
Pattern & Algebra A
Years 4  12


A Mathematics Centre Resource
You are encouraged to contribute to these notes. Comments from teachers or students, photos and student work samples will enrich the professional experience for all of us.
Picture Puzzles have multiple levels of success. They do not have to be 'finished'. They can be revisited and continued.
Odd, Even, Square and Triangle Numbers are the main study across these five investigations. The first level of each is deciding how to count the elements involved (squares in some cases, triangles in others). This usually involves the strategy of breaking the problem into parts. The next level is recognising a pattern in the count and encouraging prediction for all such cases. but the pattern needs to be checked. Can I check it another way? is asked or implied in each investigation. Further levels of challenge include representing the oral/written generalisation in symbols, seeing opportunities for introducing mathematical induction in the senior secondary school and extending the learning to related challenges such as exploring the difference between two squares.
Clearly, it is not expected that every student who begins one of these investigations will complete the whole slide show. However it is expected that every student who begins will be able to exit feeling they have successfully learnt. Perhaps more importantly, feeling that they would be able to return to the investigation to learn more.

Summary
Based on Task 108, How Many Squares?, students learn that a square made from unit squares also contains squares of other sizes. That is, the biggest square and the unit squares are obvious, but you have to learn to 'see' the other size squares. Students learn to count all the squares in a picture and discover that there is a way to predict the number of squares of each size within any large size square.
As an extra challenge, they are asked to check their prediction another way. This is not only application of a mathematician's question (see Working Mathematically), but learners who can check their work in more than one way are becoming independent. Independent learners allow teachers to change the way they use their time.
The final part of this Picture Puzzle encourages generalisation to any size square and the description of this generalisation in algebraic symbols.

Answers

 Size 1 = 9
 Size 2 = 4
 Size 3 = 1
 Total = 14


 Size 1 = 16
 Size 2 = 9
 Size 3 = 4
 Size 4 = 1
 Total = 30


 Size 1 = 25
 Size 2 = 16
 Size 3 = 9
 Size 4 = 4
 Size 5 = 1
 Total = 55


By now students might see that the total will be the sum of the first 8 square numbers:
 Total
= 8^{2} + 7^{2} + 6^{2} + 5^{2} + 4^{2} + 3^{2} + 2^{2} + 1^{2}
= 64 + 49 + 36 + 25 + 16 + 9 + 4 + 1
= 204


This slide is included to see what it brings out from the students. The outcome will depend on how they have been calculating the answers to date. Possible ways are:
 Draw and carefully count.
 Recognise and use the sum of the squares.
 Notice an addition by odd numbers pattern and apply it up to Size 8.
and there may be other approaches that come to light.

In the next section of the slide show, the 10 x 10 challenge is included to highlight the tedium of some of these methods and encourage thought about a more effective alternative.

If students, especially younger students, can say that F will equal the sum of the square numbers from 1 up to S, they have generalised and hence 'done algebra'. Creating a formula for the answer is not essential but it could be that students will see their words represented in:
 F = S^{2} + (S 1)^{2} + (S  2)^{2} + ... + 3^{2} + 2^{2} + 1^{2}
It takes considerable sophistication, worthy of Year 12 students, to show that the simplest algebraic expression for the sum of the square numbers is:
 F = ^{S(S + 1)(2S + 1)} / _{6} where S is the largest size square.
Maths300 members could explore this in Lesson 138, Pyramid Puzzle & Other Algebra Excursions. It is also important at this level to notice that none of the demonstrations in this Picture Puzzle is a proof. Asking the mathematician's question Can I check this another way? in senior secondary school can open the door to introducing Mathematical Induction and its role as a tool for proof of all cases.


To learn more about this Picture Puzzle click the photo to visit the Task Cameo. 


Summary
Based on Task 48, How Many Triangles?, students learn that an equilateral triangle made from unit equilateral triangles also contains triangles of other sizes. That is, the biggest triangle and the unit triangles are obvious, but you have to learn to 'see' the other size triangles. To add to the difficulty, some have their points up and some have their points down. Students learn to count all the triangles in a picture and discover that there is a way to predict the number of triangles of each size within any large size triangle.
As an extra challenge, they are asked to check their prediction another way. This is not only application of a mathematician's question (see Working Mathematically), but learners who can check their work in more than one way are becoming independent. Independent learners allow teachers to change the way they use their time.
The final part of this Picture Puzzle encourages generalisation to any size triangle and the description of this generalisation in algebraic symbols.

Answers
While this puzzle may have a similar structure to How Many Squares?, it is more difficult spatially. The tricky part is the Point Up and Point Down triangles and it isn't until the mathematician's strategy of breaking the problem into smaller parts is applied, that patterns are revealed. In fact, it is not only breaking the problem into different size triangles that matters, but breaking each of these further into Point Up and Point Down triangles.

Easy to start ... in fact the answer's given.
Size 1 Triangle
 Size A (1 unit per side) = 1 (1, 0)
Total = 1


But the very next level introduces two new elements  different size triangles and different orientation triangles.
Size 2 Triangle
 Size B (2 units per side) = 1 (1, 0)
 Size A (1 unit per side) = 4 (3, 1)
Total = 5


And then it becomes progressively more complicated.
Size 3 Triangle
 Size C (3 units per side) = 1 (1, 0)
 Size B (2 units per side) = 3 (3, 0)
 Size A (1 unit per side) = 9 (6, 3)
Total = 13
Students are unlikely to realise the importance yet, but they may notice that Sizes C and B are both Point Up only and Size A is made up of both Point Up and Point Down triangles. 6 up and 3 down, hence the numbers in brackets above.


Size 4 Triangle
 Size D (4 units per side) = 1 (1, 0)
 Size C (3 units per side) = 3 (3, 0)
 Size B (2 units per side) = 7 (6, 1)
 Size A (1 unit per side) = 16 (10, 6)
Total = 27


Size 5 Triangle
 Size E (5 units per side) = 1 (1, 0)
 Size D (4 units per side) = 3 (3, 0)
 Size C (3 units per side) = 6 (6, 0)
 Size B (2 units per side) = 13 (10, 3)
 Size A (1 unit per side) = 25 (15, 10)
Total = 48


Any student who has correctly counted the triangles of any size within triangles up to Size 5 has already been successful. Going further is not straightforward and it is unlikely to be possible to predict Size 8 using the data to this point. So, students might decide to get data for one more triangle.
Size 6 Triangle
 Size F (6 units per side) = 1 (1, 0)
 Size E (5 units per side) = 3 (3, 0)
 Size D (4 units per side) = 6 (6, 0)
 Size C (3 units per side) = 11 (10, 1)
 Size B (2 units per side) = 21 (15, 6)
 Size A (1 unit per side) = 36 (21, 15)
Total = 78

Now we have a little more to work with, students might notice:
 Point Up triangles follow the pattern of triangles numbers ( 0, 1, 3, 6, 10, 15, 21, ...)  zero would be the case of a Size 0 triangle.
 Size A triangles are always a total which is a square number.
 The Size A total is also the sum of two consecutive triangle numbers.
 Point Down triangles seem to 'misbehave', but closer examination suggests that:
 for even Size triangles starting from Size A they are every second triangle number back to 1 and the rest are zero.
 for odd Size triangles starting from Size A they are every second triangle number back to 0 and the rest are zero.
These insights allow us to predict Sizes 7 & 8 and beyond.
Size 7 Triangle
 Size G (7 units per side) = 1 (1, 0)
 Size F (6 units per side) = 3 (3, 0)
 Size E (5 units per side) = 6 (6, 0)
 Size D (4 units per side) = 10 (10, 0)
 Size C (3 units per side) = 18 (15, 3)
 Size B (2 units per side) = 31 (21, 10)
 Size A (1 unit per side) = 49 (28, 21)
Total = 118

Size 8 Triangle
 Size H (8 units per side) = _ (_, _)
 Size G (7 units per side) = _ (_, _)
 Size F (6 units per side) = _ (_, _)
 Size E (5 units per side) = __ (_, _)
 Size D (4 units per side) = __ (_, _)
 Size C (3 units per side) = __ (_, _)
 Size B (2 units per side) = __ (_, _)
 Size A (1 unit per side) = __ (_, _)
Total = 170 

And we could check our predictions by making or drawing triangles and counting  which would be rather tedious of course. Except that by now we can see why the Point Up triangles must follow the triangle numbers pattern. That's the way they are laid out. Concentrate on the peach from the top and you see 1, then 1 + 2, then 1 + 2 + 3, then 1 + 2 + 3 + 4 and so on, which are the triangle numbers.
Questions remain of course, such as:
 How do we explain counting only every second triangle number for Point Down triangles?
 How do we explain the zeros on the end of that sequence?
 How do we find the total for any size triangle? For example Size 1000 could be worked out using the insights above, but it would take quite a while  although a spreadsheet might help? In fact, if we plot Size against Total in a spreadsheet, or simply on paper, it appears that the points lie on a curve which suggests, a formula for the total does exist. The shape even suggests it is probably cubic.

Tackling this last question will at least involve summing triangle numbers because of the Point Down triangles. Discovering (and proving) that:
Sum of Triangle Numbers to n = ^{n(n + 1)(n + 2)} / _{6}
is a challenge for senior secondary students and doing so still doesn't solve the How Many Triangles? problem.

To learn more about this Picture Puzzle click the photo to visit the Task Cameo. 


Summary
Based on Task 111, Square Numbers, students learn that square numbers can be seen as created in two ways. The same number of rows as there are unit squares in one row, or as the addition of an odd number to the previous square. To encourage independent learning, the puzzle focuses on the mathematician's question Can I check this another way?. The additional challenge in the problem uses the idea that if a square can be made from a previous one by adding, then the amount added is the difference between the two squares. This leads to a visual demonstration of the difference of two squares rule.
For one teacher's approach to using this Picture Puzzle, read the brief article Picture Puzzles in Sweden which is based on Square Numbers being used as a whole class lesson in a BYOD school.

Answers

Students are expected to notice two things  a square number is a number multiplied by itself (which they may already know) and square numbers start at 1 and grow by the addition of odd numbers. So:
 S_{5} = 5 x 5 = 25
 S_{5} = 1 + 3 + 5 + 7 + 9 = 25
(some students might recognise the quick method of add the ends, add the second ends and so on)


 S_{20} = 20 x 20 = 40
 S_{20} = 1 + 3 + 5 + 7 + 9 + ... + 31 + 33 + 35 + 37 + 39 = 400
(and now the quick method has merit!)
Of course, if you knew the 19th square number, you would only have to add the 20th odd number to it to find the 20th square number. 

The explanation needs to recognise that because the L is being added to the sides of a square, it doesn't matter how long the sides are we will be adding the same length twice and then one extra tile is needed in the bottom right corner to complete the Lshape. So, if the side length is odd, 2xO + 1 = O and if the side length is even, 2xE + 1 = O. In fact 2N + 1 is odd for every whole number value of N, which is the definition of an odd number.
You might ask the students to sketch a starting shape that would cause an added Lshape to be an even number. 

If each square number can be made from a previous one by adding a sequence of odd numbers, then the sequence of odd numbers, the peach coloured squares in this picture, must be the difference between the two square numbers. In this case it is reasonably easy to see that S_{5} = S_{3} + (7 + 9), so the difference is 16. But suppose the two squares were S_{96} and S_{39}. Working out the difference would take a considerable amount of calculation. Could there be a quicker way?


Multiplication is quicker than addition and rectangles are the key to multiplication. If only the peach squares formed a rectangle rather than an Lshape.
Mmmm, perhaps they can.
The dissection shown in the slide sequence following this slide, turns the Lshaped difference into a rectangle with the same area, which is then easy to calculate. The new rectangle will have (5 + 3) rows with (5  3) squares in each row.

So, 5^{2}  3^{2} = 5x5  3x3 = 16 ...and... S_{5}  S_{3} = (5 + 3) x (5  3) = 8 x 2 = 16. Which method would you rather use to calculate 96^{2}  39^{2}.

To learn more about this Picture Puzzle click the photo to visit the Task Cameo. 


Summary
Based on Task 51, Staircase, students build a staircase of towers and are led to see the developing staircase as a triangle. The number representing the total of squares up to each step in the staircase is introduced as a triangle number. Students are encouraged to calculate these triangle numbers in more than one way. Then they discover links between the triangle numbers and rectangle numbers (which are the basis of multiplication). Two additional visually based challenges highlight some remarkable properties of triangle numbers.


At this point students may be thinking of the T numbers as towers and the next one after the four in the previous slides must therefore be a tower of 5. We can't predict how students will respond to the request to predict T_{5}, but one way is to sum 1 + 2 + 3 + 4 + 5 = 15.
Another might be to count all the squares one at a time (additive rather than multiplicative thinking and, perhaps, an indication of lack of confidence).


But now they are challenged to think again. Not because they are wrong, but because a mathematician asks the question Can I check this another way?. Students might:
 Notice there is a middle tower and shift squares from taller towers to smaller ones making all the towers 3 units high. This would be the calculation: 5 groups of 3 = 15.
 Notice that the 4 tower only needs 1 to make it 5 and the 3 tower only needs 2 to make it 5 thus making 3 towers of 5. This would be the calculation: 3 groups of 5 = 15.
 Notice that if the numbers are written out as the series 1 + 2 + 3 + 4 + 5 the first and last make 6. The second and second last make 6 and there is a 3 in the middle  sometimes called 'rainbow facts'. So, 2 x 6 + 3 = 15.


Asking T_{6} will encourage the students to think again about their alternatives. Now there is no middle tower. What happens to each of the approaches above in this case?


Here a new term is introduced. Triangle numbers.
Can the students explain what a triangle number is?
Their explanation should be recorded in their journals.


Calculating the 20th triangle number (210) gives students an opportunity to test their previous calculating methods again ... and perhaps think about whether there is yet another way.


This slide hints that there is.
 Make a copy of the staircase (size n).
 Join the two staircases to make a rectangle.
 Calculate the height and width of the rectangle.
 Halve the result to find T_{n}.
Which can be represented as T_{n} = ^{n(n+ 1)} / _{2}




The first extra challenge of this Picture Puzzle is an animation sequence that shows how the rectangle above can be transformed to make the shape here. The boundary is a square and it encloses a square number of small tiles. What the students might realise is that:
(8 x Triangle Number) + 1 = Square Number
Apart from being a pretty representation of a mathematical fact, this fact is critical in the deeper exploration of Task 18, Same or Different at some time around Year 10 or beyond.

In this case the transformation began with T_{3} and ended with S_{7}. The students are then asked to explore transforming other size triangle number staircases in the same way. They also find the number equivalent of their constructions, for example, in the case shown, 8 x 6 + 1 = 49 = 7^{2}.
These explorations can lead to asking the question:
 If I tell you any triangle number, can you tell me which square number it will transform to using this construction?

No doubt many things can be made with 9 triangle numbers and one square, but a gorgeous one is shown below. The blue staircase indicates the 9th copy of the chosen triangle number. The other colours are reminiscent of the 8T + 1 challenge. In fact the first challenge can be transformed in a few steps and the blue staircase added to make this new triangle number.
 Can the students explain the transformation steps?


(Hint: Cover the blue triangle number for a moment. The bottom section is almost the 8T + 1 model. How has the 8T + 1 model been transformed to get the almost model?)
If we use this model as a template, this construction shows us that triangle numbers are selfreplicating.
(Squares are selfreplicating too. Their template is to organise 4 squares into a square, then this square is organised into a bigger square and 4 of those into a bigger one and so on.)
In this case, T_{3} has generated T_{10}. Which means the number 6 generates the number 55. But T_{10} is a triangle number so 9 copies of it plus 1 will give another triangle number. 9 x 55 +1 = ??. Is the missing number a triangle number?
 What happens if we begin with a different triangle number, eg: T_{1} or T_{2} or...?
 If I tell you any triangle number can you tell me the set of triangle numbers that derives from it?
 If I tell you any triangle number can you tell me the smallest triangle number it derives from?


To learn more about this Picture Puzzle click the photo to visit the Task Cameo. 


Summary
Based on Task 186, Tetrahedron Triangles, students learn more than one way to count the unit triangles within an equilateral triangle. In the additional challenge they discover that certain sizes of these equilateral triangles of triangles can be thought of as a net that folds into a tetrahedron. They are challenged to state a generalisation linking the size of the tetrahedron and the triangles in its surface area  in effect, measuring the surface area with unit triangles.


To this point students are being encouraged to see that the number of unit triangles can be calculated by adding an odd number to the previous total, exactly the same process as in Square Numbers above. They are also likely to see that the number of triangles can be calculated by squaring the size of the triangle.


Now students can put the two processes into action. If N = number of triangles:
 N = 1 + 3 + 5 + 7 + 9 + 11 = 36
 N = 6 x 6 = 36


To this point students are being encouraged to see that the number of unit triangles is made up of the peach and the green sets. Further that each colour can be seen as a sum of natural numbers with the green (point down) set being one term less than the peach. Alternatively, if the students are familiar with triangle numbers, they can see the total as the sum of two consecutive triangle numbers.


Using this second way of looking at the pattern, again with N = the size of the big triangle:
 N = (1 + 2 + 3 + 4 + 5 + 6) + (1 + 2 + 3 + 4 + 5) = 21 + 15 = 36
 N = ^{(6 x 7)} / _{2} + ^{(5 x 6)} / _{2} = 21 + 15 = 36
The second of these uses the approach above in Staircases, which showed that the nth triangle number T_{n} = ^{n(n+ 1)} / _{2}.
Students might also add the two series like this:
1 + 2 + 3 + 4 + 5 + 6
+ 1 + 2 + 3 + 4 + 5
= 1 + 3 + 5 + 7 + 9 + 11
and hence show that they realise why the first approach above was an addition of odd numbers.


Although there are several ways to see the pattern, the simplest algebraic form for calculation is:
N = S^{2}, where S is the size, or side length, of the large equilateral triangle.
The alternative methods are important because they provide tools for students to check their own work and hence become more independent learners.


The tetrahedron has four equilateral triangle faces, so if N is the number of triangles and S is the size (or side length) of the tetrahedron, then N = 4S^{2}.
(Note: We have moved to a new problem and are now thinking in 3 dimensions. S in this problem and S in the previous paragraph are not being used to represent the same thing.)
Assuming we are building with unit triangles, for example using MiniGeofix to make the net, if I tell you any size equilateral triangle, what can you tell me about making a tetrahedron from it?
Suppose students draw and cut out an equilateral triangle that doesn't have whole number side lengths. (Many drawing programs can draw one of these in an eye blink, but can students construct one using a ruler and compass, or ruler and protractor?) Does the work above help them decide how to fold the triangle to make a tetrahedron? What will be the surface area of the tetrahedron in unit triangles?


To learn more about this Picture Puzzle click the photo to visit the Task Cameo. 

