The second post stated that I took part in International Mathematical Tournaments of Towns (IMTOT), Fall 2013. The official solution is now available on the official website:
http://www.math.toronto.edu/oz/turgor/archives.php
Have a look at the problems:
TT2013F_SOsolutions
And I was surprised by my ability of doing maths after 4-month-break: Solving all itself was not any big deal, but you would feel great if you can finish within half of the time given.
The problems below are sorted by the order of attempt.
17/12/2013: Senior O-Level.
Time given: 4 Hours
Problem 1 (3 marks). As usual, it’s easier to prove the existence of things compared to its opposite. Moreover, when the last digit is 2, 4, 6, 8, 0, 5 the number is composite (multiple of 2 or 5). Isn’t it a good strategy to move all to the right? Yes, it is!
The hard (nontrivial: this is not hard!) part is to deal with 1, 3, 7, 9.Using alternating digit argument we have 1+9=3+7, 11 divides 9317 and we can have 9317246850. However, this was not my first idea, bu I checked for divisibility of 7 using 9, 3, 1, 0, and it thankfully succeeded at 9317 😀
Problem 3 (4 marks). OK let’s start from the beginning, lcm (n, n+1)=n(n+1) is for sure, and by problem definition lcm (n, n+2)=n(n+2) is impossible, so it must be n(n+2)/2,… Ah huh! The idea is lcm (n, n+i)=n(n+i)/i (obvious, isn’t it?) This means n is a multiple of 1, 2, ,…, 35 and same goes for 36=4*9. This settles everything.
Problem 4 (5 marks). A classical trick used before: bijection. Move one column, and two rows for each rook. Oh yeah, biject the coordinates and they will be neither on same row nor same column.
Problem 2 (4 marks). Despite being the second problem, it was not as easy as problems mentioned above. How to prove something parallel? I drew few diagrams, oh wait! I saw something: YBX is similar to ABC by ratio and angles defined in the problem! What’s next? I wanted to avoid cases, and used directed angle to prove 2 lines parallel. That was a lot of work!
Great! I solved the first 4 problems in an hour. Now turn to the last challenge of the day: Nasty problem 5.
Problem 5 (6 marks). The answer turned out to be a paradox. At the first glance, we should always be able to determine the answer, right? I tried to prove that, but in vain attempt.
After a thorough thought, and knowing that 3D diagrams are harder to draw and imagine, I tried to overlap a sphere and a cube with same centre, and here comes my solution: draw six circles on each face of cube which are tangent to sides of square (for each square face) and we found the desired path. Written solution, however, was in algebra (as expected).
—
After finish writing, I had a shot on the clock: elapsed time was 1 hour and 45 minutes. Perfect! I could have a good time checking my solutions and have a rest.
After all papers have been collected, I had a survey with the other seniors: 4 of us had it all, with some others solved 4 out of 5. An easy paper, indeed! (lol I realised that I could have avoided nasty directed angle for P2 by proving AY=XZ and AZ=XY, which was what Zi Song did).
The A-Level paper was harder as usual, but overall it was a decent effort for me. I will upload it on my next post.
anzoteh96 