In this unit students explore patterns and relationships among the circle numbers and box numbers that form different arithmagons. Students identify and use rule based patterns to solve triangular arithmagons and square arithmagons and use linear equations to solve triangular arithmagons. They establish and prove algebraically the general rules for solving arithmagons.
 Devise rules based on numerical patterns to solve triangular arithmagons.
 Explain the condition for the solution of any square arithmagon.
 Form and use linear equations to solve triangular arithmagons.
 Develop proofs of rules and conditions for the solution of arithmagons.
This unit is based around exploring ‘arithmagons’. Arithmagons are polygons with a circle number on each vertex and a box number on each side such that each box number is the sum of the two circle numbers adjacent to it.
In Session 1, students must notice that the sum of the two circle numbers on either side of a box number is equal to the box number. This is the basis of all the arithmagons throughout the unit. Students are likely to notice that in Questions 1 and 2 of Task 1, the box numbers form a sequence comprising three consecutive numbers of which two are odd and one is even, (one odd and two even leads to fractional circle numbers). They are also likely to notice that the circle numbers are also consecutive but in the opposite sense to the box numbers. So when the box numbers form a clockwise sequence, the circle numbers form a counterclockwise sequence and vice versa. Noticing such features is key to generalising arithmagons.
Students are likely to use a ‘Trial and Improvement’ strategy for finding circle numbers. Encourage them to use such a strategy but to also look for a short cut as they do so. They may then notice that the circle number opposite the even box number is half that box number. This rule is developed in Task 2. Once noticed, such a rule tends to be generalised for all triangular arithmagons. However it only produces integer values when the box number is even and when the box numbers form a linear sequence. This provides a valuable lesson for all students as they seek to develop rules arising from pattern spotting. Rules that seem generalisable may not be so. For example, the box number sequence 38, 48, and 68 in Question 3c of Task 2 is not linear. The circle number opposite 48 is 29, not 24 which application of the rule may suggest. Questions 4 and 5 of Task 2 provide an opportunity to deepen students’ understanding of this first rule; when it works and when it doesn’t.
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In Session 2, students devise a rule that works for all triangular arithmagons, not just those where the box numbers form a simple linear sequence. The idea of a centre number is used to help students see links between box numbers and circle numbers in triangular arithmagons. The dotted lines below are intended to show the links.
1 + 6 = 7, 2 + 5 = 7, 4 + 3 = 7
While all triangular arithmagons can be solved using this relationship, it is also interesting to see the effect of firstly having the centre number equal to a box number and secondly having the centre number less than a box number.
When the centre number is the same as a box number, one circle number is zero. Notice that in this case, the sum of the two smaller box numbers is equal to the third box number. When the centre number is less than a box number, a negative circle number results. In this case the sum of two smaller box numbers is less than the third box number.
While it is not intended that students at this stage use algebra to write the rule for triangular arithmagons, students should be able write the rule in words. Using algebra however gives the following result.
The centre number can be represented by .
Sowhich leads to
In Session 4, students use a different approach to develop this result.
In Session 3, attention shifts to square arithmagons that have four box numbers and four circle numbers. In Question 1, students will see that the box numbers are identical but any change in a circle number leads to a different solution. In fact there is an infinite number of arithmagons that can be made since the only condition that must be satisfied is that the sum of opposite box numbers must be the same. For example, there is an infinite number of solutions with the box numbers shown in the following square arithmagon.
In Session 4, students use algebra to solve triangular arithmagons. In Question 1, the algebraic expressions for the circle numbers on either side of the box number, 30, are (25– c_{1}) and (17 – c_{1}). So the equation formed is (25– c_{1}) + (17 – c_{1}) = 30. This simplifies to 25 + 17 – 2x = 30 which gives 42 – 30 = 2c_{1 }so that the number in c_{1} is 6. The circle numbers are therefore 6, 25 – 6 = 19, and 17 – 6 = 11. In Question 3, the generalised algebraic expressions for the circle numbers on either side of the box number b_{2} are (b_{3} – c_{1}) and (b_{1} – c_{1}). So the equation formed is (b_{3} – c_{1}) + (b_{1} – c_{1}). = b_{2}. This simplifies to b_{3} + b_{1} – 2c_{1} = b_{2} which gives b_{3} + b_{1} = b_{2} – 2c_{1} so that
In the arithmagon opposite, c_{2} = b_{1} – c_{1} and c_{3} = b_{3} – c_{1}
These results are of course the same as the result derived from the ‘centre’ number approach in Session 2 (see above.)
In Session 5, students develop and use the general results for different arithmagons. The results for arithmagons that have an odd number of sides are shown below. In each arithmagon, circle numbers are in a cyclic sequence c1, c2, c3, … . Also, b1 is the box number that follows circle number c1; b2 is the box number that follows circle number c2, and so on.
Type of arithmagon 
Number of sides 
Rule for calculating 
triangular 
3 

pentagonal 
5 

heptagonal 
7 
These rules can be expressed generally as, ‘Total the values in the oddnumbered boxes beginning with the box immediately following the circle number being determined. Then subtract the total of the values in the evennumbered boxes. Divide the result by two’.
The results for arithmagons that have an even number of sides are as follows.
Type of arithmagon 
Number of sides 
Arithmagon can be solved when 
square 
4 
b_{1} + b_{3} = b_{2} + b_{4} 
hexagonal 
6 
b_{1} + b_{3} + b_{5} = b_{2} + b_{4} + b_{6} 
octagonal 
8 
b_{1} + b_{3} + b_{5} + b_{7} = b_{2} + b_{4} + b_{6} + b_{8} 
In order that these arithmagons can be solved, ‘the total of the values in the oddnumbered boxes must be the same as the total of the values in the evennumbered boxes’. Arithmagons with n sides, where n is even, and which can be solved, have an infinite number of solutions since there is an infinite number of ways that two totals that are equal can each be made from n/2 values.
Notice that the ‘centre’ number approach used for solving triangles in Session 2 can also be applied for other arithmagons. For example, the circle number, 8 in this pentagonal arithmagon can be found by subtracting the box numbers 5, and 9 from the centre number, 22. So for a pentagonal arithmagon, with circle numbers represented successively by c1, c2,… c5, and box numbers represented successively by b1, b2,… b5, the centre number can be represented by
.
The circle number c1, can then be found from
So which which is the same result derived using the approach in Task 10.
Copymaster 1 (triangular arithmagons)
Copymaster 2 (square arithmagons)
Copymaster 3 (pentagonal arithmagons)
Copymaster 4 (hexagonal arithmagons)
Copymaster 5 (heptagonal and octagonal arithmagons)
Each session in this unit comprises a sequence of tasks or challenges. It is suggested that students have an opportunity to engage in each task by themselves before they begin sharing their ideas with other students or with their teacher. The role of the teacher is to first listen carefully to students as they try to explain their thinking to others and then interact with the students only when they or their peers appear to have reached an impasse. Such interactions might then be to help clarify students’ thinking or even redirect thinking where necessary.
Notes have been included to provide mathematical background including solutions, as well as on aspects that are likely to pose some difficulty for students. The exact nature of any interactions will depend on the situation and on the thinking progress that has been made.
Session 1
In this session, students are likely to use a ‘trial and improvement’ strategy to find circle numbers. Encourage students to look for short cuts (see Task 2) that result in correct solutions. The tasks are designed to draw students’ attention to sets of box numbers for which the short cut works and doesn’t work. Identifying arithmagons that can be solved but for which the short cut doesn’t work can encourage further exploration into new and more robust short cut methods. Begin by introducing students to the rules for arithmagons, ie. that each box number is the sum of the two circle numbers adjacent to it.
Task 1

Look carefully at the completed triangular arithmagon below and then complete the unfinished one by writing the numbers that go in the circles. Show how you worked out the circle numbers.

Find the circle numbers for these arithmagons. Show how you know that your circle numbers are correct.
Task 2

Jeremy uses these arithmagons to show a rule for working out circle numbers. Show how Jeremy’s rule works.

Check that Jeremy’s rule works for the following arithmagons.

Now decide if Jeremy’s rule works for each of these arithmagons.

Use your own three box numbers to make an arithmagon (Copymaster 1) that can be solved using Jeremy’s rule.
 Now make an arithmagon that cannot be solved using Jeremy’s rule. Explain why it cannot be solved this way.
Session 2
Begin by encouraging students to look for a relationship that links the three box numbers with the centre number. If no one sees the relationship suggest they might add the three box numbers. Notice that the centre number is one half the sum of the three box numbers and is also equal to the sum of the three circle numbers. Students sometimes have difficulty reasoning that this occurs because each box numbers is the sum of the two adjacent circle numbers, so that the sum of the box numbers must be halved in order to calculate the centre number.
Whenever students are asked to see if they can find a rule, ask them to try writing the rule in their own words. Use their efforts as the basis of class discussions to clarify the meaning of the rule for everyone in the class.
Task 3

Find the circle numbers for these arithmagons.

Each arithmagon above has a centre number in the shaded circle that is related to the three box numbers and also to the three circle numbers. Figure out how these relationships work.
 Calculate the centre number for each arithmagon below. Then use it to help complete the arithmagon.


Write a rule for the centre number using b1, b2, and b3 for this arithmagon.
 Write a rule that can be used to work out the circle number, c1, for the triangular arithmagon above.
 Write a rule to work out c1, and a rule to work out c3.
 Make up several different triangular arithmagons, (Copymaster 1), with any three box numbers. Check that your rule for finding the circle numbers works.

Task 4
 Make a triangular arithmagon, (Copymaster 1), that has one box number the same as the centre number.
 What do you notice?
 Make up several different triangular arithmagons that have one box number the same as the centre number.
 Write a rule that tells you if a triangular arithmagon has one box number that is zero. Check that your rule always works my making your own triangular arithmagons.
 Make a triangular arithmagon (Copymaster 1) that has a box number greater than the centre number.
 What do you notice?
 Write a rule that tells you if an arithmagon has a circle number that is less than zero.
 Make up several different triangular arithmagons that have one box number greater than the centre number. Check that your rule works.
Session 3
When attention turns to square arithmagons, it is likely that some students will endeavour to merely extend their ideas related to triangular arithmagons. They will find however that they will need to take a fresh look at this new arithmagon. Interestingly the first square arithmagons they encounter all have multiple solutions – in fact, whenever a square arithmagon can be solved, there is an infinite number of solutions when negative or fractional values are permitted. It might be interesting to see if students spontaneously widen their perspective to include negatives etc. Square arithmagons can be solved when opposite pairs of box numbers have the same total.
Task 5

Find the missing circle numbers for these square arithmagons. One circle number in each arithmagon is shown.
 Make three more square arithmagons, (Copymaster 2), that have the same box numbers as in Question 1.
 How many different square arithmagons do you think can be made with the box numbers as above? Explain your reasoning.
Task 6
 Decide which square arithmagons below can be solved and which ones cannot.
 Val changes the box number 6 to 4. She claims that the new square arithmagon can now be solved. Check to see if she is correct.
 For any arithmagon in Question 1 that cannot be solved, change just one box number so that the new arithmagon can be solved.
 Write a rule that tells whether square arithmagons can be solved.
Session 4
In this Session, students have an opportunity to work with algebraic symbols to formulate and solve linear equations and to prove results for triangular arithmagons. Notice that rather than using algebraic symbols, x and y etc for box numbers or circle numbers, it is likely to be more helpful to use b1, b2 and b3 etc, to represent successive box numbers and c1, c2 and c3 etc to represent successive circle numbers. Students may need help to understand how this system of subscript symbols work.
It is also likely that students who will get most out of this and the next session will be those who can work accurately with symbol operations and manipulations.
Task 7
 Maraea stars with the arithmagon on the left and then puts 25 – c1 for one circle number and 17 – c1 for the other. Show how Maraea worked out these algebraic expressions.
 Maraea wrote the equation, 42 – 2c1 = 30 for her arithmagon.
 Show how she got this equation and then solve it to find the value for c1.
 Find the values for the three circle numbers in Maraea’s arithmagon.
 Use algebra to solve the following triangular arithmagons.
Task 8

In this arithmagon, the box numbers are represented algebraically by b1, b2 and b3


Write an equation for the side of the arithmagon represented as:
 Find a rule that expresses x in terms of b1, b2 and b3.
 Explain in your own words how the rule works.


Use the rule from Question 1 to find the three circle numbers for each of the following arithmagons.


Make up three triangular arithmagons with your own box numbers (use any numbers including negative numbers).
 Use your algebraic rule from Question 1 to find the circle numbers.

Session 5
In this Session, students extend their algebraic work to square, pentagonal, hexagonal, heptagonal, and octagonal arithmagons. In other words they examine the generality of solutions for arithmagons.
Task 9

Show how Albert got different algebraic expressions, b2 – (b1 – c1) and b3 – (b4 – c1) to represent one of the circle numbers in these arithmagons.

 Devise a rule for square arithmagons by simplifying the equation, b2–(b1–c1) = b3–(b4–c1).
 In your own words, explain how the rule works.

 Make up three square arithmagons with your own box numbers. (Use any numbers including negative numbers but make sure you choose numbers that allow the arithmagon to have a solution  see Task 6.)
 Use your algebraic rule from Question 2 to find circle numbers that work for the arithmagon.
Task 10
 Find the circle numbers for each pentagonal arithmagon below. One circle number is shown for each arithmagon.

Write algebraic expressions for the circles in the following arithmagons.

Devise rules to find the values of the circle numbers for pentagonal and hexagonal arithmagons given the values of the box numbers, b1, b2, b3, etc.
 Use your algebraic rules to find the values of the circle numbers for the following arithmagons.
Task 11

Write and explain a rule to find the value of the circle number immediately ahead of the first box number in a sequence of box numbers, b1, b2,… b7, for a 7sided (heptagonal) arithmagon, (Copymaster 5).
 Make a heptagonal arithmagon and check that your rule works.
Task 12
 Write and explain a condition that an 8sided (octagonal) arithmagon (Copymaster 5) can be solved.
 Make an octagonal arithmagon and check that your condition works.
Task 13
 Write and explain a rule to find the value of the circle number immediately ahead of the first box number in a sequence of box numbers, b1, b2,… bn, for an nsided arithmagon where n is odd.
 Write and explain a condition that an nsided arithmagon where n is even can be solved.