United States of America Mathematical Talent Search - Problem 1/1/24

1/1/24. Several children were playing in the ugly tree when suddenly they all fell.
• Roger hit branches Α, Β and Γ in that order on the way down.
• Sue hit branches Δ, E, and Φ in that order on the way down.
• Gillian hit branches Κ, A, and Γ in that order on the way down.
• Marcellus hit branches B, Δ, and H in that order on the way down.
• Juan-Phillipe hit branches I, Γ, and E in that order on the way down.
Poor Mikey hit every branch on the way down. Given only this information, in how many different orders could he have hit the 9 branches on his way down?

This is one of those beautiful combinatorial problems, that require a full and analytical solution through logic. No formula could give one a full solution. I solved this problem together with Demetres some day in July through Skype. Here's our solution.

From the hypothesis we can say that:

The possible configurations including Δ are:

The possible configurations including I are:

The possible configurations including Δ and H are:

We can now multiply the possible configurations including Δ and H with the first three possible configurations including I. From this we get a total of 21 possible configurations. For the last possible configuration including I we must take two cases: I will either be before or after Δ. Working similarly, we get 12 possible configurations.

In conclusion, the total amount of possible configurations is 21+12=33.

Junior International Mathematical Olympiad 1999 Problem

I found this problem in a book that states it was set as a problem in the 3rd Junior International Mathematical Olympiad held in Hong-Kong in 1999. The hypothesis is as follows:

A furniture shop has sold 225 beds during the year 1998. At first, it sold 25 beds per month, then 16 beds per month, and finally 20 beds per month. For how many months has it been selling 25 beds per month?

I really like this problem. It is quite trivial yet may look a bit difficult at first. Here's my solution.

Based on the hypothesis, considering that:

    x is the amount of months for which the shop had been selling 25 beds per month;
    y is the amount of months for which the shop had been selling 16 beds per month;
    z is the amount of months for which the shop had been selling 20 beds per month;

we can create the following system of equations:

which can be written as:

which is equivalent to the following system:

which has (x,y,z)=(1,5,6) as a solution in the set of natural numbers.

 In conclusion, the amount of months for which the shop had been selling 25 beds per month was equal to 1.

"Balkan Mathematical Olympiad 2009 - Problem 1"

Hello guys. From now on I will be joining this blog as a moderator. My name is Demetres and I am 14 years old and I am hoping to make you love Maths. Let's start with a Number Theory problem, which uses a technique named modular method.

Junior Balkan Mathematical Olympiad 2007 - Problem 3

This was the third problem of the JBMO 2007 held in Bulgaria. I find it very beautiful and I think it has a very elegant solution. Again, like all JBMO problems in my opinion, it only takes a few observations to reach a solution.

Using the Pigeon Hole Principle, we can express 50 in terms of 4. Like this:

From this statement, we can reach the conclusion that there are at least 13 points of the same colour in the plane.

Now, the key to solving this problem is to find out how many triangles we can create out of 13 points of the same colour and then reach more conclusions.

The amount of triangles we can create is equal to:

 Note that we would not be able to use this formula if it was not stated in the hypothesis that no three points are collinear.

What is left to prove, is that there exist at most 156 isosceles triangles, so that the remainder which is equal 130 will be scalene triangles.

Pay attention now, because this part is tricky.

Consider a segment created by two points. Then we can say that we can create at most 2 isosceles triangles per segment. The reason for that is that there can be at most 2 points belonging to the semi-perpendicular of that segment*. The semi-perpendicular of a segment is a straight line that bisects the segment and is perpendicular to it. Geometry tells us that the distances between any point on the semi-perpendicular of a segment and the points creating that segment are equal. For a better understanding, I have prepared a shape:

*That is due to the fact that no three points are collinear.

So now that we have proven that there can be up to 2 isosceles triangles created by a segment, we have to find the amount of segments we can create and multiply it by 2.

So the total amount of segments is 78. Should we multiply that by 2, we receive 156 isosceles triangles, which gives us 130 scalene triangles, as we wanted.

Junior Balkan Mathematical Olympiad 2012 - Problem 1

This was the first problem in last year's Junior Mathematical Olympiad, held in Veria, Greece. It was considerably easy, as it takes only a few simple observations. Let us start, shall we?

We can easily see that the inequality is cyclic, which means that it is the addition of the three following inequalities:

Since all three of them are identical, we can simply prove one and be done, since we would have to use the same method for the other two as well. So let's work on the first one.

From the hypothesis, we can say that

So the initial inequality can be written as:

Now this is the only part where the solution becomes a bit complex, so pay attention. 

Using the AM-GM inequality, we have that:

Which means that we have proven the initial inequality. Using the same method, we can prove the other two inequalities. By adding the three inequalities together, we reach the given inequality. Obviously, equality holds if, and only if:

Introductory Message

   Hello everybody! I am hoping to keep this message short enough so that it doesn't get boring. My name is Kleovoulos, I am 15 years old and I am from Greece. In this blog I will be trying to give solutions to mathematical problems from international competitions such as the Junior Balkan Mathematical Olympiad, the Balkan Mathematical Olympiad and the International Mathematical Olympiad. To whoever likes Mathematics, I am hoping that this blog will be useful, as I will be trying to provide solutions accessible by everyone. To those who don't like Mathematics, I am hoping that this blog will help you see what makes me like it and hopefully even inspire you. So, that's what I have to say for now, I hope you enjoy this blog.

 - Kleovoulos

EDIT: 2nd of August 2013

From now on, Demetres will be joining me as a moderator of this blog. He will also be posting problems from International Competitions. He has been successful in numerous competitions nation-wide and his favourite branch of mathematics is Number Theory. He hopes that you will enjoy the solutions he posts and find them useful.