The only reason I post is to show how disastrous this IMO contest is (although not participating). Here it goes, problems by problems:

1. Create four triangles using six points, but I could go no more. I turned to (b) and think: hmm the if P_(AB)A=P_(AB)B then P_(AB), P_(AC),…,P_(AZ) must be different (anyway Z here means the last point, like how we denote it as the last alphabet 😛 ) so {P_(AB), P_(AC),…,P_(AZ)}={B,C,…Z} (with some permutation) and each point represents n-1 such P’s. But if we pair PA and PB (with PA=PB and P in S) then we have (n-1)/2 pairs. OK n must be odd (and trying n=5, the star graph leads to the configuration of regular polygons HAHAHA). Back to (a). Knowing that even n fails for (b) the best is to think of equilateral triangles with a universal center. Seems doable with all points on circle except one, which is the centre O of circle. But how about AO? Here’s it: add three points A,B,C on circle with AOB=BOC=60 deg, and add two points at a time that make an angle 60 deg with the centre. Done.

2. (2,2,2) and (2,2,3) are obtained in seconds, so assume a,b,c distinct. Notice that case where a,b,c even can be happily eliminated too (why? play with powers of 2 you get two of the following is true: p(ab)>p(c) (implying p(ab-c)=p(c) ), p(bc)=p(b)+p(c)>p(a), p(ca)>p(b) where p(x) is the highest power of 2 dividing x. So if p(ab-c)=p(c) and c=d.2^k with d odd, then ab-c=2^k and we have ab-c<c, or ab<2c, so a,b<c. See the contradiction?) But no one can stay happy forever because the case 2 even and 1 odd has already confound me into the spiral of case-by-case analysis (though this implicates c=ab-1). Now replace a with 2^2k.a and b with 2^2k.b, where one of a,b odd. What do we have? 2^2k a(b^2)-(a+b) and 2^2k (a^2)b-(a+b) are powers of 2. The difference is 2^2k (a-b), and a+b cannot be odd-> a and b both odd. Take the difference 2^2k (b-a) 4 divides a+b but not b-a, and if a<b then 2^2k (a^2)b-(a+b)=2^(2k+1). We are close! Now a^2b-(a+b)/(2^2k)=2, from which we obtain (2,6,11). OK now we are left with a,b,c odd, which I haven’t done. The answer (3,5,7) was obtained fortuitously, and after some try, I surrendered.

3. Q,H,M collinear. If X is midpoint of QH then OX bisects angle QXK. Then MH.MQ=BM^2. Is that all? Yes with dilapidated compasses. I tried perpendicular bisector intersection (to locate circumcentre), play with nine point circle, draw a tangent line to circle QKH at K, hit BC at U, hoping to prove UK^2=UF.UM. All failed. I bought myself a pair of new compasses and thought of constructing another circle tangent to circle QKH and passing through FM, which turned out perfectly to be circumcircle of HKM. Now I can conveniently draw perpendicular to QM at H to meet MC at R, the radical centre of circles QKH, KFM, HFM, and now everything reduces to RXO=90 deg. But this can be bashed by using squares of length!

Well this P3 is the most beautiful problem for this olympiad (in my eye 😛 ) *Side note: IMO Problems 3 were geometry for three consecutive years since 2013, while never so for IMOs 2004-2012. What happened?

4. People said it as “ugly” but “fast killed by angle chasing”. I testified this successfully even without a decent compass.

5. *STOP! TROLLERS CROSSING!* Now f(0)=2 or 0. It used me hours for first case, but the idea was straightforward to understand: just prove f(1)=1, x+f(x) is always a fixed point (which is same for f(0)=0), and if f(k)=k then k=1 (using some substitution involving k). Now the second case. I get: all integers are fixed points, and (2^k)f(x) is fixed point for nonnegative k. But however hard I tried I couldn’t establish any of the three which would kill the problem directly: f is injective/surjective/continuous. No more progress? Game over. (An advice: always verify that your functions work. In this case they are x or 2-x.)

6. “At one point, the mean of the sequence stabilizes”. I didn’t want to try this actively on paper, considering it as a P6. But as mental exercise I thought of set {k+a_k} and concluded that only at most 2015 positive integers (including the hopeless 1) do not belong to this set. And what? Intuitively this number of such integers is the b we are searching for, and N can be taken far away from those “missing numbers”. Of course I’ll need to rigorise the proof onto paper, but I’m there (which was strange: can P6 be something solvable directly by one trick? This trick wasn’t easy to think of, but a problem stemming solely on one hard trick shouldn’t be in IMO).

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Looked like I can handle 4 problems (1,3,4,6), but how well can I do in competition? P3 took me hours, while P6 can easily be obscured by P5 under exam conditions. In the worst case, I would have 1 and 4 -> can I have 5 more points anywhere to qualify for a silver medal? I don’t know.

difficult to read as no paragraph in between.

Technical error LOL (just fixed)

cant agree more… swapping p2 with p3 (and p5 with p6) can be more desirable to many contestants…