Tournament of Towns Fall 2013, Senior A-Level

Mood of exuberance after the O-Level exam for seniors. Indeed! My result for O-Level would be an illusory of my strength of being capable to finish it in 2 hours, and the complacency feeling was ineffable!

However, the next day was the real challenge: A-Level. Some of us was quite paranoid towards it, saying that it must be harder than usual to balance the level of difficulty. I remembered how I was inundated by my first TOT A-Level for Spring 2011, where I only solved 1.5 out of 7 problems.

18/12/2013: TOT A-Level

Time given: 5 Hours

“You may start now.” Our invigilator, Mr. Loke ordered.

I turned the question paper, and stole a glance on each of them:

Hmm…where should I have my first footstep? Having an impression that TOT 2-D geometry shouldn’t be hard, I tried P3.

P3: I drew the diagram. Wow! Equilateral triangle! A pretty special case and that should trivialize the problem… What’s next? Compare angles of course! I tried the median and thought of properties of it, but in vain attempt. No!!!!!! 30 minutes passed. I gave up with a lot of confusion in my head, and tried something captivating:

P4: n=1 is obvious, but not even n=2. Though placing + and – signs for 4 consecutive cubes in a suitable way yields a linear term: 12n+6, and for 8 consecutive cubes we get a constant 48, this still got me nowhere. But choosing 2 such linear terms yields 6 isn’t it? So it suffices to prove for the case n=1,2,3,4,5 and we’re done by induction. 1:30 elapsed time. Finally solved one.

I looked at the other problems again. P6 and P7 was long-winded and I had no patience on reading it. Let’s look for easier ones. (I tried P3 again, but no progress, anyway.)

P1: Trying an error yields 4 polygons to be sufficient. How about 2 and 3? (1 polygon isn’t enough, for sure). I succumbed to the easiest problem of the paper, indeed.

P2: Looks hard. Adding two monomials and the polynomial has no real root. The leading coefficient must have even degree, so it is a must to cancel the leading coefficient for odd degree. Now see if we can add something to make it even degree: this is for sure for linear ones, but not sum of more than one monomial with odd degree. Now n=1 is possible, but not other odds. For even n it is quite simple: make the leading coefficient positive, and it has minimum value. The solution is now in front of us: add a constant to make minimum value greater than 0. We’re done!

I was not done yet! My adventurous mind asked me to have hard battle.

P6: Same pile first, different pile next, who can always win? A classical game problem, but thinking of the first pile I obtained the bijection idea. Biject it to each stone in different piles. Good. Now deal with remaining 9*10 piles. Same idea, isn’t it? I threw my pen and went to the toilet to tidy up my idea, and wrote my solutions soon. 9 points obtained by thinking 15 minutes on it. Good job, Anzo!

Now seems like I have P2, P4 and P6 in my hand, which adds up to 6+7+9=22 points. The only way to increase it was to solve P5 or P7. (Maximum three counts)

P5: (a). By having ff and gg identity, and fg (x), gf(x) greater than x what does that mean? Squeezing my brain hard itself was no-good strategy. Wait a minute! fg(x)>gg(x), with f and g both surjective (as it runs through all positive integers). This gives f(x)>g(x), and the similar argument yields a contradiction. 3 out of 8 is now in my hand.

(b). By (a), co-domain of f and g are disjoint. I kept asking myself the same question: how to construct, or, to arrive an absurd condition? Tick tock Tick tock, time was running short: 1 hours and few minutes left. Perhaps it was better to throw it aside.

Last hour: Fight or Flight?

Good question. Should I continue the battle, or to stay away from those remained unsolved and check my solutions meticulously? Knowing that my solutions for P2, 4, 6 were long, complicated and somehow naiveté (except P6 which was quite sophisticated), I chose the latter.

(Oops! I forgot to mention a single word about P7: 14 points were allocated, meaning that it could not be an apple pie.The complicated problem statement intimidated me and I never take an effort to read the problem n full.)

I read through every line of P2, and every claim was justified clearly. Looking at P4, something incongruous appeared: choosing 12n+6 for two suitable n yields 6. Oops, I fake solved P4, and it would cost me 7 points.

There no way to prevent my nonchalance from being shaken, and within the remaining 45 minutes I had to come out with a new solution. But fixing the gap wasn’t so tedious: having 4 consecutive cubes, with minus sign for the middle two we get 12n+6, and our boss, 48 (=resulting term with 8 consecutive cubes with suitable + and – sign) was facilitating us. This remains us with the small cases: 0, 1, 2, … , 11, and I had 0 to 3. Settling for 4, 5, 6 and changing -a to a we were done. After writing the solution, I stole a glance at my watch again: 15 minutes left.

5(a) solution was short, so the checking time was negligible and I took a bigger step to reach my P6 script. A lot of idea involved, but thankfully they were correct and I did not have to rewrite it. “Ding! Ding! Ding!” Time’s up. I handed my solution, with irrelevant concern of any solution being wrong.

We discussed with our peers about their performance and strategies of solving each problem. The first question by Loke after collecting the papers: “Has anyone solved P6 and P7?” As expected, nobody even read the problem statement of P7. On the other hand, 2 of us had P6 killed in our hands, and the other one had P1 to P5 all solved. Still a decent performance, and I was satisfied with the score, if I could secure full marks for the 3 problems (though solving 3.5 problems did not meet my aim, but that’s ok.)

That was the grand finale of IMO Camp 1, and the A-Level problems left me listless on the JPN van on the journey home. I had a good sleep then.

After note: I still thought that my A-Level performance was a foolhardy one: there’s some tricks which I didn’t notice of, hence could not solve some problems/took a more complicated approach on it.

P1: There’s a trap: we can draw polygon in the plane, not necessary inside the chessboard. This explains why 2 is enough.

P3: Lol. Intersect the two lines and prove that the point of intersection lies on some line, then power of point, solved!

P4: 48 is our boss, but how can we utilize it? Having (48k+1) cube gets everything congruent to 1 modulo 48, and we can add (or minus) enough 48 to get the number we want.

P5(b): Great troll! The answer is yes, but the resulting function was hard to think of…

All in all, let’s gain experience from this, and push ourselves to another level of excellence: there’s a lot of challenges in the IMO 2014 cycle: APMO 2014, TOT Spring 2014 and finally, IMO 2014!

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