# BMO1 2021

The first round of the British Mathematical Olympiad was sat on Thursday by roughly 2000 pupils in the UK, and a significant number overseas on Friday.

For obvious reasons, much of the past 18 months has been dominated by logistical rather than academic problems, so it’s refreshing to be returning to the traditional format of this exam, even if the marking has been stayed online.

I wasn’t involved in setting in this paper, so here are some thoughts on the problems. These aren’t supposed to be official solutions, and it’s entirely possible that I’m missing the most natural approach or framing. Students reading this in future years are advised, as always, that such commentaries are normally more valuable after attempting and digesting the problem yourself first.

Note that the copyright to the problems is retained by BMO, and I’m reproducing here with permission. The paper can be found in its original format here.

Problem One

If one exhibits eighteen (= 6 x 3) explicit examples of relevant sums of consecutive positive odd numbers, one has finished the problem. An exhaustive search is impractical and uninteresting. However, beyond that, we would also like to find integers N which we are convinced have this property without needing to explicitly decompose them into the desired sum format.

Let’s study such a sum. We need an even number of odd numbers to end up with an even sum. Probably the algebraically neatest way is to set up such a sum symmetrically as:

$2n-k, 2n-(k-2),\ldots,2n-1,2n+1,2n+3,\ldots, 2n+k,$

for k an odd number. The sum of this sequence of consecutive odd numbers is 2n(k+1). So, we seek integers N which can be written as 2nm, where m is an even number, in as many ways as possible. Note that if we have an integer N which can be factorised as a product of two distinct even numbers we are forced to take the larger as 2n, and the smaller as m, since we require 2n-k to be strictly positive.

It’s equivalent to finding integers M which are less than 100 and have lots of factors (then multiplying by 4 to recover an N), and there are a few ways to do this, including now potentially by exhaustive search. It is useful to remember that is M can be factorised into primes as $M=p_1^{a_1}\ldots p_k^{a_k}$ then its factors can be described as the set:

$\{1,p_1,p_1^2,\ldots,p_1^{a_1}\}\times \{1,p_2,p_2^2,\ldots,p_2^{a_2}\}\times\cdots \times \{1,p_k,p_k^2,\ldots,p_k^{a_k}\},$

In particular, the number of such factors is $(a_1+1)(a_2+1)\ldots(a_k+1)$, independently of the choice of the primes. This gives a recipe for constructing all integers M<100 with at least twelve factors. We could take $(a_i)$ to be some permutation of (11), or (5,1), or (3,2), or (2,1,1), which give the following valid M:

$M=2^5\times 3 = 96, \; M=2^3\times 3^2 = 72, \; M=2^2\times 3\times 5 = 60,$ $M=2^2\times 3 \times 7 = 84,\; M=3^2\times 2 \times 5 = 90,$

and the corresponding valid N are 240, 288, 336, 360, and 384.

For what it’s worth, this problem took me a while. I spent a long time saying things like “we need numbers that can be written as 4n and as 8n and as…” and aiming to conclude “all multiples of 24 greater than 24 x ?? will work”, but this didn’t ever generate enough examples of one class.

Problem Two

In order to have won exactly 30% of the games so far, the number of games so far must be a multiple of 10, and the same is true for 70% success rate. The other success rates have similar constraints, but the multiples are smaller. After ten games, of course Arun cannot have simultaneously won both 30% and 70% of the total! So we need at least twenty games.

It feels as if we should be able to include the other success percentages so long as we can make 30% and 70% work. But can we do this in twenty games? This would require Arun to win 3 out of the first 10, and 14 out of the first 20 (or the exact opposite), and after some head-scratching, this is impossible, since it would require winning eleven out of the ten games between these observations. This is a nice observation, and we now know the answer must be at least thirty. Given that we have made some nifty, rigorous deductions already, one might speculate that this is the true answer. To verify this, we just have to give a construction, and there are a number of ways to achieve this, for example such that Arun wins

• One of the first two
• Two of the first five
• Three of the first ten
• Twelve of the first twenty
• Twenty-one of the first thirty

To get full credit, almost any justification that a construction exists for 30 games would be fine – but one must be given. Otherwise how do you know that the true bound isn’t 40 or 50 because there’s some other reason why 30 fails?

Problem Three

Facing this problem for the first time, there’s a temptation to try values smaller than 2021, and this may well be useful. If nothing else, such experiments may well lead to a conjecture of the general form of the answer. Personally, I find it relatively easy to make mistakes when dealing with small numbers, and would turn to this as a second strategy after trying more theoretical routes first.

First note that the condition $0\le n\le 11$ could be replaced by $n\ge 0$, since we are forced to include zero pieces of weight $2^{11}=2048$, and would similarly be forced to include zero pieces of weight $2^n$ for any $n\ge 11$. (%)

When counting a potentially complex set, we should start by seeing whether there is anything which we can control effectively. For example, I don’t think it is clear how to control the number of pieces of weight 64 in a configuration. However, powers of two lend themselves well to inductive arguments (if you take the powers of two and multiply them all by two, you still have powers of two!) and to studying parity (that is, whether a number is odd or even).

In this case, note that you must have either one or three of the pieces with weight one to ensure the total weight is odd. Similarly, if the goal had been 2022, we would need either zero or two of these pieces to ensure the total weight is even.

So that settles the number of pieces of weight one, and it remains to solve the problem for total weight 2018 and 2020 with the extra condition that we are no longer allowed to use any pieces of weight one. A way to capture this is to divide everything in the problem by two. So we are now trying to capture all the ways to generate weights 1009 and 1010.

We can reformulate this more generally, and assert that the number of configurations $f(n)$ to make total weight n satisfies the relations:

$f(2n)=f(2n+1)=f(n)+f(n-1).$ (*)

It would be quite tiresome to solve backwards in a binary search fashion, but perhaps this is where trying some small examples turns useful. Either by observation on the original problem, or by thinking about (*) with some initial condition like f(0)=f(1)=1, we find that $f(n)=\lfloor n/2\rfloor +1$. (That is, $f(n)=\frac{n+2}{2}$ when n is even, and $f(n)=\frac{n+1}{2}$ when n is odd.)

As a final aside, this question is an excellent exercise in using generating functions in an olympiad context. A couple of points are worth making:

1. It is important to decide whether you’re including f(0) as an example, much more so than in the original argument. As we’ll see, it works better if you do include it, and will avoid an awkward sequence of algebra steps.
2. Figuring out the answer, for example from small-ish cases, will make a generating function argument easier unless you are very fluent with manipulating rational functions. In this case, we would be studying $\sum_{n=0}^\infty f(n)x^n$, and trying to show that it is equal to

$\sum (\lfloor n/2\rfloor +1)x^n = (1+x)(1+2x^2+3x^4+\ldots) = \frac{(1+x)}{(1-x^2)^2}.$ (*)

Many readers who’ve played around a bit with generating functions may find it easier to go from the left to the right than vice versa.

Writing down a direct generating function for f(n) and showing it matches up with (*) remains a good exercise. It is helpful to bear in mind the discussion at (%).

Problem Four

Since I last blogged regularly, I have given a series of masterclass on Euclidean geometry three times, including twice in an online format.

Part of the point of this masterclass is to get younger students to think in more general terms about what they are doing when they are solving geometry problems, even when they are straightforward. In a competition, time is tight, and anything that gets the job done is good, but there’s still value in reflecting on these ideas here. Two ideas in the first session are:

• Attacking the conclusion coherently
• Clean angle chasing

and this problem offers a good example of both of these in practice.

We’ll start with attacking the conclusion coherently. We’ve got to prove that a triangle is equilateral. How can we achieve this? There are a number of ways:

• Prove all three lengths are equal
• Prove that all three angles are equal
• Prove that two angles are equal to 60 degrees.
• Prove that one angle is equal to 60, and (any) two lengths are equal
• Prove that one angle is equal to 60, and the other two angles are equal

It’s important to keep all of these in mind as one explores the configuration. It’s genuinely unclear at the very start which of these characterisations of equilateral triangles will be most appropriate. Note though, that the configuration is symmetric in the roles of D and E, so that, for example, proving that the angle at D is 60 would be sufficient, since by symmetry the angle at E would also be 60, and we would conclude.

As regards clean angle chasing, the main moral here is to avoid angle arithmetic unless you are convinced it’s necessary. Ideally one keeps track of what could be determined in terms of other quantities, and only attempts the algebra/arithmetic once you are sure that it will work. It certainly isn’t helpful to have diagrams littered with angle measures like $240^\circ - \theta - \frac{\alpha}{2}$. Equality is the best thing to aim for. Try and demonstrate and notate equal angles wherever possible.

For example, here, even before introducing C,D,E, there is plenty of structure which might useful. Essentially every length in the configuration involving $(O_1,O_2,A,B)$ is equal, and all the angles are 60 or 120.

So we bear this in mind. I would consider proving that $\angle DO_1E=60^\circ$ with the following figure.

There are number of ways to complete the argument. Here are a couple of more obscure ones that illustrate my approach.

Firstly, let’s ignore the result that $\angle DO_1E=60^\circ$ and proceed directly to $\angle O_2DO_1=60^\circ$. It suffices to show that the blue angles are equal, which is equivalent to demonstrating that $ADO_2O_1$ is cyclic.

To achieve this, we would like to show that the angle $\angle AO_2D$ is equal to the red angle we’ve already discussed, and this follows by chasing round, using that $O_1$ is the centre of triangle ABC, and then the fact that $AO_2D$ is the external angle opposite $\angle ABC$ in cyclic quadrilateral $ABCO_2$. Indeed, note that this use of an equal external angle rather than ‘opposite angle is 180 – […]’ is the textbook example of clear angle chasing, where focusing on equality rather than arithmetic massively cleans up many diagrams.

As a second, rather overblown way to complete the argument, let’s show that the angles at D and E are equal, without calculating them. Since $O_1D,O_1E$ are angle bisectors, we have two pairs of equal angles as shown.

It would suffice to show that these measures are themselves equal. This is saying that line $ECO_2D$ is the external angle bisector of $\angle ACB$. And this is true, since $O_2$ is the arc-midpoint of AB on $\Gamma_1$!

I hope these observations are interesting. But, to reiterate, this level of complexity is unnecessary, and arguments by congruent triangles or direct angle chasing are perfectly satisfactory too!

Problem Five

The punchline of this problem is that m(N) is an integer precisely if N is a triangle number, from which it is quick to count the number of such N.

But how to arrive that this punchline? It is reasonably clear that the N-set with minimal mean must have the form $(1,2,\ldots,k, N)$. One could proceed by setting up inequalities involving the means of $(1,2,\ldots,k-1,N)$ and $(1,2,\ldots,k,k+1,N)$ to establish a relationship between k and N.

But there is a more direct route. A useful, and very intuitive result about means is the following. If you have a set of numbers with mean x, and you add into the set a new number y, then the mean of the new larger set is between x and y. In particular, if y>x, then the new mean is larger than x.

In particular, consider the mean of $(1,2,\ldots,k,N)$. If it is less than k, then removing k gives a smaller mean. If it is greater than k+1, then adding k+1 gives a smaller mean. So since the mean is an integer, it must be equal to k or k+1. That is, we have

$\frac{1+2+\ldots+k+N}{k+1}=k\text{ or }k+1.$

These lead to $N=\frac{k(k+1)}{2}\text{ or }\frac{(k+1)(k+2)}{2}$.

Problem Six

There must be some significance to the choice of 71 terms, and the 999,999 but it’s hopeless to try and see this immediately. In this scenario, it really is helpful to play around with semi-small examples.

Note that the second term is 1 regardless of whether we view this as P for PREVIOUS or S for SUM. However, if we then have an S, our sequence starts 1,1,2,2… or 1,1,2,4,… and these are bad, as once we have two even consecutive even numbers, we are doomed to see even numbers forever.

So once we have a common factor, that common factor is retained. How would we introduce a factor of, say, 11? Well, if you take 1,1,1,1,1,1,1,1,1,1,1 then take an S, you get 11, and afterwards you get either 11 or 22.

This gives a clue of the right way to set up the problem. For definiteness, we consider the second term to be an S. If you then consider a sequence of Ss and Ps, and write down the number of Ps between successive Ss (which may be zero) as a list $p_1,p_2,p_3,\ldots$, it holds that $\sum p_i=70$, and that each string of $p_i$ Ps introduces a factor of $p_i+1$, leading to

$\prod (p_i+1)=999,999=3^3\times 7\times 11\times 13\times 37.$

Note that $2\times 3 + 6+10+12+36=70$, so there exists a choice of the $p_i$s that satisfies these relations simultaneously. (And if we had replaced 71 by a smaller choice, this would have been impossible.)

The actual answer to the problem is obtained by studying how to rearrange the blocks of SPPP…s, but the main challenge lies in reducing to this form.

I thought this was a hard problem for BMO, and was impressed to see some solutions under exam conditions.

# BMO1 2018

The first round of the British Mathematical Olympiad was sat yesterday. The paper can be found here, and video solutions here. Copyright for the questions is held by BMOS. They are reproduced here with permission.

I hope any students who sat the paper enjoyed at least some of the questions, and found it challenging! The following commentaries on the problems are not official solutions, and are not always full solutions at all, but contain significant steps of solutions, so would be best saved until after you have attempted the problems, if you are planning to do so. I’ve written quite a lot about Q5 because I found it hard (or at least time-consuming) and somewhat atypical, and I’ve written a lot about Q6 because there was a lot to say. I hope at least some of this is interesting to some readers of all levels of olympiad experience.

Question 1

A list of five two-digit positive integers is written in increasing order on a blackboard. Each of the five integers is a multiple of 3, and each digit {0,1,…,9} appears exactly once on the blackboard. In how many ways can this be done? (Note that a two-digit number cannot begin with zero.)

It’s a trope of BMO1 that the first question must be doable by some sort of exhaustive calculation or listing exercise. Of course, that is rarely the most efficient solution.

However, there is normally a trade-off between eliminating all listing, and reducing to a manageable task.

The key observation here is that writing the integers in increasing order is really just a way to indicate thatÂ order of the choicesÂ doesn’t matter. Even if that seems counter-intuitive. The question wants to know how many ways to choose these five numbers. The order of choice doesn’t matter since we’re going to put them in ascending order on the blackboard anyway.

You want to make your choices with as much independence as possible. So it would, for example, be a bad idea to choose the smallest number first. How many possibilities are there where the smallest number is 24? What about 42? What about 69? These are all different, and some are zero, so will make the computation very taxing.

However, you might notice that the digits {0,3,6,9} have to go together to form two numbers, and the rest have to pair up with one digit from {1,4,7} and one from {2,5,8}. You might know that an integer is divisible by 3 precisely if its digit sum is divisible by 3, but in this context you wouldn’t lose too much time by simply listing everything! These tasks are now completely separate, so you can take the number of ways to pair up {0,3,6,9} and multiply by the number of ways to pair up {1,4,7} and {2,5,8}. You need to take care over the ordering. ItÂ does (obviously) matter which is the first digit and which is the second digit in a number!

# BMO1 2017 – Questions 5 and 6

The first round of the British Mathematical Olympiad was sat yesterday. The questions can be found here and video solutions here. My comments on the first four questions are in the previous post.

Overall, I didn’t think any of the questions on this paper were unusually difficult by the standard of BMO1, but I found everything slightly more time-consuming than typical. I thought Question 5 was a great problem, and I tried lots of things unsuccessfully, first, and so wanted to discuss it in slightly more technical language. For Question 6 I made a decisive mistake, which I’ll explain, and which cost a lot of time. But in general, my point is that the back end of the paper was a little fiddlier than normal, and required longer written solutions, and perhaps many students might have had less time than expected to attack them anyway after details earlier in the paper.

Question Five

As I said before, I thought this question was quite challenging. Not because the solution is particularly exotic or complicated, but because there were so many possible things that might have worked. In my opinion it would not have been out of place at the start of an IMO paper, because it’s perfectly possible to have enough good ideas that eliminating the ones that don’t work takes an hour, or hours. Even though it slightly spoils the flow of the solution, I’m particularly trying to emphasise the tangents that didn’t work, mostly for reassurance to anyone who spent a long time struggling.

I was thinking about this question in terms of a 2Nx2N board, where N is even, and for the given question equal to 100. I spent a while thinking that the bound was 8N-4, corresponding to taking the middle two rows and the middle two columns, but not the 2×2 square which is their intersection. If you think of a comb as a ‘handle’ of 1xN cells, with an extra N/2 alternating cells (say, ‘teeth’) bolted on, then it’s clear this construction works because there’s never space to fit in a handle, let alone the teeth.

I couldn’t prove that this was optimal though. A standard way to prove a given bound K was optimal would be to produce a tiling on the board with K combs, where every cell is included in exactly one comb. But this is clearly not possible in this situation, since the number of cells in a comb (which is 150) does not divide the total number of cells on the board.

Indeed, the general observation that if you take a comb and a copy of the comb rotated by 180, the teeth of the second comb can mesh perfectly with the teeth of the first comb to generate a 3xN unit. I wasted a moderate amount of time pursuing this route.

[Note, it will be obvious in a minute why Iâ€™m writing â€˜shadedâ€™ instead of â€˜colouredâ€™.]

But in motivating the construction, I was merely trying to shade cells so that they intersected every possible 1xN handle, and maybe I could prove that it was optimal for this. In fact, I canâ€™t prove itâ€™s optimal because it isnâ€™t optimal â€“ indeed itâ€™s clear that a handle through one of the middle rows intersects plenty of shaded cells, not just one. However, with this smaller problem in mind, it didnâ€™t take long to come up with an alternative proposal, namely splitting the board into equal quarters, and shading the diagonals of each quarter, as shown.

It seems clear that you canâ€™t fit in a 1xN handle, and any sensible tiling with 1xN handles contains exactly one shaded cell, so this shading (with 4N shaded cells) is optimal. But is it optimal for a comb itself?

Consider a shading which works, so that all combs include a shaded cell. Itâ€™s clear that a comb is contained within a 2xN block, and in such a 2xN block, there are four possible combs, as shown.

You need to cover all these combs with some shading somewhere. But if you put the shaded cell on a tooth of comb A, then you havenâ€™t covered comb B. And if you put the shaded cell on the handle of comb A, then you havenâ€™t covered one of comb C and comb D. You can phrase this via a colouring argument too. If you use four colours with period 2×2, as shown

then any comb involves exactly three colours, and so one of them misses out the colour of the shaded cell. (I hope itâ€™s clear what I mean, even with the confusing distinction between â€˜shadedâ€™ and â€˜colouredâ€™ cells.)

Certainly we have shown that any 2xN block must include at least two shaded cells. And thatâ€™s pretty much it. We have a tiling with 2N copies of a 2xN block, and with at least two shaded cells in each, that adds to at least 4N shaded cells overall.

Looking back on the method, we can identify another way to waste time. Tiling a board, eg a chessboard with dominos is a classic motif, which often relies on clever colouring. So itâ€™s perhaps lucky that I didn’t spot this colouring observation earlier. Because the argument described really does use the local properties of how the combs denoted A-D overlap. An attempt at a global argument might start as follows: we can identify 2N combs which don’t use colour 1, and tile this subset of the grid with them, so we need to shade at least 2N cells from colours {2,3,4}. Similarly for sets of colours {1,3,4}, {1,2,4}, and {1,2,3}. But if we reduce the problem to this, then using roughly 2N/3 of each colour fits this global requirement, leading to a bound of 8N/3, which isn’t strong enough. [1]

Question Six

A word of warning. Sometimes it’s useful to generalise in problems. In Q5, I was thinking in terms of N, and the only property of N I used was that it’s even. In Q4, we ignored 2017 and came back to it at the end, using only the fact that it’s odd. By contrast, in Q2, the values did turn out to be important for matching the proof bounds with a construction.

You have to guess whether 300 is important or not here. Let’s see.

I have a natural first question to ask myself about the setup, but some notation is useful. Let $a_1,a_2,\ldots,a_{300}$ be the ordering of the cards. We require that $\frac{a_1+\ldots+a_n}{n}$ is an integer for every $1\le n\le 300$. Maybe the values of these integers will be important, so hold that thought, but for now, replace with the divisibility statement that $n | a_1+\ldots+a_n$.

I don’t think it’s worth playing with small examples until I have a better idea whether the answer is 5 or 295. So the natural first question is: “what does it mean to have $(a_1,\ldots,a_{n-1})$ such that you can’t pick a suitable $a_n$?”

It means that there is no integer k in $\{1,\ldots,300\}\backslash\{a_1,\ldots,a_{n-1}\}$ such that $n\,\big|\,(a_1+\ldots+a_{n-1})+k$, which for now we write as

$k\equiv -(a_1+\ldots+a_{n-1})\,\mod n.$

Consider the congruence class of $-(a_1+\ldots+a_{n-1})$ modulo n. There are either $\lfloor \frac{300}{n}\rfloor$ or $\lceil \frac{300}{n}\rceil$ integers under consideration in this congruence class. If no such k exists, then all of the relevant integers in this congruence class must appear amongst $\{a_1,\ldots,a_{n-1}\}$. At this stage, we’re trying to get a feel for when this could happen, so lower bounds on n are most relevant. Therefore, if we get stuck when trying to find $a_n$, we have

$\lfloor \frac{300}{n} \rfloor\text{ or }\lceil \frac{300}{n}\rceil \le n-1,$ (*)

which is summarised more succinctly as

$\lfloor \frac{300}{n} \rfloor \le n-1.$ (**)

[Note, with this sort of bounding argument, I find it helpful to add intermediate steps like (*) in rough. The chance of getting the wrong direction, or the wrong choice of $\pm 1$ is quite high here. Of course, you don’t need to include the middle step in a final write-up.]

We can check that (**) is false when $n\le 17$ and true when $n\ge 18$. Indeed, both versions of (*) are true when $n\ge 18$.

So we know the minimum failure length is at least 17. But is there a failing sequence of length 17? At a meta-level, it feels like there should be. That was a very natural bounding argument for 17 (which recall corresponds to $n=18$), and it’s easy to believe that might be part of an official solution. If we achieve equality throughout the argument, that’s most of the way to a construction as well. It won’t be so easy to turn this argument into a construction for $n\ge 19$ because there won’t be equality anywhere.

We have to hope there is a construction for $n=18$. What follows is a description of a process to derive (or fail to derive) such a construction. In a solution, one would not need to give this backstory.

Anyway, in such a construction, let $\alpha\in\{1,2,\ldots,18\}$ describe the congruence class modulo 18 which is exhausted by $\{a_1,\ldots,a_{17}\}$. I’m going to hope that $\alpha=18$ because then the calculations will be easier since everything’s a multiple of 18. We haven’t yet used the fact that for a problem, we need $\alpha\equiv-(a_1+\ldots+a_{n-1})$. We definitely have to use that. There are 16 multiples of 18 (ie relevant integers in the congruence class), so exactly one of the terms so far, say $a_j$, is not a multiple of 18. But then

$0 \equiv 0+\ldots+0+a_j+0+\ldots+0,$

which can’t happen. With a bit of experimentation, we find a similar problem making a construction using the other congruence classes with 16 elements, namely $\alpha\in \{13,14,\ldots,18\}$.

So we have to tackle a different class. If $\alpha\le 12$ then our sequence must be

$\alpha,18+\alpha,2\times 18 +\alpha, \ldots, 16\times 18 + \alpha,$

in some order. In fact, let’s add extra notation, so our sequence is

$(a_1,\ldots,a_{17}) = (18\lambda_1+ \alpha,\ldots,18\lambda_{17}+\alpha),$

where $(\lambda_1,\ldots,\lambda_{17})$ is a permutation of {0,…,16}. And so we require

$k \,\big|\, 18(\lambda_1+\ldots+\lambda_k) + k\alpha,$ (%)

for $1\le k\le 17$. But clearly we can lop off that $k\alpha$, and could ignore the 18. Can we find a permutation $\lambda$ such that

$k \,\big|\, \lambda_1+\ldots+\lambda_k.$

This was where I wasted a long time. I played around with lots of examples and kept getting stuck. Building it up one term at a time, I would typically get stuck around k=9,10. And I had some observations that in all the attempted constructions, the values of $\frac{\lambda_1+\ldots+\lambda_k}{k}$ were around 8 and 9 too when I got stuck.

I became convinced this subproblem wasn’t possible, and decided that would be enough to show that n=18 wasn’t a possible failure length. I was trying to show the subproblem via a parity argument (how must the $a_i$s alternate odd/even to ensure all the even partial sums are even) but this wasn’t a problem. Then I came up with a valid argument. We must have

$\lambda_1+\ldots+\lambda_{17}=136= 16\times 8 + 8\quad\text{and}\quad 16\,\big|\,\lambda_1+\ldots+\lambda_{16},$

which means $\lambda_1+\ldots+\lambda_{16}$ must be 128 = 15×8 + 8, ie $\lambda_{17}=8$. But then we also have $15\,\big|\, \lambda_1+\ldots+\lambda_{15}$, which forces $latex\lambda_{16}=8$ also. Which isn’t possible.

If this then hadn’t wasted enough time, I then tried to come up with a construction for n=19, for which there are lots more variables, and took a lot more time, and seemed to be suffering from similar problems, just in a more complicated way. So I became convinced I must have made a mistake, because I was forced down routes that were way too complicated for a 3.5 hour exam. Then I found it…

What did I do wrong? I’ll just say directly. I threw away the 18 after (%). This made the statement stronger. (And in fact false.) Suppose instead I’d thrown away a factor of 9 (or no factors at all, but it’s the residual 2 that’s important). Then I would be trying to solve

$k\,\big|\,2(\lambda_1+\ldots+\lambda_k).$

AndÂ now if you experiment, you will notice that taking $\lambda_1=0,\lambda_2=1,\lambda_3=2,\ldots$ seems to work fine. And of course, we can confirm this, using the triangle number formula for the second time in the paper!

This had wasted a lot of time, but once that thought is present, we’re done, because we can go straight back and exhibit the sequence

$(a_1,\ldots,a_{17}) = (1, 18+1,2\times 18 +1,\ldots, 16\times 18 +1).$

Then the sum so far is congruent to -1 modulo 18, but we have exhausted all the available integers which would allow the sum of the first 18 terms to be a multiple of 18. This confirms that the answer to the question as stated is 17.

At the start, I said that we should be cautious about generalising. In the end, this was wise advice. We definitely used the fact that 18 was even in the stage I over-reduced the first time. We also used the fact that there was at least one value of $\alpha$ with an ‘extra’ member of the congruence class. So I’m pretty sure this proof wouldn’t have worked with 288 = 16×18 cards.

Footnotes

[1] â€“ If shading were a weighted (or continuous or whatever you prefer) property, ie that each cell has a quantity of shading given by a non-negative real number, and we merely demand that the total shading per comb is at least one, then the bound 8N/3 is in fact correct for the total shading. We could look at a 2xN block, and give 1/3 shading to one cell of each colour in the block. Alternatively, we could be very straightforward and apply 2/3N shading to every cell in the grid. The fact that shading has to be (in this language) zero or one, imposes meaningful extra constraints which involve the shape of the comb.

# BMO1 2017 – Questions 1-4

The first round of the British Mathematical Olympiad was sat yesterday. The questions can be found here. I recorded some thoughts on the questions while I was in Cyprus, hence the nice Mediterranean sunset above. I hope this might be useful to current or future contestants, as a supplement to the concise official solutions available. It goes without saying that while these commentaries may be interesting at a general level, they will be much more educational to students who have at least digested and played around with the questions, so consider trying the paper first. Video solutions are available here. These have more in common with this blog post than the official solutions, though inevitably some of the methods are slightly different, and the written word has some merits and demerits over the spoken word for clarity and brevity.

The copyright for these questions lies with BMOS, and are reproduced here with permission. Any errors or omissions are obviously my own.

I found the paper overall quite a bit harder than in recent years, or at least harder to finish quickly. I’ve therefore postponed discussion of the final two problems to a second post, to follow shortly.

Question One

A recurring theme of Q1 from BMO1 in recent years has been: “itâ€™s possible to do this problem by a long, and extremely careful direct calculation, but additional insight into the setup makes life substantially easier.”

This is the best example yet. It really is possible to evaluate Helen’s sum and Phil’s sum, and compare them directly. But itâ€™s easy to make a mistake in recording all the remainders when the divisor is small, and itâ€™s easy to make a mistake in summation when the divisor is large, and so it really is better to have a think for alternative approaches. Making a mistake in a very calculation-heavy approach is generally penalised heavily. And this makes sense intellectually, since the only way for someone to fix an erroneous calculation is to repeat it themselves, whereas small conceptual or calculation errors in a less onerous solution are more easily isolated and fixed by a reader. Of course, it also makes sense to discourage such attempts, which arenâ€™t really related to enriching mathematics, which is the whole point of the exercise!

Considering small divisors (or even smaller versions of 365 and 366) is sometimes helpful, but here I think a ‘typical’ divisor is more useful. But first, some notation will make any informal observation much easier to turn into a formal statement. Corresponding to Helen and Phil, let h(n) be the remainder when n is divided by 365, and p(n) the remainder when n is divided by 366. I would urge students to avoid the use of ‘mod’ in this question, partly because working modulo many different bases is annoying notationally, partly because the sum is not taken modulo anything, and partly because the temptation to use mod incorrectly as an operator is huge here [1].

Anyway, a typical value might be n=68, and we observe that 68 x 5 + 25 = 365, and so h(68)=25 and p(68)=26. Indeed, for most values of n, we will have p(n)=h(n)+1. This is useful because

$p(1)+p(2)+\ldots+p(366) - \left(h(1)+h(2)+\ldots+h(365)\right)$

$= \left(p(1)-h(1)\right) + \ldots+\left(p(365)-h(365)\right) + p(366),$

and now we know that most of the bracketed terms are equal to one. We just need to handle the rest. The only time it doesn’t hold that p(n)=h(n)+1 is when 366 is actually a multiple of n. In this case, p(n)=0 and h(n)=n-1. We know that 366 = 2 x 3 x 61, and so its divisors are 1, 2, 3, 6, 61, 122, 183.

Then, in the big expression above, seven of the 365 bracketed terms are not equal to 1. So 358 of them are equal to one. The remaining ones are equal to 0, -1, -2, -5, -60, -121, -182 respectively. There are shortcuts to calculate the sum of these, but itâ€™s probably safer to do it by hand, obtaining -371. Overall, since p(366)=0, we have

$p(1)+p(2)+\ldots+p(366) - \left(h(1)+h(2)+\ldots+h(365)\right)$

$= -371 + 358 + 0 = -13.$

So, possibly counter-intuitively, Helen has the larger sum, with difference 13, and we didnâ€™t have to do a giant calculation…

Question Two

Suppose each person chooses which days to go swimming ‘at random’, without worrying about how to define this. Is this likely to generate a maximum or minimum value of n? I hope itâ€™s intuitively clear that this probably wonâ€™t generate an extreme value. By picking at random we are throwing away lots of opportunity to force valuable overlaps or non-overlaps. In other words, we should start thinking about ways to set up the swimming itinerary with lots of symmetry and structure, and probably weâ€™ll eventually get a maximum or a minimum. At a more general level, with a problem like this, one can start playing around with proof methods immediately, or one can start by constructing lots of symmetric and extreme-looking examples, and see what happens. I favour the latter approach, at least initially. You have to trust that at least one of the extreme examples will be guess-able.

The most obvious extreme example is that everyone swims on the first 75 days, and no-one swims on the final 25 days. This leads to n=75. But we’re clearly ‘wasting’ opportunities in both directions, because there are never exactly five people swimming. I tried a few more things, and found myself simultaneously attacking maximum and minimum, which is clearly bad, so focused on minimum. Just as a starting point, let’s aim for something small, say n=4. The obstacle is that if you demand at most four swimmers on 96 days, then even with six swimmers on the remaining four days, you donâ€™t end up with enough swimming having taken place!

Maybe you move straight from this observation to a proof, or maybe you move straight to a construction. Either way, I think itâ€™s worth saying that the proof and the construction come together. My construction is that everyone swims on the first 25 days, then on days 26-50 everyone except A and B swim, on days 51-75 everyone except C and D swim, and on days 76-100 everyone except E and F swim. This exactly adds up. And if you went for the proof first, you might have argued that the total number of swim days is 6×75 = 450, but is at most 4n + 6(100-n). This leads immediately to $n\ge 25$, and I just gave the construction. Note that if you came from this proof first, you can find the construction because your proof shows that to be exact you need 25 days with six swimmers, and 75 days with four swimmers, and itâ€™s natural to try to make this split evenly. Anyway, this clears up the minimum.

[Less experienced contestants might wonder why I was worried about generating a construction despite having a proof. Remember we are trying to find the minimum. I could equally have a proof for $n\ge 10$ which would be totally totally valid. But this wouldnâ€™t show that the minimum was n=10, because that isnâ€™t in fact possible (as weâ€™ve seen), hence itâ€™s the construction that confirms that n=25 is the true minimum.]

Itâ€™s tempting to go back to the drawing board for the maximum, but itâ€™s always worth checking whether you can directly adjust the proof youâ€™ve already given. And here you can! We argued that

$450\le 4n + 6(100-n)$

to prove the minimum. But equally, we know that on the n days we have at least five swimmers, and on the remaining days, we have between zero and four swimmers, so

$450 \ge 5n + 0\times (100-n),$ (*)

which gives $n\le 90$. If we have a construction that attains this bound then we are done. Why have I phrased (*) with the slightly childish multiple of zero? Because it’s a reminder that for a construction to attain this bound, we really do need the 90 days to have exactly five swimmers, and the remaining ten days to have no swimmers. So it’s clear what to do. Split the first 90 days into five groups of 15 days. One swimmer skips each group. No-one swims in the final ten days, perhaps because of a jellyfish infestation. So we’re done, and $25\le n\le 90$.

At a general level, it’s worth noting that in the story presented, we found an example for the minimum which we turned into a proof, and then a proof for the maximum, which we then analysed to produce a construction.

Note that similar bounding argumentsÂ would apply if we fiddled with the numbers 5, 75 and 100. But constructions matching the bounds might not then be possible because the splits wouldn’t work so nicely. This would make everything more complicated, but probably not more interesting.

Question Three

It’s understandable that lots of students attempting this paper might feel ill-at-ease with conventional Euclidean geometry problems. A good first rule of thumb here, as in many settings, is “donâ€™t panic!”, and a more specific second rule of thumb is “even if you think you can calculate, try to find geometric insight first.”

Here, it really does look like you can calculate. A configuration based on a given isosceles triangle and a length condition and a perpendicular line is open to several coordinate approaches, and certainly some sensible trigonometry. Itâ€™s also very open to organised labelling of the diagram. You have three equal lengths, and a right-angle, as shown.

The key step is this. Drop the perpendicular from A to BC, and call its foot D. That alone really is the key step, as it reduces both parts of the question to an easy comparison. It’s clear that the line AD splits the triangle into two congruent parts, and thus equal areas and perimeters. So it is enough to show that triangle BMN has the same area as triangle ABD, and that their outer-perimeters (ie the part of its perimeter which is also the perimeter of ABC) are the same.

But theyâ€™re congruent, so both of these statements are true, and the problem is solved.

My solution could be as short as two or three lines, so for the purposes of this post all that remains is to justify why you might think of the key step. Here are a few possible entry routes:

• You might notice that line AD induces the required property for triangle ABD.
• You might try to find a triangle congruent to AMN, and come up with D that way.
• Thereâ€™s already a perpendicular in the question so experimenting with another one is natural, especially since the perpendicular from A has straightforward properties.
• AMN is a right angle, and so constructing D gives a cyclic quadrilateral. We didnâ€™t use that directly in the proof above, but constructing cyclic quadrilaterals is usually a good idea.
• If you were trying a calculation approach, you probably introduced the length AD, or at least the midpoint D as an intermediate step.

On the video, Mary Teresa proposes a number of elegant synthetic solutions with a few more steps. You might find it a useful exercise to try to come up with some motivating reasons like the bullet points above to justify her suggestion to reflect A in M as a first step.

Question Four

I wasn’t paying enough attention initially, and I calculated $a_2=0\text{ or }2$. This made life much much more complicated. As with IMO 2017 Q1, if trying to deduce general behaviour from small examples, itâ€™s essential to calculate the small examples correctly!

Once you engage your brain properly, you find that $a_2=0 \text{ or }3$, and of course $a_2=0$ is not allowed, since it must be positive. So $a_2=3$, and a similar calculation suggests $a_3=1\text{ or }6$. Itâ€™s clear that the set of values for $a_{k+1}$ depends only on $a_k$, so if you take $a_3=1$, then youâ€™re back to the situation you started with at the beginning. If you choose to continue the exploration with $a_3=6$, you will find $a_4=2\text{ or }10$, at which point you must be triggered by the possibility that triangle numbers play a role here.

As so often with a play-around with small values, you need to turn a useful observation into a concrete statement, which could then be applied to the problem statement. It looks like in any legal sequence, every term will be a triangle number, so we only need to clarify which triangle number. An example of a suitable statement might be:

Claim: If $a_n=T_k=\frac{k(k+1)}{2}$, the k-th triangle number, then $a_{n+1}=T_{k-1}\text{ or }T_{k+1}$.

There are three stages. 1) Checking the claim is true; 2) checking the claim is maximally relevant; 3) proving it. In this case, proving it is the easiest bit. Itâ€™s a quick exercise, and Iâ€™m omitting it. Of course, we canâ€™t prove any statement which isnâ€™t true, and here we need to make some quick adjustment to account for the case k=1, for which we are forced to take $a_{n+1}=T_{k+1}$.

The second stage really concerns the question “but what if $a_n\ne T_k$?” While there are deductions one could make, the key is that if $a_1$ is a triangle number, the claim we’ve just made shows that $a_n$ is always a triangle number, so this question is irrelevant. Indeed the claim further shows that $a_{2017}\le T_{2017}$, and also that $a_{2017}=T_k$ for some odd value of k. To be fully rigorous you should probably describe a sequence which attains each odd value of k, but this is really an exercise in notation [2], and it’s very obvious they are all attainable.

In any case, the set of possible values is $\{T_1,T_3,\ldots,T_{2017}\}$, which has size 1009.

Final two questions

These are discussed in a subsequent post.

Footnotes

[1] â€“ mod n is not an operator, meaning you shouldnâ€™t think of it as ‘sending integers to other integers’, or ‘taking any integer, to an integer in {0,1,…,n-1}’. Statements like 19 mod 5 = 4 are useful at the very start of an introduction to modular arithmetic, but why choose 4? Sometimes itâ€™s more useful to consider -1 instead, and we want statements like $a^p\equiv a$ modulo p to make sense even when $a\ge p$. 19 = 4 modulo 5 doesnâ€™t place any greater emphasis on the 4 than the 19. This makes it more like a conventional equals sign, which is of course appropriate.

[2] â€“ Taking $a_n=T_n$ for $1\le n\le k$, and thereafter $a_n=T_k$ if k is odd, and $a_n=T_{k+1}$ if k is even will certainly work, as will many other examples, some perhaps easier to describe than this one, though make sure you don’t accidentally try to use $T_0$!

# BMO1 2016 – the non-geometry

Here’s a link to yesterday’s BMO1 paper, and the video solutions for all the problems. I gave the video solution to the geometric Q5, and discuss aspects of this at some length in the previous post.

In these videos, for obvious educational reasons, there’s a requirement to avoid referencing theory and ideas that aren’t standard on the school curriculum or relatively obvious directly from first principles. Here, I’ve written down some of my own thoughts on the other problems in a way that might add further value for those students who are already have some experience at olympiads and these types of problems. In particular, on problems you can do, it’s worth asking what you can learn from how you did them that might be applicable generally, and obviously for some of the harder problems, it’s worth knowing about solutions that do use a little bit of theory. Anyway, I hope it’s of interest to someone.

Obviously we aren’t going to write out the whole list, but there’s a trade-off in time between coming up with neat ideas involving symmetry, and just listing and counting things. Any idea is going to formalise somehow the intuitive statement ‘roughly half the digits are odd’. The neat ideas involve formalising the statement ‘if we add leading zeros, then roughly half the digits are odd’. The level of roughness required is less in the first statement than the second statement.

Then there’s the trade-off. Trying to come up with the perfect general statement that is useful and true might lead to something like the following:

‘If we write the numbers from 0000 to N, with leading zeros, and all digits of N+1 are even, then half the total digits, ie 2N of them, are odd.’

This is false, and maybe the first three such things you try along these lines are also false. What you really want to do is control the numbers from 0000 to 1999, for which an argument by matching is clear, and gives you 2000 x 4 / 2 = 4000 odd digits. You can exploit the symmetry by matching k with 1999-k, or do it directly first with the units, then with the tens and so on.

The rest (that is, 2000 to 2016) can be treated by listing and counting. Of course, the question wants an actual answer, so we should be wary of getting it wrong by plus or minus one in some step. A classic error of this kind is that the number of integers between 2000 and 2016 inclusive is 17, not 16. I don’t know why the memory is so vivid, but I recall being upset in Year 2 about erring on a problem of this kind involving fences and fenceposts.

As with so many new types of equation, the recipe is to reduce to a type of equation you already know how to solve. Here, because {x} has a different form on different ranges, it makes sense to consider the three ranges

$x\in[0,1/25],\, x\in[1/25,1/8],\, x\in [1/8,\infty),$

as for each of these ranges, we can rewrite $5y\{8y\}\{25y\}$ in terms of standard functions without this bracket notation. On each range we can solve the corresponding equation. We then have to check that each solution does actually lie in the appropriate range, and in two cases it does, and in one case it doesn’t.

Adding an appropriately-chosen value to each side allows you to factorise the quadratics. This might be very useful. But is it an invitation to do number theory and look at coprime factors and so on, or is a softer approach more helpful?

The general idea is that the set of values taken by any quadratic sequence with integer coefficients and leading coefficient one looks from a distance like the set of squares, or the set $\{m(m+1), \,m\in\mathbb{N}\}$, which you might think of as ‘half-squares’ or ‘double triangle numbers’ as you wish. And by, ‘from a distance’ I mean ‘up to an additive constant’. If you care about limiting behaviour, then of course this additive constant might well not matter, but if you care about all solutions, you probably do care. To see why this holds, note that

$n^2+2n = (n+1)^2 - 1,$

so indeed up to an additive constant, the quadratic on the LHS gives the squares, and similarly

$n^2 - 7n = (n-4)(n-3)-12,$

and so on. To solve the equation $n^2=m^2+6$, over the integers, one can factorise, but another approach is to argue that the distance between adjacent squares is much more than 6 in the majority of cases, which leaves only a handful of candidates for n and m to check.

The same applies at this question. Adding on 9 gives

$n^2-6n+9 = m^2 + m -1,$

which is of course the same as

$(n-3)^2 = m(m+1)-1.$

Now, since we now that adjacent squares and ‘half-squares’ are more than one apart in all but a couple of cases, we know why there should only be a small number of solutions. I would call a method of this kind square-sandwiching, but I don’t see much evidence from Google that this term is generally used, except on this blog.

Of course, we have to be formal in an actual solution, and the easiest way to achieve this is to sandwich $m(m+1)-1$ between adjacent squares $m^2$ and $(m+1)^2$, since it is very much clear-cut that the only squares which differ by one are zero and one itself.

I really don’t have much to say about this. It’s not on the school curriculum so the official solutions are not allowed to say this, but you have to use that all integers except those which are 2 modulo 4 can be written as a difference of two squares. The easiest way to show this is by explicitly writing down the appropriate squares, treating the cases of odds and multiples of four separately.

So you lose if after your turn the running total is 2 modulo 4. At this point, the combinatorics isn’t too hard, though as in Q1 one has to be mindful that making an odd number of small mistakes will lead to the wrong answer! As in all such problems, it’s best to try and give a concrete strategy for Naomi. And it’s best if there’s something inherent in the strategy which makes it clear that it’s actually possible to implement. (Eg, if you claim she should choose a particular number, ideally it’s obvious that number is available to choose.)

One strategy might be: Naomi starts by choosing a multiple of four. Then there are an even number of multiples of four, so Naomi’s strategy is:

• whenever Tom chooses a multiple of four, Naomi may choose another multiple of four;
• whenever Tom chooses a number which is one (respectively three) modulo 4, Naomi may choose another which is three (respectively one) modulo 4.

Note that Naomi may always choose another multiple of four precisely because we’ve also specified the second condition. If sometimes Tom chooses an odd number and Naomi responds with a multiple of four out an idle and illogical sense of caprice, then the first bullet point would not be true. One can avoid this problem by being more specific about exactly what the algorithm is, though there’s a danger that statements like ‘whenever Tom chooses k, Naomi should choose 100-k’ can introduce problems about avoiding the case k=50.

I started this at the train station in Balatonfured with no paper and so I decided to focus on the case of just m, m+1 and n, n+2. This wasn’t a good idea in my opinion because it was awkward but guessable, and so didn’t give too much insight into actual methods. Also, it didn’t feel like inducting on the size of the sequences in question was likely to be successful.

If we know about the Chinese Remainder Theorem, we should know that we definitely want to use it here in some form. Here are some clearly-written notes about CRT with exercises and hard problems which a) I think are good; b) cite this blog in the abstract. (I make no comment on correlation or causality between a) and b)…)

CRT is about solutions to sets of congruence equations modulo various bases. There are two aspects to this , and it feels to me like a theorem where students often remember one aspect, and forget the other one, in some order. Firstly, the theorem says that subject to conditions on the values modulo any non-coprime bases, there exist solutions. In many constructive problems, especially when the congruences are not explicit, this is useful enough by itself.

But secondly, the theorem tells us what all the solutions are. There are two stages to this: finding the smallest solution, then finding all the solutions. Three comments: 1) the second of these is easy â€“ we just add on all multiples of the LCM of the bases; 2) we don’t need to find the smallest solution â€“ any solution will do; 3) if you understand CRT, you might well comment that the previous two comments are essentially the same. Anyway, finding the smallest solution, or any solution is often hard. When you give students an exercise sheet on CRT, finding an integer which is 3 mod 5, 1 mod 7 and 12 mod 13 is the hard part. Even if you’re given the recipe for the algorithm, it’s the kind of computation that’s more appealing if you are an actual computer.

Ok, so returning to this problem, the key step is to phrase everything in a way which makes the application of CRT easy. We observe that taking n=2m satisfies the statement â€“ the only problem of course is that 2m is not odd. But CRT then tells us what all solutions for n are, and it’s clear that 2m is the smallest, so we only need to add on the LCM (which is odd) to obtain the smallest odd solution.

# BMO1 2016 Q5 – from areas to angles

For the second year in a row Question 5 has been a geometry problem; and for the second year in a row I presented the video solution; and the for the second year in a row I received the question(s) while I was abroad. You can see the video solutions for all the questions here (for now). I had a think about Q5 and Q6 on the train back from a day out at Lake Balaton in Western Hungary, so in keeping with last year’s corresponding post, here are some photos from those sunnier days.

I didn’t enjoy this year’s geometry quite as much as last year’s, but I still want to say some things about it. At the time of writing, I don’t know who proposed Q5, but in contrast to most geometry problems, where you can see how the question might have emerged by tweaking a standard configuration, I don’t have a good intuition for what’s really going on here. I can, however, at least offer some insight into why the ‘official’ solution I give on the video has the form that it does.

The configuration given is very classical, with only five points, and lots of equal angles. The target statement is also about angles, indeed we have to show that a particular angle is a right-angle. So we might suspect that the model approach might well involve showing some other tangency relation, where one of the lines AC and BC is a radius and the other a tangent to a relevant circle. I think it’s worth emphasising that throughout mathematics, the method of solving a problem is likely to involve similar objects to the statement of the problem itself. And especially so in competition problems â€“ it seemed entirely reasonable that the setter might have found a configuration with two corresponding tangency relations and constructed a problem by essentially only telling us the details of one of the relations.

There’s the temptation to draw lots of extra points or lots of extra lines to try and fit the given configuration into a larger configuration with more symmetry, or more suggestive similarity [1]. But, at least for my taste, you can often make a lot of progress just by thinking about what properties you want the extra lines and points to have, rather than actually drawing them. Be that as it may, for this question, I couldn’t initially find anything suitable along these lines [2]. So we have to think about the condition.

But then the condition we’ve been given involves areas, which feels at least two steps away from giving us lots of information about angles. It doesn’t feel likely that we are going to be able to read off some tangency conditions immediately from the area equality we’ve been given. So before thinking about the condition too carefully, it makes sense to return to the configuration and think in very loose terms about how we might prove the result.

How do we actually prove that an angle is a right-angle? (*) I was trying to find some tangency condition, but it’s also obviously the angle subtending by the diameter of a circle. You could aim for the Pythagoras relation on a triangle which includes the proposed right-angle, or possibly it might be easier to know one angle and two side-lengths in such a triangle, and conclude with some light trigonometry? We’ve been given a condition in terms of areas, so perhaps we can use the fact that the area of a right-angled triangle is half the product of the shorter side-lengths? Getting more exotic, if the configuration is suited to description via vectors, then a dot product might be useful, but probably this configuration isn’t.

The conclusion should be that it’s not obvious what sort of geometry we’re going to need to do to solve the problem. Maybe everything will come out from similar triangles with enough imagination, but maybe it won’t. So that’s why in the video, I split the analysis into an analysis of the configuration itself, and then an analysis of the area condition. What really happens is that we play with the area condition until we get literally anything that looks at all like one of the approaches discussed in paragraph (*). To increase our chances, we need to know as much about the configuration as possible, so any deductions from the areas are strong.

The configuration doesn’t have many points, so there’s not much ambiguity about what we could do. There are two tangents to the circle. We treat APC with equal tangents and the alternate segment theorem to show the triangle is isosceles and that the base angles are equal to the angle at B in ABC. Then point Q is ideally defined in terms of ABC to use power of a point, and add some further equal angles into the diagram. (Though it turns out we don’t need the extra equal angle except through power of a point.)

So we have some equal angles, and also some length relations. One of the length relations is straightforward (AP=CP) and the other less so (power of a point $CQ^2 = AQ\cdot BQ$). The really key observation is that the angle-chasing has identified

$\angle PAQ = 180 - \angle \hat C,$

which gives us an alternative goal: maybe it will be easier to show that PAQ is a right-angle.

Anyway, that pretty much drinks the configuration dry, and we have to use the area condition. I want to emphasise how crucial this phase in for this type of geometry problem. Thinking about how to prove the goal, and getting a flavour for the type of relation that comes out of the configuration is great, but now we need to watch like a hawk when we play with the area condition for relations which look similar to what we have, and where we might be going, as that’s very likely to be the key to the problem.

We remarked earlier that we’re aiming for angles, and are given areas. A natural middle ground is lengths. All the more so since the configuration doesn’t have many points, and so several of the triangles listed as having the same area also have the same or similar bases. You might have noticed that ABC and BCQ share height above line AQ, from which we deduce AB=BQ. It’s crucial then to identify that this is useful because it supports the power of a point result from the configuration itself. It’s also crucial to identify that we are doing a good job of relating lots of lengths in the diagram. We have two pairs of equal lengths, and (through Power of a Point) a third length which differs from one of them by a factor of $\sqrt{2}$.

If we make that meta-mathematical step, we are almost home. We have a relation between a triple of lengths, and between a pair of lengths. These segments make up the perimeter of triangle APQ. So if we can relate one set of lengths and the other set of lengths, then we’ll know the ratios of the side lengths of APQ. And this is excellent, since much earlier we proposed Pythagoras as a possible method for establish an angle is a right-angle, and this is exactly the information we’d need for that approach.

Can we relate the two sets of lengths? We might guess yes, that with a different comparison of triangles areas (since we haven’t yet used the area of APC) we can find a further relation. Indeed, comparing APC and APQ gives CQ = 2PC by an identical argument about heights above lines.

Now we know all the ratios, it really is just a quick calculation…

[1] â€“ I discussed the notion of adding extra points when the scripts for the recording were being shared around. It was mentioned that for some people, the requirement to add extra points (or whatever) marks a hard division between ‘problems they can do’ and ‘problem they can’t do’. While I didn’t necessarily follow this practice while I was a contestant myself, these days the first thing I do when I see any angles or an angle condition in a problem is to think about whether there’s a simple way to alter the configuration so the condition is more natural. Obviously this doesn’t always work (see [2]), but it’s on my list of ‘things to try during initial thinking’, and certainly comes a long way before approaches like ‘place in a Cartesian coordinate system’.

[2] â€“ Well, I could actually find something suitable, but I couldn’t initially turn it into a solution. The most natural thing is to reflect P in AC to get P’, and Q in BC to get Q’. The area conditions [AP’C]=[ABC]=[BCQ’] continue to hold, but now P’ and B are on the same side of AC, hence P’B || AC. Similarly AQ’ || BC. I see no reason not to carry across the equal length deductions from the original diagram, and we need to note that angles P’AC, ACP’, CBA are equal and angles Q’AB and BAC are equal. In the new diagram, there are many things it would suffice to prove, including that CP’Q’ are collinear. Note that unless you draw the diagram deliberately badly, it’s especially easy accidentally to assume that CP’Q’ are collinear while playing around, so I wasted quite a bit of time. Later, while writing up this post, I could finish it [3].

[3] â€“ In the double-reflected diagram, BCQ’ is similar to P’BA, and since Q’C=2P’C = P’A, and Q’B=AB, you can even deduce that the scale factor is $\sqrt{2}$. There now seemed two options:

• focus on AP’BC, where we now three of the lengths, and three of the angles are equal, so we can solve for the measure of this angle. I had to use a level of trigonometry rather more exotic than the Pythagoras of the original solution, so this doesn’t really serve purpose.
• Since BCQ’ is similar to P’BA and ABQ’ similar to CP’A, we actually have Q’BCA similar to AP’BC. In particular, $\angle CBP' = \angle ACB$, and thus both are 90. Note that for this, we only needed the angle deductions in the original configuration, and the pair of equal lengths.
• There are other ways to hack this final stage, including showing that BP’ meets AQ’ at the latter’s midpoint, to give CP’Q’ collinear.

The first round of the British Mathematical Olympiad (BMO1) takes place tomorrow. Last year I wrote a brief note to my mentoring students about the exam. Most of the advice is fairly obvious, but I guess it never does any harm to be reminded. In particular, while it is tempting to give lots of mathematical guidance, under exam pressure good deductive ideas either will or won’t come, and there’s relatively little to be done about it in advance to help. However, especially for students for whom this is their first experience of a long olympiad style paper, there are a few practical and general points to be made, so you have the best chance of turning good ideas into good solutions during the time allowed.

DON’T waste time. 3.5 hours is a long time, but it will pass quickly when you have lots to think about. Obviously, you will inevitably spend some time just thinking vaguely about the problems, or even daydreaming, just to give your brain a break. Don’t worry about that, but do try not to waste time pursuing methods which don’t look like they are working. If you have made 6 algebraic substitutions and the expression now takes up an entire line, ask yourself whether you’re going anywhere. If your geometrical diagram now has dozens of extra points, or if you are trying to solve a polynomial in n variables where n is large, question yourself. Maybe you’re missing something more obvious?

On the subject, DO flit between questions. The rubric says that full solutions are better than partial solutions. However, if moving to another question allows you to take a fresh stab at the first one in 15 minutes or whatever, that is a good thing.

Also, DO take food or drink (within reason and so long as whoever is invigilating doesn’t mind), if you think it will help. 3.5 hours of concentration can be draining! The 200g value pack of Dairy Milk was my preference back in the day…

On a more mathematical note, DON’T draw rubbish geometrical diagrams. DO use a compass and a ruler. These geometry problems normally want you to spot similar triangles or something like that. These will be much much easier to find if they actually look similar on your diagram! Markers also like seeing good diagrams.

DO write up relevant bits of your rough. It’s a good way to grab small marks, and you never know, you might have had all the right ideas, just missed the final crucial step. It sometimes says not to hand in rough: so make sure what you hand in looks vaguely neat, and has key steps or results you’ve proved underlined or in a box, so that they are as visible as possible to the marker. Checking small cases explicitly will be useful to your understanding of the problem, and so may gain credit.

DON’T wait until the end to write up bits of your rough. The temptation to keep working on them will be too strong, and you might have forgotten what seemed interesting an hour ago. Crucially, deciding carefully what the most important steps of your working are may very well help you to finish the problem.

DO read the question properly. Trying to prove something false will waste your time; trying to prove something simpler than the actual question will cost you marks. Things to consider include:

• If the question says ‘If and only if’, you have to prove it both ways. Similarly if it asks for a converse.
• Check what the domains are. Does n have to be an integer or is it a real number? Can it be zero?
• In a counting question, does order matter?
• Is the triangle allowed to be obtuse? Does this change anything important in the argument?

DON’T waffle. If you are writing a massive load of text, have a think about whether that’s a good idea. It is very easy, especially for fiddly combinatorics questions, for a simple equation to turn into a sprawling essay. Keeping sentences very short (no long subordinate clauses) and leaving space between displayed maths and words will help. Remember that whether or not you know what you are doing, you want to GIVE THE IMPRESSION that you know what you are doing!

DO be clever. Sometimes the questions are hard but routine, sometimes they require clever ideas. If your current method isn’t making any progress and you have a crazy idea, try it – it might be just the thing.

However, DON’T be too clever. It’s very tempting, especially to new mentoring students, to try to use every bit of theory you’ve recently learned. Remember that not every geometry question requires the Angle Bisector Theorem, and you don’t always need to deploy Fermat’s Little Theorem or even modular arithmetic on every problem about integers. In particular, avoid applying anything you don’t properly understand – under the pressure of an exam, it’s easy to forget the details, and end up assuming something that is false!

DO relax. I know that is easier said than done, but this is an academically stressful time of life, so enjoy the fact that this is a rare exam where doing well is not of huge importance to the rest of your life. I haven’t seen this year’s paper, but the questions are normally interesting, and should bring out the best in a strong young mathematician. As with many things, if you stop worrying about the outcome, you often do better than you might expect.

Best of luck to everyone sitting the exam tomorrow!