#### Selection on Hong Kong IMO Team

The selection process of the Hong Kong Team for IMO starts annually with the Hing Kong preliminary selection contest in May a year before. The contest consists of 20 short answered within 3 hours. The 100 most outstanding contestants are selected for an eight-month training programme, begins at mid-July and consist three-hour weekly class and four selection tests, three of which are prepared by the committee and held in August, October and December, with the last being Hong Kong China Mathematical Olympiad (CHKMO). All proof based, and 3 hours given to solve each of test. And the last selection test is the Asian Pacific Mathematical Olympiad serves as the last selection test. After APMO…

Wait, Why I am saying this?

Ah, yes. The first selection test, known as the August Test, are prepared to be held 2 days later.

#### Badly prepared status

However I am in a very badly prepared status, I have a running nose right now and I haven’t memorized the formulas of Complex numbers and barycentric coordinates. Oh for god sake, Look I have nothing. Geometry? I literally haven’t solve a single problem of geometry. Hence coordinate geometry, complex numbers please. Combinatorics? I have nothing but an unconvincing confidence. Algebra and Number Theory is likely okay. Oh holy god, I haven’t learnt primitive roots, quadratic reciprocity… Also I am quite angry that I can’t hand in the problems solved in Mathematical Excalibur(Today is the deadline), oh I am really sorry to Dr. Kin Y Li…I thought to myself, I won’t hand in any solutions if I can’t solve it all. Hence in physically and psychologically, I am absolutely nothing compare to the team member going to IMO. Fine, 2 days left, right?

#### Argentina 2006 Team Selection Test

Tonight I select the Argentina 2006 TST, not only because it should not be hard, but more it have solutions in Art of Problem Solving Community, and it have 2 days each have 3 problems, and august test have 6 problems in 3 hours too. Hence it makes me a perfect choice… I hope they are not hard…math ahead

Problems are:

Problem 1. Let be a set of natural numbers in which if , belong to () then either or belong to ( both cases may be possible at the same time). Decide whether there is or not a set consisting on exactly elements which has four elements , , , ( not necessarily distinct) that satisfy and

Problem 2. Given 365 cards, in which distinct numbers are written. We may ask for any three cards, the order of numbers written in them. Is it always possible to find out the order of all 365 cards by 2000 such questions?

Problem 3. In a circumference with center we draw two equal chord and if then and

We consider and such that

Prove that the chord determined by extending has the same as length as both and

Problem 4.Find all integers solutions for

Problem 5. Let be a prime with , and let . Prove that contains two elements and such that and .

Problem 6. Let be a natural number, and we consider the sequence where If we make the sum of consecutive members of the sequence, starting from one with an odd index and finishing in one with and even index, the result is and

How many sequence are there satisfying this conditions?

After an hour killing problems (I gave up after a hour, I should note it down). Only 2 problems solved (And I didn’t write it down formally).

My first thought is to do problems 4 first. Turn out it’s quite simple, I used up 10 minutes trying lots of inequality bonding, don’t work apparently. But when I tried , and I substitute , and finished.

Problem 5 stuck me around, here is the solution.

We show that the smallest possible works.

Let . If , then we set and . This works, because surely is odd, so giving .

In all other cases we can set . There are three cases:

a) : Then we set . Additionally gives .

b) : this implies , impossible.

c) : Then we set . Additionally gives . If , then we would have giving , impossible. So again .

In all possible cases, we now set and get , being the result.

Instead I solved Problem 1, I got frustrated considering the two conditions, it’s too messy considering either or belong to , so I construct the so call image set a subset of integers, where contains only elements in and adding minus sign to elements in . Hence we can eliminate one of the condition. After proving 2 is an element in , boom done.

Problem 2 is like something bubble sort. However I failed (My method is to do induction on n, I think my combinatorical argument need bush up) The answer is also like that: Suppose we have cards. We divide the cards into groups and sort each of them independently, so that we obtain one group , another and the last group . So I ask a question about , and whichever is the smallest must be the smallest in the entire list, WLOG , then I ask a question about so whichever is the smallest out of these will be the next smallest in the entire list, and I can keep adding element in order, and with questions the entire list will be in order (since for the last I add them all at once). Now just keep applying the above strategy and you get the recursion

We shall count for after noticing that cards you can do it in questions

For it would take , and so for we can do it in less than questions for sure.

I actually don’t know what is , oh my geometry sucks.

Let and cut line segment at and respectively.And let and be projections of to and respectively. and -cyclic implies that is isosceles trapezoid.. and implies that and implies that which implies that and -cyclic and similarly we get that is cyclic and from we get that which implies that is cyclic and implies that if cuts circle at and . Let angle bisector of cut at and let lies between

and

which is contradiction thus we get that is isosceles trapezoid which implies that .

And I have no idea in problem 6.I search AoPs and just someone thought.

“I didn’t have time to make all the calculations but here’s some thoughts:

Since I couldn’t find any nice combinatorial argument I thought of a “brute force approach”. We see that each pair has a sum in but we have three types of , two types of and one type of .

Let be the number of sequences we are looking at. Let denote the number of sequences of length with entries in (taking different types into account) so that all consecutive sums have absolute value , so that we have

Now let be respectively the number of sequences of length with entries in so that the absolute value of the sum of consecutive terms doesn’t exceed 2 and so that the sums of the first terms () lies in respectively.

Then we have

If we let and we have

Now everything can be computed recursively and with some algebra we should be able to get the answer.”

Oh my holy Chloe.