# 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…)

# Implicit Differentiation

Normally, I don’t have trouble teaching implicit differentiation. However, I’m never satisfied with what I do. I’m fairly certain that I have taught it four different ways in the past four years. But what’s common is that we do a lot of algebra. By the end, they can find $\frac{dy}{dx}$ for a relation like $\sin(xy)+y^3=2x+y$. Or something like that. But we lose the meaning of what we’re doing.

I realized we can do all this algebra, but it’s all procedure. And so there’s no real depth.

So today, after introducing implicit differentiation (including some visual motivation), I assigned 5 basic problems from the textbook. Each of the problems has an equation like $3y^3+x^2=5$ and students are asked to find $\frac{dy}{dx}$. My kids are going to go home today and struggle with it. We’ll spend about 20 or 25 minutes in our next class going over their solutions, talking about things, whatever.

And then… then… I’m going to hand out this sheet I wrote today.

[.doc, .pdf]
[if you’re wondering, the graphs were made by the fabulous winplot which I adore… it can do implicit plotting!]

My kids found $\frac{dy}{dx}$ for homework. Now in class, my kids are going to interrogate what that means.

I am not sure yet how I’m going to structure the class. I think I might have us all work together on the first problem (#9), and then assign pairs to work on two of the remaining problems. And then I’ll pick one problem for each pair to present to the class. But what I’m truly happy about is that each problem gets kids to relate implicit differentiation to a graphical understanding of the derivative. It forces my kids to look at the derivative equation, and make connections to the original graph.

Although I’m proud of it, I’m honestly just not sure if this investigation is beyond the scope of my kids’s abilities. It pulls together a lot of concepts. I think it’ll work for them. This year I have a really really strong crew so I have faith. However, it’s an activity I’m going to have to give my kids time to do, and room to struggle. I know me, and I’m going to want to rush it, and I’m going to want to help them in ways that aren’t good for them. The struggle is where they’re going to learn in this, so I have to give it time and stay out.

I am in the middle of a hellish week, but if I have time, I’ll try to report back how it goes after we do it in class.

# Taking a Moment… in Calculus

In calculus, I’ve historically asked kids to take the derivative of: $f(x)=\frac{2x^2+\sqrt{x}}{\sqrt{x}}$

and students will immediately go to the quotient rule. OBVIOUSLY! There’s a numerator and denominator. Duh. So go at it!

Unfortunately, this is VERY UNWISE because it leads to a lot more work. And I was sick of my kids not taking a moment to think: what are my options, and what might be the best option available? Also, kids generally found it hard to deal when we started mixing the derivative rules up!

So I came up with a sheet to address this and paired kids to work on it.

(I’ve also had kids think they can do some crazy algebra with $g(x)=\frac{x^2+1}{x+1}$. This sheet also helped me talk with kids individually about that.)

For a little context, my kids have only learned the power rule, the product rule, the quotient rule, and that the derivative of $e^x$ is $e^x$. They have not yet been formally exposed to the chain rule.

[.pdf, .doc]

# Believe it or not… a log question which was briefly stumping us

Hi all,

A teacher approached me with the following question.

The function $\ln(x^{-2})$ has a graph that looks like: It makes sense that the function exists for all negative x values, because when you raise a negative number to the -2 power, you’re going to get a positive number. And you can take the natural log of a positive number.

Then the teacher said to consider the following function: $-2 \ln(x)$, and the graph looks like: Notice that you can’t input negative x values, because the domain of natural log doesn’t allow for it.

Here’s the question.

According to the log rules/properties, we know that: $\ln(a^b)=b\ln(a)$ (obviously).

So $\ln(x^{-2})=-2\ln(x)$. But the graphs are different.

We went a little crazy trying to figure out what’s going on… For about 3 minutes, we were having a great conversation. But we quickly converged on the little text that accompanies the log rules in any textbook… and this text says that these rules work but are only valid for $a>0$.

I kinda love this as an in-class exercise (I’ll probably forget this when I get to logarithms, but maybe posting it here will prevent me from forgetting it). Because it will force kids to (a) be confuzzled, (b) talk through ideas, (c) go back to the definition and qualifications for the log rules, and (d) see that these rules are indeed valid (we didn’t break math), but they are a bit more restrictive that we might have thought.

What I love is that $\ln(x^{-2})=-2\ln(x)$ isn’t actually an identity. But we are so used to using the rules blindly, robotically, that we never think about it. But for it to be a good mathematical statement, you need to qualify it! You need to say this is only an equivalence for $x>0$. This was a good reminder for us.
{"ft_lang_label":"__label__en","ft_lang_prob":0.9427882,"math_prob":0.93922764,"size":2752,"snap":"2021-43-2021-49","text_gpt3_token_len":638,"char_repetition_ratio":0.12117904,"word_repetition_ratio":0.0,"special_character_ratio":0.23255815,"punctuation_ratio":0.10482529,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9817058,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100,101,102],"im_url_duplicate_count":[null,10,null,5,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-17T00:47:08Z\",\"WARC-Record-ID\":\"<urn:uuid:a2b62466-523a-4554-ae54-509acab85b1d>\",\"Content-Length\":\"137827\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2d208da3-f5cd-47bb-b50d-de90eac84536>\",\"WARC-Concurrent-To\":\"<urn:uuid:a2d627df-9754-492b-8e4f-6cdcf08e988d>\",\"WARC-IP-Address\":\"192.0.78.25\",\"WARC-Target-URI\":\"https://samjshah.com/tag/good-math-problems/page/2/\",\"WARC-Payload-Digest\":\"sha1:XCYMS5DOAUTVG7IKQ6CL2V752LN2JBE7\",\"WARC-Block-Digest\":\"sha1:UZQG4MG3QDWEIX2T52IGNNJF2BTVMXVF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585045.2_warc_CC-MAIN-20211016231019-20211017021019-00465.warc.gz\"}"}