Problem Set 1

I TRIED all 7 Problems 

I am posting them in order although I did not solve them in order. I solved problem 1 first, problem 2 second, then problem 5, problem 3 (attempt), problem 4, problem 6, and problem 7 last. I selected them in order of what I thought was their diffculty but I am still struggling with 3 and 7. I will post below the problem prompts, pictures of my solutions, and my thoughts on how I solved each problem.

Problem 1. An Equation 

Consider the equation (x^2) + (y^2) = (z^2) and respond to the following prompts:

• What are some answers? 

• What are some pictures that could be described by this equation? 

• What else can you do with this equation?

Solution

I kept thinking of ideas for this problem as I was working on other problems. I added the triangle after 'attempting' problem 7. I think there is more solutions I just can not think of them yet.

Problem 2. Handshakes at a Party 

A man and his partner invite five other couples to dinner. As they enter the house, and accept pre-dinner drinks, some of the people at the party (but not all of them) shake hands. Naturally, no guest shakes hands with his or her own partner. At dinner, the host brings up the subject of the handshakes. Going round the table, he asks each of the people at the party how many hands he or she shook. He finds that he obtains 11 different answers. How many hands did the host’s partner shake? 

Solution 

      

I feel that there was no one answer to this problem. I decided to figure out how many possible solutions there are, "how many hands did the host partner shake?" I did not think there was enough information made available to solve the problem to find a definite number of hand shakes. He could have shaken 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0 hands, he would not shake his partners hand or his own so these are the only possible outcomes. Any two of these 11 numbers could be repeated so I used the 12! method, 12*11*10*9*8*7*6*5*4*3*2*1. So I determined there would be 479, 001, 600 possible solutions for all of the hand shaking going on. (I just noticed a typo in my calculating of 12! so the answer picture above is not correct, the 479 value in my description here should be correct.)

Problem 3. Parallelepiped

Find the length of the diagonal of a rectangular parallelepiped of which the length, the width, and the height are known. 

Solution

I feel like I still am not sure how to solve this problem. I am actually no sure if I should be solving for the diagonal through the entire parallelepied or on the surface of one of the faces. I am stuck.

Problem 4. Forest for the Trees 

(I drew the tree diagram best as I could on my solution)

Imagine standing at the corner of a forest with trees planted in equally spaced horizontal and vertical rows. Which trees are visible from your position at the “origin”? As you explore this question, keep a running list of any conjectures you make, questions you have, patterns/extensions that you see, and techniques, tools, or extra information that you use (or wish you had available to use). a. Suppose you can see as far as the third row and column (call this “depth 3” in the forest). How many trees can you see? b. Suppose you can see to depth 4 in the forest. How many trees do you see? c. Think about a way to represent each of the visible trees. Find at least two different ways to do this. d. Suppose you can see to depth n in the forest. Can you predict how many trees you’ll be able to see? 

Solution

I tried my best, but I could not figure out if I was the bottom left corner dot or if I was behind the corner dot. I solved it both ways. I think I represented b correctly but I had trouble thinking of another way to solve for "think of another way to represent the visual trees." I was trying to think of an algorithm but now I am wondering if this just meant how you would draw them or illustrate them. I tried squares in my solution so I think this would work for another representation. I would also use numbers like in my second drawing in the photo solution to the right.

Problem 5. The Squareness Task

Below is a collection of rectangles.

 a. Which of the rectangles is the “squarest”?

 b. Arrange the rectangles in order of “squareness” from most to least square.

 c. Devise a measure of “squareness,” expressed algebraically, that allows you to order any collection of rectangles in order of “squareness.”

 d. Devise at least one more measure of “squareness” and discuss the advantages and disadvantages of each of your measures. 

Solution

I first defined what it means to be a square, or to be the most "squareness". I then measured each rectangle, recorded their length and widths in the chart in the left photograph. I then I divided the length by the width to get as close to a 1:1 ratio as possible as this I determined would be the most "squareness." I decided through my measuring and calculating that rectangle C is the most square followed by E, B, D, A, and F.

Problem 6. The Wet Box

Part 1 A shipper is shipping large crates of iPhones from Shanghai. Each large crate measures 36” × 18” × 16”. It is filled completely with iPhones, which are in boxes that are 3” × 4” × 6”. One crate falls of the boat and submerges completely, but is pulled out of the water almost immediately. Upon talking with Apple, the shipper concludes that all iPhones that were touching the outside of the wet crate will have to be returned to Apple to check if they still work. How many iPods will have to go back? Be sure to explain how you arrived at your answer, and use words, symbols, and/or diagrams to support your explanation.

Part 2 Upon hearing the bad news, Apple decides that they need to ship their iPhones in a larger crate. (Why would that be a good idea?) Specifically, they want to design a crate that has twice the volume of the original and that minimizes the number of iPhones that would be damaged if a similar accident were to occur. What are the dimensions of a crate that would serve this purpose? Is there more than one answer to this question? How do you know? If Apple’s box supplier charges by the square inch of surface area, how much more will the new box cost?

Solution

Part 1: I found how many they will need to ship back by finding the directionality of the phones first. Then figuring out how many phones fit in the box, it was "filled" so this was possible. I then found out if the phones were situated in the way I determined how many were touching the outside. I subtracted two phones from each of the directionality dimensions so as to not count the phones on the top and bottom or far right and far left edges. I found the volume of this smaller box and subtracted this from the larger box to find the number of phones only on the edges, or the ones they need to ship back because they are or could be wet.

Part 2: I found all of the solutions for calculating double volume of the box. I did this by again finding the volume by number of phones, doubled this value then came up with 3 dimension possibilities. 

12 x 6 x 4

8 x 6 x 6

8 x 9 x 4

I then used these values in various orders to calculate which directions the phones were facing, I determined if they had to be facing up down, right or left based on how the dimensions of the phones fit into these boxes. All fit since the boxes are twice the size. I multiplied the dimensions by the phone dimensions 6 x 4 x 3 to convert the phones into real inches to be used in measuring the crates. I then used these measurements and looked for duplicates, I crossed off duplicates and then calculated the surface area of each set of the 11 total possible options for crate dimensions to see which would be the cheapest. I multiplied each of the length, width, and height by each other so I had 3 seperate face (surface) areas which I then multiplied by 2 because there are two sides to every face (a front and a back, a top and a bottom, and a side and a side). I added each of these surface area calculations together to find the cheapest one. 

Original Crate

Measure: 36 x 18 x 16 inches

Volume: 10,368 inches cubed

Surface Area: 3168 inches squared 

New Crate  

Measure: 24 x 24 x 36 inches

(Double) Volume:  20,736 inches cubed 

(Cheapest) Surface Area: 4608 inches squared 

Problem 7. Squares and Triangles (Oh My!)

In the diagram below, the four corners of a square are each connected to a midpoint of one of the square’s sides, forming a second (smaller) square inside. What fraction of the area of the original square is the area of the shaded square?

Solution

I am not sure where to go with this problem. I drew the square. I know the answer is 1/5. I am confused how to prove the answer is 1/5. I know that each small triangle and each trapezoid makes up the same square that the shaded (center) square makes up. I think the proportions are correct so that there are four tiny triangles and four trapezoids to make 4 Hodge-Poge squares plus the one in the middle for a total of 5 squares. The shaded one being 1 of 5 of the squares taking up 1/5 of the original squares area.