The scale problem

When working through this problem, I came across the solution of the weights being 1g, 3g, 9g, 27g. To demonstrate how I solved this problem, we need to think of an equal-arm weight scale like Figure 1. I will also denote the left side of the scale with a red x. When we have x+(some amount), then the items are on the same side of the equal-arm weight scale and when we have x-(some amount), then we have items on opposite sides of the equal-arm weight scale.

Figure 1: An equal-arm weight scale

As the vendor has four different weights to measure herbs from 1 to 40g, the first weight must be 1g, to measure 1g herbs.

For the second weight, we need to be able to measure 2g of herbs. If the vendor has a 2g weight, then we can measure 2g of herbs, 3g of herbs, or also 1g of herbs by setting 2g on the left and 1g weight + 1g herbs on the right. If we had a 2g weight, we need a 4g weight to measure 4g of herbs. If we had a 4g weight, we need to satisfy combinations of x-1, x-2, x+1, x+2, x-1-2, x-1+2, x+1-2, x+1+2. Therefore, we can measure 3g, 2g, 5g, 6g, 1g, 5g, 3g, 7g herbs. As we can measure the same herb amounts in several ways, this may not be the most effective weight combination and we would need our last weight to be 40g-1g-2g-4g = 33g.

If our second weight was x=3g, then we can measure x-1=2g or x+1=4g herbs. Hence now we can measure 1, 2, 3, 4 grams of herbs with our two weights. As noted before, to find our third weight, we need to satisfy x-1, x-3, x+1, x+3, x-1+3, x+1-3, x-1-3, x+1+3. 

We conclude that the third and fourth weights must be 9g and 27g. When we have weights 1g, and 3g, we cannot have a third weight larger than 9g as we would be unable to measure 5g herbs at the minimum. 

I believe this is the only solution. I noticed that 1, 3, 9, 27 is part of the geometric sequence for 3^y, where y is the set of whole numbers. An extension for this activity could be to examine geometric sequences backwards in the form of half-life, which is the time it takes for a quantity to reduce to half of its initial value. Half-life is commonly observed in biology or nuclear physics where substances or atoms decay. 

The giant soup can of Hornby Island

Figure 1: The giant soup can of Hornby Island

When trying to answer the question on the actual size of the painted water tank in the photo, it is important to keep in mind that the water tank has the exact same proportions of a Campbell’s Soup can. We can use the dimensions of a can of Campbell’s soup and scale up to estimate the size of the water tank using the bicycle as a reference. 

Figure 2: A can of Campbell’s Vegetable Soup

From Figure 2, the Campbell vegetable soup I will be measuring from has 284mL. For this exercise, I will assume the volume of the can is as the equation for the volume of a cylinder: pi*r^2*h. Taking measurements of our can, I find that it has a diameter of 6.6cm (radius is 3.3cm) and height of 10.2cm. Entering these in our equation for the volume of a cylinder, I get 348.96cm³.

I googled ‘dimensions of a bicycle’ for diagrams and found that most bikes had a length of around 990mm (99cm) to 1080mm (108cm) as the distance between the centre of the rear wheel to hub of front wheel. Bicycle wheels have a diameter ranging between 419mm (41.9cm) to 633mm (63.3cm) according to PandaEbikes. If this bike had a 520mm diameter (52cm) wheel, then the radius would be 260mm (26cm). For this estimate, I will take the length of the bike to be 108cm + 52cm = 160cm (adding 26cm to rear wheel radius and 26cm to front wheel radius). 

Figure 3: Example of bike dimension found on google images

I then imagined how many bikes would fit length-ways across our water tank. Figure 4 shows that approximately 3.5 bikes fit lengthways, giving our height of the water tank; h = (3.5)(160) = 560cm. Comparing this to the height of the Campbells soup can of 10.2cm, this is 54.9 times larger. I can then estimate the radius of the water tank to also be 54.9 times larger, which equates to 181.17cm. Therefore the volume of water that the water tank can hold is pi*(181.17)^2(560) = 57744479.14cm³ or in terms of liquids to be 57744479.14mL or 57744.48 litres of water.  

Figure 4: Estimating height of water tank

An extension I have for an activity similar to this for secondary students, is to estimate the volume of places such as how much water could fill up the classroom, or ask students to scale up objects such as their phones and estimate the surface area it would cover. 

 

Source:

Pandaebikes. Ebike-motor-wheel-rim-size-guide. Accessed November 9, 2020 from https://www.pandaebikes.com/what-is-my-wheel-size/ebike-motor-wheel-rim-size-guide/. 

Geometric/numerical puzzle

Question: Thirty equally spaced points on the circumference of a circle are labelled in order with the numbers 1 to 30. Which number is diametrically opposite to 7?

To solve this geometric/numerical puzzle, I drew a circle and made diametric lines across from numbers 1 to 30. Using my image, I find that the number diametrically opposite to 7 was 22. After drawing my image, I noticed that there are 15 ‘spaces’ between the starting number and the number diametric from it. 

Figure 1: My drawing 

A puzzle related to this can be to rearrange the numbers so that they are not organized in ascending order around the circle and ask students to find numbers diametric and next to a number, etc. If this were a word problem, it would be lengthy just to set up the problem alone. An extension of this is to use numbers 1-24 to correspond with letters of the alphabet and students can use it to create coded messages.

I feel like all problems can be logic problems, but a problem is considered “geometric” when there are aspects that allow for visualization of the problem geometrically. When I read this problem, I immediately thought of it as a geometric problem because I began visualizing a circle with the numbers arranged around it. I drew diametric lines to help me think of the solution and did not make the logical connection between half of 30 and properties of a circle.

I think there is more value to give students problems with no single correct answer (in an academia, these are considered wicked problems). This encourages discussion and conversation amongst students to share their reasoning and logic when arriving at their answer. 

The dishes problem

Problem: "How many guests are there?" said the official. "I don't know," said the cook, "but every 2 used a dish of rice, every 3 used a dish of broth, and every 4 used a dish of meat between them". There were 65 dishes in all. 

The way I solved this problem without algebra is to consider how many dishes are used following a lowest common multiple of people, ie. 12. So 12 guests minimum use 13 dishes total. 13 dishes per 65 total dishes equals 5. Taking 5 by 12 guests gives the result 60 guests. 

I think offering examples of math problems to students from different cultures can be worthwhile as it encourages students to think beyond their current cultural lens. Math history from different cultures can be especially important as it offers an opportunity to see how different cultures come up with different solutions to the same problem. I enjoy having imagery to math word problems and puzzles as it gives it allows me to make a realistic connection. This dishes problem is extremely relevant in Asian cultures as Asian cuisines often involve large banquets around a table consisting of family and friends. 

Goldberg Polyhedrons

In geometry, a polyhedron is a 3D shape with polygon faces, straight edges, and sharp corners or vertices. It derives from the Greek word “many-base.” Polyhedrons are convex, meaning that their faces are curved outward. A Goldberg polyhedron is a convex polyhedron made of pentagon or hexagon faces, ie. five or six sided polygons and exactly three faces meet at each vertex.

 Goldberg Polyhedrons have notation of the form GP(x,y), where x is the number of steps in one direction, and y is the number of steps taken after turned 60^o. We can use x, y to calculate the triangulation number, T, where T = x² + xy + y² = (x+y)² – xy. Multiplying the triangulation number and the number of original faces the polyhedron has will determine the number of triangles the new polyhedron will have. This process is a type of subdivision known as principal polyhedral triangle (PPT), which breaks down the structure of a polyhedron. By subdividing the Goldberg polyhedron and slicing the spherical shape in half, we create a geodesic dome. The triangular shapes are structurally rigid and create a sturdy and durable structure. Any pressure applied to one triangle is evenly distributed to other triangles.  

Figure 1:  A GP(5,5) Goldberg Polyhedron (left).

Figure 2: A GP(5,5) subdivided by triangulation number (left). 

For this project, I tried to recreate creating several different kinds of Goldberg polyhedrons by colouring golf balls. The divots in the golf balls are all hexagon shaped. A problem I ran into while working on this project, was that divots were not entirely uniform, ie. same size. Nonetheless, it was an interesting opportunity to try and draw my own Goldberg polyhedron. An extension of this activity in the classroom could be to give students toothpicks and marshmallows and challenge them to first create a Goldberg polyhedron, then subdivide the faces to create a geodesic dome and compare strengths between their creations. This activity teaches students teamwork and cooperation, but also lets them explore and experiment on creating integral structures resistant to falling. There is another interesting resource online that is found on https://levskaya.github.io/polyhedronisme/, which generates different polyhedrons. It would be useful to integrate this resource in the digital classroom as it is a great visualization tool that demonstrates easily how polyhedrons change when different operators are applied such as truncating, orthogonality, duality, reflecting and snubbing. 

Figure 3: A GP(5,0) created from a golf ball. 

Figure 4: Virtual recreation of Fathauer's Goldberg Polyhedron using polyHedronisme. 

This was my first group project since starting the teacher education program at UBC, and I had a blast. I’ve learned so much through creating Goldberg polyhedral patterns on golf balls and researching polyhedrons. My favourite part of this project would be the interactive session using polyHedronisme. As a teacher, I hope to integrate many different digital tools in the classroom for learning. We are living in such a technologically advanced generation, I feel that incorporating these technologies into the classroom can be opportunities for students to engage with mathematics on a deeper level than equations seen in a textbook. 

Sources:

Fathauer, R. 2019. Robert Fathauer. Retrieved September 27, 2020 from http://gallery.bridgesmathart.org/exhibitions/2020-bridges-conference/fathauer. 

Hart, G.W. n.d. A Twelve-Part Puzzle Based on the (4,2) Goldberg Polyhedron “Tectonic Plates.”  Retrieved September 27, 2020 from https://georgehart.com/puzzles/GBpuzzle-4-2/goldberg-puzzle-2.html. 

Levskaya, A. 2019. polyHedronisme. Retrieved September 27, 2020 from https://levskaya.github.io/polyhedronisme/.

The 1000 locker problem

When attempting to solve this problem, I began by rewording the information provided. I then drew pictures to help visualize the problem. I defined ‘O’ to be an open locker and ‘X’ to be a closed locker. I also defined student 0 to be the initial state of the lockers before student 1 closes them. After drawing locker changes up to student 4, and extending the lockers altered to locker 15, I began to write out patterns I saw. For example:

I saw that locker 2 was opened 1000 times (student 0, 2, 3, …, 1000) and closed once (student 1).

I saw that locker 4 was open twice (student 0 and student 2) and closed everywhere else (student 1, 3, 4, 5, …, 1000) for a total of 999 times.

I saw that locker 9 was opened 7 times (student 0, 3, 4, 5, 6, 7, 8) and closed 994 times (student 1, 2, 9, 10, 11, …, 1000).

I saw that locker 15 was opened 989 times (student 0, 3, 4, 15, 16, 17, …, 1000) and closed 12 times (student 1, 2, 5, 6, 7, …, 14).

From here, I saw a pattern and came up with the conjecture which states: “lockers opened an odd number of times will be open at the end [after the 1000^th student].” At the time of working on the problem, I wanted to conduct a few more tests to verify the conjecture:

Consider locker 18. It was opened 988 times (student 0, 2, 6, 7, 8, 18, 19, …, 1000) and closed 13 times (student 1, 3, 4, 5, 9, 10, 11, 12, … ,17). 

This counterexample showed that the conjecture was false as locker 18 was opened an even number of times and remains open after the 1000^th student. As I am writing this blog now, I immediately see that this conjecture is false by the counterexample of locker 9 as the locker was opened an odd number of times but remains closed.

I then noticed that factors of the locker numbers were related to whether they were opened or closed, so I began listing out factors. I discovered that the lockers with an even number of factors would be open and the lockers with an odd number of factors would be closed. The only time a number would have an odd number of factors is if it was a perfect square. As the first 31 perfect squares are below 1000, that led me to conclude that there would be 31 lockers closed and 969 lockers open after all 1000 students had gone through opening and closing.