OSN 2010 Proposal

Source: majalahmatematik.wordpress.com

In 2010, I proposed several problems to Indonesia National Mathematical Olympiad (called OSN). Apparently there are 3 problems accepted by the jury team as the test problem. All of them in one day! I think I had set up fair standard for OSN 2010 but of course it is boring if everything is fair. So, I think OSN 2010 problems are not really suprising at all. In this post, I will tell about some stories behind the problems.

Problem 6. Find all positive integers n > 1 such that
\tau(n)+\varphi(n)=n+1
where \tau(n) denotes the number of positive integers that divides n, and \varphi(n) denotes the number of positive integers less than n that are relatively prime to n.

This is actually a nice problem. The idea of creating this pretty easy. Well, I just know that if p is prime and \tau(p)=2 and \varphi(p)=p-1. The we can easily see that \tau(p)+\varphi(p)=p+1. Then, I asked what are all n such that the equation \tau(n)+\varphi(n)=n+1. A little bit working on inequalities, one may obtain that the only non-prime number satisfying this equation is n=1 and n=4. I created this problem like 5 or 6 months before OSN; when I was in 2nd stage of Indonesia IMO Camp. I wrote the problem on the board but only some people want to see, lol. I actually I wonder how to generalize this problem. After the OSN, someone from Mathlinks gave the link to a similar problem: find all n such that |\tau(n)+\varphi(n) -n| \le 1, which is pretty nice! But for the sake of Internet Explorer, I didn’t plagiarize!

Now, let’s take a look at another problem.

Problem 7. Let a and b be positive real numbers. F(x) = x^2 + ax + b and G(x) = x^2 + bx + a are polynomials such that the roots of polynomials F(G(x)) and G(F(x)) are all real numbers. Prove that both a and b are greater than 6.

This problem does not have spirit of mathematics; it is ugly! I do not have intention to make a tighter bound of a and b.. it will be a little bit interesting I think, but it does not make significance impact on the beauty. So that’s it, the problem I proposed with hope that this was accepted as easiest problem of the day. Dang, it is weird that this problem appears as problem 7. Well, I don’t really care, then. Heheh, let’s see how to hack this one. Well, since G(x) must be real, then F(x) must have real roots. Then let’s say F(x)=(x-x_1)(x-x_2) and thus, G(x)=x_i must have real roots. It appears that b^2 \ge 4(a-x_i) and thus b^2 \ge 2(2a+a)=6a. Similarly, you can get a^2 \ge 6b. Yeah, it’s then easy to see the rest.

The last one..

Problem 8. Given ABC an acute triangle that is not an equilateral, with O as the circum-center and H as the orthocenter. Let K be any point in the interior of ABC different from O and H. The points L,M are outside of ABC such that AKCL and AKBM are parallelograms. Lastly, let BL and CM intersect at N, and let J be the midpoint of HK. Show that KONJ is a parallelogram.

Dang! This problem is made with complex numbers! I asked Feri Priatna [IMO 2010 st. 3, FE UI] to give me synthetic solution for this problem. It is pretty nice, because his solution used similarity, rotation, and other stuff. So I proposed the problem! Another solution is to use well-known results in geometry (that I apparently not-so-well-known to me, haha). I don’t know if this problem is hard or not, but complex numbers overkill.

So, that is what it is. I have never proposed problem anymore, but I really wish I can propose some problems to IMO, APMO, USAMTS, and other nice competitions! This, year one of my friend proposed from South Africa proposed a nice functional equations problem to IMO 2012. Whoa, that is cool!

Source: imo-official.org

 

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