Characterising fixed points in geometry problems

There’s a risk that this blog is going to become entirely devoted to Euclidean geometry, but for now I’ll take that risk. I saw the following question on a recent olympiad in Germany, and I enjoyed it as a problem, and set it on a training sheet for discussion with the ten British students currently in contention for our 2017 IMO team.

Given a triangle ABC for which AB\ne AC. Prove there exists a point D\ne A on the circumcircle satisfying the following property: for any points M,N outside the circumcircle on rays AB, AC respectively, satisfying BM=CN, the circumcircle of AMN passes through D.

Proving the existence of a fixed point/line/circle which has a common property with respect to some other variable points/lines/circles is a common style of problem. There are a couple of alternative approaches, but mostly what makes this style of problem enjoyable is the challenge of characterising what the fixed point should be. Sometimes an accurate diagram will give us everything we need, but sometimes we need to be clever, and I want to discuss a few general techniques through the context of this particular question. I don’t want to make another apologia for geometry as in the previous post, but if you’re looking for the ‘aha moment’, it’ll probably come from settling on the right characterisation.

At this point, if you want to enjoy the challenge of the question yourself, don’t read on!

Reverse reconstruction via likely proof method

At some point, once we’ve characterised D in terms of ABC, we’ll have to prove it lies on the circumcircle of any AMN. What properties do we need it to have? Well certainly we need the angle relation BDC = A, but because MDAN will be cyclic too, we also need the angle relation MDN = A. After subtracting, we require angles MDB = NDC.

Depending on your configuration knowledge, this is all quite suggestive. At the very least, when you have equal angles and equal lengths, you might speculate that the corresponding triangles are congruent. Here that would imply BD=CD, which characterises D as lying on the perpendicular bisector of BC. D is also on the circumcircle, so in fact it’s also on the angle bisector of BAC, here the external angle bisector. This is a very common configuration (normally using the internal bisector) in this level of problem, and if you see this coming up without prompting, it suggests you’re doing something right.

So that’s the conjecture for D. And we came up with the conjecture based on a likely proof strategy, so to prove it, we really just need to reverse the steps of the previous two paragraphs. We now know BD=CD. We also know angles ABD = ACD, so taking the complementary angles (ie the obtuse bit in the diagram) we have angles DBM = DCN, so we indeed have congruent triangles. So we can read off angles MDB = NDC just as in our motivation, and recover that MDAN is cyclic.

Whatever other methods there are to characterise point D (to follow), all methods will probably conclude with an argument like the one in this previous paragraph, to demonstrate that D does have the required property.

Limits

We have one degree of freedom in choosing M and N. Remember that initially we don’t know what the target point D is. If we can’t see it immediately from drawing a diagram corresponding to general M and N, it’s worth checking some special cases. What special cases might be most relevant depends entirely on the given problem. The two I’m going to mention here both correspond to some limiting configuration. The second of these is probably more straightforward, and was my route to determining D. The first was proposed by one of my students.

First, we conjecture that maybe the condition that M and N lie outside the circumcircle isn’t especially important, but has been added to prevent candidates worrying about diagram dependency. The conclusion might well hold without this extra stipulation. Remember at this stage we’re still just trying to characterise D, so even if we have to break the rules to find it, this won’t damage the solution, since we won’t be including our method for finding D in our written-up solution!

Anyway, WLOG AC < AB. If we take N very close to A, then the distances BM and MA are c and b-c respectively. The circumcircle of AMN is almost tangent to line AC. At this point we stop talking about ‘very close’ and ‘almost tangent’ and just assume that N=A and the so the circle AMN really is the circle through M, tangent to AC at A. We need to establish where this intersects the circumcircle for a second time.

To be clear, I found what follows moderately tricky, and this argument took a while to find and was not my first attempt at all. First we do some straightforward angle-chasing, writing A,B,C for the measures of the angles in triangle ABC. Then the angle BDC is also A and angle BDA is 180-C. We also have the tangency relation from which the alternate segment theorem gives angle MDA = A. Then BDM = BDA – MDA = 180 – C – A = B. So we know the lengths and angles in the configuration BDAM.

At this point, I had to use trigonometry. There were a couple of more complicated options, but the following works. In triangle BDM, a length b is subtended by angle B, as is the case for the original triangle ABC. By the extended sine rule, BDM then has the same circumradius as ABC. But now the length BD is subtended by angle DMB in one of these circumcircles, and by DAB in the other. Therefore these angles are either equal or complementary (in the sense that they sum to 180). Clearly it must be the latter, from which we obtain that angles DMA = MAD = 90 – A/2. In other words, D lies on the external angle bisector of A, which is the characterisation we want.

Again to clarify, I don’t think this was a particularly easy or particularly natural argument for this exact problem, but it definitely works, and the idea of getting a circle tangent to a line as a limit when the points of intersection converge is a useful one. As ever, when an argument uses the sine rule, you can turn it into a synthetic argument with enough extra points, but of the options I can currently think of, I think this trig is the cleanest.

My original construction was this. Let M and N be very very far down the rays. This means triangle AMN is large and approximately isosceles. This means that the line joining A to the circumcentre of AMN is almost the internal angle bisector of MAN, which is, of course, also the angle bisector of BAC. Also, because triangle AMN is very large, its circumcircle looks, locally, like a line, and has to be perpendicular to the circumradius at A. In other words, the circumcircle of AMN is, near A, approximately line perpendicular to the internal angle bisector of BAC, ie the external angle bisector of BAC. My ‘aha moment’ factor on this problem was therefore quite high.

Direct arguments

A direct argument for this problem might consider a pairs of points (M,N) and (M’,N’), and show directly that the circumcircles of ABC, AMN and AM’N’ concur at a second point, ie are coaxal. It seems unlikely to me that an argument along these lines wouldn’t find involve some characterisation of the point of concurrency along the way.

Do bear in mind, however, that such an approach runs the risk of cluttering the diagram. Points M and N really weren’t very important in anything that’s happened so far, so having two pairs doesn’t add extra insight in any of the previous methods. If this would have been your first reaction, ask yourself whether it would have been as straightforward or natural to find a description of D which led to a clean argument.

Another direct argument

Finally, a really neat observation, that enables you to solve the problem without characterising D. We saw that triangles DBM and DCN were congruent, and so we can obtain one from the other by rotating around D. We say D is the centre of the spiral similarity (here in fact with homothety factor 1 ie a spiral congruence) sending BM to CN. Note that in this sort of transformation, the direction of these segments matters. A different spiral similarity sends BM to NC.

But let’s take any M,N and view D as this spiral centre. The transformation therefore maps line AB to AC and preserves lengths. So in fact we’ve characterised D without reference to M and N ! Since everything we’ve said is reversible, this means as M and N vary, the point we seek, namely D, is constant.

This is only interesting as a proof variation if we can prove that D is the spiral centre without reference to one of the earlier arguments. But we can! In general a point D is the centre of spiral similarity mapping BM to CN iff it is also the centre of spiral similarity mapping BC to MN. And we can find the latter centre of spiral similarity using properties of the configuration. A is the intersection of MB and CN, so we know precisely that the spiral centre is the second intersection point of the two circumcircles, exactly as D is defined in the question.

(However, while this is cute, it’s somehow a shame not to characterise D as part of a solution…)

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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 the 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. The 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 – ratios of areas works – but the rest of the equivalences are straightforward. Later I’m going to talk about the difference between an exercise and a 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 to 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’ is 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’ the symmedian of triangle ABC. Using the first set of equivalences, the symmedian of triangle ABC bisects 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 the 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 ABC in green. Two angles 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, the 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’ is an isosceles trapezium, and arguing that E’ lies on the given circle, and the circle through BME, and thus must coincide with E, in a 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’s similar in spirit to lots of things that are 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 often 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 important 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 a harmonic range.

If I’d wanted, I could instead have written the prelude about harmonic ranges, 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. 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_is. 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.

bmo1-2016-q5aI 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.

bmo1-2016-q5bThe 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 – 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.

bmo1-2016-q5cNow we know all the ratios, it really is just a quick calculation…

[1] – 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] – 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] – 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.

Pencils, Simson’s Line and BMO1 2015 Q5

When on olympiad duty, I normally allow myself to be drawn away from Euclidean geometry in favour of the other areas, which I feel are closer to home in terms of the type of structures and arguments I am required to deal with in research. For various reasons, I nonetheless ended up choosing to present the solution to the harder geometry on the first round of this year’s British Mathematical Olympiad a couple of weeks ago. The paper was taken a week ago, so I’m now allowed to write about it, and Oxford term finished yesterday so I now have time to write up the notes I made about it during a quick trip to Spain. Here’s three gratuitous photos to remind us all what a blue sky looks like:

And here’s the statement of the problem:

BMO1 Q5 a

and you can find the video of the solution I presented here (at least for now). Thanks to the AV unit at the University of Bath, not just as a formality, but because they are excellent – I had no right to end up looking even remotely polished.

As so often with geometry problems, the key here is to find an entry point into the problem. There are a lot of points and a lot of information (and we could add extra points if we wanted to), but we don’t expect that we’ll need to use absolutely all the information simultaneously. The main reason I’m going to the trouble to write this blog post is that I found an unusually large number of such entry points for this problem. I think finding the entry points is what students usually find hardest, and while I don’t have a definitive way to teach people how to find these, perhaps seeing a few, with a bit of reverse reconstruction of my thought process might be helpful or interesting?

If you haven’t looked at the problem before, you will lose this chance if you read what follows. Nonetheless, some of you might want to anyway, and some of you might have looked at the problem but forgotten it, or not have a diagram to hand, so here’s my whiteboard diagram:

BMO1 Q5 b

Splitting into stages

A natural first question is: “how am supposed to show that four points are collinear?” Typically it’s interesting enough to show that three points are collinear. So maybe our strategy will be to pick three of the points, show they are collinear, then show some other three points are collinear then patch together. In my ‘official solution’ I made the visual observation that it looks like the four points P,Q,R,S are not just collinear, but lie on a line parallel to FE. This is good, because it suggests an alternative, namely split the points P,Q,R,S into three segments, and show each of them is parallel to FE. We can reduce our argument by 1/3 since PQ and RS are symmetric in terms of the statement.

BMO1 Q5 c

So in our reduced diagram for RS, we need an entry point. It doesn’t look like A is important at all. What can we say about the remaining seven points. Well it looks like we’ve got a pencil of three lines through C, and two triangles each constructed by taking one point on each of these lines. Furthermore, two pairs of sides of the triangles are parallel. Is this enough to prove that the third side is parallel?

Well, yes it is. I claim that this is the natural way to think about this section of the diagram. The reason I avoided it in the solution is that it requires a few more lines of written deduction than we might have expected. The key point is that saying BF parallel to DR is the same as saying BFC and DRC are similar. And the same applies to BE parallel to DS being the same as saying BEC similar to DSC.

We now have control of a lot of angles in the diagram, and by being careful we could do an angle chase to show that <FEB = <RSD or similar, but this is annoying to write down on a whiteboard. We also know that similarity gives rise to constant ratios of lengths. And this is (at least in terms of total equation length) probably the easiest way to proceed. FC/RC = BC/DC by the first similarity relation, and EC/SC=BC/DC by the second similarity relation, so FC/RC = EC/SC and we can reverse the argument to conclude FE || RS.

So, while I’m happy with the cyclic quadrilaterals argument in the video (and it works in an almost identical fashion for the middle section QR too), spotting this pencil of lines configuration was key. Why did I spot it? I mean, once A is eliminated, there were only the seven points in the pencil left, but we had to (actively) make the observation that it was a pencil. Well, this is where it becomes hard to say. Perhaps it was the fact that I was working out of a tiny notebook so felt inclined to think about it abstractly before writing down any angle relations (obviously there are lots)? Perhaps it was because I just knew that pencils of lines and sets of parallel lines go together nicely?

While I have said I am not a geometry expert, I am aware of Desargues’ Theorem, of which this analysis is a special case, or at least of the ingredients. This is not an exercise in showing off that I know heavy projective machinery to throw at non-technical problems, but rather that knowing the ingredients of a theorem is enough to remind you that there are relations to be found, which is certainly a meta-analytic property that exists much more widely in mathematics and beyond.

Direct enlargment

If I’d drawn my board diagram even more carefully, it might have looked like FE was in fact the enlargement of the line P,Q,R,S from D by a factor of 2. This is the sort of thing that might have been just an accidental consequence of the diagram, but it’s still worth a try. In particular, we only really need four points in our reduced diagram here, eg D,E,F,R, though we keep in mind that we may need to recall some property of the line FR, which is really the line FC.

Let’s define R’ to be the enlargement of R from D by a factor 2. That is, we look along the ray DR, and place the point R’ twice as far from D as R. We want to show that R’ lies on FE. This would mean that FR is the perpendicular bisector of DR’ in the triangle FDR’, and would further require that FR is the angle bisector of <DFR’, which we note is <DFE. At this stage our diagram is small enough that I can literally draw it convincingly on a post-it note, even including P and P’ for good measure:

BMO1 Q5 d

So all we have to do is check that FC (which is the same as FR) is actually the angle bisector of DFE, and for this we should go back to a more classical diagram (maybe without P,Q,R,S) and argue by angle-chasing. Then, we can reverse the argument described in the previous paragraph. Q also fits this analysis, but P and S are a little different, since these lie on the external angle bisectors. This isn’t qualitatively harder to deal with, but it’s worth emphasising that this might be harder to see!

I’ve described coming at this approach from the observation of the enlargement with a factor of 2. But it’s plausible that one might have seen the original diagram and said “R is the foot of the perpendicular from D onto the angle bisector of DFE”, and then come up with everything useful from there. I’m not claiming that this observation is either especially natural nor especially difficult, but it’s the right way to think about point R for this argument.

Simson Lines

The result about the Simson Line says that whenever P is a point on the circumcircle of a triangle ABC, the feet of the perpendiculars from P to the sides of the triangle (some of which will need to be extended) are collinear. This line is called the Simson line. The converse is also true, and it is little extra effort to show that the reflections of P in the sides are collinear (ie the Simson line enlarged from P by factor 2) and pass through the orthocentre H of ABC.

It turns out that this can be used to solve the problem quite easily. I don’t want to emphasise how to do this. I want to emphasise again that the similarity of the statement of the theorem to the statement of this particular problem is the important bit. Both involve dropping perpendiculars from a single point onto other lines. So even if it hadn’t worked easily in this case, it would still have been a sensible thing to try if one knew (and, crucially, remembered) the Simson line result.

I was working on this script during an evening in Barcelona, and tapas culture lends itself very well to brief solutions. Whether it was exactly between the arrival of cerveza and the arrival of morcilla or otherwise, this was the extent of my notes on this approach to the problem:

BMO1 Q5 e

And this makes sense. No computation or technical wizardry is required. Once you’ve identified the relevant reference triangle (here HEC), and have an argument to check that the point playing the role of P (here D) is indeed on the circumcircle (it’s very clear here), you are done. But it’s worth ending by reinforcing the point I was trying to make, that considering the Simson line is an excellent entry point to this problem because of the qualitative similarities in the statements. Dealing with the details is sometimes hard and sometimes not, and in this case it wasn’t, but that isn’t normally the main challenge.