# 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

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
• A broken egg must be discarded
• The effect of a fall is the same for all eggs
• If an egg breaks when dropped, then it would break if dropped from a higher window
• If an egg survives a fall, then it would survive a shorter fall

You only have two eggs with which to test the drops. What is the least number of egg-droppings that is guaranteed to work in all cases?

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