Date: 8,9 July
Yup, these were the VVID all the teams looking for, and prepared for. The two days would be determining who would create a new record, who would achieve the lofty aim of gold medal, and who would be holding the title of absolute winner.
On top of that, mental equanimity was vital for everyone. I woke up at 6 with mind ready for challenge for both days, and nutritious breakfast provided me the maximum energy that I need for both days. Zi Song made joke of himself again by adding excessive amount of tomato sauce into his meals, which should be ascribed as his dietary habit. Nevertheless, a good laugh mollify the extreme nervous of us.
This time, the organiser did not specify our seat number in our name tag like 3 previous years, but we were given permission to enter our contest area for searching our seat number at around 8:30. From the clue that I was in block 1, I rushed to the area and spotted 1D10 for mine.
In attempt of maintaining ebullience, the Kangaroo team members shouted their Aussie slogan like some other teams last years. In contrast, Mr. Iqbal detected apprehension on me, and advised me to work on changing my beta brain waves to delta waves. It worked after some efforts of meditation, a gift from Justin’s father to us. Now I just had to wait for time to past. In the meantime, some thermal support was necessary for my hands: these shivering specialists would mar my work if there was no proper treatment.
“You may open the folder now,” the chief invigilator ordered 1 minute before the exam. With heart accelerating on its rhythm, I unsealed the white closed folder and looked for the problem sheet. “Start your work,” she added when the clocked stroke 9.
At the instinct, Problem 1 looked rather non-traditional, Problem 2 was typical TOT-style problem, and Problem 3 was a geometry with bizarre angle condition. When the history repeated itself with a hard geometry, wasn’t that a golden opportunity to score a perfect 21 for the day? “Conquer!” shouted the General of Brain.
P1. The problem suggested an experiment of the typical sequence of a_i=i+1, and experimenting of small values gives validity to n=1 but not others, since this leads to violation of first inequality. This immediately gave a motivation of proving that: if for some n, the second inequality was true, then the first inequality was false for all m>n. Analogously, the first inequality for some n implies the infringement of second inequality for numbers less than n. This was evident enough to solve the uniqueness part of the problem, no?
This excitement led me to write when I was thinking (with hands still quivering) : a taboo for contestants because of the saying of “more haste, less speed”. This was true especially when I had to figure out the existence part of the problem, which was not that easy. However, we must have a_0<a_0+a_1. Since it’s impossible to have smaller numbers violating second inequality, and bigger numbers violating first inequality with no gap between, the facts yielded we can assume that second inequality was invalid. Ironically, the step to contradict this wasn’t that easy, and I raised my toilet card to find some inspiration.
The toilet was itself a miracle when I figured out that f(n)=a_0+…+a_(n-1)-(n-1)a_n was decreasing function (why? consider f(n+1)-f(n).) This settled everything, and I quickly finished my solution, which turned out to be really elaborated. The clock displayed 9:45 in the end, which wasn’t satisfactory.
P1(Solution for P1)
P2. Considering the main diagonals we had approximately n/2 as our desired answer, and k=1 for n=2, 3 verifies this. However, for n=4 there was a traitor configuration of (1,1),(2,3),(3,2),(4,4), resulting me no choice but to play with n=5. This play was fruitful, however, since if k=1, the first 4*4 box to have exactly 4 rooks, leaving the bottom-right corner to have a rook. Didn’t that suggest rotation of chessboard? I was near to bull’s-eye now! For the upper boundary we ought to imitate our method of 4*4, where for 9*9 I experimented the configuration of (1,1), (2,4), (3,7), (4,2), (5,5), (6,8), (7,3), (8,6), (9,9), and to my surprise it worked magically. I had no reason not to grab my pen and write my work.
This writing, however, was full of friction: the lower boundary’s proof was settled with ease, but for the opposite direction I was posed with challenge of verifying that the configuration could not allow any better bound. Oops! Time to go to toilet, again. And……thanks to the “toilet magic” again, the notion of separating into 4 cases depending on the location of the k*k square came across my mind! That’s tedious enough, and it ended me up to 4 pages long. Fortunately the clock generously showed 11:15, implying I had half of IMO full-time to vanquish the boss of the day.
P3. Question at sight: how were we going to use the given angle condition into the solution? Well, first idea that I could squeeze of was to fix C, H, S and line BS, then construct B, A and D, accordingly. Since the problem condition could be changed to “circumcentre of CHS lies on AH”, we may as well fix the cicumcentre, and point T would be apparent by then (just for sake of construction). Good thing was, the 90 degree difference implied that circumcentre of CHS lies on AB (name K), and analogous condition holds for the other side (name L).
However, that’s all I could do with limited capability of brain twisting, scarce knowledge, and quick draining of mind battery. The proposition that bisectors of angles HKS and HLT should meet at AH was bright in the middle of the stage, but I couldn’t see the inner layer of that suggesting the method of Apolonious circle (or simply HK/KA=HL/LA). I raised my green card again, and the toilet apologized to me for being unable to create magic as what it did for P1 and P2 before. I put it aside and double-checked my solution on P1 and P2, and thankfully there was no erroneous argument.
It’s after 12:30 when I got over my devastation and try some hard approach on trigonometry (since this was my master tool in solving most sufficiently hard geometry problems). With triangle AKL in mind with points H and C this gave me rooms of brute-forcing, until the nick of time where I discovered something I missed it before (using this ugly method): reduction of the problem to CH perpendicular to ST, which could be equated as HK/KA=HL/LA. It’s 1:25, but I couldn’t go any further than this. I could only leave my last sentence “this being obvious by H as the reflection of H’ (perpendicular from C to BD) in midpoint of BD.”
“Stop your work and put your work properly into the folder,” the chief invigilator ordered sternly. I checked my folder and all written papers, just to confirm that no sheet of paper was misplaced. We were then asked to leave row by row, and I left my seat despondently, thanks to my terrible defeat to P3.
Reports from my teammates were marvelous: all of us solved P1, and 5 of us had P2 done. Moreover, Justin managed to complete P3 using 11 pages, hence victoriously submitted a perfect paper for the day. I was glad that he assuaged my broken heart by pointing that P3 was more brutal than last year.
Studying after the contest was neither productive nor useful, even when you considered spotting the topic combination based on the first paper. While there were many guesses (based on instinct or what they hope so?) on the topics for P1 to P6 a month before that (with some insensible claim of cut-off points at the same time), we derided at it. Knowing this fact, Mojalefa decided to take us out for another walk–this time slightly beyond just the campus, and it aimed to refresh our mind after the grueling mind fight for the morning, besides hoping that we could forget all moments in it.
(Above: some extraordinary buildings and statues that we came across)
Taking back into AoPS again, the problems were posted onto the website, and as before there were some users use ridiculed the difficulty of P3 and said “this is a relatively easy one”. Moreover, this, along the somewhat easy P2, invited claims on high boundary for medals. Well, that hurt.
On the positive side, it doesn’t seem to be a challenge for us to raise the Malaysian benchmark in IMO again, based on the exceptional performance of the day. Nevertheless, let’s hope for a brighter day tomorrow!
1 thought on “IMO 2014 (IV)-The climactic section (First Half)”
Nice Blog, thanks for sharing this kind of information.