LESSON STUDY VIGNETTE: THE CATHEDRAL PROBLEM
For this lesson study, we decided to focus on one specific problem that was provided to the teachers as an opening problem for the day, the cathedral problem that is adapted from Burns, S. (2003):
Text Box 9.3 A Math Happening 9b: The Cathedral Problem
While building a medieval cathedral, it cost 37 guilders to hire 4 artists and 3 stonemasons, or 33 guilders for 3 artists and 4 stonemasons. What would be the expense of just 1 of each worker?
The following sections include the data analysis of teacher thinking that went into solving this problem. Each group was asked to create a poster representing their solution and each teacher was also asked to reflect on how they participated in the problem-solving process and how they would take this problem back to their classroom to present to their students. Photographs of all of the posters created by the teacher groups of the cathedral problem were taken and arranged in order on the day of completion.
Topics included reasoning up and down, direct and inverse thinking, unitizing, and, ratios and proportional thinking. The data from the posters were analyzed for content, connections between concepts, and any possible differences related to the time already spent in the seminar. Data were also analyzed from the teacher reflections for common themes as well as individual perspectives. First, we will present the analysis of the poster artifacts based on what the teachers in the respective groups shared followed by our analysis of the corresponding teacher reflections.
Poster Proofs to Document Visible Thinking
The first group, Figure 9.4 (left), argued that if one artist and one stonemason together made $11, then the total for three of each would be $33. However, they reasoned that because we know that $33 is enough to pay those six workers plus another stonemason, then one stonemason and one artist must together make less than $11. This group then used the guess-and-check method. They first assumed that the total for one artist and one stonemason was $8. They tried the combination of $1 for the cost of one artist and $7 for the cost of one stonemason ($8 total); however, they discovered that the total for 3 artists and 4 stonemasons was less than the needed $33.
Figure 9.4 Poster artifacts showing guess and check, tabular/pictorial, linear addition. Source: Authors.
They tried other combinations but saw that the total cost was decreased; so, they abandoned the idea of an $8 total. They then assumed that the total for one artist and one stonemason was $9. However, their starting guess for the cost of one artist was $2, not $1. They found that the combination of three artists at $3 each and four stonemasons at $6 did total the needed $33. However, when they used these amounts in the second scenario, they found that it did not work: four artists at $3 each and three stonemasons at $6 did not total the needed $37. They then assumed that the total for one artist and one stonemason was $10. Using the same logic, starting at $1 per artist and $9 per stonemason, then $2 per artist and $8 per stonemason, etc., they arrived at a solution of $3 per artist and $7 per stonemason, which they demonstrated would satisfy both requirements.
In all three guess-and-check calculations, they assumed that the artists would earn less than the stonemasons would; so, they arrived at the correct figures for the solutions but had the assignments to the two types of workers backward. They showed that four artists at $3 each and three stonemasons at $7 each would total $33. However, the original question stated that the cost of $33 applied to three artists and four stonemasons. So, although their logic was correct, they made a minor error in the interpretation.
The second group, Figure 9.4 (middle), presented tabular and pictorial representations of the two scenarios. There are also several indications that they chose the values of seven and $3 for the costs of the two types of workers, but there is no clear explanation of how they arrived at that conclusion. At the top left of the poster, four rows of seven marks each are made to represent the artists; each group of seven is circled, showing that the cost of each of four artists is $7.
To the right, there are three rows of three marks each, representing that each of three stonemasons earns $3 each. There is no indication that any values other than the correct solution were considered. There is also no indication of exactly how the correct values were calculated. However, at the bottom of the poster, the expressions written seem to indicate that an algebraic solution using two simultaneous equations was employed.
The third group, Figure 9.4 (right), presented a linear addition solution using symbols to balance two equations. The top of the poster shows the two scenarios and the bottom of the poster shows the addition of these two. The top left of their poster shows 3 tens (squares) and 7 units (circles) to represent $37. On the right of the equal sign, there are four A’s, for artists, and three S’s, for stonemasons. Directly below that is a similar configuration to represent a $33 cost for three artists and four stonemasons. The lower left portion of the poster shows 6 tens (squares) and 10 units (circles) to represent $70. This is the addition of the tens (squares) and units (circles) from the two equations on the top of the poster.
On the right side, there are seven A’s and seven S’s, which are the sum of the A’s and S’s from the two equations. This group reasoned that they now had a total of seven artists and seven stonemasons and a total of $70. They circled one artist, one stonemason, and the group of ten units (circles) to show that one of each worker would cost $10. This was one of the few groups who answered the question as written. They made no attempt to determine the individual costs for one artist or one stonemason. Members of this group were not unanimous about whether or not they should do so;
however, several members of this group were confident that the question merely asked for the cost for one artist and one stonemason together and that individual costs were not required.
The fourth group (not shown) listed two scenarios and then depicted the artists making $7 each and the stonemasons making $3 each. This seemed to be working backward. They reason that $33 and $37 added together equals $70; simultaneously, they reason that three artists and four stonemasons added to four artists and three stonemasons results in seven of each type of worker. If seven artists and seven stonemasons cost $70, the group reasons that one artist and one stonemason cost $10.
An interesting approach to finding the individual cost for each type of workers follows. First, the group realizes that both scenarios have three artists and three stonemasons. One scenario has an extra artist, and the other scenario has an extra stonemason. Based on their conclusion that one artist and one stonemason cost $10, they derive that three artists and three stonemasons cost $30. Using this baseline, they argue that the scenario, which has the extra artist, is $37, which is $7 more than their baseline. Therefore, the artist must cost $7. And, the scenario which has an extra stonemason costs $33, which is $3 more than their baseline. Therefore, the stonemason must cost $3.
While each poster seemed to represent a different thinking behind the solution strategy, teachers were able to make important connections between their respective poster proofs. All five of the possible representation strategies were used by the groups: tables, pictures, graphs, numbers and symbols, and verbal descriptions. A review of the artifacts from the other three groups (B, C, D) revealed a similar variety of approaches. Some groups used a strictly algebraic strategy with the simultaneous equations 4a + 3s = 37 and 3a + 4s = 33.
However, there were a variety of other interesting approaches that were employed by the teachers in solving these equations beyond the traditional textbook approaches including substitution, linear addition, or matrix approaches. One group started off with finding ways to arrive at partitioning the number 37, including 25 + 12, 26 + 11, and 27 + 10, even though none of these contain one value which is divisible by 7 and another value which is divisible by 3. Their last attempt, though, 28 + 9, did factor correctly. They then used the same strategy to arrive at partitioning 33, using 19 + 14, 20 + 13, and finally arriving at the correct 21 + 12.
Another group, first wanted to do a comparison to determine who made more. They reasoned that the cost with an extra artist was greater than the cost with an extra stonemason, concluding that artists cost more. Using algebra, they found that the cost for one artist was $4 greater than the cost of a stonemason. Their poster used this idea to show that an artist earns $7 and a stonemason earns $3. Another interesting observation involving parity was made by one of the groups because the total cost in either scenario was odd and the number of total workers in each scenario was odd, then the individual pay for each type of worker must be odd. If there are four workers of the same type, then their total pay will be an even integer. But, the three remaining workers must have an odd wage or else the total cost would be an even integer. Using the same logic in the second scenario, their poster showed that both types of workers must have an odd value for their daily pay.
As these poster illustrations clearly indicate, the teachers exhibited a wide variation in their thinking. Specifically, the comparison strategy generated a lot of interesting conversation that helped the instructors to bring a nice closure to this problem using proportional reasoning. Since this group demonstrated that one artist must cost $4 more than a stonemason, starting from the first setup of four artists and three stonemasons that costs $37, it became clear to the teachers that adding an artist and taking away a stonemason to this will yield five artists and two stonemasons that costs $41.
Continuing this process, we have six artists and one stonemason that costs $45 and doing this one more time yields seven artists that costs $49 which immediately helps to solve for the cost of an artist which was $7 and from that obtain the cost of a stonemason which was $3. It was interesting to see such profound thinking from teachers for a problem that most students would normally do using a textbook approach of solving system of linear equations.