Week 7

 

{Apply and prove the standard circle theorems concerning angles, radii, tangents and chords, and use them to prove related results} (4)

Our look at the circle theorems continues. Here we consider perhaps the most fundamental of them all, Angle at the centre, but approach it by comparing arcs, which will seem novel to many of us in the UK. This is an intriguing approach, which I hope will interest teachers, even if they feel it could be too arcane for some of their students. [I wrote an article about this in Mathematics in School (2005, 34, 2): Angle at the centre, transformed.] 

We look at a more familiar approach in next week's tasks, namely the one which appears in Euclid's Elements and is based on isosceles triangles.

The theorem, known formally as Proposition III.20, states that
The angle which an arc of a circle subtends at the centre is double that which it subtends at any point on the remaining part of the circumference.

The first three tasks take us step-by-step through a (partial) proof of the theorem, using arcs. (To keep things a bit simpler, we work with specific rather than general angles.) You might want to start with the third task (Wednesday's), to give students a better sense of where we're going, and then use the earlier tasks where necessary. Thursday's and Friday's tasks provide opportunities to consolidate some of the earlier ideas and we then provide two extra tasks where the use of arcs can be applied.

MONDAY: Here we rotate a chord so that it is parallel to another chord. The fact that the centre of rotation is not on the object being rotated, will make this quite challenging for some students. Paradoxically, this may especially be the case if students try to visualise the rotation. Others, who adopt a more general approach about the properties of rotation, may find the task trivial.

An obvious way to visualise the rotation is to join chord AB to the centre, O. The shortest line that does this is the perpendicular from O to AB (OP in the diagram below, left). But having rotated P to P', does this visualisation really help? How easy is it to determine the size of the angle of rotation POP'?

A more salient, but less direct, visualisation is to consider a line segment through the centre parallel to our chord AB (DE in the diagram below, right). Here it is far more obvious that an anticlockwise rotation of 50˚ will make this line segment (and hence AB) parallel to CB.

 

TUESDAY: Having rotated the chord AB, we now focus on arcs, especially the arcs along which the points A and B have travelled.

Points A and B are the same distance from the centre of the circle. So they will travel on identical arcs when rotated about that centre. So arc AA' = arc BB'. However, arc BB' is also identical to arc A'C (by symmetry!), and so arc AA' = arc BB' = arc A'C.

WEDNESDAY: This is where the proof begins to crystalise (perhaps!), though it is not quite the final step of a (specific) proof of Angle at the centre - having established that angle AOA' = 50˚ = angle A'OC, we will still need to make explicit that angle A'OC = 100˚ and that this is twice as big as angle ABC, but we have done all the work needed to establish this.

To help students appreciate the general nature of the proof, you might want to challenge students to produce a diagram similar to that of the present task, but with B in a slightly different position (as in the diagram below, left) or with a different angle at B (as in the diagram below, right). In each case, it is interesting to consider which features of the original diagram have changed and which have stayed the same.

 

THURSDAY: Here is an opportunity to consolidate some of the ideas from the earlier tasks, although we can also solve the task using pre-existing knowledge.

We can solve the task from first principles, as in the diagram below. Chord HC becomes parallel to NC when its end-point C moves to F (and the other end-point H moves to K). So the end-point C has rotated about O through the angle COF. As there are 18 equally spaced points on the circle, and hence 18 equal length arcs between adjacent points, each such arc subtends an angle of 20˚ at the centre (360˚÷18). So angle COF = 3×20˚ = 60˚. Notice, by the way, that K lies midway along the arc HN, which is the arc that subtends the angle at C.

It is also possible to find the angle of rotation by applying the Angle at the centre theorem. This tells us that angle NCH, which we know is the angle that chord HC turns through, is half the angle NOH. This angle is equal to 6×20˚ and so angle NCH = 3×20˚ = 60˚.

 

FRIDAY: Here is another chance to consolidate earlier ideas. This time we start with a parallel line which we rotate.

When chord HE is mapped onto IF by a rotation about the centre of the circle, radius OE is mapped onto radius OF and so turns through the angle EOF. This is the angle subtended at the centre by the arc EF and so has an angle of 360˚÷18 = 20˚.

We know from the first step of our proof (Task 07A) that the angle between IF and LA is equal to the angle of rotation, namely 20˚. Alternatively, we can find this angle by finding the angle between IF and ID, where ID is parallel to LA (see below). This is the angle DIF whose size, we know from the Angle at the centre theorem, is half of angle DOF, where DOF = 2×20˚.

 

EXTRA 1: Here is the first of three Extra tasks where we can apply the earlier thinking about arcs and rotations. However we can also solve them in other ways - for example, by using our previously existing knowledge of the Angle at the centre theorem, or by using more basic geometry.

Arc AB subtends an angle of 40˚ at the centre, so arcs BC and CD will also subtend angles of 40˚. This means angle BOD = 40˚ + 40˚ = 80˚. We can now use the theorem to state that angle BPD is 80˚÷2 = 40˚.
Alternatively, we can now apply the External angle theorem to the triangle OPD to state that angle BOD = angle OPD + angle ODP = 2 times angle OPD as triangle OPD is isosceles; and so angle BPD = angle OPD = 80˚÷2 = 40˚.

 

EXTRA 2: Again, this involves a straightforward application of Angle at the centre and Equal chords subtend equal angles.

The angle subtended at the centre by arc BE is 2×45˚ = 90˚. So the angle subtended at the centre by an arc that is one third the length of BE, such as arc AB, is 90˚÷3 = 30˚.

EXTRA 3: This is less straightforward than the previous two tasks, which makes it much more interesting! It turns out that it can be solved in quite a simple way, but it may take some time to hit upon a solution (as I discovered on revisiting the task some weeks after having written it!). This means that the task gives students the opportunity to experience the joy of achieving the sudden insight that leads to a successful resolution.

The task is based on the item below (Item 34, p 346) that I found in A New Geometry for Schools by C V Durell (1939, reprinted 1945). Students might find it intriguing that the result does not depend on where the chords cross: translating one of the chords, so that one arc gets bigger but the other gets smaller doesn't, it seems, affect the total length of the two arcs.

It looks as though about one third of the circumference is taken up by arcs u and v. Might it be exactly one third? If so, how can we be sure? The chords seem to meet at an arbitrary point inside the circle. What if they met at the centre (below, left) or what if chord CD was further to the left so that points C and D coincide and so the chords meet on the circumference (below, right)?

If the chords meet at the centre, then u and v both take up 60˚/360˚ = ⅙ of the circumference, so together they take up one third. If the chords meet at A on the circumference, whereby C and u disappear, then arc v subtends an angle of 2×60˚=120˚ at the centre and so takes up 120˚/360˚ = ⅓ of the circumference.

So the hunch is confirmed, but does it always hold, that is when chords AB and CD intersect at an angle of 60˚ at any point inside the circle? A neat way to confirm this is to join points A and D (or B and C), as in the diagram below. Then arcs u and v subtend the angles d and a at the circumference, whose sum is 60˚ (Exterior angle of a triangle theorem). So the sum of the angles subtended by u and v at the centre is 2×60˚ = 120˚ and so together the arcs cover 120˚/360˚ = ⅓ of the circumference.

A condition that we haven't yet considered is where the chords meet at an angle of 60˚ but outside the circle (below). This time the difference between the lengths of arcs u and v is one third of the circumference (or their sum is still ⅓ of the circumference if we now regard u as negative).