Three hours gone in the morning. Three more hours to fight. Thankfully we had two-hour-break in between but I desperately needed coffee. Our proctor/Putnam coach (Prof. Stephen New) asked if anyone in the room needs more coffee and I was among the very few people who raised their hands, “you gonna finish most of them”, he joked to me.

Jokes aside, here’s section B. Assuming that the methodology of trying the first few problems actually works, I first looked at B1. “Easy”, I thought, “just use my trigonometric specialization that I had in my IMO journey back then right?” Unfortunately, there are just too many cases involved (point inside/outside the angle domain that gives positive or negative ratio), so I put it on hold and turn to B2.

B2 was another problem that looked routine: the first question we should ask ourselves was: what are the numbers that can be represented by sum of k consecutive positive numbers? Removing the constraint that *a* has to be positive, N has to be divisible by k if k is odd, and N-(k/2) divisible by k if k is even. Adding the “positive *a*” constraint simply means N is at least 1+2+…+k = k(k+1)/2. But what’s next? Hmm…well now N=2017m with m>=1008. If m is divisible by some odd prime p then k=p would work (come on, 2017m>=p(p+1)/2 shouldn’t be hard to prove). Otherwise, consider m=1024 and the smallest k>1 that is not 2017 is k=2048, but 2048*2049/2 > 1024*2017. This gives our desired form, and hence the answer a=16.

Turning back to B1, I wasn’t convinced that I could make my trigonometric argument work. But the next thing came into my mind: what if, we rotate a line around P and consider the ratio? As it turns out, this ‘covers’ all possible lambdas except zero, as desired.

Time for B3. First thought: prove that the sequence c_i must be eventually periodic for f(1/2) to be rational. There came my machinery: Fermat’s little theorem (every number 1/q with q odd can be written as p/(2^k-1) for some k and p integers, and 1/(2^k-1) is 0.000…1000…1… in binary form, with period k). Still…this looks messy, and I had to skip one final computation part and straight claim this identity directly. The rest wasn’t too hard to complete: if the series c_i is eventually periodic, then f(2/3) has odd denominator and cannot be 1/2. Yay, no? Oh well, not really (considering the minute details that ate up substantial amount of my time).

I couldn’t remember how much time left by then, but the aim was to kill another problem. Curiosity on how the pattern of sums for B4 dominated my mind, so I tried that first. On its first sight, rearrangement of the terms in the right-hand-side looks tempting because this trick simplifies computations substantially most of the times, according to my experience. No. I couldn’t do this arbitrarily: it doesn’t convergence absolutely so I had to do the rearrangement judiciously. But how? 😥

After the futile experimentation on B4, I proceeded with B5 instead (geometry ❤ ). So there were a few things to keep in mind: if we fix the line to partition the triangle into equal perimeters and vary it, then using extremal arguments each triangle must have at least one equalizer. Having that in mind, I continued to play with such lines, looking for the critical cases of the lines intersecting one of the three vertices of the triangle, and anything interesting in between. But I got myself into the whirlpool of cases, the most distinctive of which was whether the shortest side is at least a quarter of the perimeter of the triangle. Nervous. Can’t continue. Time’s up. Devastated. I couldn’t even squeeze time to *look* at B6.

In the short post-contest discussion, many peers claimed the first three, and concurred with me that B4 was hard (except the Putnam coach, Prof. Stephen New, who vanquished B4 rather instantaneously). There was one who additionally claimed B5, which gained him a bucket list of 8 problems solved. *Oh well, I just had to hope that all my 7 solutions were good enough to earn full/near full credits*.

Ps: B4 actually wasn’t that hard. *If only I was a bit braver and try to avoid the convergence issue by thinking of partial sums…*