Honestly, I don’t expect it. There are a lot of other people better than me. Let me recap everything… There are 17 people in Training/Selection Camp 2, including me. 9 people were from last year’s camp (automatically advance here) and 8 are new from Camp 1 in October last year. I’m one of the new. In the tean of six that made it, at least one is from Camp 1 (guaranteed spot). The resulting team has two from Camp 1, I’m one of them. Among the new ones, I would rank myself fighting for Top 5, but would be somewhere around #3. So that’s surprising that I got a spot, and neither of those that I project above me made it.

In case youre wondering, here’s the list from the rumors:
Tobi Moektijono
Ahmad Zaky
Stefanus
Bivan Alzacky
Ivan Adrian Koswara
Fransisca Susan

There. Now you get my real name down to six possibilities. Ouch. Meh whatever.

Assuming it’s true, then I would be focusing to IMO a lot. That means this blog will be significantly more…dead than it was. Is something can be “more dead”? Whatever. Don’t expect puzzles and non-math contents to pop up often, at least until IMO 2012 finished.

Speaking of math contents…

IMO 2009-5
Let f be a function from the positive integers to itself, satisfying the property that a, f(b), f(b+f(a)-1) always form the sides of a non-degenerate triangle for all positive integers a,b. Find all such f.

This took three incorrect attempts (as in incorrect conclusions taken from what I know at the time), my third one wasn’t spotted by myself until a day after I found that “solution”.

Also, if you’re craving for some inductive logic puzzle, here’s one that might take you some time to figure out:

1, 10, 9, 2, 4, 3, 7, 19, 8, 11, ???, ???, ???
The sequence contains all positive integers. One trillion (10^12 aka 1,000,000,000,000) appears before the 50th term.

And if you wonder about the answer, highlight: 90, 5, 6