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…)

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!

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.

Popoviciu’s Inequality

I’ve just returned to the UK after an excellent stay at the University of British Columbia. More about that will follow in some posts which are being queued. Anyway, I flew back in time to attend the last day of the camp held at Oundle School to select the UK team for this year’s International Mathematical Olympiad, to be held in Cape Town in early July. I chose to give a short session on inequalities, which is a topic I did not enjoy as a student and do not enjoy now, but perhaps that makes it a particularly suitable choice?

We began with a discussion of convexity. Extremely occasionally in olympiads, and merely slightly occasionally in real life, an inequality arises which can be proved by showing that a given function is convex in all its arguments, hence its maximum must be attained at a boundary value in each variable.

In general though, our main experience of convexity will be through the medium of Jensen’s inequality. A worthwhile check is to consider one form of the statement of Jensen’s inequality, with two arguments. We are always given a convex function f defined on an interval I=[a,b], and x,y\in I, and weights \alpha,\beta which sum to 1. Then

\alpha f(x)+\beta f(y)\ge f(\alpha x+\beta y).

How do we prove this? Well, in fact this is the natural definition of convexity for a function. There had initially been vague murmurings that convexity should be defined as a property of the second derivative of the function. But this is somewhat unsatisfactory, as the function f(x)=|x| is certainly convex, but the second derivative does not exist at x=0. One could argue that the second derivative may not be finite at x=0, but is nonetheless positive by defining it as a limit which happens to be infinite in this case. However, I feel it is uncontroversial to take the case of Jensen given above as the definition of convexity. It is after all a geometric property, so why raise objections to a geometric definition?

The general statement of Jensen’s inequality, with the natural definitions, is

\sum_{i} \alpha_i f(x_i)\ge f(\sum_{i}\alpha_ix_i).

This is sometimes called Weighted Jensen in the olympiad community, with ‘ordinary’ Jensen following when the weights are all 1/n. In a probabilistic context, we write

\mathbb{E}[f(X)]\ge f(\mathbb{E}X),

for X any random variable supported on the domain of f. Naturally, X can be continuous as well as discrete, giving an integral version of the discretely weighted statement.

Comparing ‘ordinary’ Jensen and ‘weighted’ Jensen, we see an example of the situation where the more general result is easier to prove. As is often the case in these situations, this arises because the more general conditions allow more ‘elbow room’ to perform an inductive argument. A stronger statement means that assuming the induction hypothesis is more useful! Anyway, I won’t digress too far onto the proof of discrete ‘weighted’ Jensen as it is a worthwhile exercise for olympiad students.

What I wanted to discuss principally was an inequality due to Tiberiu Popoviciu:

\frac13[f(x)+f(y)+f(z)]+f(\frac{x+y+z}{3})\ge \frac23[f(\frac{x+y}{2})+f(\frac{y+z}{2})+f(\frac{z+x}{2})].

We might offer the following highly vague intuition. Jensen asserts that for sums of the form \sum f(x_i), you get larger sums if the points are more spread out. The effect of taking the mean is immediately to bring all the points as close together as possible. But Popoviciu says that this effect is so pronounced that even with only half the weight on the outer points (and the rest as close together as possible), it still dominates a system with the points twice as close together.

So how to prove it? I mentioned that there is, unsurprisingly, a weighted version of this result, which was supposed to act as a hint to avoid getting too hung up about midpoints. One can draw nice diagrams with a triangle of points (x,f(x)), (y,f(y)), (z,f(z)) and draw midpoints, medians and centroids, but the consensus seemed to be that this didn’t help much.

I had tried breaking up the LHS into three symmetric portions, and using weighted Jensen to obtain terms on the RHS, but this also didn’t yield much, so I warned the students against this approach unless they had a specific reason to suppose it might succeed.

Fortunately, several of the students decided to ignore this advice, and though most fell into a similar problem I had experienced, Joe found that by actively avoiding symmetry, a decomposition into two cases of Jensen could be made. First we assume WLOG that x\le y \le z, and so by standard Jensen, we have

\frac13[f(x)+f(y)]\ge \frac23 f(\frac{x+y}{2}).

It remains to show

\frac13 f(z)+f(\frac{x+y+z}{3})\ge \frac23[f(\frac{x+z}{2})+f(\frac{y+z}{2})].

If we multiply by ¾, then we have an expression on each side that looks like the LHS of Weighted Jensen. At this point, it is worth getting geometric again. One way to visualise Jensen is that for a convex function, a chord between two points on the function lies above the function. (For standard Jensen with two variables, in particular the midpoint lies above the function.) But indeed, suppose we have values x_1<x_2<y_2<y_1, then the chord between f(x_1),f(y_1) lies strictly above the chord between f(x_2),f(y_2). Making precisely such a comparison gives the result required above. If you want to be more formal about it, you could consider replacing the values of f between x_2,y_2 with a straight line, then applying Jensen to this function. Linearity allows us to move the weighting in and out of the brackets on the right hand side, whenever the mean lies in this straight line interval.

Convex function chordsHopefully the diagram above helps. Note that we can compare the heights of the blue points (with the same abscissa), but obviously not the red points!

In any case, I was sceptical about whether this method would work for the weighted version of Popoviciu’s inequality

\alpha f(x)+\beta f(y) + \gamma f(z)+f(\alpha x+\beta y+\gamma z)\ge

(\alpha+\beta)f(\frac{\alpha x+\beta y}{\alpha+\beta})+(\beta+\gamma)f(\frac{\beta y + \gamma z}{\beta+\gamma})+(\gamma+\alpha)f(\frac{\gamma z+\alpha x}{\gamma+\alpha}).

It turns out though, that it works absolutely fine. I would be interested to see a solution to the original statement making use of the medians and centroid, as then by considering general Cevians the more general inequality might follow.

That’s all great, but my main aim had been to introduce one trick which somewhat trivialises the problem. Note that in the original statement of Popoviciu, we have a convex function, but we only evaluate it at seven points. So for given x,y,z, it makes no difference if we replace the function f with a piece-wise linear function going through the correct seven points. This means that if we can prove that the inequality for any convex piece-wise linear function with at most eight linear parts then we are done.

(There’s a subtlety here. Note that we will prove the inequality for all such functions and all x,y,z, but we will only use this result when x,y,z and their means are the points where the function changes gradient.)

So we consider the function

g_a(x)=\begin{cases}0& x\le 0\\ ax & x\ge 0\end{cases}

for some positive value of a. It is not much effort to check that this satisfies Popoviciu. It is also easy to check that the constant function, and the linear function g(x)=bx also satisfy the inequality. We now prove that we can write the piece-wise linear function as a sum of functions which satisfy the inequality, and hence the piece-wise linear function satisfies the inequality.

Suppose we have a convex piecewise linear function h(x) where x_1<x_2<\ldots<x_n are the points where the derivative changes. We write

a_i=h'(x_i+)-h'(x_i'),\quad a_0=h'(x_1-),

for the change in gradient of h around point x_i. Crucially, because h is convex, we have a_i\ge 0. Then we can write h as

h(x)=C+ a_0x+g_{a_1}(x-x_1)+\ldots+g_{a_{n}}(x-x_n),

for a suitable choice of the constant C. This result comes according to [1] as an intermediate step in a short paper of Hardy, Littlewood and Polya, which I can’t currently find online. Note that inequalities are preserved under addition (but not under subtraction) so it follows that h satisfies Popoviciu, and so the original function f satisfies it too for the values of x,y,z chosen. These were arbitrary (but were used to construct h), and hence f satisfies the inequality for all x,y,z.

Some further generalisations can be found in [1]. With more variables, there are more interesting combinatorial aspects that must be checked, regarding the order of the various weighted means.

[1] – D. Grinberg – Generalizations of Popoviciu’s Inequality. arXiv

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