seemed a bit too straightforward.

## Regular tetrahedron – computing various quantities in terms of the side-length

Sometimes one needs to find certain quantities related to the regular tetrahedron in , like volume, radius of the circumscribed sphere, radius of inscribed sphere, distance between opposite sides, etc. in terms of the side-length which we’ll note in the following with . In the past I needed to find the angle under which every side is seen when looking from the center of the regular tetrahedron.

Here’s a trick which can help you find rather easily everything you need related to the regular tetrahedron: *just embed it into a cube*. We can see rather easily that when drawing certain diagonals of the faces of a cube, like in the figure below, we can recover a regular tetrahedron. Now it becomes rather easy to solve all questions above. We note that the ratio between the side of the cube (denoted by ) and the side of the embedded tetrahedron (denoted by ) is : .

Here are a few ideas:

**1. Finding the volume of the tetrahedron in terms of its sides.**

The volume of the cube is . The volume of the tetrahedron can be obtained by cutting four corner pyramids with volumes (recall that volume of a pyramid is (area of base) (height) ). Therefore the volume of the regular tetrahedron is . Replacing we get that the volume of the tetrahedron is .

**2. Finding the circumradius .**

It is not difficult to see that the sphere passing through the vertices of the tetrahedron also passes through the vertices of the cube. Therefore its radius is a long diagonal of the cube divided by . This gives . Replacing we get that the circumradius is .

**3. Finding the inradius .**

Once we have the volume and the circumradius, finding the inradius is not that difficult, since the symmetry of the figure shows that , where is the distance from a vertex to the opposite face, also called the height. We can find the height from the formula of the volume (recalled above), and then find . Also note that since the center of the tetrahedron is also the centroid, we must have , so we have another quick finish solution.

However, let’s use the cube to do this. We can choose a system of coordinates putting the origin at the center of the cube (and tetrahedron). Put the cube so that its vertices have coordinates and suppose that the tetrahedron corresponds to the vertices , , , . All we need to do is to compute the distance from the origin to the plane . This is immediate if we know the equation of this plane. Fortunately, it is really easy to see that the coordinates of verify (if not, then note that the normal to is the vector and figure out the remaining translation constant). We know that if a plane is defined by the equation then the distance from to this plane is

Apply this to our problem and the distance from the origin to becomes . Replace and we get .

**4. Find distance between opposite sides.**

This is particularly easy with the cube. The distance between the opposite sides is exactly the distance between two parallel faces of the cube and that is .

**5. Find angle made by two rays connecting the center with vertices.**

Use the coordinate system introduced in **3.** and just compute the angle between vectors and , for example. If is the angle between and we get that

Therefore .

## Putnam 2017 A3 – Solution

**Problem A3.** Denote . Then is continuous and . We can see that

Now note that on we have so . Furthermore, on we have so multiplying with we get . Therefore

To prove that goes to we can still work with . Note that the negative part cannot get too big:

As for the positive part, taking we have

Next, note that on

We would need that the last term be larger than . This is equivalent to . Since is continuous on , it is bounded above, so some delta small enough exists in order for this to work.

Grouping all of the above we get that

Since we get that goes to .

## Putnam 2017 A2 – Solution

**Problem A2.** We have the following recurrence relation

for , given and . In order to prove that is always a polynomial with integer coefficients we should prove that divides somehow. Recurrence does not seem to work very well. Also, root based arguments might work, but you need to take good care in the computation.

A simpler idea, which is classic in this context, is to try and linearize the recurrence relation. In order to do this, let’s write two consecutive recurrence relations

We add them and we obtain the following relation

which leads straightforward to a telescoping argument. Finally, we are left with a simple linear recurrence with integer coefficient polynomials, and the result follows immediately.

## A hint for Project Euler Pb 613

The text for Problem 613 can be found here. The hint is the following picture 🙂

## IMC 2017 – Day 2 – Problems

**Problem 6.** Let be a continuous function such that exists (finite or infinite).

Prove that

**Problem 7.** Let be a nonconstant polynomial with real coefficients. For every positive integer let

Prove that there are only finitely many numbers such that all roots of are real.

**Problem 8.** Define the sequence of matrices by the following recurrence

where is the identity matrix.

Prove that has distinct integer eigenvalues with multiplicities , respectively.

**Problem 9.** Define the sequence of continuously differentiable functions by the following recurrence

Show that exists for every and determine the limit function.

**Problem 10.** Let be an equilateral triangle in the plane. Prove that for every there exists an with the following property: If is a positive integer and are non-overlapping triangles inside such that each of them is homothetic to with a negative ratio and

then