Math Camp

DAY 1

Spirolaterals and Dragon Curves

The first day of math camp is done. One participant was absent, but she will join us tomorrow. The finger math lesson would have been useful for the puzzle, but Robert breezed through it anyway. Gabriel is having trouble getting the cobwebs out of his mental math machine, but he completed most of it. Generating all the spirolaterals was a lot more work than I thought it would be. Most of them require 36 iterations before the pattern starts repeating. We did them for +1 through +9 on both graph paper and isometric graph paper. The regular graph paper spirolaterals were definitely more interesting. We only briefly talked about the dragon curves. After four folds, it gets really tricky.

Puzzle Winner: Robert

Posing in front of their spirolaterals
Puzzle work

DAY 2

Fibonacci Numbers

We had a fractions review that generated a great deal of conversation. I introduced the pizza's per boy model of fractions. For example, 1/2 is one pizza shared among two boys. The kids found it very funny when I explained dividing fractions by considering problems like one pizza per half a boy: 1/1/2 = 2. They thought the idea of cutting a boy in half and giving each half of him a pizza gross. When we got to notion of perfect sequence, analogous to perfect numbers like 6 = 1 + 2 + 3 (note, 1,2,and 3 are all the factors of 6 excluding 6 itself), we invented our own perfect sequences before getting to Fibonacci's sequence. We had the Lucky Sequence (7,7,14,21,35,56,...), the Meri Sequence (5,5,10,15,25,40,...), the Gabriel Sequence (6,6,12,18,40,...), and the Robert Sequence (2,2,4,6,10,16,26,...) which could also be called the TV sequence because of the 16:10 aspect ratio. Our projects were composing using the golden ratio (~1 5/8) or golden spiral. They each designed a company logo. We had two wolf companies (?) and one called Muscle Cat. Finally, they each photographed a subject and composed the picture using three methods: centering, rule of thirds, and the golden spiral.

Puzzle Winner: Gabriel

Most Beautiful (person whose ratio of height to belly button is closest to golden ratio): Meri - 1.67, Robert - 1.73, Gabriel - 1.76

Posing with their work (shown below)
Kids working diligently on their daily puzzle.

Robert

Revolutionary War

Gabriel

T-Rex Circuit

Meri

Planets

DAY 3

Figurate Numbers

We used mm's to construct square and triangular numbers before switching to Cuisenaire rods. Surprisingly, we didn't run out of mm's, which I guess means the teacher can eat the leftovers. It was a lot of fun showing relations for the pentagonal and hexagonal numbers, but the most fun, yet frustrating, was the tetrahedral (pyramid) numbers. We used double-sided tape to secure the base, but several times after the third or fourth level the pyramid would collapse. Fortunately, we had just enough time to get a picture. They did not find the formula for tetrahedral numbers but they did find a relationship using the Cuisenaire rods. Finally, they each made a poster illustrating one of the theorems from the lesson. Robert did the theorem of Theon of Smyrna, Gabriel showed that the rectangular numbers are the sum of the even numbers, and Meri showed that the square numbers are the sum of the odd numbers.

Puzzle Winner: Robert and Gabriel tied.

Square Numbers

NxN

Triangular Numbers

1/2 N(N+1)

Rectangular Numbers

N(N+1)

Pentagonal Numbers

1, 5, 12, 22, ...

Pentagonal Numbers

Nth square number + (N-1)th triangular number

Hexagonal Numbers

1, 6, 15, 28, ...

Hexagonal Numbers

Nth square number + 2 x (N-1)th triangular numbers.

5th Tetrahedral Number

Pyramid bases are the triangular numbers (1 + 3 + 6 + 10 + 15 = 35)

5th tetrahedral Number

Sum of (N-3)th and (N-1)th square numbers and the Nth triangular number (4 + 16 + 15 = 35)

On day 4, you can derive that these are equal to (N+2)-choose-3. In this case we have 7-choose-3 = 7!/4!3! = 7x5 = 35

Meri

The sum of the odd numbers is a square number.

Robert

The sum of two consecutive triangular numbers is a square number.

Gabriel

The sum of the even numbers is a rectangular number.

DAY 4

Pascal's Triangle

Today we brought together many of the new topics learned earlier in the week—Fibonacci numbers, Triangular numbers, Tetragonal numbers—by use of Pascal's triangle. We first briefly touched on basic combinatorics, counting: counting the number of ensembles given three shirts and two pairs of pants, the number of ways to rearrange the letters in a three letter word, etc. The latter served as the basis for committee problems (called n-choose-k by mathematicians). How do you choose the number of ways to choose a team of three from a class of five students? The answer is 5-choose-3. It is also the solution to the number of ways to rearrange the letters in the word OOOXX (5!/3!2! = 10). Finally we got to the meat of the lesson and talked about a coin toss game in which one player gets a point if a coin lands heads and the other gets a point if it lands tails. The first person to 10 points wins $100. How do we split the winnings if the game stops at a score of 9-9? Since they both have an equal chance of winning, then they should split the winnings equally: $50-$50. What if the game is ended prematurely at a score of 8-9? At most, they could play 2 more games. The possibilities for those two games are AA (A wins both), AB, BA (they both win one), and BB (B wins both). If A had a score of 9 and B a score of 8, then as you can see A has three ways to win, but B has only one. So A should get 3/4 of the winnings and B only 1/4: $75-$25. Notice that AA is the number of ways in which A can win two out of two games. AB and BA is the number of ways in which A can win only one of the two remaining games. Finally, BB is the number of ways in which A loses both games. These are all n-choose-k problems: 2-choose-2, 2-choose-1, and 2-choose-0. If there were three games remaining then we would estimate the number of ways for A to win as 3-choose-3, 3-choose-2, 3-choose-1, and 3-choose-0. These are the second and third rows of Pascal's triangle. At once, the students can notice the relationship between the entries in the subsequent rows. The next column is the sum of the two columns above it:

1

1 1

121

1331

14641

And so on. The final art project was to make the biggest Pascal's triangle they could and color the even numbers one color and odd numbers another resulting in a beautiful Sierpinski triangle. We actually got a print out of a Pascal's triangle to do this because we could not fit three and four digit numbers in the small triangles on the isometric graph paper.

Puzzle Winner: Gabriel

The kids posing with their Sierpinski triangles.
Sierpinski triangles

DAY 5

Binary Math and Computers

This is the last day of math camp. The kids learned how to count in binary and convert from binary to base-10. We also briefly talked about Hexadecimal (base-16) as well as other historical bases like base-12 (Egyptians and clocks) and base-60 (Sumerians and circles). I augmented the lesson with a video of Nobel Prize winning physicist Richard Feynman's lecture on computers which makes the link between a base-2 number systems and computers. I provided the animations for the video. The link on vimeo and the password are in the DIY Math Camp 2019 Booklet. After watching the video we made a human computer (see below), but our computations were limited after our third bit dropped out and decided not to play any more. We finished up by watching the movie War Games. The movie touches on themes from math camp (computers on day 5 and game theory on day 4). Mainly, though, I thought it was a nice reward after a long week of running math camp!

Puzzle Winner: Gabriel, who now gets the title of puzzle master.

With 3 bits, we could have calculated up to 8.

00 (0 base-10)

00 + 01 = 01 (1 base-10)

When hands touch, they move together.

01 + 01 = 10 (2 base-10)