# Conception of the good

### Insights into our current education system

This blog post is the fourth of a series ‘Reattempting Circle Theorems’. We have looked at the following circle theorems and a non-circle theorem:

Circle Theorem #1: A right angle triangle in a semi-circle

Show that: Two radii and a chord form an Isosceles Triangle

Circle Theorem #3: A cyclic quadrilateral’s opposite angles add to 180 degrees

We will now look at the next circle theorem:

Circle Theorem #4: Angles at the circumference in the same segment are equal

The booklet attached below follows the same structure:

1. Starting with the circle and outlining its features to deduce the circle theorem
1. the circle theorem in isolation
2. interleaved basic angle facts
3. interleaved previously taught circle theorems
3. Algebraic section (In this case, I haven’t attached it because I would do it differently now)

Can you imagine your weakest children being able to find all the missing angles?

Introducing the Circle Theorem: Angles in the same segment are equal

Here is a circle:

Mark the first point on the circumference:

Mark a second point on the circumference:

Mark a third point on the circumference:

Draw a chord from the first point to the third point:

Draw a chord from the second point to the third point, and this forms the first angle:

We can draw a chord between the first and second point, so we form two different segments. The angle is in the bigger segment. We can call this the major segment.

Draw a fourth point in the same segment, and draw a chord from the first point to the fourth point:

Draw a chord between the second and fourth point on the circumference, and this forms the second angle in the same segment:

We can see that the angles are equal. We can see that using the first and second point on the circumference we have created two equal angles in the same segment.

Mark a fifth point on the circumference and draw a chord between the first and fifth point:

Draw a chord between the fifth and second point, to form an angle at the circumference in the same segment:

We can see that three equal angles made by the first and second point are in the same segment.

What if we mark a sixth point on the circumference that is the other segment?

And we draw a chord between the second and sixth point to form an angle in the other segment:

We can see that the angle in the other segment is not the same size even though it is made by the first and second point on the circumference. It is not the same size because the angle is not in the same segment.

Pedagogical Quandary

As a side note, I’ve come to realise that there are a few different ways that teachers describe what’s happening here.

The first, which I was using, was to explain that any angles in the same segment, subtended by the same arc, are equal.

The second, is that any angles in the same segment, subtended by the same chord, are equal.

So, which is correct; which should we use?

Well, a chord necessarily forms an arc… and each unique chord forms a unique corresponding arc… so, if we know the chord, we know the arc, and if we know the arc, we know the chord.  In other words, its redundant to use both, and we can choose either and be completely correct in what we’re saying.

For me, I used to use the language of ‘arc,’ but have chosen to now switch to the talking about the angles as subtended by the chord instead, since it’s the chord that splits the circle into different segments, not the arc.

You can’t avoid drawing in a chord at some point, to create the two segments, so discussing the arc as well, right up front, is probably just adding more unnecessary, extraneous cognitive load.

Even for me as a teacher, I was struggling to understand the point about the angles being in the same segment, since I wasn’t drawing in any chords… I felt like I was just saying the words, just getting kids to parrot back ‘angles in the same segment are equal,’ when even I couldn’t see where the segment was!  I’m questioning the value using this whole ‘arc’ language now. This is something I saw yesterday at Mathsconf which confirmed that I felt needed addressing.

Find the missing angle using the Circle Theorem

At this point, I would state the steps to find the missing angle before looking at a few examples. Pupils can follow the steps if they lose track during the example explanation.

Here are the steps:

• Circle known angle
• Highlight the two lengths that form that angle
• Circle the two points at the end of each length of that circled angle
• Draw a dashed chord between the two points to show the segment that the angle is made
• Highlight the pair of lengths that form another angle in the same segment
• The angles at the circumference made in the same segment are equal

Here are the pupil-attempt (We do) examples to find a missing angle or angles. The variation between the examples isn’t evident here. I try to change one change at a time between the ‘I do’, and the ‘We do’ example. I’ll explain the thinking behind choosing each example:

1. State that angle ‘a’ is 60 degrees because the angles at the circumference are equal in the same segment.

2. State that angle ‘a’ and ‘b” is 60 degrees because the angles at the circumference are equal in the same segment. Here we are showing that more than two angles are equal in size at the circumference from being in the same segment. I have interleaved the first circle theorem – an angle in a semi-circle is 90 degrees. If we total ‘b’ and 90 degrees to then subtract from 180 degrees then we can find ‘e’.

3. State that angle ‘e’ is 28 degrees because the angles at the circumference in the same segment are equal. Find ‘a’ using the basic angle fact that angles on a straight-line sum to 180 degrees. Find ‘b’ using the basic angle fact that angles in a triangle sum to 180 degrees. State that ‘f’ is the same angle size as ‘b’ because of the circle theorem we have just learnt.

4. This is the same problem type as the third example but now I’ve added a chord to form a triangle with angle ‘f’, ‘h’ and ’48 degrees’. We can find ‘f’ to be 88 degrees because it is vertically opposite to it. We can find ‘h’ using the basic angle fact that angles in a triangle sum to 180 degrees.

5. Pupils can see that there are two equilateral triangles meaning ‘a’, ‘h’, ‘e’ and ‘I’ are all 60 degrees. I would mention that angle ‘h’ and ‘I’ are equal because those angles at the circumference are in the same segment.

1. State ‘f’ is 37 degrees because angles at the circumference made in the same segment are equal. Find ‘g’ because the total of ‘f, ‘g’ and 29 degrees is 90 degrees because the triangle is right angle triangle. We know this because the hypotenuse of the triangle is also the diameter of the circle. Find ‘a’ using the basic angle fact of angles in a triangle sum to 180 degrees. State that ‘a’, ‘b’ and ‘e’ are all angles at the circumference made in the same segment. They are equal to 60 degrees.

Booklet attached below:

This blog post is the third of a series ‘Reattempting Circle Theorems’. We have looked at the following circle theorem and a non-circle theorem:

Circle Theorem #1: A right angle triangle in a semi-circle

We will now look at the next circle theorem:

Circle Theorem #3: A cyclic quadrilateral’s opposite angles add to 180 degrees

The booklet attached below follows the same structure:

• Starting with the circle and outlining its features to deduce the circle theorem
• Show a series of examples and non-examples of the circle theorem

1. the circle theorem in isolation
2. interleaved basic angle facts
3. interleaved previously taught circle theorems
• Algebraic section (In this case, I haven’t attached it because I would do it differently now)

Introducing the Circle Theorem: A cyclic quadrilateral’s opposite angles add up to 180 degrees

Here is a circle:

Then we draw four points on the circumference of the circle:

We draw a straight line between one point and the next point on the circumference. We can go clockwise or anti-clockwise when choosing which points to connect:

And keep going:

All four corners of the four-sided shape lie on the circumference. This is a cyclic quadrilateral.

Here we have a four-sided shape. A quadrilateral where all four corners of the shape lie on the circumference of the circle. Here is another example:

All four corners of the four-sided shape lie on the circumference. This is a cyclic quadrilateral.

Here is an example of a quadrilateral but NOT all its corners lie on the circumference. Here one of the corners is the centre of the circle; this isn’t on the circumference of the circle. This is not a cyclic quadrilateral.

All four corners of the four-sided shape DO NOT lie on the circumference. This is NOT a cyclic quadrilateral.

Here we do have four corners of a shape on the circumference of the circle, but the shape doesn’t have four sides. This is not a cyclic quadrilateral.

##### Examples and Non-Examples of the Circle Theorem

Specify the language to state the correct circle theorem examples and its non-example.

For example:

“All four corners are on the circumference then we have a cyclic quadrilateral.”

For a non-example:

“Not all four corners on the circumference then we don’t have a cyclic quadrilateral.”

“The shape doesn’t have four side lengths even though all four corners are on the circumference. We don’t have a cyclic quadrilateral.”

##### Find the missing angle using the Circle Theorem

At this point, I would state the steps to find the missing angle before looking at a few examples. Pupils can follow the steps if they lose track during the example explanation.

Here are the steps:

1. Check that the shape is a four-sided shape and all four corners lie on the circumference
2. Identify the opposite angles in a cyclic quadrilateral
3. If you have one the opposite angles, then subtract from 180oto find the unknown missing angle

Here are the teacher examples to find a missing angle or angles. The variation between the examples isn’t evident here. I try to change one change at a time between the ‘I do’, and the ‘We do’ example. I’ll explain the thinking behind choosing each example:

1. Find the opposite angle in the quadrilateral by subtracting the angle you have from 180 degrees

2. Find the opposite angle in the quadrilateral by subtracting the angle you have from 180 degrees. Two incomplete pairs of opposite angles.

3. Find the opposite angle in the quadrilateral by subtracting the angle you have from 180 degrees

4. Using basic angle facts. Find one of the opposite pair of angles in the quadrilateral to find the second missing opposite angle.

5. Interleaving the first circle theorem – A right angle triangle in a semi-circle

• Identify ‘a’ and ‘b’ as a right angle of 90 degrees.
• Find ‘c’ by adding 90 and 43 degrees and then subtracting from 180 degrees.

6. Interleaving the first circle theorem – A right angle triangle in a semi-circle. Interleave the ‘show that’ feature of two radii and a chord form an Isosceles triangle

• Identify ‘h’ as a right angle of 90 degrees.
• Identify the total of ‘a’ and ‘f’ as 90 degrees.
• Find ‘a’ as 29 degrees as it is one of the two equal angles in an Isosceles triangle
• Find ‘b’ by adding the two equal size angles and subtracting the result from 180 degrees.
• Find ‘e’ by subtracting ‘b’ from 180 degrees using the basic angle fact that angles on a straight-line sum to 180 degrees.
• Find ‘g’ and ‘f’ by subtracting ‘e’ from 180 degrees and dividing the result by 2.

7. Interleaving the first circle theorem – A right angle triangle in a semi-circle

• Identify ‘a’ and ‘f’ as a right angle of 90 degrees
• Find ‘e’ by adding 90 and 43 degrees and then subtracting from 180 degrees
• Find ‘b’ by adding 90 and 47 degrees and then subtracting from 180 degrees
• Different approach: Interleaving angles on parallel lines to determine ‘b’ as an alternate angle to 43 degrees. And ‘e’ as an alternate angle to 47 degrees.

8. Interleaving basic angle facts to

• Find ‘a’ since the triangle is an Isosceles triangle
• Find ‘f’ and ‘g’ as they are corresponding angles to ‘a’
• Find ‘b’ by subtracting ‘a’ from 180 degrees since angles on a straight-line sum to 180 degrees
• Find ‘e’ since its equal to ‘b’ because the shape is an Isosceles trapezium.

Booklet attached:

This blog post is a follow up of the ‘Reattempting Circle Theorems’ blog post series. Last time we looked at this circle theorem:

Circle Theorem #1: A right angle triangle in a semi-circle

We will now look at a potential non-circle theorem. I’m opening this up for debate. I’ve had a look around online and in articles whether ‘Two radii and a chord form an Isosceles triangle’ is considered a circle theorem or not. Right now, I’m going to state it is not a circle theorem but instead the start of a proof about the perpendicular bisector of a chord. Thank you to Ed Southall for clarifying. Follow him – he’s my geometry go to! I am flexible to change my mind.

I deliberately included this after the first circle theorem so I could interleave it into new problems.

If I have a triangle where two lengths are radii and the third length is a chord, then it will be an Isosceles triangle. Two radii connected by straight line at both endpoints form an Isosceles triangle. The two of the side lengths are of equal measure.

I hope all pupils will be able to access a mathematical problem like this:

##### Show that: Two radii and a chord form an Isosceles Triangle

Here is a diagram for you to see this circle theorem form:

I have a circle where I draw one radius:

I draw a straight line between the two endpoints of each radii which are on the circumference of the circle:

I then would show pupils that I can mark the two radii to show they are of equal length. I would show that the two equal angles of the triangle are of same size. And even mark the angles with identical values. I would do this live rather than with static images:

Now, there are two types of angles you can find:

1) One of the two equal angles

2) One non-equal angle

To find one of the equal angles, you’ll have the non-equal angle. Here are the steps:

1. 180 degrees – non-equal angle
2. Divide the result by 2

To find the non-equal angle, you’ll have one of the equal angles. Here are the steps:

1. Double one of the equal angles
2. 180 degrees – (double one of the equal angles) = Non-equal angle

The structure of the booklet is the same as outlined in the previous blog post. There are ‘I do, We do’ examples and then a practice exercise at the end. Since this isn’t a circle theorem, I’m not going to show a selection of examples and non-examples. But, I’m interleaving the first circle theorem into the sequence.

##### Finding the missing angle

Here are the teacher examples to find a missing angle or angles. The variation between the examples isn’t evident here. I try to change one thing at a time between the ‘I do’ and the ‘we do’ example. I’ll explain the thinking behind choosing each example:

1. Show that two radii and a chord form an Isosceles triangle, only. Given the non-equal angle, find the one of the two equal sized angles.

2. Show that two radii and a chord form an Isosceles triangle, only. Given one of the equal sized angles, find the missing non-equal angle.

3. Interleaving the angle fact: angles around a point sum to 360 degrees. Find the non-equal angle to then find one of the two equal sized angles.

4. Two Isosceles triangles sharing a length. One triangle you are given one of the two equal sized angles to find the non-equal angle. Once you’ve calculated ‘m’ you can use the basic angle fact that angles on a straight-line sum to 180o to find ‘p’.

5. Same as Example 4 – angles needed to find before ‘u’ are not marked. Same logical steps. Also, can use the fact that angles on a straight-line sum to 180 degrees to then find ‘u’.

6. Interleaving the first circle theorem – A right angle triangle in a semi-circle

1. Identify ‘a’ as a right angle of 90 degrees.
2. Find ‘e’ by adding 48 and 90 degrees and then subtracting from 180 degrees.
3. Find ‘b’ by finding the non-equal angle of the triangle using 180 – 58 degrees, and dividing the result by 2

7. Three Isosceles triangles.

1. Find ‘m’ and ‘n’ by doubling one of the two equal sized angles of each triangle and subtracting from 180 degrees.
2. Find ‘k’ by using the fact that angles around a point sum to 360 degrees.
3. Use ‘k’ to find angle ‘p’ by subtracting the non-equal angle and dividing the result by 2.

8. Two Isosceles triangles that are not sharing a side length:

1. Find ‘u’ using the one of the two equal angles of the triangle
2. Find ‘w’ by doubling the triangle’s two equal angles and subtracting from 180 degrees.
3. Find ‘x’ by doubling 64 degrees and subtracting from 180 degrees.
4. Find ‘y’ by using the angle fact that angles around a point sum to 360 degrees.
##### Algebraic Application

Another reason why I wanted to include this feature because it allows pupils to apply prior knowledge from the topic of solving equations. There are two problem types that we will have to deal with:

• Two equal angles labelled – solve equations with unknowns on both sides

In this case, the variable used for the two equal angles represented as expression will be the same. It has to so we can form an equation with unknowns on both sides to find the value of the unknown. I then labelled the non-equal angle with a different variable. I did this because I didn’t want to make an error in choosing an algebraic expression which shared the same variable as the two equal sized angles.

• All angles are an algebraic term, expression (same variable, not different) or number. Form equation and equate to 180 degrees

In the next problem type, I deliberately labelled the two equal sized angles with the same algebraic term or expression. This re-iterated that the two angles of the Isosceles triangle are equal. Given more time I would create more examples where the equal sized angles are labelled with different expressions sharing the same variable.

##### Is it possible for a mathematical problem be incorrect?

Thankfully, Tom Francome highlighted that I mislabelled one angle in a triangle with a different variable so the possibilities for that variable were endless:

I changed ‘4c’ to ‘4f’ so that all angles in this triangle share the same variable. We can then collect the algebraic terms and equate to 180o to find the value of the unknown. We can substitute the value of the unknown back into the algebraic term to find the size of each angle.

Booklet attached:

In January 2017, I wrote two blog posts on a selection of circle theorem resources. I was in my fourth year of teaching and I loved creating curriculum resources. I made the problems using activinspire because I hadn’t mastered Geogebra by that point. There were so many mistakes in the questions.

Some lovely teachers got in touch via DM and gave their friendly feedback. Others wielded their axe and punched me with mean tweets. Some others pretended to care by writing blog posts about the situation. As if they knew the cause behind my errors.

It wasn’t the best of times. But, I learnt that it is OK to make mistakes that you didn’t even know were mistakes. It highlighted that subject knowledge development is never ending.

What have I learnt from the experience:

1. Using GeoGebra meant that I was able to draw the angles, so a 26 degrees angle looked like one. The largest angle was opposite the longest side length etc.
2. I was able to check that all the angles within a triangle and a straight line totalled to 180s. And that angles in a quadrilateral or around a point totalled to 360 degrees.
3. I was able to learn about construction invariance – something I didn’t know about. It is also something taught after A level Mathematics. My knowledge of school taught mathematics goes this far.
4. Always have your work reviewed by a fresh pair of eyes. Ask a friend, a colleague etc.

I am familiar with Geogebra and enjoy using the application. Last year I re-created the circle theorem resources into a prepared booklet. I will release each circle theorem’s resources one blog post at a time.

Here are a few things to know about the resources that you are about to see:

1. The first section is showing pupils how the circle theorem comes into existence. Starting with the circle and outlining its features to deduce the circle theorem. If I showed this in class, then I would do this live rather than using static images.
2. I then show a series of examples and non-examples of the circle theorem. This helps pupils to identify a circle theorem on complicated problems.
3. I start with examples which model the circle theorem in insolation. So, a pupil’s attention is on the theorem alone and identifying the missing angle.
4. I then start interleaving basic angle facts in the circle theorem problems. E.g: angles in a triangle, straight line, right angle, isosceles triangle etc.
5. In future circle theorem resources, you will see me interleave learnt circle theorems. This is to build gradual difficulty within the examples.
6. There is algebraic section included at the end where angles are expressions.

One common question I get from the resources I make: which sets are the resources appropriate for?

This isn’t a popular answer but it’s true. When you are teaching a concept for the first time the teaching should be the same for all sets. The only difference between high and low attaining pupils is the time taken to learn the concept. High attaining pupils recall prior knowledge and using it with fidelity compared to weaker pupils. I would teach a new concept explicitly with all pupils across the attainment spectrum. Can you imagine your low-attaining pupils finding the missing angles on this problem?

##### Introducing the Circle Theorem: A right angle triangle in a semi-circle

Here is a circle:

Draw a line from one point on the circumference, through the centre of the circle marked ‘C’ through to the opposite end on the circumference. This line is the diameter:

Teacher Language: Edge – Centre – Edge – check it is a straight line!

Draw a straight line to another spot on the circumference in one half of the circle from one end of the diameter:

Complete the three-sided shape (triangle) by drawing a straight line between the two unconnected points:

We have a triangle that is within a circle, and all the triangle’s vertices are on the edge of the circle. In other words, all the triangle’s vertices are on the circumference on the circle. The angle on the circumference, not touching the diameter is 90 degrees. It is a right angle. This is a right-angle triangle.

##### Introducing Examples and Non-Examples

Specify the language to state the correct circle theorem example and its non-example.

For example:

“If we have the longest side of the triangle as the diameter of the circle, then we have a right-angled triangle.”

For a non-example:

“If the longest side of the triangle is NOT the diameter of the circle, then we don’t have a right-angled triangle.”

The right angle is always touching the circumference! ‘a’ in the examples mark the right angle:

##### Finding the missing angle using the circle theorem

At this point, I would state the steps to find the missing angle before looking at a few examples.  Pupils can follow the steps if they lose track during the example explanation.

Here are the steps:

1. Identify the right angle of the triangle
2. Calculate the remaining interior angles by
4. Subtracting the result from 180 degrees

I usually do a ‘I do, We do’ set up when going through teacher examples. I will go through one example, and then pupils will go through a similar example. Same problem just different numbers. Completed on mini-whiteboards. Here is an example:

Example 1: Find the missing angles:

Here is the teacher sequence series only (‘I do’ only):

Example Breakdown:

1. Circle theorem in Isolation – Find right angle and second missing angle
2. Circle theorem in Isolation – Find right angle and second missing angle
3. Interleave an Isosceles right-angled triangle
4. Find the missing angles across two right-angle triangles. One being an Isosceles triangle.
5. Same as example 4 but both triangles are in the same semi-circle
6. Interleaving angles on a straight line

I included Example 5 to relate to a future circle theorem: Angles in the same segment are equal.

The algebraic section is incomplete. I thought I’d include it to see the variation theory applied in the examples.

A month ago, I was resourcing for the topic ‘Expressions and Substitution’ for the Y9 Curriculum. A common approach is to substitute values to then solve a linear equation. I was keen to create a problem type when you substitute values you then have to solve a quadratic equation. Something like this:

On the surface you can see that the sequence of examples have one aspect varied at a time – value for ‘u’. This is inspired from Ference Marton’s Variation Theory (VT). One theory out of a body of educational research underpinning the United Learning Mathematics resources.

We use VT because perceived variation generates expectations for pupils. When these expectations are confirmed pupils’ perceptions are relevant mathematically. VT guarantees more clarity and honesty about mathematical ideas. Anne Watson and John Mason explain the meaning of ‘clarity of mathematical ideas’:

“the structure of examples through which learners encounter mathematically significant variation…attention can be focused on the use of variation to reveal the patterns and generalities which result from the techniques. This is the clarity.” ¹

This blog post will look at how VT is applied on the surface by changing the value of ‘u’, but also resulting in one change in the procedure of factorising a quadratic expression. This post is about the thinking behind creating an exercise like this, teaching it would be different post. To appreciate what follows here is the working out for part (a) and part (b)

On the surface, you can see that only initial velocity’s value changes. This is deliberate. Under the surface, here are the quadratic equations for each part to show that only one thing is changing between each part:

Pupils overgeneralise the procedure of substituting values to then solve a quadratic equation.

This example shows variation on the surface, but also with the quadratic equations solved so pupils develop mathematically relevant expectations that they will be solving a quadratic equation.

Here is the thinking process behind making this exercise:

Step 1:The kinematic equation where t is unknown results in solving a quadratic equation:

Step 2: I then thought what would have to be the unknown variable and the known variables? For the following formula to result in a quadratic equation t is unknown.

Step 3: In Y9, pupils can factorise a quadratic expression where all coefficients and constants are integers, and where the coefficient of ‘a’ in a quadratic equation of ax² + bx + c is equal to 1, only, not greater than or less than 1.

I have to be selective with each known variables’ values. Some are permissible and where some aren’t. The values that aren’t permissible would mean pupils factorising a quadratic expression where ‘a’ in ax² + bx + c is greater than 1.

Step 4: I looked at a sequence of quadratic expressions that can be factorised. Variable t will be the unknown rather than x to correspond to the kinematics equation:

So, here is the sequence I made:

There are a few expressions that I can’t factorise because of the context of the kinematics. During the solving process, I can’t have two values of  t being equal to a negative value. I can’t have this situation below:

In the context of kinematics, I can’t have negative values for time. Deciding which one to pick will confuse pupils.

I can have either: a quadratic expression where t would be both positive, or one value of  would be positive and the other negative. If there are two positive values for time, then that means the case is true at two different times. If one value of time is positive and the other is negative, then we accept the positive value for time only. In sum:

I can’t have a quadratic expression where the constant is positive AND the value of ‘b’ in ax² + bx + c is also positive.

I can have a quadratic expression where the constant is negative AND the value of ‘b’ in ax² + bx + c is either positive or negative.

Step 5: Back to the condition stated in Step 3 where the coefficient of ‘a’ in a quadratic expression of ax² + bx + c is equal to 1, only.

For this to be the case, I know that

The value of ‘a’ cannot vary, because pupils will have to factorise a quadratic expression where ‘a’ in ax² + bx + c is greater than 1.

Step 6: If we look back at the sequence of quadratic expressions, I have kept the value of the constant the same -24. And since,

The value of ‘a’ and ‘s’ remain the same throughout so the only varying value is for ‘u’ . Initial velocity can be positive or negative because it is a vector quantity.

In summary, on the surface of the exercise only the value of initial velocity varies at a time. Beneath the surface, one coefficient of the quadratic expression varies at a time. We are varying on the surface and deep within the mathematical problem. The two controlled dimensions encourages the learner to find the value of t:

• By overgeneralising the process of substituting to then solve a quadratic equation.

Pupils focus on this overgeneralisation because they can factorise the quadratic expressions used. They learn the mathematical structure to substitute and solve a quadratic equation.

1. References:

¹ Anne Waton and John Mason (2006) Variation and Mathematical Structure MT194

This exercise is under review. If you spot an error or have any feedback then please feel free to get in touch.

Chapter 21, in Engelmann’s Magnus Opus, Theory of Instruction has changed the way I sequence worked examples to communicate a concept. This change took place greatly whilst I was at Michaela Community School when I started experimenting how to teach a UK Maths Challenge After school club. This is written about in more detail, here.

What I noticed, which may sound obvious, is that if you teach using examples which demonstrate the explicit features of a concept first and then progressively make those features implicit then the probability of success for each child in the learning process is greater. Engelmann refers to this as Covertization, “instruction that involves prompt shifts progressively from highly-prompted examples to unprompted examples.”

One example which will be detailed below, would look like this:

So a sequence of examples with explicit features transitioning to implicit features provides a process where pupils take overt steps to go from the first line of working to the last line of working. This means that the first example pupils encounter is structured within the simplest context.

Covertization results in examples that communicate a concept, or part of a concept, to be communicated in its most explicit and misconception-proof form.

The best way to appreciate covertization is to look at some examples which aren’t explicit.

Here are examples of fractions:

Here pupils can develop the following misconceptions:

• all fractions have a one at the top
• all fractions increase by ‘1’ in the bottom value
• fractions get bigger when the bottom value is greater

Another example:

Here are examples of expressions commonly known as a ‘difference of two squares’

Here pupils can develop the following misconceptions:

• The first term is always positive and a
• The second term is always a constant
• The second term is always an integer
• The second term is always negative

What are the examples that I would use instead? To communicate an expression as a ‘difference of two squares’ I would start with the following, by stating that an expression must meet the following conditions:

1. The coefficient of the variable is a square number
2. The power of the variable must be even
3. The power of the variable is never an odd number
4. One term will be negative AND the other term will be positive
5. The constant must be a value that can be square rooted.

Here are the following examples that I would use to reiterate the conditions stated above:

There are other examples that I could include here, however, what I’ve done is limit the examples to only meet the conditions I’ve stated. But look at the complexity of what pupils can identify as an expression known as a difference of two squares by selecting the examples that I have.

More importantly, I’ve started with an example which is explicit with the features that make an expression a difference of two squares. With the initial set of examples, you can see that the coefficient of 1 for isn’t obvious because pupils can’t see it, and ‘1’ as a constant isn’t an obvious example of a square number. Explicit features allow pupils to build a clear and misconception proof understanding of a concept, which is what makes an example powerful.

My final point, we can see that the process of covertization has highly prompted examples allowing pupils to respond successfully to writing the product of two expressions because the coefficients and powers provide them that structure. Starting the other way around with implicit features now looks unstructured and poorly guided for a pupil to know how to rewrite the expression as a product of two expressions.

On February 15th, the world lost an educator who spent his life developing an approach to accelerate the learning of disadvantaged pupils.

Engelmann was a Marketing Director turned Professor Emeritus of Education at the University of Oregon. He co-authored the famous ‘Theory of Instruction’ with Douglas Carnine and co-developed the term ‘Direct Instruction’ while working with Carl Bereiter. Through grant funding, they set up the Bereiter-Engelmann Pre-school which demonstrated the extent to which disadvantaged pupils could accelerate their learning in comparison to the performance of middle-class pupils.

I have spent the last couple of years becoming familiar with Engelmann’s work, taking aspects of his Theory of Instruction and applying it in my resource creation.

So, what have I learnt from Engelmann?

Answer: That a learner’s inability to respond appropriately to a form of instruction may not be the fault of the child; instead, it can be a problem with what she’s being taught.

This means it’s possible to teach a syllabus in a way she can respond to appropriately without dumbing it down. Here are the four things I keep in mind when creating resources that allow the highest percentage of pupils to understand the course content on the first attempt.

Atomisation

When I used Engelmann’s Connecting Maths Concept textbook series with my Year 7 and Year 8 Intervention pupils, I saw that Engelmann had taken a concept and broken it down into several sub-tasks. A sub-task is a small aspect of a concept. For example, A sub-task of how to add fractions with denominators would be finding the lowest common denominator. For a pupil to develop a flexible understanding of a concept, she needs to be taught as many sub-tasks of the concept and then shown the connections between each of the sub-tasks.

Atomising does exactly this. When I plan a unit of work, I take a concept and break it down into its sub-tasks, and I explicitly teach even the most nuanced aspects of the concept. For example, before I teach pupils how to factorise an expression, I teach pupils how to divide an algebraic expression by an integer, or by an algebraic term. Before I even do this, I teach pupils whether we can also divide an algebraic expression by a number or an algebraic term. An example is shown below:

Here are some examples, of where we can simplify the algebraic fraction:

Here are some examples of where we CANNOT simplify the algebraic fractions because we cannot divide ALL the terms in the fraction:

The value of this exercise is two-fold:

1)      Pupils are taught the most nuanced aspects of a concept which are usually the most difficult parts of the unit being taught. If the most challenging part of the concept isn’t taught explicitly then how can we expect pupils to attempt the most complex applications of the concept? We need to be more thorough and comprehensive than you might think and teach the most complex elements of a concept as well as the most basic.

2)      Pupils develop a flexible understanding of the concept because they can see the big picture. If you plan an entire unit rather than isolated lessons parts, you are more likely to teach as many sub-tasks as possible and not miss anything that’s essential to a student’s understanding. Missing out sub-tasks inevitably means you have to re-teach. Engelmann set up his textbook series to avoid the need to re-teach. If re-teaching is required, Engelmann provides appropriate correction and reinforcement exercises for each unit of work.

Sequencing the learning in the most effective manner

Engelmann’s Connecting Maths Concept textbook structures the content of a unit of work in just that sequence where the learning can be delivered most effectively. Engelmann believed that all future learning is dependent on prior learning and that there is an optimal sequence for each concept. Provided the lessons are sequenced in the most effective way, pupils always have the knowledge required to access the topic they are about to learn. At United Learning, scheme of work is structured and resourced with the same philosophy in mind. The underlying idea is that how effectively the pupils learn depends on the sequence in which they learn about a particular concept.

Scripting the lessons – Pedagogy

Scripting how you communicate the concept is essential. Now, many teachers despise scripted lessons, and some with good reason, e.g. the script they’re expected to follow is sub-optimal. Another reason for their scepticism is the belief that there is more than one optimal way to teach pupils about a particular concept. However, Engelman persuasively argues that there is only one optimal way to teach a particular concept – and his scripted lessons were field tested with tens of thousands of pupils and constantly being refined in response to feedback. Consequently, he was confident that the scripted lessons he and his colleagues developed embodied the most optimal learning sequences.

When I created my resources at Great Yarmouth Charter Academy, I started scripting how I would communicate concepts, to ensure pupils received the most effective and efficient form of instruction.

Then, I would think carefully about what method to communicate.  For example, I didn’t want to teach pupils how to add fractions using a method which was limited to only a few problem types, and then create a different method for another set of problem types. Instead, I tried to create methods that could be applied consistently to as many problem types as possible. This allowed pupils, especially the weakest, to master each concept in all its myriad complexity; evidenced by ever increasing scores in weekly quizzes.

Lastly, my scripted lessons were designed to give pupils the grounding they needed to articulate their understanding. Here is a video showing how a pupil using this knowledge to subtract negative fractions:

Low-stake quizzing and providing appropriate corrections and reinforcements

Engelmann’s Connecting Maths Concept textbook has many opportunities for pupils’ understanding to be tested. The script includes hundreds of questions for teachers to ask. Pupils are given exercises to try with the teacher, as well as independent exercises. Similarly, after every ten lessons, there are also small quizzes recapping what pupils have learnt, not only in the last ten lessons but in the previous 20, even 30.

At Charter, one visitor tallied the number of questions I asked pupils in a single lesson, and they totted up 76 questions in about 25 minutes. I learnt from Engelmann’s teacher scripts how to ask pupils’ questions which test their ability to recall prior knowledge, articulate their knowledge of a concept, to explain a misconception, etc.

In summary, I believe that Engelmann is one of the most important educators of the 20th and 21st Century. I think his work will stand the test of time. By applying his teaching principles to resource creation, I have helped my pupils learn more, and remember it for years to come.  My experience confirms, for me, that teacher quality is a function of the resources they have access toChildren are more likely to be successful with a teacher, who has access to exceptional resources, than a teacher who doesn’t, and never has.

Engelmann’s work has taught me more than any educator that I studied with during my PGCE and MA. My next post will look into the evidence for the effectiveness of Engelmann’s approach and the reasons why his work hasn’t been more influential.

After my podcast with Craig Barton, I have received many emails asking to share more booklets. I have attached the booklets that I made during my time at Charter. They aren’t perfect, and with my current workload, I am not in a position to refine them. However, I do think they are useful for teachers who want to start designing their own booklets. I used each booklet with all my classes. I hope they are helpful.

There will inevitably be mistakes in the booklets. I take full responsibility for any errors that you see.

In February 2018, I heard Andrew Blair speak at a CPD event about Inquiry Maths. Andrew is a Head of Maths and founder of his website Inquiry Maths where teachers can gain access to ideas about teaching Mathematics using the Inquiry Maths model.

Andrew and I tend to disagree over the best pedagogical approach when teaching Mathematics, but we agree in teaching many of the problems that he has shared on his website. When I create resources, I check out different sites for ideas, and I regularly check Andrew’s website. It’s fantastic! When I saw Andrew speak at this CPD event, he showed a mathematical inquiry task that I made a mental note to include in my resourcing for the following year. The problem is below:

40% of 70=____% of _____

I love this type of mathematical problem. There is so much mathematics that needs to be communicated, and I would recommend readers to check out Andrew’s page on this inquiry. At United Learning I’ve finished the Fraction, Decimals and Percentage booklet for the Y8 curriculum and I’ve made resources for this type of mathematical inquiry to be shared with the kids.

Here is an example of how I broke down the teaching sequence so all pupils can access a problem like this but also support all pupils to attempt more complex forms as well.

The teaching sequence is designed to increase the probability that all pupils can be successful in learning the subject content.

1) Showed a multiplication model for the following problem types

I have listed some of the potential decimal multiplication calculations and then demonstrated how to draw the model for the equivalent decimal multiplication calculation:

Pupils are only shown the first line because they are taught that the mathematical model of rewriting the decimal multiplication calculation.

A few points: I’ve deliberately kept the position of the mixed number or decimal and the integer in the same place within the calculation. In the second example, the first term of the calculation is a mixed number and the second is an integer, it is the same on the right-hand side of the equal sign. This is because I want pupils to focus on the digits changing position, and that only. Everything else is kept constant.

This is how I would communicate the model in the classroom to pupils:

I am multiplying a two decimal places decimal by an integer, so my equivalent calculation will have a two decimal places decimal and an integer.

I would get kids to draw the underlines to show me how their equivalent calculation would look. I have done this because learning the mathematical structure of the calculation is another thing a pupil needs to learn. This also reduces the likelihood of a future error caused by pupils by not knowing whether the calculation will have an integer and a decimal, or an integer or a decimal, and how many decimal places the decimal will have.

2) Introduce the steps to go from the problem to the equivalent calculation

a) Rewrite decimal place structure
b) Fill in the digits from the right side
c) CHECK: Multiply out both sides of the calculation to see that the result is the same

In my teaching, I write the generalised steps for the calculation pupils are about to do because they can see that even if I am rewriting a multiplication calculation with

• an integer and decimal or;
• two decimals or;
• decimals with the same number of decimal places or;
• decimals with a different number of decimal places

the generalised steps remain the same.

Generalising method stops is something that Siegfried Engelmann calls ‘Logically Faultless Communication’ which has been blogged about here.

A logistical thing, it also allows pupils to follow the steps better if they can flicker their eyes from the live worked example demonstrated by the teacher and the steps already written on the board.

3) Introduce the Example-Pupil attempt sequence

At this point, I then design the teacher-led worked examples and the pupil attempt questions. The problem type that the teacher goes through and the pupil goes through after seeing the teacher demonstrate the same problem type.

Here is the sequence:

Pupils aren’t evaluating the calculation. They are rewriting the calculation so their answers would look like this:

I have deliberately started with explicit examples meaning that the 2 decimal places decimal and two-digit integer have digits more than 0. Only from example 4, I have introduced a single figure of 5; this is because pupils must write the digit 5 as 0.05 where there is a zero between the decimal point and the 5. Starting with explicit features such as digits more than 0 in decimals and integers makes the change from one calculation to the equivalent one more clear.

Pupils then will complete a practice exercise where they will rewrite a decimal calculation but ending in the same result as the first decimal calculation.

They will then transition onto a filling in the blank exercise like shown below:

4) Equating Percentages: Introduce the Example-Pupil attempt sequence

I believe that I’ve now taught all the prior knowledge pupils need to complete this statement:

40% of 70=____% of _____

The prior knowledge is listed below:

• Writing a percentage in its equivalent decimal form
• Rewriting a decimal multiplication calculation in a visually different form but where the result is the same as the original calculation
• Decimal place value
• Evaluating decimal multiplication calculations

Given that the prior knowledge has been taught and committed to long term memory, there is only one new thing pupils need to learn to complete the statement. They need to write the percentage as a decimal and replace the ‘of’ with a multiplication sign. Then every other line of working is prior knowledge that has already been covered.

Here is the working out, and the new line of working is in bold:

40% of 70 = ____% of _____
0.40 × 70 = __.__ __×____
0.40 × 70 = 0.70 × 4

5) Comparing Percentage Increase/Decrease

What has been learnt here can also be used to complete statements like this below, and place the correct symbol between the statements <, > or =

6) Equating Percentage Increase/Decrease

The next level of difficulty that is accessible because of the prior knowledge taught is that pupils can evaluate the complete statement and equate the result to the incomplete statement.

Here the new knowledge being taught is dividing the result by the value in the incomplete percentage calculation.

However, what makes this more complex is that you must evaluate the complete calculation and equate the result to the incomplete calculation to then find the missing percentage. So pupils must:

1) Convert the decimal into a percentage
2) If the calculation is a percentage decrease, then subtract the percentage from 100%
3) If the calculation in a percentage increase, then subtract the percentage from 100%

Summary

I appreciate all the beautiful problems and mathematical tasks that Andrew provides on his website. They are incredibly rich in knowledge and can create a valuable mathematical conversation between teachers and pupils. There is a great deal of importance in teaching inquiry tasks because it is an opportunity to develop a flexible understanding of the subject. However, where I would disagree with the Inquiry Maths model would be how the mathematical task would be used and how the content would be taught to children. I think there is a great deal of prior knowledge that needs to be sequenced, so all future learning is supported by prior learning, and that only one new thing is being taught at each part of the sequence. This is to avoid cognitive overload.

I hesitate to have pupils taking responsibility for directing the lesson because each child will focus on a different aspect of the mathematical problem in front of them. More importantly, they will focus on various aspects depending on what knowledge they already have. It is inevitable that in any learning experience, each pupil will be starting at a different point based upon how much knowledge they have. However, when pupils direct the lesson that will result in more knowledgeable peers leading the lesson, and weaker pupils struggling to identify or learn aspects of mathematics that they need even to attempt the mathematical problem. By laying out the prior knowledge and sequencing it, you are allowing the more knowledgeable pupils to consolidate any existing knowledge they may already have, but you are letting the weaker pupils close any gaps they may have. Overlearning is an essential part of the learning process; it helps the more knowledgeable pupils to learn something to the point that they never make a mistake.

This attempt, I believe allows all pupils regardless of their knowledge gaps to have any missing gaps filled, but also allows all pupils to be able to access the mathematical problem.

#Mathsconf18: Atomisation Pt 2

Atomisation: Breaking down your teaching as you have never seen before…

On Saturday 9th March I delivered a workshop at the La Salle Mathematics Conference in Bristol. This blog post is a summary of some of the points made in the session.

For this workshop, I chose to look at an uncontroversial topic such as Angles on parallel lines. I think it’s uncontroversial because teachers know that it is an undeniable part of geometry that is commonly assessed. Also, I think it’s a topic which is taught with a poor sequence of examples. Lots of angle problems on parallel lines feels like an angle chase – when you find one angle then how can you find the other angle. However, in the process, pupils can’t have the rich mathematical discussion between the relationships of different angle facts on parallel lines. Part of teaching this topic effectively is dependent on the sequence in which the examples are organised.

When I started making the worked examples for this topic, I thought about the simplest application of angles on parallel lines for each angle fact and the most complex application. What I realised is that I could have spent hours or even days making lots of different worked examples. To avoid this, I thought of how I could cover all the myriad of complexities for each angle fact within the fewest number of worked examples.

The first step was to write out all the sub-tasks that I planned to teach:

–   Vertically Opposite Angles are Equal

–   Alternate Angles are Equal

–   Co-interior Angles sum to 180o

–   Corresponding Angles are Equal

–   Basic Angle facts on parallel lines

–   Angles on parallel lines – Algebraic

o   Simplified expressions equal to 180o or 360o

o   Simplified expressions equal to form an equation with unknowns on both sides.

After I listed the sub-tasks, I realised that I wanted to try a different pedagogical approach from the status quo approach. Historically, pupils are told that one unknown and one known are equal, and they are to accept it, and then identify the unknown and known angle pair which are equal in a similar looking example. Instead, I showed a selection of worked examples where the angles were of equal size, and I would state that these two angles are equal. I used Geogebra which is an online graphing programme where I would have an interactive set up so if I moved one of the parallel lines or the traversing line, the equal-sized angles would change from what they were before, but they would still be equal.

So, using a sequence of worked examples, I would

1)   Show the relationship with the position of angles and the angle fact

2)   Find the missing angle

3)   Find the missing angle by using a basic angle fact

NOTE: Diagrams aren’t drawn to scale here.

Vertically Opposite Angles are equal

Here is an example sequence for Vertically Opposite angles being equal

Show the Relationship

At this point, I transitioned to asking pupils in a whole class discussion the size of the unknown angles because they had seen the relationship between the position of two vertically opposite angles.

I deliberately used more challenging examples for pupils to identify one missing angle which is vertically opposite to one known angle. This is because I knew it is these type of angle problems, they would struggle with the most so I went through it with them so they would be successful when they would attempt similar issues independently. I felt that showing the relationship was explicit enough for pupils to attempt the simplest applications of identifying vertically opposite angles being equal.

The third section is using basic angle facts from the list below:

1. Angles in a triangle sum to 180o
2. Angles in a quadrilateral sum to 360o
3. Angles on a straight-line sum to 180o
4. Angles around a point sum to 360o

To either use the angle fact to determine one of the two vertically opposite angles or find one of the vertically opposite angles to then find another angle using one of the basic angle facts.

Here are some worked examples of this with an explanation:

I made Example 4 deliberately to highlight that the small triangle FCB and the large triangle KCE have the same angles.

In Example 5 and 6 I have included 2 parallel lines and 1 parallel line segment to allow Example 5 to include an opportunity to use the ‘Angles around a point sum to 360o’ fact. Similarly, in Example 6 I deliberately didn’t label the vertically opposite angle because I want pupils to start finding angles that aren’t labelled but are required to find another unknown angle.

In summary, I followed the sequence structure of:

1. Show the relationship
2. Find the missing angle using the angle fact given
3. Find the missing angle fact using basic angle facts to then determine angles using the new angle facts learnt.

Alternate Angles are equal

Here is a worked example sequence showing the relationship in positioning of Alternate Angles being equal

Now using the same geometric structure as the worked examples used to show the relationship of alternate angles, I’m asking pupils to find the size of the missing angle:

Here is a sequence of worked examples where basic angle facts have been interleaved:

Co – Interior Angles sum to 180o

Here is a worked example sequence for Co-interior angles summing to 180o

In Example 7 and 8, pupils can see that each triangle formed by both traversing lines and one of the parallel lines all have the same size angles.

Here are examples where pupils are asked to use the fact that co-interior angles sum to 180o to find the missing angle:

In the last two examples we can explore so many interesting mathematical patterns between co-interior angles and parallel lines. In Example 6, A = I and G = C

Here is the next section of worked examples where basic angle facts are being interleaved:

Corresponding Angles are Equal

Here is a worked example sequence for Corresponding angles being equal

Here is a worked example sequence where pupils are using the angle fact that corresponding angles are equal to find the missing angle:

Here is a worked examples sequence where basic angle facts are being interleaved with the angle fact that corresponding angles are equal:

Atomisation: Breaking down your teaching as you have never seen before…

On Saturday 9th March I delivered a workshop at the La Salle Mathematics Conference in Bristol. This blog post is a summary of some of the points made in the session.

This presentation was on the topic of atomisation. Recently, atomisation has created a lot of conversation on Twitter and at a couple of conferences. What is atomisation and is it the next fad?

Atomisation is the process of breaking down a topic into its sub-tasks.

Atomisation is a term coined by Bruno Reddy who taught me about it during my second school placement in 2014.

Atomisation is the starting point to every booklet I create in my role as Curriculum Advisor at United Learning. I sit down and list all the sub-tasks that I need to teach that is within the topic. An example of this for Perimeter is available here.

What are the benefits of Atomisation in respect to the big picture?

Atomisation is a process in which teachers can collaboratively identify the specific and detailed knowledge that pupils must know to be academically successful.

Identifying this specific knowledge means that pupils can learn as close to 100% of the domain of knowledge that they need to know. Exam boards and the national curriculum, truthfully, may provide high-level specificity of what needs to be taught, but not the finer details or detailed knowledge that goes into making pupils able to do a task.

For example, pupils are expected to know the following fact below:

Fraction x Fraction’s Reciprocal = 1

And upon reflection, this seems understandable in respect to multiplying fractions, index notation, ratio, proportion etc. but it is not necessarily explicitly stated in exam board specifications or Curricula.

The lack of detailed knowledge outlined by exam boards or textbooks results in teachers reducing the subject content being taught. And the source of information of what needs to be taught from children usually comes from exam tasks, and then the curriculum becomes an endless repetition of exam materials. Similar thoughts have been shared on Daisy Christodoulou’s blog.

Mathematics as a subject is vast. There is so much to teach a child in the space of their academic career. Here is a visual:

If the bubble represents 100% of the domain of knowledge that needs to be taught. It is the case that we teach small samples of that domain. We teach concepts and how they overlap with other concepts, and we also teach connections between different concepts. However, if we use exam boards and textbooks as our source of information of what needs to be taught in the curriculum, then we inevitably miss out teaching other parts of the domain.

The process of atomisation enables teachers to focus on the concept and identify the finer details that aren’t readily available. Atomisation allows the teacher to teach as close to 100% of the domain of knowledge.

What are the benefits of Atomisation in respect to the teaching process?

Identifying and sequencing all the sub-tasks and specific knowledge that needs to be taught for a concept increases the probability each child will be successful in the learning process.

This means that pupils will be able to appropriately and accurately respond to planned questions, or that they will remember what has been taught at that moment after a long period.

Teaching aspects of a concept that are usually overlooked undermine how successful each pupil will be in respect to their future learning.

For example, not teaching pupils that:

Fraction x Fraction’s reciprocal = 1

undermines a child’s ability to attempt questions like this:

Show that the ratio in the form 1:n can be written as 1: 1.5

If pupils get these questions wrong, then the incorrect inference is made. Instead, they must be retaught the topics of ratios from the beginning, but they only need to be taught this fact. And this is a fact has been mentioned several times in the Year 8 Curriculum so far. I’ve resourced a section on problem types like this for the most recent ratio booklet coming out.

Teaching all the sub-tasks of a concept and sequencing it in a logical sequence prevents pupils from being cognitively overloaded because each sub-task taught is being used or covered in future learning. In the teaching process, pupils will have committed prior knowledge in their long term memory so any future learning will occupy space in their limited working memory.

What are the costs of Atomisation?

The cost of atomisation is a valid one. The initial stages of teaching a concept take more time to cover the content. If we explicitly teach more sub-tasks than we would normally then we would have more time available to teach the rest of the curriculum. However, by not explicitly teaching all the sub-tasks of a concept it will result in the following consequences:

1. Reduce the probability of a child learning parts of a concept on the first attempt
2. Undermine a child’s ability to access future learning

Atomisation avoids these inadvertent consequences by:

1. Guaranteeing a greater likelihood a pupil will learn the sub-task on the first attempt
2. Increasing the probability of success when learning future content or more complex applications of the concept
3. Saving time in the future which would be inevitably spent re-teaching
4. Revealing sub-tasks that need to be taught that are usually overlooked in the curriculum.

In summary, Atomisation has improved my teaching and as a result my pupils learning more and remembering it a half term later, a year later etc. More importantly, it benefits the children that struggle with learning the most. Atomisation has allowed my weakest pupils learn more in less time, and in my small world, I believe that this is a path that I will continue down when creating resources.