It’s almost 140 days from the contest date (11/3/14), and I wrote it as I promised it in my previous post–how can a man break his promise? Better late than never.

—

That was the third camp of IMO 2014 cycle, and we took it after all the lecture sessions had ended. Let’s have a look on my previous achievement:

APMO 2011: 77000 (14), Bronze

APMO 2012: 77770 (28), Silver

APMO 2013: 77717 (29), Bronze

The aim was clear: to fight for a gold. To do that, not only I had to meet the cut-off points, but also had to beat *all* other Malaysia contestants (check daryn.kz/apmo for award rule). That was never an easy job-I failed the second requirement for the previous two years.

—

**On The Day**

Thus that’s the plan for the day when I entered the exam hall in Permata Pintar at 8:35am of the day: make myself as compose as possible, and try to tackle all 5 problems to get a perfect score (note: we had 3 perfect scores in Malaysia for APMO 2013, which pulled me down to a bronze).

A short briefing by chief invigilator, Mr. Suhaimi embarked 10 minutes later, and some rules specifically highlighted to reach the maximal integrity. At the end of the briefing, which was just before the clock struck 9, he accentuated his last statement: “Good luck, and you may turn the question paper and start your work”.

**P1:** the problem didn’t look traditional, and at the first glance, “Hmm, is that wasn’t easy”. Some observation gives S(a_i) and P(a_i) has prime divisors in set {2,3,5,7}. Let’s experiment with 9^k, 9^{k+1},… where k is some suitable number. However, it seems that it was easier to use 2^k, 2^{k+1},…, and, Bingo! I got it at 9:15. The writing of solution was hurdled with a flaw, but that’s mistaking P(a_1)<P(a_2)<…<P(a_n) instead of S(a_1)<S(a_2)<…<S(a_n). Having known that these S and P are cycle, the gap is no longer hard to fix. 9:30, done.

**P3: **that’s short and cute, so I skipped the wordy P2 and began with this. Simple experiment for primes gives that was invalid for 4k+1 (for p divides a^2+1 has a solution), and 2,7,11 but valid for 3. How? Let’s try n=9, and we get a positive answer! OK! Let’s see a^3+a=b^3+b (mod n), which in turn equals to (a-b)(a^2+ab+b^2+1). Simple try gives the second term cannot be divisible by 3, and we get 3^k as valid solution!

How about other prime? It’s 4 months from now, but I remembered that my brain began to blurred by that: I fake solved the problem and wrote a 2-page crap, with 3^k being the only useful point (not even mentioning case p=4k+1). The work was fallacious: the work seems correct for p not equals to 3, but the reasoning applies to p=3 too, hence there’s mistake in between (I forgot exactly what detail as that, but it was something to do with quadratic reciprocities). The best thing was, I was still in quandary whether my solution was correct! With lots of question mark in my head, I returned to this:

**P2:** For S any member x can be chosen, similar reasoning applies for S\{x}, and if we choose y the set, similar reasoning is valid for S\{x, y}, and so on…hence we have 2014! possible choices now. Moreover x is chosen for any set of at most 2012 elements and containing x, as you can choose 2 more non-empty set that together forms a disjoint union as S, while for those with 2013 elements, since the above is not valid, anything can be chosen. By continuing S\{x}, S\{x,y},…, we arrived with this:

Unfortunately, I met difficulties in verifying that all of them are valid: in fact this is wrong. That’s the time I realized that I spent 2 pages and 30 minutes in writing *rubbish* (as shown above)–“well done”! Hence, the 2013-element thing is wrong: only one way to choose them. This is true since for any such set take 2011+1+1 several times you get that x must be chosen. (In fact, take 2010+1+1, 2009+1+1,…,3+1+1 this is true). However, for {a,b,c,d} and let a be chosen: 3-element subset can have any element chosen, and for {b,c,d}, among it any 2-element number is good. This gives another 3^3 times 2^2 choices, and we have 108*2014!……strange, isn’t it? The answer is not even beautiful.

Another 2 pages and 30 minutes spent…and let’s see the last two (still didn’t see why P3 is wrong? **Stupid**!)

**P4:** Why? For part (a) we can solve it for 7 element (obvious! Take 3^k, k=0-6) and for part (b) it’s easy for 10 element (as in IMO 1972, since there are 1024 subset and total sum of each subset is at most 99+98+…+90<1000, which is pigeonhole principle!) Now they make both problem to have slightly tighter bound, but significant increase in difficulty. Hence for part (a) here it went: I tried Fibonacci+1 (failed), and something like 4,5,10,…,79 (still failed). It took me a while to see that 3^k (k=1-6), 97,98 is good. That’s good to be happy of, isn’t it?

**P5:** Some knowledge of Casey’s theorem would help: use 3 vertices of the triangle along with the length of tangent from each of them with respect to the second circle, and it is true the tangent point must lie opposite the intersection of AB and l_p. But…was it even easy to see how the lengths relate to each other? I did some algebra and got nowhere, then instantly gave up…

Time flies, and I only saw what’s wrong with P3 10 minutes before the end of the competition. Luckily I checked my P1, P2 and P4(a), and for this very moment I stopped my mind from thinking P4(b) and P5. Oops! I still couldn’t fix P3 when Mr. Suhaimi announced time’s up. I made final sorting of my papers, and stapled it with *frustration,* feeling defeated.

—

Upon submission, I was quite curious on the performance of my peers. Justin hinted to me that he didn’t do well too, and Si Yu helplessly told me that he submitted nothing for 3 problems. 3 of us had 2 problems out of 5, and that’s the most we could offer: many claimed a solution for P1, but apart from that no other problem was well-answered (2 of us solved P2, 1 for P3, none for complete P4 and P5, but two of us has first part of P4). Although it’s obvious that the award boundary with be lower, not only when compared to last year but also to that of 2010-2012, my hope was still slim: how can you get it with only 40% of score?

—

**13/5/2014**

Mr. Suhaimi sent an email to notify us about our results: 3 Silver (I’m one of them), 3 Bronze and 3HM for Malaysia. Ranked 17/36 and team score plummeted from 213 to 94. A silver was pretty much expected when I knew my score was 77120=17, but the rage didn’t arise until I saw the complete result with cut-offs below:

Bronze=8 pts (just 1 point more than a complete solution!)

Silver=12 pts (Not even 2 complete solutions!)

Gold=*18 pts*

With the fact that I was #1 in Malaysia, you know the feel: 1 point away from gold award! Plus I could have obtained an extra point for P3 for writing the case for p=4k+1, ARGH!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! That’s remorseful…

Hence that’s the way I release my anger:

You should expect people to say you “show-off” when you post this–pretty arrogant for those who see it as telling others how many silver medals you have won, but I posted it as I had yet to win *any* gold medal in Mathematical Olympiad competition at international level.

Thankfully I received a compliment from my high school teacher:

Hey! That was just few minutes after my original post! This showed appreciation of CLB even after my graduation…it’s still my school, right?

On overall, we weren’t too bad, since those strong countries are damaged by this brutal set of paper. The champion, as always, is Korea, followed by USA and Russia. But that’s fair to some contestants compared to last year: from the last IMO I heard that in Korea, there were 23 perfect scorers that some of them didn’t win anything despite submitting a perfect paper.

So I knew what I should do: make myself into the IMO 2014 team, and fight hard for a gold!