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…)

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.