Glass Champagne Tower Project
By Ryan Tang, Junsung Lim, Junseo Song
Often at elegant parties and ceremonies, we can find champagne glasses stacked together that form a tower. Then, a person with a champagne bottle pours the liquid into the champagne glass that is located at the very top of the tower, causing the liquid to flow down to every single cup. Champagne towers typically look like this:
In the picture above, we can recognize a simple pattern of 1,4,9,16… cups as you go down each level. In other words, we can see a pattern of an =1,4,9,16,25,36 … ,n².
So then we asked ourselves, what other patterns are there that can work as a champagne glass tower? Adding on, what can we use from what we learned in Honors precalc unit 9 to determine the total number of champagne glasses required without having to count them individually?
This project incorporated recognizing patterns, mathematical induction, and finding partial sums.
Question:
Junsung Lim is considering using champagne towers at his wedding. However, he cannot decide what sequence the tower should be constructed in. The ballroom can only fit up to 6 levels of the tower. You are hired as an event organizer to advise Junsung on how to construct his champagne tower so that it works and uses the optimal number of glasses to satisfy the number of guests at his party:
a) Explore the various combinations possible to build such towers and determine how many designs will be practical if the first level of the tower can only accommodate 1 glass and it must be centered. Find the sequence(s) and calculate how many glasses will be needed for each design if there are 6 levels by deriving a partial sum formula. Use mathematical induction to prove your derivation.
b) 100 guests have agreed to participate in Junsung’s wedding. You may utilize multiple designs/multiple towers to meet this criterion. Then, determine the minimum number of towers that will add up to exactly 100 champagne glasses.
Our Designs
We experimented with various combinations to construct the champagne tower and reached the conclusion that only two designs were feasible. The criteria for a champagne tower is to start off with a single glass, in which liquid is poured, then as the glass fills, the liquid would flow down to the bottom layers. We concluded that only two types of combinations for the second layer of the tower are possible: 3 glasses and 4 glasses. This means that only two types of sequences will be practical, fewer or greater number of glasses in the second row will cause the top glass to topple or be misplaced from the center, not allowing liquid to flow properly to the levels below. Therefore, only two types of design can be constructed.
Design 1:
For our first design, we decided to use sequences of 1, 4, 9, 16, 25, 36… as you go down the levels. We conducted this design to the 6th row. We first found the general formula:
We determined that the sequence represented a quadratic series since the second difference was a constant, 2. The calculations eventually boiled down to a system of equations with 3 variables. We then solved for the variables a, b, and c using our knowledge of unit 7 in Honors PreCalculus, by the method of substitution. We substituted known values such as the sequence, and the first and second difference.
We eventually arrived at the formula:
After we derived the general formula, we found the partial sum of this sequence to determine the total number of glasses needed.
We discovered that the summation sequence of this is a cubic series since the third difference was the constant, 2. The calculations boiled down to a system of equations with 4 variables. We then solved for the variables a, b, c, and d using our knowledge of unit 7 in Honors PreCalculus, by the method of substitution. We substituted known values such as the sequence, and the first, second, and third differences.
We then reached the summation formula:
Mathematical Induction
In order to prove the validity of the formula derived above, we proved it by using mathematical induction.
From the mathematical induction above, we were able to prove true for the partial sum of this sequence.
Design 2:
For our second design, we decided to use the sequence 1,3,6,10,15,21… We conducted this design to the 6th row which is 21. We first found the general formula of this sequence through the following steps.
We first determined that the sequence represented a quadratic series since the second difference was a constant, 1. The calculations eventually boiled down to a system of equations with 3 variables. We then solved for the variables a, b, and c using our knowledge of unit 7 in Honors PreCalculus, by the method of substitution. We substituted known values such as the sequence, and the first and second difference.
We eventually arrived at the formula:
After we derived the general formula, we found the partial sum of this sequence to determine the number of glasses needed through the following steps.
We discovered that the summation sequence of this is a cubic series since the third difference was the constant, 1. The calculations boiled down to a system of equations with 4 variables. We then solved for the variables a, b, c, and d using our knowledge of unit 7 in Honors PreCalculus, by the method of substitution. We substituted known values such as the sequence, and the first, second, and third differences.
We then reached the summation formula:
Mathematical Induction
In order to prove the validity of the formula derived, we proved it by using mathematical induction.
From the mathematical induction above, we were able to prove true for the partial sum of this sequence.
Solution — Part A
Design 1:
Design 2:
To answer the question, Junsung Lim would need 91 champagne glasses if he chooses design 1 while he needs 56 champagne glasses if he chooses design 2 in a 6 level tower.
Solution — Part B
To answer the question, we decided to first find the partial sums of each level.
Design 1:
Design 2:
Then, using the partial sums of each level in each design, we explored different ways of arranging these glasses to create a perfect combination with 100 glasses.
To answer the question, Junsung Lim would have to build a minimum of three towers to exactly use 100 champagne glasses. Below are some different combinations of towers he could use:
Combination 1:
- One 3 level tower using Design 1 (14 glasses)
- One 4 level tower using Design 1 (30 glasses)
- One 6 level tower using Design 2 (56 glasses)
- 14+30+56=100 glasses, therefore satisfying all the guests with no extra glasses remaining.
Combination 2:
- One 5 level tower using Design 1 (55 glasses)
- One 3 level tower using Design 2 (10 glasses)
- One 5 level tower using Design 2 (35 glasses)
- 55+10+35=100 glasses, therefore satisfying all the guests with no extra glasses remaining.
Combination 3:
- One 2 level tower using Design 1 (5 glasses)
- One 6 level tower using Design 1 (91 glasses)
- One 2 level tower using Design 2 (4 glasses)
- 5+91+4=100 glasses, therefore satisfying all the guests with no extra glasses remaining.
Combination 4:
- One 4 level tower using Design 1 (30 glasses)
- Two 5 level towers using Design 2 (35 x 2 glasses)
- 30+35+35=100, therefore satisfying all the guests with no extra glasses remaining.
The prompt asks for the minimal number of towers, hence there are only 4 combinations possible to use exactly 100 glasses.
Conclusion
This real-life application project allowed us to explore different sequences and series, thus deepening our understanding of unit 9. Adding on, we added restraints to this question challenging us to go one step further than what we learned in class.
As we continue on with the champagne glasses project, we ask the readers: Is there a method that could be used to determine the combination to best fit the restraints when there is a given number of guests without having to individually find the sum of each level and combining them?
This is a project done by the Honors PreCalculus curriculum at Concordia International School Shanghai
