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:
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.
]]>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.
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There are some basic rules one need to respect to use parfor efficiently:
The most restrictive requirement is the fact that one cannot use the same variables in the computations for different processors. In order to do this, the simplest way I found was to use a function for the body of the loop. When using a matlab function, all variables are local, so when running the same function in parallel, the variables won’t overlap, since they are local to each function.
So instead of doing something like
parfor i = 1:N commands ... array(i) = result end
you can do the following:
parfor i=1:N array(i) = func(i); end function res = func(i) commands...
This should work very well and no conflict between variables will appear. Make sure to initialize the array before running the parfor, a classical Matlab speedup trick: array = zeros(1,N). Of course, you could have multiple outputs and the output array could be a matrix.
There is another trick to remember if the parpool cannot initialize. It seems that the parallel cluster doesn’t like all the things present in the path sometimes. Before running parfor try the commands
c = parcluster('local'); c.parpool
If you recieve an error, then run
restoredefaultpath c = parcluster('local'); c.parpool
and add to path just the right folders for your code to work.
]]>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 .
]]>I’ll briefly describe below how to use Paraview to make some nice pictures of level-sets. First of all, you’ll need your data in some format that Paraview can understand. I use vtk file format for which there is a nice automated interface in the software I use for the optimization (FreeFem++). In the vtk file you need to have a set of points and a scalar value attached to them.
If you want to create level-sets associated to certain values, follow the steps below:
If your level-set cuts the boundary of the domain, Paraview will draw a hole there. If you want to have a closed region instead, you need to use the IsoVolume filter instead of the Contour one. The difference is that you need to specify two values and Paraview will draw the surface enclosing the points corresponding to these values. Many other features are directly available: you can color the level set following another scalar value, you can set the lighting, etc. You can also symmetrize your geometry using the Reflect filter. Below you can see a result built from my work.
You can also create animations in a pretty straightforward way. Just go to View and select the Animation box. Then you’ll see the animation options. Add a Camera object with Orbit field selected. You’ll be presented with multiple options, like the center of rotation, direction of the vertical and initial position. Once everything is set, click the play button to see the animation. Then go to File/Save Animation to save it to a file.
I heard that Paraview could to many things when dealing with visualization aspects, but I hesitated to use it until now since the interface is not straightforward. The use of Filters is not clear in the beginning, but after playing with some examples, everything becomes really easy to use. The next step is to automatize all this using scripts.
Happy New Year!
]]>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 .
]]>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.
]]>Next, let’s note what elements cannot be in . Note that taking square roots and squaring cannot change a non-zero remainder modulo into a zero remainder. Therefore, starting from one could never get a multiple of following the allowed operations. Thus we can safely say that multiples of are not in the minimal set .
Furthermore, could only be obtained as a square root of itself with the allowed operations. Starting from , one could never get below by performing square roots or . Therefore, the minimal set does not contain and multiples of .
Now, we show that it contains all the rest. The general idea is as follows: it is enough to find which is the smallest element in a class of remainders modulo to deduce that all larger elements are there (recall the operation ). Now in order to obtain small elements of , one would need to take successive square roots. So if we prove that for some we have for some then we get that .
Now let’s start from the beginning. We have so . Since is in for every , we get that all squares of the form greater than are in . Moreover, so all numbers of the form greater than are in . Since it follows that . Moreover, ends in and is greater than so . Next, we have which ends in and is greater than so it is also in . Therefore .
Finally, we have that , and , . This means that the minimal set is .
]]>Problem A1. Let be the smallest set of positive integers such that
Which positive integers are not in
(The set is “smallest” in the sense that is contained in any other such set.)
Problem A2. Let , and
for all Show that, whenever is a positive integer, is equal to a polynomial with integer coefficients.
Problem A3. Let and be real numbers with and let and be continuous functions from to such that but For every positive integer define
Show that is an increasing sequence with
Problem A4. A class with students took a quiz, on which the possible scores were Each of these scores occurred at least once, and the average score was exactly Show that the class can be divided into two groups of students in such a way that the average score for each group was exactly
Problem A5. Each of the integers from to is written on a separate card, and then the cards are combined into a deck and shuffled. Three players, and take turns in the order choosing one card at random from the deck. (Each card in the deck is equally likely to be chosen.) After a card is chosen, that card and all higher-numbered cards are removed from the deck, and the remaining cards are reshuffled before the next turn. Play continues until one of the three players wins the game by drawing the card numbered
Show that for each of the three players, there are arbitrarily large values of for which that player has the highest probability among the three players of winning the game.
Problem A6. The edges of a regular icosahedron are distinguished by labeling them How many different ways are there to paint each edge red, white, or blue such that each of the 20 triangular faces of the icosahedron has two edges of the same color and a third edge of a different color?
Problem B1. Let and be distinct lines in the plane. Prove that and intersect if and only if, for every real number and every point not on or there exist points on and on such that
Problem B2. Suppose that a positive integer can be expressed as the sum of consecutive positive integers
for but for no other values of Considering all positive integers with this property, what is the smallest positive integer that occurs in any of these expressions?
Problem B3. Suppose that
is a power series for which each coefficient is or . Show that if , then must be irrational.
Problem B4. Evaluate the sum
(As usual, denotes the natural logarithm of )
Problem B5. A line in the plane of a triangle is called an equalizer if it divides into two regions having equal area and equal perimeter. Find positive integers with as small as possible, such that there exists a triangle with side lengths that has exactly two distinct equalizers.
Problem B6. Find the number of ordered -tuples such that are distinct elements of and
is divisible by
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