Week 6

{Apply and prove the standard circle theorems concerning angles, radii, tangents and chords, and use them to prove related results} (3)

This is another set of tasks where we look at properties derived from the circle theorems. This time we focus on the angle at the centre theorem, as in the question below, which is from a 1959 GCE examination paper. Indeed, the tasks are all variations of this one, in which we are presented with two circles, one of which passes through the centre of the other.

 

MONDAY: This task provides an introduction to the week's scenario. It is perhaps more immediately accessible than later tasks in that it involves a specific angle - the 90˚ interior angle of a square. However, students still have work to do, to decide how to use this and other implicit angles in the given figure.

One way to approach a classic task of this sort is to start with what we know (eg the angles of the square are 90˚) and hope that this can lead us to where we need to end up. So, with luck and/or experience, one might produce a chain of reasoning like this:
Angle ADC = 90˚, so angle ADB = 45˚, so angle AOB = 45˚, so angle APB = 22½˚.
Of course, one would need to justify each step.

Another approach is to work backwards, at least for a while, as here:
I could find the desired angle APB, if I knew angle AOB, which I could find if I knew angle ADB, which I can find because it is half of angle ADB which is 90˚.

TUESDAY: This is the original GCE question, but without the initial hint about the angle at the centre theorem.

This is a subtle and demanding task, even with the angle at the centre hint. But it has the potential of giving students a hugely important experience: of perhaps struggling with a task, but having a sudden insight that solves it!

We can solve the task if we can show that the triangle BQP (which is not shown explicitly in the diagram) is isosceles. But what do we know about its base angles? - ah, that their sum is equal to the exterior angle AQB. And we potentially know a lot about this angle as it is related to two other angles in the diagram (albeit angles that are also not shown explicitly!). 

You might decide to scaffold the task for students, for example by drawing the line-segment BP, perhaps also AO and BO, although that would hugely reduce the insight that students could at some point gain and take pleasure from!

WEDNESDAY: This is essentially the same situation as before, except that Q has moved to a position on the smaller circle that takes it outside the larger circle and takes P inside the smaller circle. Can we use the same or similar arguments as before?

THURSDAY: Here Q has moved further along the smaller circle, leaving it still outside the larger circle but with P back outside the smaller circle. Can we still use our earlier arguments?

FRIDAY: Here Q has moved further still, until it coincides with point A and essentially disappears. How does this affect our earlier arguments?

Week 5

{Apply and prove the standard circle theorems concerning angles, radii, tangents and chords, and use them to prove related results} (2)

This is another set of tasks where we look at properties derived from the circle theorems. In fact, we continue our look at the properties of the cyclic quadrilateral. Who knew that it could be so interesting! 

MONDAY: This is quite a straightforward task, though it might seem odd that the two quadrilaterals don't turn out to be similar.

   

As corresponding sides of the two quadrilaterals are parallel, their corresponding angles will be equal. So, just as in the smaller quadrilateral, opposite angles of PQRS will be supplementary, which is a sufficient condition for it to be cyclic.

However, as we can plainly see, the two quadrilaterals are not similar, despite their corresponding angles being equal. One might have casually thought that the particular way in which the second quadrilateral is generated from the first, might cause them to be similar, and it is perhaps quite difficult to construct an argument to show that in this situation they don't have to be. On the other hand, it is quite easy to see that, in general, given a cyclic quadrilateral, we can translate one of the sides to construct another quadrilateral (A'B'CD, below) that is plainly not similar but which is still cyclic because the angles are unchanged.

TUESDAY: Here we again have two cyclic quadrilaterals where corresponding angles are equal. Are the quadrilaterals similar this time?

 

We know that in our two cyclic quadrilaterals, the angles at A and A', and at B and B', are equal, because of the parallel lines. Do students see that this means that the other corresponding angles are equal too? And can they explain why?

It turns out that the two quadrilaterals are not similar, despite having equal corresponding angles. Why? One way to justify this is to consider the segments formed by the chords AB and A'B' (see Task 03A). The segments would need to be similar, but plainly they are not (or if they happened to be similar, we could nullify this by translating one of the chords).

WEDNESDAY: Here we have a trapezium (which takes us back to Week 4) and a quadrilateral with the same corresponding angles as a cyclic quadrilateral but whose shape is rather extreme - can it nonetheless be cyclic?

 

Quadrilateral FADE is a trapezium. It can't be cyclic as it is not isosceles (only then would its opposite angles be supplementary). We explored this situation in Week 4.

The angles of quadrilateral FBCE are the same as the corresponding angles of the cyclic quadrilateral ABCD, because of the parallel lines, so it too is cyclic (its opposite angles are supplementary). This may come as a surprise, given its extreme shape and it is worth asking students whether they can visualise the circle that goes through its vertices. Where roughly is its centre? It is also interesting to consider what happens if we extend BA produced and CD produced further to the left. What happens when the lines cross so that E and F 'flip over'? Is quadrilateral FBCE (or EBCF?) still cyclic?

THURSDAY: Here we have four circles each containing a cyclic quadrilateral (which might be worth making more visible by drawing the chords JC, ID and HE). What can we say about the angles of these quadrilaterals, and hence about the slopes of these chords?

 

It turns out that the four cyclic quadrilaterals all have equal corresponding angles, though their shapes are quite different. A good place to start is by comparing the angles of JABC and IJCD. We can show that angle A of JABC is equal to angle C of IJCD (Why?) and similarly that angle B of JABC is equal to angle J of IJCD. Continuing with this kind of argument, we can soon establish that AB is parallel to ID, which in turn means that ID is parallel to GF, and so AB is parallel to GF.

FRIDAY: At first, this task looks rather different, as there are no cyclic quadrilaterals in sight. However, we can think of this as involving two overlapping circles, having cyclic quadrilaterals with a common chord, that have been pulled apart until the circles only touch and the common chord has become a single point - or a tangent.

 

If we think of the circles as overlapping (below, left) we get the situation in Task 05D, where we have shown that the brown (formerly green) lines are parallel. Alternatively, we can add a tangent to the given diagram (below, right) and use the theorem about the angle between a tangent and chord to show the brown lines are parallel.

The use of 'tangent and chord' is shown below, left, while a nice alternative proof is shown on the right. Both were posted on Twitter.

EXTRA 1: Here is a simple variant on Friday's task. The circles still touch but this time one circle is inside the other. Can we apply the same arguments as before?

  

EXTRA 2: Finally, a somewhat different task about cyclic quadrilaterals. Here one circle goes through the centre of another, which is a situation that we focus on in next week's tasks.

 

Week 4

{Apply and prove the standard circle theorems concerning angles, radii, tangents and chords, and use them to prove related results} (1)

This is the first of several Weekly sets of tasks looking at properties derived from the circle theorems. In this Week (and the next) we focus on the cyclic quadrilateral, in particular the property that opposite angles of a cyclic quadrilateral are supplementary (ie, their sum is 180˚).

MONDAY: The given quadrilateral is not cyclic as opposite angles do not sum to 180˚. Could we move D (to a point D') in such a way that the angle at D increases from 95˚ to 100˚? If so, this transformed quadrilateral ABCD' would be cyclic: the angles at D' and B would sum to 180˚ (and so of course would the other pair of angles at A and C, as the total sum of the interior angles of a quadrilateral is 360˚).

We can increase the angle at D by moving D 'inwards', towards B. So D lies outside the circle through A, B and C (and D').

TUESDAY: This is very much a visualising task. What sort of size is the circle through A, D and C? Roughly where is its centre? (Note that the given quadrilateral is a trapezium.)

It is not essential to know the location of the circle's centre, though some students will find it helpful: it lies on the perpendicular bisector of the chords AD and DC. The key, perhaps, is to realise that the circle is symmetrical about the perpendicular bisector of DC: point B is closer to this line of symmetry than point A, so B lies inside the circle.

WEDNESDAY: Here we meet the same trapezium again. The task is quite challenging and needs calm and careful thought! You might prefer to go straight to Thursday's task, though that has challenges of its own!

 

One way to approach this task is to build on Tuesday's visualisation. The circle through P, D and C will go through a point P' that lies to the right of CB, with PP' parallel to DC, and so the circle will intersect BC at a point that is 'lower' (further away from DC) than P (see diagram, below-left). Similarly, the circle through P, A and B will go through a point P'' that lies to the left of BC, with PP'' parallel to AB, and so the circle will intersect BC at a point that is 'lower' (nearer to AB) than P (see diagram, below-middle). Could it be that these two points on BC are the same (see diagram, below-right)?! 

THURSDAY: This is a much more open version of Wednesday's task, involving the same trapezium. Initially, students might try to split the trapezium with a steepish line, ie one that goes through AB and DC. This can be quite intriguing, though it doesn't lead to a solution. It produces quadrilaterals where two pairs of adjacent angles are supplementary, which would allow one of the resulting quadrilaterals to be cyclic, but only if it were an isosceles trapezium. In turn, this would mean that the other quadrilateral had opposite angles of 60˚ and 80˚, and 100˚ and 120˚, but these pairs don’t sum to 180˚ and so this other quadrilateral is not cyclic.

The key is to draw a flatish line, ie one that goes through AD and BC.

Any line such as the green line (below) will do, where angle DPQ = 80˚ and where P is between A and D and Q is between B and C. (What happens if, say, P is between A and D but Q is on CB produced?)

FRIDAY: This interrogates the idea that may have arisen in Thursday's task, of splitting our given trapezium with a steepish line through DC and AB. The two resulting points can't both be the points where the two circles intersect. More specifically, if the two circles intersect at U and V, with U on DC, then V cannot be on AB.

If V were on AB, then the circles can't go through U, C, B and V, and through U, D, A and V, as UCBV and UDAV can't both be cyclic. In the case of UCBV, say, the adjacent angles C and B would be supplementary, as would U and V. If in addition, opposite pairs were supplementary, the shape would have to be an isosceles trapezium, in which case the other quadrilateral would be a (non-rectangular) parallelogram and thus not cyclic. 

The two circles will look something like this (below). Point V is not on AB.

EXTRA: This is a bonus question. It does not fit perfectly into this Week's set of task, but is interesting in its own right!

How do students solve this? A promising grounded approach is to think of the radius OA (where O is the centre of the circle) as the hand of a clock, and to consider what happens to the angles in, for example, triangle OAD as the hand turns.

Here is an interesting diagram (posted on Twitter by @aburazi1956), where the argument is that the angle at A doesn't change but AO rotates through 6˚, so angle AOA' = 6˚ and so the angles ABA' and ADA', also on the arc AA', will both be 3˚.