Harmonic ranges and Balkan MO 2018 Q1

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

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

Balkan MO 2018 Problem One

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

An alternative synthetic approach

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

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

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

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

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

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

Eventually Almost Everywhere

A blog about probability and olympiads by Dominic Yeo

Harmonic ranges and Balkan MO 2018 Q1

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