# Revisiting Mathematics Competitions (Hotseat: AIME II, 2015)

I remember I had a pretty good time doing the Haskell January Test earlier this year, and thought it might be both enjoyable and useful to revisit some of these past mental exercises. Hotseat is intended to be a series that discusses my thoughts, ideas and reflections as I take or re-take examinations or competitions; many of these will end up being related to mathematics or computer science in some way, though I’ll try to also include a few that go against that grain.

#### Background

I didn’t actually participate in the American Invitational Mathematics Examination (AIME) as I don’t think my high school offered it (the only maths competitions I remember taking myself are the Singapore Math Olympiad and the Australian Mathematics Competition). This is typically the “second round” of the main mathematical competition in the US (the first being the AMC, and the third being the USAMO), and features 15 problems, all of which have integer answers from 0 to 999. The point of this is to facilitate machine scoring. While the questions might indeed have integer answers, they can be very difficult especially towards the end.

2015 II was supposed to be an easy paper, though I didn’t find it to be very much easier than what I recall these to be. You can find the paper here.

I managed to fight my way through 12 out of 15 questions, leaving out two hard problems towards the end and a not-so-difficult one in the middle (mainly because I struggle somewhat with geometry in three dimensions). I think I’m not used to doing these as quickly as I would have in the past as well; problem 14, for instance, would probably have taken under 15 minutes in the past while it now took 32. 8 minutes on problem 1 is rather bad, as well.

The values below indicate the time spent on each problem; here green means the problem was answered correctly while grey means no answer was given. (In practice, in a contest you would take the 1-in-1000 shot and guess, since there are no penalties for guessing!). I’ve also outlined the subject areas – you can see that I tend to be stronger with combinatorics and probability, and struggle somewhat with geometry!

I think I remember mentioning in a post a very long time ago that when I look at data, I tend to initially look for things that seem unusual. Bearing in mind that the difficulty of these papers tends to rise, we could say the 32 minutes spent on question 14 or 15 on question 7 is perhaps not too far from expectations (the target “pace” here is 12 minutes per question, if you’re trying to finish). Bearing that in mind, we can look at a couple of outliers:

• Problems 1, 6 and 11 took unduly large amounts of time.
• Problems 2, 3, 8 and 12 were done well above pace.

Some of these were actually pretty interesting, which is one reason I tend to enjoy these competitions as well.

#### Selected Problems in Depth

Problem 1: Let $N$ be the least positive integer that is both 22 percent less than one integer and 16 percent greater than another integer. Find the remainder when $N$ is divided by 1000.

For this question you were actually supposed to find $N$; the “divided by 1000” shenanigans was so that the answer would be an integer from 0 to 999. The challenge here was really formulating a requisite equation from the problem statement, which would be

$\dfrac{39}{50} k_1 = N = \dfrac{29}{25} k_2$

and then reasoning that $N$ must be divisible by both 29 and 39, such that $k_1$ and $k_2$ were both integers. Since these are relatively prime $N = 29 \times 39 = 1131$ and the answer was 131. I guess I hadn’t really warmed up yet, and wanted to be very careful to set up the expression correctly, so this took rather long.

6 was a bash with the Vieta formulas, though my brain temporarily switched off when I somehow thought a cubic equation should have four roots. 11, on the other hand, was a rather messy geometry question which I initially found tough to visualise. The solution on Art of Problem Solving involves constructing perpendiculars from the centre of the circle to the two lines, but I didn’t think of that (and for that matter wouldn’t have known that was a property of a circumcentre), and instead had a rather painful trigonometry bash to find the answer.

On the brighter side, 2 and 3 were rather simple questions (an elementary probability question, and a rather quick modulo-arithmetic one) and I cleared them off very quickly. I dealt with 8 using some rather quick-and-dirty casework that yielded the desired result, though it wasn’t too interesting. I’d say 12 was a bit more of a fun one.

Problem 12There are $2^{10} = 1024$ possible 10-letter strings in which each letter is either an A or a B. Find the number of such strings that do not have more than 3 adjacent letters that are identical.

I thought in terms of recurrences and dynamic programming, and realised that valid strings (in the long case) must end in BA, BAA, BAAA or their flipped versions (alternatively, with a blank space in front). To end a string with BAA, it must previously have ended with BA, and so on for BAAA. Thus, the following equations drop out:

$BA(n) = AB(n - 1) + ABB(n - 1) + ABBB(n - 1)$

$BAA(n) = BA(n - 1); BAAA(n) = BAA(n - 1)$

Since the AB and BA cases are symmetric, we can simplify the first equation above to

$BA(n) = BA(n - 1) + BAA(n - 1) + BAAA(n - 1)$

And we have of course the base cases; $BA(1) = 1, BAA(1) = 0, BAAA(1) = 0$. Computing $2 \left( BA(10) + BAA(10) + BAAA(10) \right)$ is then not difficult, and yields $548$.

In mathematical terms I would say problem 14 required some insight, though for seasoned competitors it would probably still not have been too difficult.

Problem 14Let $x$ and $y$ be real numbers satisfying $x^4y^5 + y^4x^5 = 810$ and $x^3y^6 + y^3x^6 = 945$. Evaluate $2x^3 + (xy)^3 + 2y^3$.

I initially tried squaring the resultant expression, and combining the two given results in various ways. This went on for about 20 minutes with little meaningful progress. I then took a step back, attempted and finished up with problem 11, and came back to this. I suddenly had the idea of combining the two givens to force a cubic factorisation:

$x^3y^6 + 3x^4y^5 + 3x^5y^4 + x^6y^3 = 810 \times 3 + 945$

And that simplifies down to

$x^3y^3 (x+y)^3 = 3375 \leadsto xy(x+y) = 15$

We then proceed by factoring the first result:

$x^3y^3 (xy(x+y)) = 810 \leadsto x^3y^3 = 54$

We can get $17.5$ for $x^3 + y^3$ by factoring the second given to $x^3y^3(x^3 + y^3)$. The final result is then just $35 + 54 = 89$.

I subsequently figured out solutions for 13 (my brief intuition about complex numbers was kind of on the right track, but I didn’t have time) and then 9 (not good with 3D geometry), and had a look at the rather painful solution for 15.

#### Meta-Analysis

It’s worth taking a look at how this performance was relative to the actual score distribution, which is available on the AMC statistics page. A score of 12 would be in the 96.09th percentile, which is interesting; I did the 2015 AIME I a few weeks ago and scored 11, but on that contest that was the 97.49th percentile!

There’s also a report on item difficulty (i.e. the proportion of correct responses to each problem). I’ve graphed this below (items I answered correctly are in green, while those I did not are in red):

It’s interesting that problem 12, a problem I found to be one of the easier ones (assuming you sort by success and then time, I found it to be the 8th hardest) turned out to be the second “hardest” problem on the paper! I’d attribute some of this to a programming background; if I took this contest when I was in high school I think I would probably struggle with 12.

13 and 15 were hard as expected; the next hardest problem in general after that was 10 (just a notch above 12 in difficulty for me) though again it relies on tricky combinatorics. We then have 6, 11 and 8 before 14; I found 6 and 11 relatively harder as well, as mentioned above, though 8 was fairly trivial for me. It turns out that my geometry weakness is fairly real, as 9 was completed by more than half of the participants! We can also observe problem 1 being challenging relative to its location in the paper.

#### Conclusion

This was a good mental workout, and it felt nice to be able to hack away at the various problems as well. I think a 12 is pretty solid (the 96th percentile is that of AIME qualifiers, which itself is something like the top 5 percent of high school students taking the AMC, so that would be the 99.8th percentile – and you could argue that considering only students taking the AMC already introduces a selection bias towards those with greater aptitude). I’d probably have been able to solve 9 and maybe 15 if I was doing this back in high school, though I’m not sure I would have managed 12.

I can certainly feel a loss of speed, though I think my problem-solving ability is for the most part intact, which is good.