CAMP is now waitlisted
We’ve experienced strong interest in our 2016 CAMP program, and have moved the registration into waiting list mode.
* We are still processing current applications, and expect to make a first round of invitations very soon.
* If you are interested in attending CAMP, please do complete the application form to be placed on the waiting list. We will make additional invitations as space becomes available.
Bard Math Circle
2016 C.A.M.P. – Save the Date!
We’ve set the date for our 2016 Creative and Analytical Math Program (C.A.M.P.): August 22-26, 2016.
If you would like to receive the announcement for when registration opens, please join our mailing list and indicate your preference in the Summer C.A.M.P.
We look forward to welcoming our next cohort of middle school students (current 5th-8th graders) to campus at the end of the summer!
CAMP is over. For now!
Our 2015 summer C.A.M.P. (Creative and Analytical Math Program) is over, but our 2015–2016 program will be announced in the next few weeks. Please join our email list (Sign Up) if you are interested in our library programs, competition programs or C.A.M.P.
CAMP is over. For now!
Our 2015 summer C.A.M.P. (Creative and Analytical Math Program) is over. If you are interested in next summer’s program, please make sure to join our email list: Sign Up
Day 5
Today was our last day of math CAMP! Fortunately, this didn’t mean that students didn’t continue to solve interesting problems.
We began our day by playing “Liar’s Bingo,” a game that utilized sequences of six different numbers that were chosen in order to follow a very mysterious rule. The process can be described as follows:
Each group of students sat together with some of these six number sequences, attempting to find any obvious patterns between them. The numbers were either contained in a red box, or a black box, suggesting some relevant relationship between color and the number contained. Since no student found any predictable pattern, we were able to safely have volunteers come to the front of the room to Eliana where the student needed read the sequence of colors on their strip, but also lie about precisely one of the colors in their sequence. from this information, Eliana was able to deduce the exact value of the number that was lied about.
After successfully guessing many of the numbers, students then attempted to guess what the “trick” was behind Eliana’s odd ability to read their minds! One student in the back row suggested that the colors were actually being assigned values such as “0” for red, while another student suggested that perhaps the sequence of numbers could be subdivided into groups of three. Both of these suggestions lead to an entire class that had the answer on the tip of their tongue! However, the morning logic warm up concluded and students shuffled out of the room and into their other classes speaking to each other about what the trick might be.
Hint: Consider the fact that every number in liar’s bingo had either one or two digits. Now take a look at two distinct groupings of three numbers whose color seems to represent a particular value. Keep place value in mind! Hmm…
Here is a link where students can find their own “Liar’s Bingo” cards:
http://sigmaa.maa.org/mcst/documents/U.pdf
In geometry, Group A started with a hexagon template and were assigned to create their own design within the confines of the template. Here, the expectation was that students would use geometric principles in order to make shapes within the hexagon using only a compass and straight edge. Thankfully they had also been working on this technique in art class and were able to create many interesting designs.
Group B on the other hand used all of the tools they had gathered during the week in order to construct a “Pi Sandwich.” Here, the goal was to understand how ancient civilizations were able to calculate Pi to an arbitrary number of decimal places! The stock answer for many middle schoolers in terms of how one might obtain the value of Pi is by dividing the circumference of a circle by its diameter. However, Group B should now understand that you can’t actually measure an irrational circumference to any degree of accuracy using this method. Instead, the group worked together with Frances in order to show that one can “sandwich” the circumference of a circle in between two hexagons (one that circumscribes the circle, and the other is inscribed.) This means that students were able to find lower and upper bounds for the exact value of Pi. Unfortunately, the Hexagon method seems to be insufficient as the bounds for Pi are still quite large with this method.
With this understanding, students then worked in smaller groups to implement a similar method with a dodecagon (A shape whose perimeter is closer to the circumference of a circle.) While some groups didn’t come to an exact numerical estimation of Pi with the dodecagon, many of them were able to see that one could use this method for an arbitrarily large “n-gon” and get the value of pi to as many decimal places as they had the patience for.
In graph theory, students were given an assortment of problems concerned with “tree graphs” and “bipartite graphs.” Unbeknownst to many students, these are actually very applicable forms of mathematical organization. A bipartite graph is a graph that can be partitioned into two different categories in which each “vertex” from one category is related to a respective vertex in the other. A natural formulation of a bipartite graph is to think of a soccer game. Imagine one category that is the team to which each player belongs, and another category that consists of the players themselves. Draw an edge between a player and their team if and only if the player is in that team, and you will have a bipartite graph!
Here is another natural example of a bipartite graph:
As you guys can probable imagine, this can get very complicated very fast! Try doing the graphs for some of the other polyhedra we worked with this week. More importantly though, try to take your own shape and draw a graph of it.
**Challenge problem: Use some of the basic polyhedra that you know and draw their graphs. See if you can relate the number of faces, edges, and vertices in each graph.
In computer science, the students also completed their work with graphics and animation. As is apparent to anyone who saw their work, many of the students worked very intensively in order to create works of sophistication. More impressive is that students were able to learn an entirely new language and create something with it. Hopefully all of the projects have been shared with parents, and the students will use all of the language (as well as the downloadable software) they have learned about this week and continue to work on their craft.
All in all, it was a very successful day, and showed how valuable even a week can be in the development of any academic pursuit. I hope that all of the students continue to work independently but also strive to stay in touch with their like-minded peers. I also hope to see many of you back next year for another week of exploration and growth!
Good luck to everyone,
Andres
P.S If anyone has questions about any of these problems (or just want to talk about any of the math they’re doing during the academic year!) you guys can all reach me at andresnayeem@gmail.com.
have fun with these interesting questions and all of the other great questions you can get your hands on this year.
Day 4
The mathematicians began their day as greedy pirates, swash-buckling, mathematically of course, to get the most gold coins in a famous game theory conundrum.
Here’s the problem for those who haven’t seen it before:
There are 5 rational pirates A, B, C, D and E. They find 100 gold coins. They must decide how to distribute them.
Five pirates vying for 10 coins is an interesting problem but what if there were one hundred pirates each trying to get the most loot out of 100 coins? What if there were more pirates than coins? If there were more pirates than coins would the first pirate be able to get any coins? Is there a strategy where the first pirate would be able to stay onboard? Can you devise a formula for the number of coins any given pirate in such a situation would get where the number of pirates (N) is greater than the number of coins (G)? Use the backwards induction model like we used with the first version of the game!
Speaking of formulating your answer using variables, we continued to derive the pythagorean theorem in Geometry, replicating the work of the indian mathematician Bhakarsa in his famous proof. Walking the same path he did in 1114 CE!
In graph theory we built off of our definitions of days one through three to find isomorphic graphs amidst a maelstrom of interconnected nodes and edges. Mathematicians then used this fresh skill set to draw their own planar graphs.
During lunch I noticed a young mathematician who was playing a game of Cat’s Cradle. I wondered, if they were to chop up the string could they make a planar graph with the same number of nodes and edges as we see in the picture below?
If you were to play cat’s cradle without chopping up the string, how many different non-planar graphs can you make? How many of these are identical?
In computer science we talked about Boolean logic and the special definitions of and or or in programming.
Here’s a joke about the inclusive or:
A logician’s wife is in labor. The logician is in the waiting room.
The doctor comes out of the delivery room and says to the logician, “Congratulations, your wife gave birth to a beautiful baby!”
“Is it a boy or a girl?” asks the logician.
“Yes”, says the doctor.
In art, mathematicians continued to use their budding architectural acumen to make platonic solids. We’re eagerly shedding the templates that we made in the beginning of the week as tetrahedrons, octahedrons, dodecahedrons and more take form.
We’re disappointed that C.A.M.P. is discrete and not continuous. The last update (until next summer of course) is tomorrow, stay tuned!
— Eliana
Day 3
The warm up on Day 3 was slightly different than usual. Instead of doing one of the usual logic puzzles, the mathematicians had to figure out how to draw a perpendicular lines using a compass. Then they moved on to figuring out how to draw a pentagon using a compass, building on what they learned in art.
In Geometry, the young mathematicians saw the “Chinese Proof” of the Pythagorean Theorem, and figured out how the diagram proved it.
In Graph Theory, we defined what a complete graph is (a graph with an edge between every pair of vertices), and then the mathematicians found a formula for the number of edges in a complete graph with with n vertices.
We figured out that you can represent the number of edges in Kn with the series:
Then we used the Pigeon Hole Principle to prove that every graph has at least two vertices of the same degree. The Pigeon Hole Principle states that if you have n pigeons, and n-1 pigeon holes, then at least one pigeon hole will have two or more pigeons.
For those that were wondering, pigeon holes look like this:
Can you think of any other scenarios where the Pigeon Hole Principle would apply?
During activities we split off in to groups, and some of us went to the waterfall!
To keep safe, we used the buddy system to keep an eye on each other. Since there was seven of us, there were two pairs of two buddies and one group of three buddies. How would you represent this with a graph? Would it be a connected graph? Would it be a simple graph?
In art the mathematicians continued the work they did yesterday and started cutting and assembling their polyhedron outlines.
That’s all for today. Check back again for more exciting mathematical adventures!
Day 2
We began our day with a logic puzzle that goes by the moniker, “Impossible Elevators.” Although, the bright mathematicians got to work making it possible to get to any floor in a seven floor building with six elevators granted each elevator could only stop at three floors and you’re not allowed to switch elevators! Can you figure out how?
What if each elevator could stop at 6 floors, you still have six elevators, and you can still get to any floor from any other floor granted you find the right elevator — how many floors could this building have? Can you use your answer from the first part to guide you?
As we’ve seen in our Geometry Module oftentimes numerical patterns have geometric representations. In this case, the nifty Projective Plane of Order 2 also known as a “Fano Plane” may help you muddle through this elevator enigma.
Speaking of geometric representations, in Geometry mathematicians transformed into architects to make tessellations out of the triangles we made on Day 1. These tessellations led to new revelations. We wrote proofs for the triangle inequality and created a derivation of the area of a triangle; centuries of mathematical research in just one day!
On the topic of math research — Graph Theory took us to Leonard Eulors theorem for the Eulerian Path, and our young mathematicians began constructing their own algorithm’s for a Eulerian path.
Is there a Eulerian path in the Projective Plane of Order 2 shown above? Can you write your own algorithm for traversing any Eulerian path?
In art we continued to use the tool of the week, the compass, to construct elaborate shapes. Today we drew tetrahedrons, also known as pyramids.
Using your knowledge of drawing triangles with a compass can you figure out how to construct an octahedron, a dodecahedron or an Icosahedron?
Octahedron |
Icosahedron |
Dodecahedron |
Mathematicians made nets (outlines) to later create polyhedrons just like the ones above!
Lastly, I promised to share with you the answer to the pesky Egg Drop Problem, if you haven’t solved it yet — steer clear of what’s next.
Here is the complete problem for reference. Take a moment to solve it if you haven’t already:
You are assigned the task of determining from which windows in a 36-story building it is safe to drop eggs from. We make a few assumptions:
-
An egg that survives a fall can be used again
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A broken egg must be discarded
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The effect of a fall is the same for all eggs
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If an egg breaks when dropped, then it would break if dropped from a higher window
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If an egg survives a fall, then it would survive a shorter fall
I talked to a few mathematicians at C.A.M.P. who cleverly figured out that they didn’t have to drop the egg from every single floor to know exactly where it would break. They could split the floors in half. They would drop the first egg from the 18th floor, and then if it broke they would start dropping the second egg on the 1st floor, dropping each floor up until the 17th floor. If the egg survived they would drop it on the 36th floor. If it broke there they would repeat the same strategy with the second egg, except start on the 19th floor and drop all the way up to the 37th floor. They would know exactly where the egg broke while saving themselves 18 unnecessary drops! Not bad, using this strategy, in the worst case scenario they could accomplish the task in 19 drops.
Some mathematicians went even further, they figured they could split the floors into thirds and drop the egg every 12 floors. Using the same strategy as with every 18th floor, they would drop the first egg a maximum of three times (once on the 12th floor, once on the 24th floor and once on the 36th floor) and then they would drop the second egg a maximum of 11 times (up from the last interval where it survived to the interval where it broke).
What if we split the floors up into intervals of 4, 5, 6, 9, 12 and 18? Using the same strategy as above and imagining the worst case scenario, we get the following chart to describe the maximum number of egg drops for each interval.
Floors Skipped In between Drops
|
Maximum # of First Egg Drops
|
Maximum # of Second Egg Drops
|
Maximum Number of Total Egg Drops
|
18
|
2
|
17
|
19
|
12
|
3
|
11
|
14
|
9
|
4
|
8
|
12
|
6
|
6
|
5
|
11
|
4
|
9
|
3
|
12
|
3
|
12
|
2
|
14
|
2
|
18
|
1
|
19
|
What do you notice? It seems that two numbers are important. The maximum number of times you can drop the first egg and the maximum number of times you can drop the second egg. It appears that the closer those two numbers are together, the smaller the total number of drops is.
So, while intervals of 6 (with 6 and 5 as the two numbers) is a great answer — giving you a maximum of only eleven drops! It would be better if the two numbers are exactly the same!
But how can we do that?
What if we didn’t divide the intervals equally?
Is there a series where the number in the series is equal to the difference between that number and the last? There is, the triangular numbers have this pattern!
What triangular number series adds up to 36?
8!
8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 36
How would a series of drops that travels along the triangular series of 8 look? You would drop from the 8th floor then hop up 7 floors to the 15th floor then hop up 6 floors to the 21st floor, ect.
If you get up the 8th floor and it breaks then you have to try floors 1-7 giving you a total of eight drops.
If you get up to the 15th floor and it breaks, you’ve already used two drops and then you have to try floors 7-15 giving you a total of 8 drops.
If you get up to the 21st floor and it breaks, you’ve already used three drops and then you have to try floors 15-21 giving you a total of 8 drops.
So the maximum number is always 8 drops! That’s the least that we can possibly do it with, since the number of intervals traversed is always the same as the number in-between the intervals
That’s all for now! More to come soon!
— Eliana
Day 1
The day began with our young mathematicians looking at a puzzle about dropping eggs from tall buildings. Here’s a variation of the problem that you can mull over.