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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