# Waiters, Waiters, everywhere…

Today I was nerdsniped in the math office. My department head and two colleagues were working on this problem. I don’t know where it came from. But golly, did I enjoy it!

Imagine you have a row of waiters all facing forward. Each waiter has a beautiful silver platter that they are carrying. They have to choose: will they hold it directly in front of them, or on their left side, or on their right side? Here’s a diagram showing the three options (I imagine I’m looking down on the waiter.)

Okay, so there is one constraint. Remember the waiters are all standing in a row. So you can’t have the platters crash into each other. So here’s an example of an OK way the waiters could hold their platters, and then a NOT OK way the waiters could hold their platters.

So here’s the question… If you have n waiters standing in a row, how many different ways could they hold their platters?

I am not going to post the answer here, because I like to nerdsnipe! But if you want to check your answer, for 20 waiters, I calculate 267,914,296 different positions!

I bet you will have a lot of fun with this problem. One person in our office came up with many pages of work, and had a very complex approach which yielded some deep insights. She was super psyched about the intricate superstructure she was building. Another person got to review solving a particular type of thingie using matrices (I want to keep things vague so I’m going to use the word thingie to avoid giving anything away). I and another person had the same approach that led to a quick and elegant solution, but left me with rich conceptual questions to pursue. And as I started doing that, I realized that I had accidentally stumbled on the complex approach that the first person had taken.

# Spiral Challenge

Megan Schmidt is obsessed with spirals. Her obsession got me hooked — for hours — on a math problem. I thought it would take maybe an hour or two, but I’m still at it and I’ve probably been working four or five hours.

I’ve been having so much fun with it.

Here’s the problem. Look at the spiral below…

We see that 1 is located at (0,0).
We see that 2 is located at (0,1).
We see that 8 is located at (-1,0).

If we continue this spiral in this manner, can you come up with a formula for the coordinates of the kth number?

So what I want to know is if we consider the number 2016, can we come up with a way to precisely define where it is? What about 820526487?

One easy way around this is to write a computer program that just brute forces our way through it. So here’s the constraint: I want a closed formula for the x-coordinate and y-coordinate. That means no recursion! No if/then statements! Just an equation that relies on k only.

You know one of the most frustrating things? Going down a path and feeling good about it, even though it is pretty complicated. And then having a new insight on how to attack the problem (which *just* happened to me now as I typed up the problem and look at the image I created for this post) [1]. And realizing that approach might yield it’s secrets so much easier!

In any case, I thought I’d share the problem because it’s given me so much enjoyment thus far. If you do get obsessed and solve it, please feel free to put your answer in the comments. I have a feeling there are a variety of valid solutions which look very different but yield the same answer.

[1] What this reminds me is how slight changes in representations can lead to new insights! Before I was using this image that Megan sent me:

## UPDATE!: I solved it!

If you want to see that I did solve it, check out this Geogebra sheet. It won’t give away how I solved it (unless you download it, look at how I defined each cell, and then reverse engineered it).

https://www.geogebra.org/m/cXhwYh3P

So yeah… 2016 is at (-22,13), and 820526487 is at (-14322, 4784).

I am so proud of myself! I came up with a closed form solution!!!

I am going to put a “jump” below here, and then show what my solution is, and write a little about it. So only read below the jump (meaning: after this) if you want some spoilers.

# A New Insight on the Famous Painted Block Problem

There is a famous, well-known problem in the world of “rich math tasks” that involves taking an nnn cube and painting the outside of it. Then you break apart the large cube into unit cubes (see image below cribbed from here for n=2 and n=3):

Notice that some of the unit cubes have 3 painted faces, some have 2 painted faces, some have 1 painted face, and some have 0 painted faces.

The standard question is: For an nnn cube, how many of the unit cubes have 3 painted faces, 2 painted faces, 1 painted face, and 0 painted faces.

[In case you aren’t sure what I mean, for a 3 x 3 x 3 cube, there are 8 unit cubes with 3 painted faces, 12 unit cubes with 2 painted faces, 6 unit cubes with 1 painted face, and 1 unit cube with 0 painted faces.]

Earlier this year, I worked with a middle school student on this question. It was great fun, and so many insights were had. This problem comes highly recommended!

Today we had some in house professional development, and a colleague/teacher shared the problem with us, but he presented an insight I had never seen before that was lovely and mindblowing.

Spoiler alert: I’m about to give some of the fun away. So only jump below / keep reading if you’re okay with some some spoilers.

# Clock Puzzle

In our last department meeting, one teacher presented a puzzle/problem for us to figure out.

At 3:00, the hour and minute hands on a clock form a right angle. What is the next time that happens?

The presenting teacher had a pretty darn elegant solution. But I enjoyed working it out using brute force. (That’s pretty much my go-to.) I’m going to type my solution down below the jump.

# Stuffing Sacks

Matt Enlow (math teacher in MA) posted a fascinating problem online today, one he thinks of when storing all those plastic bags from the grocery store. You shove them so they all lie in a single bag, and throw that bag under the sink. Here’s the question: how many different ways can you store these bags?

For 1 bag, there is only 1 way.
For 2 bags, there is still only 1 way.
For 3 bags, there are 2 ways.

Here is a picture for clarification:

Can you figure out how many ways for 6 bags? 13 bags?

You are now officially nerdsniped.

A number of people had trouble calculating 4 bags correctly, so I’ll post the number of ways 4 bags could be stored after the jump at the bottom, so you can at least see if you’re starting off correctly…

Additional Information: Matt and I figured the solution to this problem together on twitter. It was an interesting thing. We didn’t really “collaborate,” but we both refined some of our initial data (for 5 bags, he undercounted, and I overcounted). It seemed we were both thinking of similar things — one idea in particular which I’m not going to mention, which was the key for our solution. What blew my mind was that at the exact time Matt was tweeting me his approach that he thought led to the solution, I looked at my paper and I had the exact same thing (written down in a slightly different way). I sent him a picture of my paper and he sent me a picture of his paper, and I literally laughed out loud. We both calculated how many arrangements for 6 bags, and got the same answer. Huzzah! I will say I am fairly confident in our solution, based on some additional internet research I did after.

Obviously I’m being purposefully vague so I don’t give anything away. But have fun being nerdsniped!

Update late in the evening: It might just be Matt and my solution is wrong. In fact, I’m now more and more convinced it is. Our method works for 1, 2, 3, 4, 5, and 6 bags, but may break down at 7. It’s like this problem — deceptive! I’m fairly convinced our solution is not right, based on more things I’ve seen on the internet. But it is kinda exciting and depressing at the same time. Is there an error? Can we fix the error, if there is? WHAT WILL HAPPEN?!

The number of ways 4 bags can be stored is… (after the jump)

# Doodling in Math

A few years ago, I blogged about this fun little doodle that students often make — and how another teacher and I found out the equation that “bounds” the figure. I honestly can’t remember if I ever posted how I got the answer. If I did and this is a repeat, apologies.

Tonight I wanted to see if I could re-derive it like I did before — and lo and behold I did. I’m curious if any of you have done it the way I did it, or if there are other ways you’ve learned to approach this problem. (There is a student who I had last year who created this amazing 3-d version of this using the edges of a cube and some string. I love the idea of asking — for this 3-d figure — what surface is generated by the intersections of these strings.)

We start out by having these lines which form a family of curves. But of course we’re not graphing all the lines. If we were, we’d get something more dense like this.

The main idea of what I’m going to do to find that curve… I’m going to pick two of those lines which are infinitely close to each other and find their point of intersection. That point of intersection will lie on the curve. (That’s the big insight in this solution.) But I’m not going to pick two specific lines — but instead keep things as general as possible. Thus when I find that point of intersection for those two lines, it will give me all the points of intersection for all the lines.

Watch.

First we pick two arbitrary lines.

We’ll have one line move down on the y-axis $k$ units (and thus over on the x-axis $k$ units). And the second line will be moved down on the y-axis just a tiny bit more (down an additional $e$ units). Yes, we are going to have that tiny bit, that $e$, eventually go to zero.

The two lines we have are:

$y=\frac{k-1}{k}(x-k)=\frac{k-1}{k}x-(k-1)$

$y=\frac{k+e-1}{k+e}(x-(k+e))=\frac{k+e-1}{k+e}x-(k+e-1)$

A little bit of algebra is needed to find the point of intersection. Setting the y-values equal:

$\frac{k-1}{k}x-(k-1)=\frac{k+e-1}{k+e}x-(k+e-1)$

And then doing some basic algebra:

$k^2+ke=x$

Now solving for $y$ we get:

$y=\frac{k-1}{k}(k)(k+e)-(k-1)$

$y=k^2+ke-2k-e+1$

So the point of intersection is:

$(k^2+ke, k^2+ke-2k-e+1)$

Here’s the kicker… Remember we wanted the two lines to be infinitely close together, right? So that means that we want $e$ to go to zero. Thus, our point of intersection of these infinitely close lines will be:

$(k^2, k^2-2k+1)$ or $(k^2,(k-1)^2)$.

Beautiful! And recall that we picked the lines arbitrarily. By varying $0\leq k \leq 1$ and plotting $(k^2,(k-1)^2)$, we can get any two lines on our doodle.

But I want an equation.

Simple. We know that $x=k^2$. Thus $x=\sqrt{k}$.*

Since $y=(k-1)^2$, we have $y=(\sqrt{x}-1)^2$

Let’s graph it to check.

Huzzah!!! And we’re done!

I wonder if I can do something similar with this cardioid:

I think I must (for funsies) do some investigation of “envelopes” this summer. I mean, Tina at Drawing on Math even introduces conics with these envelopes!

An extension for you. Do something with this 3d string-art.

*Of course you might be wondering why I don’t say $x=\pm \sqrt{k}$. Since $k$ is between 0 and 1, we know that $x$ must be positive.

# An Animal Problem

One of our math club leaders gave out this problem as the final problem of math club for the year. I had never seen it before, and after she handed it out, a number of math teachers were in a tizzy about finding the solution. So instead of planning for classes, we enjoyed working on this problem. But we got it! HUZZAH!

Here’s the problem:
In how many ways, counting ties, can eight horses cross the finishing line?

So we fully understand the problem, let me list all possibilities for three horses: Adam, Beatrice, and Candy. No, wait, those are better names for unicorns:

1st: A   2nd: B   3rd: C

1st: B   2nd: A   3rd: C

1st: A   2nd: C   3rd: B

1st: C   2nd: A   3rd: B

1st: B   2nd: C   3rd: A

1st: C   2nd: B   3rd: A

1st: AB (tie)   2nd: C

1st: AC (tie)   2nd: B

1st: BC (tie)    2nd: A

1st: A   2nd: BC (tie)

1st: B   2nd: AC (tie)

1st: C   2nd: BC (tie)

1st: ABC (tie)

That comes out to 13 different ways these horses unicorns can finish the race.

That’s the answer for 3 unicorns. What’s the answer for 8 unicorns?

(FYI: If you want to know if you’re on the right track… I have 541 for 5 unicorns…)