BMO2 2022

The second and final round of this year’s British Mathematical Olympiad took place on Thursday.

Here are some thoughts on the problems. I wasn’t involved in choosing the problems, although I did write Q4. I’ll say a bit more about this geometry problem than about the other three, so I’ve put this first.

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 Four

Let me explain how I devised this problem. Let A’ be the point diametrically opposite A on the circumcircle of triangle ABC. (Sometimes called the antipode of A.) Now consider any line L through A’, and let P be the point where it meets the circumcircle again. I was particularly thinking about lines L where P lay on the same side of BC as A’, as depicted.

Because AA’ is a diameter, there is a right angle $\angle APA'$. So I considering intersecting L with the lines $\ell_B,\ell_C$ given in the question, since this would create two cyclic quadrilaterals, denoted $BDPP_B, DCP_CP$ in the figure.

Let’s just focus on $BDPP_B$. This cyclic quadrilateral gives us two equal angles in many different ways. But a particularly nice pair is the one shown, because ACPB is also cyclic, which confirms that the angle measure is $\angle C$. In particular, by extending $P_BD$ to E on AC, we get a third cyclic quadrilateral $BECP_B$, and hence find out that $P_BD,P_BE\perp AC$.

So four lines go through D: AP, BC, and the perpendiculars from $P_B,P_C$ to the opposite sides of the triangle.

How to turn this into a question? The steps outlined above are relatively clear because

• the right angle at P is essentially given;
• there are four other right angles, two involving $\ell_B,\ell_C$ which are given, then one involving $P_BD,P_BE$ being perpendicular to AC, and the same for $P_C$ and AB, which are to be deduced,

so the cyclic quadrilaterals just pop out directly. This would then be too straightforward for BMO2, and would also waste the fact that four lines meet at D. It would suffice to work with $P_B$ as we did above, and assert that the corresponding result also held for $P_C$.

To move towards the form of the BMO2 question, we consider taking P=A’, so that line L is actually the tangent to the circumcircle. We still have the quadruple concurrency at D, but now D lies on AA’. It seems almost certain that we wouldn’t have this concurrency if we built the figure using the tangent from any point other than A’. So this will be the configuration for the problem.

As the writer, I know that the four lines meet when T=A’, so I know I can force T=A’ by insisting that three lines meet, and then set the problem asking to prove that the fourth goes through the same point. One could have stated that $P_BE,P_CF,AT$ meet, and ask to prove that the meeting point lies on BC. I preferred the one I chose. Reflecting on the difficulty of the problem:

• It is hard to draw a diagram which looks right! Essentially you have to draw a few, and if T ends up close to A’ on one, the diagram will look right.
• Crucially, we now only have four right angles in the figure, so only one pair of (symmetric) cyclic quads can be read off immediately

I think the cleanest way to attack the problem is to extend AQ to meet the circumcircle at T’, with the goal of showing T’=T. Then $BECP_B, ACT'B$ cyclic gives $BQT'P_B$ cyclic (in an exact reversal of the argument in the prelude earlier). What we don’t know is that $P_B,T',P_C$ are collinear, but this follows since now we derive $\angle QT'P_B=\angle P_CT'Q=90^\circ$ from this new cyclic quad. Consequently T’=T.

There are many routes for coming up with olympiad problems in geometry and other topics, and the scenario I’ve described isn’t particularly typical for myself and certainly isn’t necessarily standard for other problem setters. Nonetheless, I hope that’s an interesting insight into the origin story of a problem like this.

Problem One

This question reduces to establishing whether

$m(m+k)+k=n^2$

has infinitely many solutions in (m,n). The technique for proving via ‘square sandwiching’ that this is normally not possible has been discussed previously on this blog here:

and here:

This is a good opportunity to reflect in more detail. Let $P(x)$ be a polynomial with integer coefficients, and we consider whether the image $P(\mathbb{Z}):= \{ P(n)\,:\, n\in \mathbb{Z}\}$ contains infinitely many squares.

Clearly if $P(x)=Q(x)^2$ for Q another polynomial with integer coefficients, then this holds. Square sandwiching establishes that the answer is no when P(x) is a monic quadratic (meaning that the coefficient of $x^2$ is one) unless P(x) is the square of a linear polynomial. The same holds if the leading coefficient is a square. Note that the general case where P(x) is negative except on a finite range (eg if P has even degree and negative leading coefficient) is also immediate.

But there are several cases where the answer is yes:

• For example, when $P(x)=2x^3$, we have P(n) a square whenever $n=2k^2$.
• Indeed, note that $P(x)=x$ has this property too!
• Amongst non-monic quadratics, note that $P(x)=nx^2+1$ works when n is not a square because of the theory of Pell’s equation.

We could have the alternative question: if $P(n)$ is always a square, does this imply $P(x)=Q(x)^2$. Indeed, one could generalise this and ask: if $P(\mathbb{Z})\subset R(\mathbb{Z})$ does this imply that $P(x)=R(Q(x))$ for some choice Q?

Our original setting with $R(x)=x^2$ is classical, due to Polya, and is a good problem for any students reading this. The case $R(x)=x^k$ is handled by Polya and Szego. The case of general R is more challenging, and though it can be treated without heavier concepts, is probably best left until one knows some undergraduate algebra. Readers looking for a paper may consult Davenport, Lewis and Schinzel Polynomials of certain special types (Acta Arith. 1964).

Problem Two

Rearranging to get

$\frac{f(2xy)}{2y} = \frac{f(f(x^2)+x)}{f(x+1)}$

is a good first step, as y appears only on the LHS. As y varies, the LHS is therefore constant, and so we deduce that $f(2y)=Cy$, ie f is linear on the even integers.

There were then a number of ways to proceed, bearing in mind one had to study the odd integers and solve for C. Taking x even in the given equation is a good next step, because (after some careful argument about C and divisibility by powers of 2) $f(f(x^2)+x)$ is already known, leading to an expression for f(x+1), ie some control over the odd integers.

There were a few possibilities for how to piece the odds and evens together, and extract the constant C. It’s worth mentioning that if the problem had studied functions from the non-negative integers to the non-negative integers, then C=0 would have been a possibility. You might like to consider whether this could lead to extra functions satisfying the given equation.

Problem Three

I personally found this quite difficult. We have a large number of 50-dimensional integer vectors with entries bounded by [0,2022] such that their sum is (n,n,…,n) and we are tasked to show that we can split the sum to obtain (m,m,…,m) and (n-m,n-m,…,n-m).

It seemed helpful to replace each vector $(x_1,x_2,\ldots,x_{50})$ with the 49-dimensional vector $(x_2-x_1,x_3-x_1,\ldots,x_{50}-x_1)$. We now have a large number of 49-dimensional integer vectors with entries bounded by [-2022,2022] with sum equal to the zero vector. Our task is to show that we can partition into two smaller sums with sum zero. (*)

It seems likely that the role of 2022 is not important, and will just determine how large n must be. I was hoping there might be a quick proof along the following lines: Let $\mathcal V$ denote the collection of vectors, and $A,B\subset \mathcal V$ some subsets, inducing sums $\Sigma_A,\Sigma_B$ respectively. Clearly $\Sigma_A=-\Sigma_B$ if $B=A^c$, and we might expect to be able to find $A\ne B$ such that $\Sigma_A=\Sigma_B$. I was hoping one could then move some vectors around (eg from $B\mapsto A^c$) to produce disjoint sets A,B for which $\Sigma_A=-\Sigma_B$, but I couldn’t see any way to achieve this.

This doesn’t mean it’s impossible though of course! And perhaps a contestant will find a successful argument along these lines during the olympiad.

In fact, it seems that a careful induction (on the number of cards, ie the dimension of the vectors) is the best way to proceed, and I will leave that there.

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.

BMO2 2019

The second round of the British Mathematical Olympiad was taken on Thursday by the 100 or so top scoring eligible participants from the first round, as well as some open entries. Qualifying for BMO2 is worth celebrating in its own right. The goal of the setters is to find the sweet spot of difficult but stimulating for the eligible participants, which ultimately means it’s likely to be the most challenging exam many of the candidates sit while in high school, at least in mathematics.

I know that lots of students view BMO2 as something actively worth preparing for. As with everything, this is a good attitude in moderation. Part of the goal in writing about the questions at such length is because I think at this level it’s particularly easy to devote more time than needed to preparation, and use it poorly. This year time is tight at the end of semester, and so what follows is closer to a set of complete solutions than usual, for which apologies, although I hope it is still possible to get a sense of how one might have come across the solutions yourself. Of course, this means that what follows will certainly spoil the problems for anyone who hasn’t tried them by themselves already.

The copyright for the problems is held by BMOS, and reproduced here with permission.

Question 1

As if often the case in geometry questions, what you’ve been asked to prove here isn’t the most natural property of the configuration. A good first step would be to see if there are stronger statements which are true.You are asked to show that triangle BPE is isosceles, but you aren’t told which of the three vertices is the apex. In fact, the task is to show that $BP=EP$ or, alternatively, $\angle BEP=\angle PBE$. It’s not in general true that BE is equal to BP=EP. Unless you’re very unlucky, you can establish this from one diagram.

Now, you don’t immediately know whether it’s going to be easier to show that two lengths are equal, or that two angles are equal. However, you know that P lies on the perpendicular bisector of BC, hence BP=CP, which is a big clue. In particular, this means that P would be the centre of the circle through BCE. This clearly implies the given result, so deciding to prove this instead is a good strategy.

There are now a number of ways to prove this. Note that D lies on the altitude from A, and the feet of the perpendiculars from D to sides AB and AC are both present in the configuration so (just as for the orthocentre diagram) we can calculate most of the angles involving {A,B,C,D,E}.

For example, ABDE is cyclic, so $\angle BED=\angle BAD = 90-\hat{B}$, hence $\angle AEB=\hat{B},\,\angle EBA=\hat{C}$. This shows that AB is tangent to the circumcircle of BCE. But then the line L is a radius of this circle, and so its centre must be P, the unique point on L which is equidistant from B and C.

Alternatively, we could directly calculate $\angle BEC=180-\hat{B}$ and $\angle CBP=90-\hat{B}$. But BPC is isosceles so $\angle BPC=2\hat{B}$. In general, the converse of ‘angle at centre is twice angle at circumference’ does not hold, but when we know P is equidistant from B and C this does hold, and so the angle relations precisely confirm that P is the centre of the circle through BPE.

My intention had been that the triangle would be acute-angled, to reduce the number of diagram options based on the magnitude of $\hat{B}$. If pursuing this second approach, one would need to be careful to account for whether P is on the same side or the opposite side of BC to E. That said, unless you do something very exotic, it should be exactly the same argument or calculation, and such a case distinction probably isn’t very important.

Question 2

First, a short remark. As stated, if n=5, a piece could move 3 squares to the left then 4 squares up, by Pythagoras. Handling all such options is likely to be quite annoying, since some values of n can be written in this Pythagorean form, and others cannot. This brings us to some good general principles for olympiad problems which look like this one:

• A construction, when one exists, will probably be possible using simple versions of the allowed moves / structures.
• An argument why a construction is impossible should probably be based on ideas which treat the simple moves similarly to the more complicated moves.

The setup of the problem encourages you to think about dividing the board into $n^2$ sub-boards, each with dimensions $n\times n$. Continue reading

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!

Harmonic ranges and Balkan MO 2018 Q1

A discussion of the non-geometry questions {Q2,Q3,Q4} on the Balkan MO 2018, held in Serbia, may be found here.

A blog post about the UK team’s experience is here, and a more formal pdf report is here.

Balkan MO 2018 Problem One

A quadrilateral ABCD is inscribed in a circle $\Gamma$, where AB>CD, and AB is not parallel to CD. Point M is the intersection of the diagonals AC and BD and the perpendicular from M to AB intersects the segment AB at the point E. If EM bisects the angle CED, prove that AB is a diameter of $\Gamma$.

I do not think that this was the hardest question on the paper, but I have the most to say about it, so it gets its own post. The section entitled ‘Step One’ contains (including the exercise at the end) a complete solution which only uses familiar material. The remaining sections have to quote some more obscure material, and may be of less interest to inexperienced readers, for whom many other Balkan and IMO geometry problems might be more appropriate.

Although I’ve been working hard to improve my geometry over the past couple of years, my attitude to the subject remains recreational. I prefer problems with a puzzle-like quality rather than this sort of question, whose statement is, after a little thought, not so surprising, even if most proof methods are either complicated (but elementary) or exotic. I feel most approaches to this problem require three steps: it’s easy to read a solution and forget that the first step really is a step!

I’m fairly vigorously opposed to software diagrams, as at least for me they discourage exactly the sort of insights one is generally hoping for. If you are reading this section carefully, almost certainly the most useful approach is to draw your own diagram, moderately accurate. There are only five points, though you might like to peek at Step Zero to inform drawing an accurate enough diagram without needing to apply the condition by eye.

Step Zero: Introduce X, the intersection of AD and BC.

To follow through any synthetic approach, it’s essential to have a good perspective on what the diagram means, and you will almost certainly need to introduce X to get such a perspective. Here are a couple of reasons why you might think to introduce X:

• If the conclusion is true, then $\angle ADB=\angle ACB=\pi/2$, and so M lies on two altitudes, and thus is the orthocentre of some triangle. Which triangle? It’s triangle AXB.
• Alternatively, the corresponding altitude is an angle bisector of the pedal triangle, and so the given diagram might remind you very strongly of this. Which triangle has pedal triangle CED? It’s AXB again.
• If your diagram was accurate enough (and since part of the statement is a ‘given…’ this is not so easy) you might have noticed that AD, ME and BC were concurrent. Where? At X:= AD n BC, obviously.
• In a similar vein, if the conclusion is true, then ADME and BEMC are both cyclic, and we are given ABCD cyclic. The radical axes of these three circles are AD, ME, and BC, so it is reasonable to guess that X, the (hypothesised) point of concurrence is relevant. See later.
• You are given part of a complete quadrilateral (since M is one of the intersection points of quadrilateral ABCD$– it might well be useful to complete it! • Random luck. It’s not unreasonable to consider arbitrary intersections, though this can be a low-reward strategy in general. If you did introduce X for no reason, you then had to guess, observe or realise that X, M and E should be collinear. Step One: Proving X, M, and E are collinear. This is harder than Step Two I think, so is postponed. Step Two: showing the result, given X,M,E collinear The official solution proposes introducing the reflection of A in E, which is certainly a good way to get lots of equal angles into useful places rather than not-quite-useful places. However, probably one didn’t spot this. Whether or not this was your motivation in the first place, once X is present, it’s natural to look for an argument based on the radical axis configuration. Our conclusion is equivalent to showing that ADME or BEMC are cyclic, and obviously ABCD is given as cyclic. However, motivated by the radical axis configuration (Which you can look up – but I recommend not getting distracted by what radical axis means at this stage. It’s a theorem concerning when three pairs of points form three cyclic quadrilaterals, and it has a valid converse! I also recommend not drawing any circles when thinking about the diagram.) let E’ be second intersection of circles ADM and BMC. We know that E’ lies on line XM, and so it suffices to show that E’=E. But by chasing angles in the cyclic quadrilaterals involving E’, we find that if $E\ne E'$, then $\angle EE'A=\angle BE'E$, and so $\triangle AEE'\equiv \triangle BEE'$, which after a bit of thought implies triangle AXB is isosceles, which contradicts the given assumptions. Step One: Proving X, M, and E are collinear By introducing enough extra notation and additional structure, one can prove this part by similar triangles. I think a natural approach in a question with significant symmetry is to use the sine rule repeatedly. This has pros and cons: • Disadvantage: it’s easy to get into an endless sequence of mindless calculations, which don’t go anywhere and leads more towards frustration than towards insight. • Advantage: one can plan out the calculation without actually doing it. Imagine, to give a completely hypothetical example, trying to plan such an approach in a lurching Serbian minibus with only one diagram. You establish which ratios can be calculated in terms of other ratios, and wait until you’re back in a quiet room actually to do it. You might try to show that $\angle ADB=\angle ACD=\pi/2$ directly by such a method, but I couldn’t make it work. I could plan out the following though: • Start with some labelling. I write $\alpha,\beta$ for $\angle XMD, \angle CMX$, and a,b for $\angle DME,\angle EMC$. The goal is to prove that $(a,\alpha)$ and $(b,\beta)$ are complementary by showing that $\frac{\sin \alpha}{\sin \beta}=\frac{\sin a}{\sin b}$. Will also refer to $\hat{A}$ for $\angle BAD$ when necessary. • The first ratio of sines is the easier one. Using the equal length MX in triangle DXM, triangle CMX, and then the sine rule in triangle DXC, obtain $\frac{\sin\alpha}{\sin\beta}=\frac{DX}{CX} = \frac{\sin \hat{A}}{\sin \hat{B}}$. • We can obtain $\frac{\sin a}{\sin b}=\frac{DE/DM}{CE/CM}$, but this could get complicated. However, by exploiting the equal angles $\angle DEA=\angle BEC$, we can derive $\frac{DE}{CE}=\frac{AD}{BC}\frac{\sin \hat{A}}{\sin\hat{B}}$. But of course, ABCD$is cyclic, and so there are relevant similar triangles, from which $\frac{AD}{BC}=\frac{DM}{CM}$. So in fact we have shown $\frac{\sin a}{\sin b}=\frac{\sin \hat{A}}{\sin \hat{B}}$, as we wanted since now we know:

$\frac{\sin \alpha}{\sin \beta}=\frac{\sin a}{\sin b}.$ (1)

• We need to be careful as this doesn’t immediately imply $\alpha=\pi-a$ and $\beta=\pi-b$. (For example, we need to exclude $\alpha=a$! It’s useful to exploit the fact that both a and b are obtuse here. For this type of thing, it’s more useful to focus on showing uniqueness (we definitely know one solution!) rather than finding all solutions. We are essentially asked to show uniqueness of a solution to an equation like

$\frac{\sin(\theta-x)}{\sin x}=z,$ (2)

• where $\theta<\pi$. After suitable rearranging, (2) determines $\tan x$, and so certainly has at most one solution in any interval of width less than $\pi$. This is a standard issue when using this type of argument and it’s important to know how roughly how to resolve such issues, as you wouldn’t want to waste significant competition time on such technicalities.

As an exercise, you can try to prove Step Two using this method. A hint: suppose M is not the orthocentre of triangle AXB. Introduce points C’,D’ such that $\angle AD'B=\angle AC'B=\pi/2$. Now AE bisects $\angle DEC$ but also $\angle D'EC'$. Can you use this to find two congruent triangles which can’t possibly actually be congruent?

An alternative synthetic approach

UK student Alex started with the following observation. Simple angle-chasing in cyclic quadrilateral ABCD reveals that

$\pi/2-\angle AME = \angle EAM=\angle MDC,\quad \pi/2 - \angle EMB=\angle MBE=\angle DCM.$ (3)

But we are given that M lies on the angle bisector of $\angle CED$. So we make the following claim.

Claim: the only point M which lies on the angle bisector and satisfies (3) is the incentre of triangle CED.

Remark: This claim is false. However, it is true that such a point can only be the incentre or E-excentre of triangle CED. One could salvage the original by restricting M to lie inside the triangle.

Remark: As was heavily discussed, this claim is certainly not well-known. It is very believable, but it is also not obvious either. An approach by ratios of sines, for example, as in the solution given above, seems rather tricky. Aron’s argument below is lovely, but again briefÂ is not equal toÂ easy’!

Proof of claim (Aron):Â Write $\theta:= \angle MDC$ and $\varphi:=\angle DCM$. Consider the altitude MX in triangle MDC. This is isogonal in this triangle to line ME, because the angles $\pi/2-\theta$ and $\pi/2-\varphi$ are interchanged at M. This means that the circumcentre of triangle MDC lies on ME. (Perhaps you are more familiar with the stronger statement that the orthocentre and circumcentre – eg of triangle MDC – are isogonal conjugates.) But the circumcircle of triangle MDC also lies on the perpendicular bisector of CD, and this meets the angle bisector on the circumcircle of triangle CED. Indeed, this intersection point is the arc midpoint of CD, and this really is well-known to be the circumcentre of $\odot ICI_ED$, the circle which includes the incentre and the E-excentre, and so this characterises the two possibilities for M, as required.

Harmonic ranges

In the end, the most straightforward approach to this question was to use harmonic ranges. Personally, I would use this to complete what I referred to as Step One, namely showing X,M,E collinear. I feel the radical axis argument given above is a more natural way to handle the second step, though one can also deploy projective theory for this too in relatively few steps.

This is not the place for an in-depth introduction to harmonic ranges. However, I think less experienced students are often confused about when they should consider looking for them, so I’ll try to focus on this.

Symmedians and Balkan MO 2017 Q2

While I was away, I wrote about my latest approach to teaching geometry at olympiad camps. This post will end up being about Q2 from the Balkan MO which took place yesterday in Macedonia, but first there is quite a long prelude. My solution, and probably many solutions, to this problem made use of a standard configuration in triangle geometry, namely theÂ symmedian. I want to introduce the configuration, give some simpler examples in practice, and along the way talk about my slightly patched-together philosophy about the merits of practising Euclidean geometry.

The symmedian

Draw a triangle ABC, with A at the top of the page, and extend the rays AB and AC.Â TheÂ median is the line from A through M, the midpoint of BC. Now take points D and E on AB and AC respectively. The following properties are equivalent:

• DE is parallel to BC;
• triangle ADE is similar to triangle ABC;
• the median of ABC passes through the midpoint of DE, and thus is also the median of ADE.

I think it’s a little awkward to prove either of the first two from the third –Â ratios of areas works – but the rest of the equivalences are straightforward. Later I’m going to talk about the difference between anÂ exercise and aÂ problem. These are all, at best, exercises.

Now take B’ on the ray AC, and C’ on the ray AB such that triangle AB’C’ is similar to triangle ABC. One way to achieve this is toÂ take B’ and C’ to be the reflections in the angle bisector of A of B and C respectively (so then AB’=AB and AC’=AC). We say the line B’C’ isÂ antiparallel to BC, as is any other line DE parallel to B’C’. (Probably this should say ‘with respect to triangle ABC’ or similar, but the context here is very clear, and I want this to seem natural rather than opaque.) Note that DE is an antiparallel line iff BCED is a cyclic quadrilateral. We should remember that, as cyclic quadrilaterals are the signposts for progress in both exercises and problems.

The median of triangle AB’C’ obeys the same equivalences as described above, and so bisects any antiparallel segment. We call the median of triangle AB’C’ theÂ symmedian of triangle ABC. Using the first set of equivalences, the symmedian of triangle ABC bisectsÂ any line antiparallel to BC. Furthermore, by construction, the symmedian is the image of the median of ABC under reflection in the bisector of the angle at A. We sometimes say that the symmedian is theÂ isogonal conjugate of the median.

That’s my definition. Note that there was essentially one definition then a couple of easy equivalent definitions. At no point again will I discuss the equivalence of these definitions – we have to take that for granted if we want to get on to more interesting things.

Intersection of tangents + concurrency

Now, in triangle ABC, draw the tangents to the circumcircle at B and C. These meet at P. It turns out that AP is the symmedian. This could have been our definition of a symmedian, but it wasn’t, so let’s quickly prove this.

Trigonometric arguments are very accessible, but I’ll give a Euclidean argument. Draw the antiparallel DE through P, as shown. Our task is to show that EP=PD. At this point, I would again say that this is an exercise.

We colour the angle ABCÂ in green. Two anglesÂ around point C share this measure by the alternate segment theorem. The angle at E shares this measure because DE is antiparallel. Therefore CPE is isosceles, and so EP=CP. But CP=BP, so by applying the same argument for the orange angles, we get EP=CP=BP=DP as required.

Pause to regroup. Proving this wasn’t hard, but it was perhaps surprising. If this was all new to you, and I told you to consider the reflection of the median in the angle bisector, you probably wouldn’t instantly exclaim “it goes through the tangent intersection!” So this is a useful piece of knowledge to have gained, in case we ever have to work with the intersection of two tangents like this. Maybe it won’t be useful, but maybe it will. Maybe the statement itself won’t but some extra insights from the proof will be useful, like the fact that we actually showed P is the centre of the circle BCED, and thus angles ECD=EBD=90.

A second property is that in a triangle ABC, the symmedian from A, the symmedian from B and the symmedian from C intersection at, naturally, theÂ symmedian point, which is usually denoted K. This comes from the fact that each symmedian is the isogonal conjugate of the respective median, and the medians are known to concur at the centroid. I’m not going to get into this now.

Configurations – an example

Here’s a problem. Take an isosceles trapezium ABCD as shown (ie throughout I don’t want to worry about alternative diagrams).

Let M be the midpoint of AD, and let E be the point on CM such that angle DBM = EBA. Prove that ABCDE is cyclic.

Well, certainly ABCD is cyclic. So we just need to show E also lies on that circle. And we have two equal angles, but they aren’t in the right place to conclude this immediately. However, we have angle MCA = DBM = EBA, so ABCE is cyclic, and the result follows.

Why is angle MCA = DBM? Well, the isosceles trapezium has an axis of (reflective) symmetry, and MCA is the is image of DBM under that reflection. Simple. If we wanted to do it with congruent triangles, this would all be a bit more laborious. First have to show BD=AC using one set of congruent triangles, then CM=BM using another, finally finishing using DM=MA. This is much less interesting. The symmetry of the configuration is a higher-level observation which could be proved from the axioms of geometry if necessary, but gives us more information more quickly. When we use a configuration like the symmedian configuration, we are really doing a higher-again-level version of this.

Anyway, that problem is fine, but it’s not especially difficult.

Consider instead the following problem. (I saw this online, possibly with slightly different notation, a few days ago and can no longer find the link. If anyone can help, I will add the link.)

Let AB be a chord of a circle, with midpoint M, and let the tangents at A and B meet at P. Consider a line through P which meets the circle at C and D in that order. Extend CM to meet the circle again at E. Show DME is isosceles.

Here’s a diagram, though it includes some clues.

I thought this was a fun problem, and for a while I couldn’t do it because despite lots of equal angles and equal lengths, I couldn’t conjure any congruent triangles in the right places, and I didn’t care enough about solving it to get involved in trigonometry. Then came the moment of insight. We have a midpoint, and also the intersection of the tangents. So DP is the symmedian of triangle DAB, and DM is the median. This gives us the two equal orange angles. Cyclicity gives us an extra equal angle at E as well.

Note now that the situation is very very similar to the previous question (after changing some of the labels), only this time we know ACBDE is cyclic, but don’t know that ABDE is an isosceles trapezium. If ABDE is an isosceles trapezium, we are clearly finished, as then by the same symmetry argument, EM=DM. This direction is probably harder to prove than the direction of the previous problem. Again there are a couple of ways to proceed, but one way is to consider the point E’ such that ABDE’Â isÂ an isosceles trapezium, and arguing that E’ lies on the given circle, and the circle through BME, and thus must coincide with E, in aÂ reverse reconstruction argument.

Anyway, this is all slightly a matter of taste, but I would say the second problem is much much more fun than the first problem, even though the second part of the solution is basically the first problem but in a more awkward direction. If you’re going to do Euclidean geometry at all (very much another question), I think you should do questions like the second question wherever possible. And the enjoyable ‘aha moment’ came from knowing about the symmedian configuration. Is it really plausible that you’d look at the original diagram (without the dashed orange lines) and think of the antiparallel to AB in triangle DAB through point P? Probably not. So knowing about the configuration gave access to the good bit of a nice problem.

‘Philosophy of this sort of thing’

If the goal was to solve the second problem in a competition, knowing about the symmedian configuration would be a big advantage. I’ve tried to justify a related alternative view that knowing about the configuration gave access to an enjoyable problem. The question is how many configurations to study, and how hard to study them?

We tend not to think of cyclic quadrilaterals as a special configuration, but that is what they are. We derived circle theorems from the definition of a circle so that we don’t always have to mark on the centre, every single time we have a cyclic quadrilateral. So becoming familiar with a few more is not unreasonable. In particular, there are times when proofs are more important than statements. In research (certainly mine), understanding how various proofs work is the most important aspect, for when you try to extend them or specialise. And in lots of competition problems, the interesting bit is normally finding the argument rather than basking in wonder at the statement (though sometimes the latter is true too!).

To digress briefly. In bridge, I don’t know enough non-obvious motifs in bidding or gameplay to play interesting hands well. I trust that if I thought about some of it very very carefully, I could come up with some of them, especially in gameplay, but not in real time. And it is supposed to be fun right?! Concentrating very very hard to achieve a basic level of competence is not so enjoyable, especially if it’s supposed to be a break from regular work. The end result of this is that I don’t play bridge, which is a shame, because I think the hurdles between where I am currently and a state where I enjoy playing bridge are quite low. If I knew I was going to play bridge regularly, a bit of time reading about conventions would be time well spent. And obviously this applies equally in pursuits which aren’t directly intellectual. Occasionally practising specific skills in isolation broadens overall enjoyment in sport, music, and probably everything. As anyone who’s played in an orchestra knows, there are standard patterns that come up all the time. If you practise these occasionally, you get to a stage where you don’t really need to concentrate that hard in the final movement of Beethoven 5, and instead can listen to the horns, make funny faces at the first violins, and save your mental energy for the handful of non-standard tricky bits. And of course, then move on to more demanding repertoire, where maybe the violas actually get a tune.

This is highly subjective, but my view is that in all these examples are broadly similar to configurations in geometry, and in all of them a little goes a long way.

How? In lots of the geometry configurations you might meet in, for example, a short session at a training camp, most of the conclusions about the configurations have proofs which, like in our symmedian case, are simple exercises. Once you’ve got over some low initial experience hurdles, you have to trust that you can normally solve any simple exercise if required. If you can’t, moving on and returning later, or asking for help is a good policy. The proof shown above that symmedians pass through tangent meet points (and especially a trigonometric alternative) really isn’t interesting enough to spend hours trying to find it. The statements themselves are more useful and interesting here. And it can often be summarised quite quickly: “symmedians are the isogonal conjugates of the medians, so they bisect antiparallels, meet at K, and pass through the alternate tangent meeting points.” Probably having a picture in your mind is even simpler.

There’s a separate question of whether this is worthwhile. I think solving geometry problems occasionally is quite fun, so I guess yes I do think it is worthwhile, but I understand others might not. And if you want to win maths competitions, in the current framework you have to solve geometry problems under time pressure. But from an educational point of view, even though the statements themselves have no real modern research value, I think a) that’s quite a high bar to set, and there’s no a priori reason why they should – >99.9% of things anyone encounters before university have no value to modern research maths; b) in terms of knowledge acquisition, it’sÂ similar in spirit to lots of things thatÂ areÂ relevant to later study. I don’t have to solve PDEs very often, but when I do, I hope they are equivalent or similar to one of the small collection of PDEs I do know how to solve. If I worked more with PDEs, the size of this collection would grow naturally, after some initial struggles, and might eventually match my collection of techniques for showing scaling limits of random processes, which is something I need to use often, so the collection is much larger. Maybe that similarity isn’t enough justification by itself, but I think it does mean it can’t be written off as educationally valueless.

Balkan MO 2017 Question Two

An acute angled triangle ABC is given, with AB<AC, and $\omega$ is its circumcircle. The tangents $t_B,t_C$ at B,C respectively meet at L. The line through B parallel to AC meets $t_C$ at D. The line through C parallel to AB meets $t_B$ at E. The circumcircle of triangle BCD meets AC internally at T. The circumcircle of triangle BCE meets AB extended at S. Prove that ST, BC and AL are concurrent.

Ok, so why have I already written 1500 words about symmedians as a prelude to this problem? Because AL is a symmedian. This was my first observation. This observation is then a route into non-Euclidean solutions. It means, for example, that you can describe the point of concurrency fairly explicitly with reference to triangle ABC. If you wish, you can then proceed using areal coordinates. One member of the UK team, whom I know is perfectly capable of finding a synthetic solution, did this. And why not? It’s a competition, and if you can see a method that will definitely work, and definitely take 45 minutes (or whatever) then that’s good.

I was taking a break from work in my office, and had no interest in spending the time evaluating determinants because that isn’t enjoyable at any level, so I focused on the geometry.

I think there’s a good moral from the diagram above, which is the first moderately correct one I drew. I oftenÂ emphasise that drawing an accurate diagram is important, as it increases the chance that you’ll spot key properties. In this case though, where you’re trying to examine a known configuration, I think it’s more importantÂ what you choose to include on your diagram, than how accurately you draw it. (In a moment, we’ll see why it definitely wasn’t very accurate.)

In particular, what’s not on the diagram? E is not on the diagram, and S got added later (as did the equal length signs in TB and CS, which rather spoil what’s about to happen). My first diagram was wildly incorrect, but it also suggested to me that the line ST was hard to characterise, and that I should start by deducing as much as possible about S and T by themselves. So by symmetry, probably it was enough just to deduce as much as possible about T.

Label the angles of triangle ABC as <A, <B, And therefore TB is an antiparallel in triangle ABC. (Note this doesn’t look antiparallel on my diagram at all, but as I said, this didn’t really matter.) Obviously you then guess that CS is also an antiparallel, and on a different diagram I checked this, for essentially the same reasons.

We haven’t yet made any use of the symmedian, but this is clearly where it’ll be useful. Note that if we didn’t know about everything in the prelude, we might well have deduced all of this, but we might not have thought to prove that AL bisects TB unless we’d drawn a very accurate diagram.

At this point, we have to trust that we have enough information to delete most of the diagram, leaving just {A,B,C,S,T} and the line AL. There are a few ways to finish, including similar triangles if you try very hard or trigonometry if you do it right, but again knowledge of some standard configurations is useful. Probably the quickest way is to use Ceva’s theorem in triangle ACS. You can also use Menelaus’ theorem in ABC, so long as you know a little bit about where the symmedian meets the opposite side.

An alternative is the following. We have a complete quadrilateral here, namely BTCS, and the intersection of all its diagonals. One is A, one is the proposed point of concurrency, and one is the point at infinity, since TB || CS. You can chase that, but I found it more clear to let P be the intersection of ST and BC (which we want to prove lies on AL), then look at the complete quadrilateral ATPB. Then AT and BP meet at C, and AB and TP meet at S. So if we look at where the diagonals of ATPB meet the line CS, we have aÂ harmonic range.

If I’d wanted, I could instead have written the prelude about harmonic ranges,Â but I had fewer ideas how to set these up in a slick Euclidean way. Also, it feels better to talk about the start, rather than the end of a proof, especially when there were alternative endings. Anyway, a harmonic range is a collection of two pairs of points on a line (A, B; C, D), satisfying the following ratio of directed lengths:

$\frac{AC}{BC} = -\frac{AD}{BD}.$

A classic example is when D is the point at infinity, the RHS is -1, and so C is the midpoint of AB. Being happy about using the point at infinity is a property of projective geometry, of which this is a good first example. Anyway, returning to the problem, we are looking at where the diagonals of ATPB meet line CS, and this pair of points forms a harmonic range with (C,S). TB meets CS at the point at infinity, and so AP meets CS at the midpoint of CS. But from the symmedian configuration, AL bisects CS, so AP and AL are in fact the same line, and so P lies on AL as required.

I think was a brilliant example of when knowing a bit of theory is enjoyable.Â It wasn’t at all obvious initially how to use the symmedian property, but then the observation that TB is antiparallel felt like a satisfying breakthrough, but didn’t immediately kill the problem.

EGMO 2017 – Paper One – Geometric subconfigurations

I’ve recently been in Cambridge, running the UK’s annual training and selection camp for the International Mathematical Olympiad. My memories of living and studying in Cambridge are very pleasant, and it’s always nice to be back.

Within olympiad mathematics, the UK has traditionally experienced a weakness at geometry. By contrast to comparable nations, for example those from Eastern Europe, our high school curriculum does not feature much Euclidean geometry, except for the most basic of circle theorems and angle equalities, which normally end up as calculation exercises, rather than anything more substantial. So to arrive at the level required to be in with a chance of solving even the easier such questions at international competitions, our students have to do quite a lot of work for themselves.

I’ve spent a bit of time in the past couple of years thinking about this, and how best to help our students achieve this. The advice “go away and do as many problems as you can, building up to IMO G1, then a bit further” is probably good advice, but we have lots of camps and correspondence training, and I want to offer a bit more.

At a personal level, I’m coming from a pragmatic point of view. I don’t think Euclidean geometry is particularly interesting, even though it occasionally has elegant arguments. My main concern is taming it, and finding strategies for British students (or anyone else) to tame it too [1].

Anyway, I’m going to explain my strategy and thesis as outlined at the camp, then talk about Question 1 from EGMO 2017, a competition held in Zurich this year, the first paper of which was sat earlier today (at time of writing). The UK sent a strong team of four girls, and I’m looking forward to hearing all about their solutions and their adventures, but later. I had intended to talk about the other two questions too, but I can’t think of that much to say, so have put this at the end.

My proposed strategy

Before explaining my proposed strategy, let me discuss a couple of standard approaches that sometimes, but rarely, work at this level:

• Angle chase (or length chase) forwards directly from the configuration. Consider lots of intersection points of lines. Consider angles and lengths as variables, and try to find relations.
• Exactly as above, but working back from the conclusion.
• Doing both, and attempting to meet in the middle.

The reason why this doesn’t work is that by definition competitions are competitive, and all participants could probably do this. For similar reasons competition combinatorics problems tend not to reduce instantly to an exhaustive search.

It’s also not very interesting. I’m certainly unlikely to set a problem if it’s known to yield to such an approach. When students do try this approach, common symptoms and side-effects involve a lot of chasing round conditions that are trivially equivalent to conditions given in the statement. For example, if you’re given a cyclic quadrilateral, and you mark on opposing complementary angles, then chase heavily, you’ll probably waste a lot of time deducing other circle theorems which you already knew.

So actually less is more. You should trust that if you end up proving something equivalent to the required conclusion, you’ll notice. And if you are given a cyclic quadrilateral, you should think about what’s the best way to use it, rather than what are all the ways to use it.

On our selection test, we used a problem which essentially had two stages. In the first stage, you proved that a particular quadrilateral within the configuration was cyclic; and in the second stage, you used this to show the result. Each of these stages by themselves would have been an easy problem, suitable for a junior competition. What made this an international-level problem was that you weren’t told that these were the two stages. This is where a good diagram is useful. You might well guess from an accurate figure that TKAD was cyclic, even if you hadn’t constructed it super-accurately with ruler and compasses.

So my actual strategy is to think about the configuration and the conclusion separately, and try and conjecture intermediate results which might be true. Possibly such an intermediate result might involve an extra point or line. This is a standard way to compose problems. Take a detailed configuration, with some interesting properties within it, then delete as much as possible while keeping the properties. Knowing some standard configurations will be useful for this. Indeed, recognising parts of the original diagram which resemble known configurations (possibly plus or minus a point or line) is a very important first step in many settings.

Cyclic quadrilaterals and isosceles triangles are probably the simplest examples of such configurations. Think about how you often use properties of cyclic quadrilaterals without drawing in either the circle or its centre. The moral is that you don’t need every single thing that’s true about the configuration to be present on the diagram to use it usefully. If you know lots of configurations, you can do this sort of thing in other settings too. Some configurations I can think up off the top of my head include: [2]

• Parallelograms. Can be defined by corresponding angles, or by equal opposite lengths, or by midpoint properties of the centre. Generally if you have one of these definitions, you should strongly consider applying one of the other definitions!
• The angle bisector meets the opposite perpendicular bisector on the circumcircle.
• Simson’s line: the feet of the three perpendiculars from a point to the sides (extended if necessary) of a triangle are collinear precisely when the point is on the circumcircle.
• The incircle touch point and the excircle touch point are reflections of each other in the corresponding midpoint. Indeed, all the lengths in this diagram can be described easily.
• The spiral similarity diagram.
• Pairs of isogonal conjugates, especially altitudes and radii; and medians and symmedians.

Note, all of these can be investigated by straightforward angle/length-chasing. We will see how one configuration turned out to be very useful in EGMO. In particular, the configuration is simple, and its use in the problem is simple, but it’s the idea to focus on the configuration as often as possible that is key. It’s possible but unlikely you’d go for the right approach just by angle-analysis alone.

EGMO 2017 Question 1

Let ABCD be a convex quadilateral with <DAB=<BCD=90, and <ABC > <CDA. Let Q and R be points on segments BC and CD, respectively, such that line QR intersects lines AB and AB at points P and S, respectively. It is given that PQ=RS. Let the midpoint of BD be M, and the midpoint of QR be N. Prove that the points M, N, A and C lie on a circle.

First point: as discussed earlier, we understand cyclic quadrilaterals well, so hopefully it will be obvious once we know enough to show these four points are concyclic. There’s no point guessing at this stage whether we’ll do it by eg opposite angles, or by power of a point, or by explicitly finding the centre.

But let’s engage with the configuration. Here are some straightforward deductions.

• ABCD is cyclic.
• M is the centre.

We could at this stage draw in dozens of equal lengths and matching angles, but let’s not do that. We don’t know yet which ones we’ll need, so we again have to trust that we’ll use the right ones when the time comes.

What about N? If we were aiming to prove <AMC = <ANC, this might seem tricky, because we don’t know very much about this second angle. Since R and Q are defined (with one degree of freedom) by the equal length condition, it’s hard to pin down N in terms of C. However, we do know that N is the midpoint opposite C in triangle QCR, which has a right angle at C. Is this useful? Well, maybe it is, but certainly it’s reminiscent of the other side of the diagram. We have four points making up a right-angled triangle, and the midpoint of the hypotenuse here, but also at (A,B,D,M). Indeed, also at (C,B,D,M). And now also at (C,Q,R,N). This must be a useful subconfiguration right?

If you draw this subdiagram separately, you have three equal lengths (from the midpoint to every other point), and thus two pairs of equal angles. This is therefore a very rich subconfiguration. Again, let’s not mark on everything just yet – we trust we’ll work out how best to use it later.

Should we start angle-chasing? I think we shouldn’t. Even though we have now identified lots of potential extra pairs of equal angles, we haven’t yet dealt with the condition PQ=RS at all.

Hopefully as part of our trivial equivalences phase, we said that PQ=RS is trivially equivalent to PR=QS. Perhaps we also wrote down RN=NQ, and so it’s also trivially equivalent to PN=NS. Let’s spell this out: N is the midpoint of PS. Note that this isn’t how N was defined. Maybe this is more useful than the actual definition? (Or maybe it isn’t. This is the whole point of doing the trivial equivalences early.)

Well, we’ve already useful the original definition of N in the subconfiguration (C,Q,R,N), but we can actually also use the subconfiguration (A,P,S,N) too. This is very wordy and makes it sound complicated. I’ve coloured my diagram to try and make this less scary. In summary, the hypotenuse midpoint configuration appears four times, and this one is the least obvious. If you found it, great; if not, I hope this gave quite a lot of motivation. Ultimately, even with all the motivation, you still had to spot it.

Why is this useful? Because a few paragraphs earlier, I said “we don’t know very much about this second angle <ANC”. But actually, thanks to this observation about the subconfiguration, we can decompose <ANC into two angle, namely <ANP+<QNC which are the apex angle in two isosceles triangles. Now we can truly abandon ourselves to angle-chasing, and the conclusion follows after a bit of work.

I’m aware I’ve said it twice in the prelude, and once in this solution, but why not labour my point? The key here was that spotting that a subconfiguration appeared twice led you to spot that it appeared a further two times, one of which wasn’t useful, and one of which was very useful. The subconfiguration itself was not complicated. To emphasise its simplicity, I can even draw it in the snow:

Angle-chasing within the configuration is easy, even with hiking poles instead of a pen, but noticing it could be applied to point N was invaluable.

Other questions

Question 2 – My instinct suggested the answer was three. I find it hard to explain why. I was fairly sure they wouldn’t have asked if it was two. Then I couldn’t see any reason why k would be greater than 3, but still finite. I mean, is it likely that $k=14$ is possible, but $k=13$ is not.

In any case, coming up with a construction for $k=3$ is a nice exercise, and presumably carried a couple of marks in the competition. My argument to show $k=2$ was not possible, and most arguments I discussed with others were not overwhelmingly difficult, but didn’t really have any key steps or insight, so aren’t very friendly in a blog context, and I’ll probably say nothing more.

Question 3 – Again, I find it hard to say anything very useful, because the first real thing I tried worked, and it’s hard to motivate why. I was confused how the alternating turn-left / turn-right condition might play a role, so I ignored it initially. I was also initially unconvinced that it was possible to return to any edge in any direction (ie it must escape off to infinity down some ray), but I was aware that both of these were too strong a loosening of the problem to be useful, in all likelihood.

Showing that you can go down an edge in one direction but not another feels like you’re looking for some binary invariant, or perhaps a two-colouring of the directed edges. I couldn’t see any way to colour the directed edges, so I tried two-colouring the faces, and there’s only one way to do this. Indeed, on the rare occasions (ahem) I procrastinate, drawing some lines then filling in the regions they form in this form is my preferred doodle. Here’s what it looks like:

and it’s clear that if the path starts with a shaded region on its right, it must always have a shaded region on its right. As I say, this just works, and I find it hard to motivate further.

A side remark is that it turns out that my first loosening is actually valid. The statement remains true with arbitrary changes of direction, rather than alternating changes. The second loosening is not true. There are examples where the trajectory is periodic. I don’t think they’re hugely interesting though, so won’t digress.

Footnotes

[1] – “To you, I am nothing more than a fox like a hundred thousand other foxes. But if you tame me, then we shall need each other. To me, you will be unique in all the world. To you, I shall be unique in all the world,” said the Fox to the Little Prince. My feelings on taming Euclidean geometry are not this strong yet.

[2] – Caveat. I’m not proposing learning a big list of standard configurations. If you do a handful of questions, you’ll meet all the things mentioned in this list several times, and a few other things too. At this point, your geometric intuition for what resembles what is much more useful than exhaustive lists. And if you’re anxious about this from a pedagogical point of view, it doesn’t seem to me to be a terribly different heuristic from lots of non-geometry problems, including in my own research. “What does this new problem remind me of?” is not unique to this area at all!

RMM 2017 – Problems 2, 3 and 6

In the previous post, I discussed Problems 1, 4 and 5 from this year’s Romanian Master of Mathematics competition. In this post, I discuss the harder problems (modulo my subjective appreciation of difficulty).

Problem 2

Determine all positive integers n satisfying the following condition: for every monic polynomial P of degree at most n with integer coefficients, there exists a positive integer $k \leq n$, and (k+1) distinct integers $x_1,\ldots,x_{k+1}$ such that

$P(x_1) + P(x_2) + \cdots + P(x_k) = P(x_{k+1}).$

Parsing this question deserve at least a moment. Straight after a first reading, I find it worth writing down any key quantifiers which I might forget later. Here, it’s the words at most. If you want to show the statement holds for n=2, you need to investigate monic polynomials with degree zero, one and two. You should also make sure that any instances of $x_i$ really are always distinct.

This matters in competitions! Two of our contestants failed to get the mark for showing n=2 works, precisely because of not checking the linear case, and a third could have lost it for using examples which are sometimes not distinct. On hard papers, one mark actually is the difference between triumph and frustration. And of course it matters outside competitions too, since small cases are exactly what your reader might examine first, to check they understand the problem posed, so it’s not a good place for awkward errors.

I started by trying to show that it couldn’t possibly happen that every polynomial with degree at most n had this property, for some combinatorial reason. For example, that if every set of distinct integers could only be a solution set for a small number of polynomials, then we would end up with not enough polynomials. But I couldn’t make this work at all; every bound ended up heavily in the wrong direction.

The next natural question is, does a typical polynomial of degree at most n have this property? But choosing a typical polynomial is hard, so in fact I asked, do the simplest polynomials of degree at most n have this property? I think the simplest polynomials of degree at most n are $\{1,x,x^2,\ldots,x^n\}$. Under what circumstances does

$x_1^m + \ldots x_k^m = x_{k+1}^m,$ (1)

have solutions in distinct integers? Famously, when k=2 and $m\ge 3$ this is a very very hard problem indeed. So the first point is that it though it might be useful to use Fermat’s Last Theorem, it would be foolish to pursue a strategy which, if successful, would have a proof of FLT as a sub-problem. At least, it would be foolish if the aim was to finish this strategy within a few hours.

So my main comment on this question is meta-mathematical. If lots of attempts at general arguments don’t work, there must be some special example that does it. And what properties do I want this special example to have? Maybe one might have thought of this from scratch, but my motivation came from (1) in the case m=p-1. Then, by Fermat’s Little Theorem, all the summands are equal to 1 or 0 modulo p. If k>p, then after discounting any uniform factors of p, we obtain a congruence equation which is, in informal terms,

$\left(0\text{ or }1\right)+\ldots+\left(0\text{ or }1\right) \equiv \left(0\text{ or }1\right).$

This looks really promising because it’s quite restrictive, but it’s still just a bit annoying: there are quite a few solutions. But it does give us the right idea, which is to find a polynomial P for which $P(x)\equiv 1$ modulo n. The equation $1+\ldots+1\equiv 1$ modulo n has solutions only if the number of summands on the LHS is 1 modulo n. So in this context, this reduces to showing that P is, additionally, injective on the integers, ie that P(x)=P(y) only when x=y.

It’s a nice exercise to show the existence of polynomials which are constant modulo n, and a good problem to work out how to force injectivity. If a polynomial is increasing everywhere, then it is certainly injective, and so the problem ends up being slightly easier in the case where the degree is odd than when the degree is even, but this is a nice conclusion to a nice problem, so I’ll save it for any interested readers to finish themselves.

Problem 3

Let n be an integer greater than 1 and let X be an n-element set. A non-empty collection of subsets $A_1,\ldots, A_k$ of X is tight if the union $A_1 \cup \dots \cup A_k$ is a proper subset of X and no element of X lies in exactly one of the $A_i$s. Find the largest cardinality of a collection of proper non-empty subsets of X, no non-empty subcollection of which is tight.

Note. A subset A of X is proper if $A\neq X$. The sets in a collection are assumed to be distinct. The whole collection is assumed to be a subcollection.

By Neel Nanda:

If |X|=n, there are $2^n$ possible subsets, so at first glance the answer could be a variety of things, from a linear to an exponential function of n, each of which would suggest a different approach. So the first step is to conjecture an answer, and by examining small cases it seems impossible to do better than 2n-2. There are several natural constructions for this bound, such as n subsets of size (n-1) and (n-2) subsets of size 1, so we guess this to be our answer (which later turn out to be right!).

From here, a solution is deceptively simple, though empirically the five full solutions in the contest show that it was by no means easy to find. We proceed by induction on the size of X, and want to show that any collection of subsets S has size at least (2n-2). By assumption all subcollections are not tight, so if the union of a subcollection is not the whole set X, then there is an element which appears in exactly one subset. This is a useful result, so we’d like to force a subcollection whose union is not the whole set X.

One way to guarantee that the union of a subcollection is not X is by taking the subcollection of all subsets not containing some element b. So there is some element c which appears in only one subset not containing b. If we choose b so that it’s the element contained in the fewest subsets of S, c is in at least as many subsets of S, but in only one subset not containing b. This means that at most one subset containing b doesn’t contain c. This is useful, because after removing at most 2 subsets (the coefficient of n in 2n-2, importantly!), we now have that every subset in S either contains both b and c or neither. This means that we can replace the pair (b,c) with a new element d, to get a new collection of subsets S’ of a set X’, of size n-1, so by induction $|S| \le |S'|+2\le 2n-2$.

There is also the case where all subsets contain b, but we can create an equivalent collection of subsets of X \ {b} by removing b from all subsets. So again by induction we are done.

Problem 6

Let ABCD be any convex quadrilateral and let P, Q, R, S be points on the segments AB, BC, CD, and DA, respectively. It is given that the segments PR and QS dissect ABCD into four quadrilaterals, each of which has perpendicular diagonals. Show that the points P, Q, R, S are concyclic.

I thought this problem was extremely hard. The official solution starts with a ‘magic lemma’, that isn’t quite so magic if you then read how it’s used. The overall claim is that PQ, RS and AC are concurrent (or parallel), and this is proved using the fact that the radical axis of the two circles with diameters PQ and RS also passes through this point of concurrency. Hunting for key properties of subsets of points in the diagram is an important skill in hard olympiad geometry, since it exactly reflects how problem-setters produce the problems. All the more so when there is lots of symmetry in the construction. But this is a hard example – there are a lot of potentially relevant subsets of the configuration.

When you’re really stuck with how to get involved in a synthetic configuration, you might consider using coordinates. Some of the UK students have been reading some chapters of a book (Euclidean Geometry in Mathematical Olympiads by Evan Chen. I’ve only had my own copy for a couple of days, but my initial impression is very positive – it fills a gap in the literature in a style that’s both comprehensive and readable.) focusing on various analytic approaches, so James and I felt it was safer to make sure we knew what the best settings were, and how far we could take them.

You almost certainly want the intersection of PR and QS to be your origin. I wanted to set up the configuration using the language of vectors, referenced by (P,Q,R,S). This was because $PQ\perp BO$ and so on, hence $\mathbf{b}\cdot (\mathbf{q}-\mathbf{p})=0$ and so on. An alternative is to use complex numbers, which makes this condition a bit more awkward, but is more promising for the conclusion. Concyclity is not a natural property in vectors unless you can characterise the centre of the circle, but can be treated via cross-ratios in $\mathbb{C}$. You also have to decide whether to describe the collinearity of A, B and P by expressing $\mathbf{p}=\lambda_{\mathbf{p}} \mathbf{a}+(1-\lambda_{\mathbf{p}})\mathbf{b}$, or via something more implicit. There definitely are not four degrees of freedom here, since specifying A certainly defines at most one valid set of (B,C,D), so one is mindful we’ll have to eliminate many variables later. We also have to account for fact that $\mathbf{r}$ is a negative scalar multiple of $\mathbf{p}$, and it’s not clear whether it’s better to break symmetry immediately, or use this towards the end of a calculation.

The point of writing this was that if your initial thought was ‘this looks promising via coordinate methods’, then I guess I agree. But there’s a difference between looking promising, and actually working, and there are lots of parameterisation options. It’s certainly worth thinking very carefully about which to choose, and in this case, challenging though they were, the synthetic or synthetic-trigonometric methods probably were better.

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.