July has been highly mathematical for me. The release of IMO 2015 shortlist, my unstoppable penchant for trying out problems, the endless journey of searching for mathematical topics to teach,…all drive me to a unprecedented craze for mathematics. Above all, however, was the annual mathematical festival that captivated me–the International Mathematical Olympiad.
Here they came: the problems.
Problem 1: Too many points clustered into a single diagram, which makes the problem rather unconventional. Never mind, just draw the semicircle circumscribing triangle BCF and all is set, because X and D will be on this circle as well (I used trigonometry to established that, only to discover an elementary solution using similar triangle a week after). Moreover, E, X, D will be on the same line parallel to CF, and EM parallel to DC (easy to prove: EA=MX=MC=MF, the circumradius of BCF). If we let DC intersect FX at R then ME bisects RF (by parallel+midpoint argument above). Simple angle chasing yields BD bisects FX, too. Done.
Comment–not an easy geometry; I believe there should have been prettier geometry problems on the IMO 2016 shortlist.
Problem 2: 3 divides n (don’t ask me why), but obviously 3 doesn’t work–I initialized the middle square as M and quickly filled up the diagonals, only to have this diagram:
Wait a minute, can’t you see … that I (key to construction) gonna fall from the stars, straight into your arms… I quickly understood the construction for 9*9 with nine 3*3 similar grid above, only to modify the middle grids of each 3*3 squares.
It turned out that 9 must divide n: let n as 3k. First count the entries in each column and each row, then count the entries in diagonals. Now split them into k^2 3*3 grids, and we have those in the centre of grids counted 4 times, corner: 3 times and sides 2 times. If m of the k*k entries are M, then counting 2nd, 5th, 8th,…3k-2th column and 2nd, 5th, 8th, …, 3k-2th column gives all sides and centered grids, so: 2(k^2-c) M’s on the sides, leaving k^2+c M’s on corners. But each M is counted 8k*k times! Summing up, c is k*2/3, and k is a multiple of 3.
Comment: Beautiful puzzle, with difficulty extremely suitable for Problem 2. I’m actually amazed by how fast I tackled the problem.
Problem 3: Attempted, but no work. It turned out to be the toughest nut to crack in the entire IMO paper, and I’m convinced that it will remain a conundrum until I read the solution.
This significantly redefined my view on number theory problems: I had always thought that I have decent mastery in Olympiad number theory, but this NT problem doomed me. Can the unconventionality of the problem be an excuse? No.
Problem 4: Let’s see primes that potentially divide P(n) and P(n+k) for at least one n. k=1, none. k=2…only 7 (take n=7x+2), k=3…only 3 (take n=3x+1), and 19 is good for k=4. (Take n=19x+7). Easily seen that b=2,3,4 can’t work (just take the middle elements and you can’t ‘pair’ all of them), with b=5 more work to do. b=6 can be constructed, however: 19x+7, 7y+2, 3z+1, 7y+4, 19x+11, 3z+4 — they exist by Chinese Remainder Theorem.
Comment: Cute puzzle in number theory, and I enjoy the problem.
Problem 5: Wanted to construct total of 2016 linear factors, but my thinking wafted to an awry position. x-1, x-4, x-5, x-8,… on one side and x-2, x-3, x-6, x-7 on the other side?
No no no my mind said it can’t work (disproved on AoPS forum) so I mistakenly ditched it. I experimented on things like x-1 to x-c (c odd), then x-2016 to x-2016+c-1 on one side, and the rest on the other side. Sadly, this pushed me into the black hole of futile verification, and time’s up (4.5 hours mark) before I could tell whether this works.
As it turned out, the modulo 4 construction that wafted through my mind can be simply verified using product telescoping (remember, that, (x-1)(x-4) and (x-2)(x-3) always differ by 2 !)
Comment: Another non-traditional polynomial problem with slight combinatorial flavour.
Problem 6: I did (a), out of the contest time frame. To do this, find three lines and form a triangle, then set a direction for one line. For the other lines, identify the rays which has even and odd points on sides with a reference line, and if the direction of travel is from the even side for a reference point, do the opposite for this new point (and vice versa). There will be no contradiction between different sets of triangles (use Menelaus’ theorem), and this is of course valid (for if the frogs land on the intersection points at the same time, then they had travelled the same number of points before, which means coming from both odd or even sides!) How about 6(b)? Maybe not within my capability, lol. (I intended to show that there are two lines whose ‘midpoints’ coincide, but an AoPS user gave a counterexample 😦 )
Trivia: It’s hilarious to see the different names (instead of Geoff) used in papers of different language: for Malay version, it’s Suhaimi. 😛
Overall, this try wasn’t particularly great when the performance is roughly at the borderline silver/bronze (worst case scenario: 770700; should have managed 770772).