Harmonic ranges and Balkan MO 2018 Q1

Posted on May 15, 2018 by dominicyeo

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

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

Balkan MO 2018 Problem One

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

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

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

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

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

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

If the conclusion is true, then $\angle ADB=\angle ACB=\pi/2$ , and so M lies on two altitudes, and thus is the orthocentre of some triangle. Which triangle? It’s triangle AXB.
Alternatively, the corresponding altitude is an angle bisector of the pedal triangle, and so the given diagram might remind you very strongly of this. Which triangle has pedal triangle CED? It’s AXB again.
If your diagram was accurate enough (and since part of the statement is a ‘given…’ this is not so easy) you might have noticed that AD, ME and BC were concurrent. Where? At X:= AD n BC, obviously.
In a similar vein, if the conclusion is true, then ADME and BEMC are both cyclic, and we are given ABCD cyclic. The radical axes of these three circles are AD, ME, and BC, so it is reasonable to guess that X, the (hypothesised) point of concurrence is relevant. See later.
You are given part of a complete quadrilateral (since M is one of the intersection points of quadrilateral ABCD$ – it might well be useful to complete it!
Random luck. It’s not unreasonable to consider arbitrary intersections, though this can be a low-reward strategy in general. If you did introduce X for no reason, you then had to guess, observe or realise that X, M and E should be collinear.

Step One: Proving X, M, and E are collinear.

This is harder than Step Two I think, so is postponed.

Step Two: showing the result, given X,M,E collinear

The official solution proposes introducing the reflection of A in E, which is certainly a good way to get lots of equal angles into useful places rather than not-quite-useful places. However, probably one didn’t spot this. Whether or not this was your motivation in the first place, once X is present, it’s natural to look for an argument based on the radical axis configuration. Our conclusion is equivalent to showing that ADME or BEMC are cyclic, and obviously ABCD is given as cyclic.

However, motivated by the radical axis configuration (Which you can look up – but I recommend not getting distracted by what radical axis means at this stage. It’s a theorem concerning when three pairs of points form three cyclic quadrilaterals, and it has a valid converse! I also recommend not drawing any circles when thinking about the diagram.) let E’ be second intersection of circles ADM and BMC. We know that E’ lies on line XM, and so it suffices to show that E’=E. But by chasing angles in the cyclic quadrilaterals involving E’, we find that if $E\ne E'$ , then $\angle EE'A=\angle BE'E$ , and so $\triangle AEE'\equiv \triangle BEE'$ , which after a bit of thought implies triangle AXB is isosceles, which contradicts the given assumptions.

Step One: Proving X, M, and E are collinear

By introducing enough extra notation and additional structure, one can prove this part by similar triangles. I think a natural approach in a question with significant symmetry is to use the sine rule repeatedly. This has pros and cons:

Disadvantage: it’s easy to get into an endless sequence of mindless calculations, which don’t go anywhere and leads more towards frustration than towards insight.
Advantage: one can plan out the calculation without actually doing it. Imagine, to give a completely hypothetical example, trying to plan such an approach in a lurching Serbian minibus with only one diagram. You establish which ratios can be calculated in terms of other ratios, and wait until you’re back in a quiet room actually to do it.

You might try to show that $\angle ADB=\angle ACD=\pi/2$ directly by such a method, but I couldn’t make it work. I could plan out the following though:

Start with some labelling. I write $\alpha,\beta$ for $\angle XMD, \angle CMX$ , and a,b for $\angle DME,\angle EMC$ . The goal is to prove that $(a,\alpha)$ and $(b,\beta)$ are complementary by showing that $\frac{\sin \alpha}{\sin \beta}=\frac{\sin a}{\sin b}$ . Will also refer to $\hat{A}$ for $\angle BAD$ when necessary.
The first ratio of sines is the easier one. Using the equal length MX in triangle DXM, triangle CMX, and then the sine rule in triangle DXC, obtain $\frac{\sin\alpha}{\sin\beta}=\frac{DX}{CX} = \frac{\sin \hat{A}}{\sin \hat{B}}$ .
We can obtain $\frac{\sin a}{\sin b}=\frac{DE/DM}{CE/CM}$ , but this could get complicated. However, by exploiting the equal angles $\angle DEA=\angle BEC$ , we can derive $\frac{DE}{CE}=\frac{AD}{BC}\frac{\sin \hat{A}}{\sin\hat{B}}$ . But of course, ABCD$is cyclic, and so there are relevant similar triangles, from which $\frac{AD}{BC}=\frac{DM}{CM}$ . So in fact we have shown $\frac{\sin a}{\sin b}=\frac{\sin \hat{A}}{\sin \hat{B}}$ , as we wanted since now we know:

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

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

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

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

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

An alternative synthetic approach

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

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

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

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

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

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

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

Harmonic ranges

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

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

What is it? Study four points A,B,C,D on a line $\ell$ , grouped into two pairs (A,B),(C,D)$ Then define the cross-ratio to be

$(A,B ; C,D ) := \frac{\overrightarrow{CA}}{\overrightarrow{CB}}\,\div \,\frac{\overrightarrow{DA}}{\overrightarrow{DB}}.$ (4)

We say that (A,B;C,D) form a harmonic range (or harmonic bundle, harmonic system etc etc.) if their cross-ratio is -1. This certainly implies that one of (C,D) lies between A and B, and the other lies outside. Note that this is a property of two pairs of points, not of four points! (A,B;C,D) harmonic does not imply (A,C; B,D) harmonic and so on. Crucially, there is an analogous definition for two pairs of points lying on a given circle.

What can you do with harmonic ranges? There are two reasons why they are useful in solving geometry problems:

They often appear in standard configurations and given configurations!
Given one harmonic range, there are natural ways to generate other harmonic ranges.

We’ll discuss both of these in a second, but a rough outline of a typical proof using harmonic ranges is as follows. First, identify a harmonic range in the configuration, perhaps using a standard sub-configuration; then, project this first harmonic range around to find some new, perhaps less obvious, harmonic ranges; finally, use some converse result to recover a property about the diagram from your final harmonic range.

We need to discuss the two useful reasons given above in more detail:

Take a triangle ABC, and consider the intersection points D,E of the internal and external A-angle bisectors with the opposite side BC. Can you prove (for example using a theorem about lengths in the angle bisector configuration…) that (B,C; D,E) is harmonic?

A related example occurs when you have both Ceva’s configuration and Menelaus’s transversal present in a given triangle, as you then have a harmonic range too. (See the suggested notes.)

One of the points may be the point at infinity on $\ell$ . Without getting into philosophy, can you see how to choose C so that $(A,B; C,\infty)$ is harmonic? This is a very very useful example.

There are plenty of good examples for cyclic ranges too, which you can explore yourself.

Harmonic ranges live in the world known as projective geometry. What this means in general is not relevant here, but it’s a good mnemonic for remembering that one can project one harmonic range to acquire another. The most simple example is this.

Given A,B,C,D on a line $\ell$ , let P be some point not on $\ell$ . The set of lines (PA,PB,PC,PD) is often referred to as a pencil. Now, consider intersecting this pencil with a different line $\ell'$ (again not through P) to obtain a new set of points (A’,B’,C’,D’). The key fact is that if (A,B; C,D) is harmonic, then (A’,B’; C’,D’) is also harmonic!

Not only does this give a new harmonic range, it establishes that the harmonic property really depends on the pencil of lines, rather than the choice of $\ell$ . Letting $\ell$ vary, we get an infinite collection of harmonic ranges. So if your diagram has a suggestive pencil of four lines, this is a promising sign that harmonic ranges may have value.

One can also project between lines and circles and from circles to circles, and typically you will need to do this.

How do you prove the results? If you proved the first example above using the angle bisector theorems, you might ask `how do you prove the angle bisector theorem’? Well, there are elegant synthetic methods, but the sine rule is a fail-safe mode of attack too. Essentially, almost all results about harmonic ranges can be proved using the sine rule, perhaps with a bit of help from other standard length-comparison results, in particular Menelaus, Ceva, and trigonometric Ceva.

As we’ve seen in the first attempt at Step One, sine rule calculations can be arduous. Projecting harmonic ranges can be a shortcut through such calculations, provided you know enough examples.

How do I know when to use them? This is really just a reiteration:

If you are given a configuration and you recognise part of the diagram as a harmonic range, it might well be worth pursuing this. If you can’t project it into any useful other harmonic range (even after, for example, introducing one extra intersection point), this might lead nowhere, but you’ll probably find something.
If you see that part of the diagram is well-suited for projecting harmonic ranges into other harmonic ranges, this is relevant. For example, if there are several lines through one point, particularly if that point also lies on a relevant circle.
Similarly, if you require some sort of symmetric result like ‘points $\mathcal{A}$ have some tangency condition iff points $\mathcal{B}$ have the same tangency condition’, then consider whether the condition has a harmonic range interpretation, and whether $\mathcal{A}$ can be projected onto $\mathcal{B}$ .
If it feels like the problem could be solved by a giant sine rule calculation comparing various ratios, it might be amenable to harmonic range analysis, so long as you find a first example!

Where can I find actual details? Because this is a report on a contest, rather than a set of lecture notes, the level of detail given here is intentionally very low. Though I hope it gives a useful overview of why such approaches might be useful, perhaps especially for those students who have a passing familiarity with harmonic ranges, but are not yet fluent at successfully applying the methods in actual problems.

The detail is important though, and I recommend these resources, among many articles on the internet:

Alexander Remorov’s sheet on Projective Geometry, which also includes a discussion of polars. My own knowledge of the subject is particularly indebted to this source. I like Question 4.
Sections 9.2–9.4 of Evan Chen’s recent book Euclidean Geometry in Mathematical Olympiads includes an ideally compact repository of useful statements. Problems, some of which veer into more challenging territory, are at the end of the section.

Characterising fixed points in geometry problems

Posted on May 20, 2017 by dominicyeo

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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Harmonic ranges and Balkan MO 2018 Q1

Characterising fixed points in geometry problems