# OSN 2010 Proposal

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 such that

where denotes the number of positive integers that divides , and denotes the number of positive integers less than that are relatively prime to .

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 and be positive real numbers. and are polynomials such that the roots of polynomials and are all real numbers. Prove that both and are greater than .

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 an acute triangle that is not an equilateral, with as the circum-center and as the orthocenter. Let be any point in the interior of different from and . The points are outside of such that and are parallelograms. Lastly, let and intersect at , and let be the midpoint of . Show that 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!