Tag problem solving

Paula Beardell Krieg’s 72 degree question

A few weeks ago I got this question from Paula Beardell Krieg on Twitter:

Today I went through this problem with the boys – the difficulty of this exercise surprised me a bit. They really struggled to see how you could tell if an angle was 72 degrees.

Here’s the introduction. The boys noticed a few things about the picture and got some ideas with how to proceed:

Next we drew the two squares on a piece of paper and I let the boys explore the question. Here they struggled to make much progress beyond the things that they noticed in the first part of the project:

The thing giving them trouble was that they didn’t know any relationships between angles in a right triangle with a 72 degree angle. That left them completely stuck. Eventually they decided to measure the squares and found that they had something that looked like a triangle with side 1, 3, and \sqrt{10}.

Next we explored some of the ideas around 1, 3, \sqrt{10} triangles. After a little nudging from me they decided to measure the angles with a protractor.

Now I showed them my solution and let them see where the 1, 3, \sqrt{10} triangle comes up:

Finally, I let them play with two sets of triangles that I printed overnight. Two of these triangles are right triangles with 72 and 18 degree angles, and two of them are 1, 3, \sqrt{10} triangles. The question is -> are all 4 triangles the same?

Here are pictures (to scale) of the two triangles. You can see how similar they are.

First, the right triangle with a 72 degree angle:

72 degree picture

Second, the 1, 3, \sqrt{10} triangle:

one three root 10 picture.jpg

Tomorrow we’ll explore a second similarity between these two triangles. I found it playing around while I was making the triangles yesterday 🙂


Working through an AMC 8 geometry problem

My younger son was working through the 1989 American Junior High Mathematics Exam this morning and got stuck on this problem:

Screen Shot 2017-07-14 at 8.33.41 AM copy.jpg

Here’s a link to the entire exam on Art of Problem Solving’s website:

The 1989 AJHME on Art of Problem Solving’s website

I thought this problem would make for a nice project since there are a couple of good mathematical ideas in it, so we sat down to talk about it. My younger son talked through his approach first:

My older son went next and had a different approach:

To wrap up we talked about how the answer would change if the problem was set up with a slightly different arrangement of the cubes and the boys found their way to an important idea in geometry:

I’m really happy that the old AMC problems are available – they are a wonderful resource to use to find challenging but accessible problems for kids.

A terrific example for calculus students from Nassim Taleb and Alexander Bogomolny

I saw a wonderful exchange on twitter yesterday on a problem posted by Alexander Bogomolny:

At first this problem didn’t really jump off the page as a good first year calculus example, but then I as the solution that Nassim Taleb posted:

I’m a tiny bit time constrained this morning and can’t get the Taleb tweet to embed right, so here’s the solution a second time just in case the embedding remains broken:


So, Taleb reduces the difficult-looking limit and sums to two integrals. The ideas underlying this reduction are both beautiful on their own and fundamental in calculus.

A few questions that I think would be worth discussing with calculus students are:

(1) [this one was discussed in the twitter thread] Why did the integrals start at 0, and does that matter?

(2) Why is ratio of the integrals equal to the ratio of the sums? This answer to this question is related to the answer to (1). It is also an excellent way to reinforce some of the main ideas behind Riemann sums.

(3) Probably less mathematically interesting, but a good challenge exercise for students is evaluating Taleb’s integral formulation of the problem using l’Hospital’s rule. I say “less mathematically interesting” because you have to evaluate the same integrals in both approaches, but the approach via l’Hospital’s rule allows you to discuss the Fundamental Theorem of Calculus and also review the chain rule. The arithmetic here requires you to be extra careful, but I think the other ideas outweigh the annoying arithmetic.

Too bad that the school year is over – but this is a great example to keep in your back pocket for next year’s calculus classes!

A night with Cut the Knot, Nassim Taleb, and some Supernova

Please note the correction at the bottom of the post

A further correction – there is still an error. Ugh. This approach may not work, unfortunately . . .

Saw a neat problem from Alexander Bogomolny earlier today:

I actually missed the problem when it was initial posted, but saw it via Nassim Taleb’s solution:

The problem sort of gnawed at me all day and I figured it was in the maybe 1 in 10 problems that Bogomolny posts that I might be able to solve.

So, tonight I poured a glass of Supernova and gave it a go

One thing on my mind all day with this problem was Jensen’s inequality. What I would *love* to be able to do is say that by Jensen’s inequality:

(1/3) \sqrt{x^2 + 3} + (1/3) \sqrt{y^2 + 3} + (1/3) \sqrt{xy + 3}

\geq \sqrt{ (1/3)( x^2 + y^2 + xy) + 3}

Which is easily seen to be \geq 2 because of the constraint x + y = 2. That work would show the original inequality was \geq 6.

The approach has a tiny bit of merit since \sqrt{x^2 + 3} is concave up for x between 0 and 2 -> here’s a little Mathematica plot showing that the second derivative is indeed positive on 0 to 2:


But . . . the problem is that folding in the 3rd term in the sum is stretching the rules of Jensen’s inequality a bit, I think, since it is not of the form \sqrt{a^2 + 3}.

With the first two terms, though, applying Jensen’s inequality seems ok, but I now need (1/2)’s instead of (1/3)’s since there are only two terms. So, I’ll use Jensen’s on the first two terms only and try to show that

(1/2) \sqrt{x^2 + 3} + (1/2) \sqrt{y^2 + 3} + (1/2) \sqrt{xy + 3} \geq 3

By Jensen’s inequality this new sum is

\geq \sqrt{ \frac{x^2 + y^2}{2} + 3} + (1/2) \sqrt{xy + 3}

A bit of algebra and the fact that x + y = 2 allows us to simplify this expression to:

\sqrt{5 - xy} + (1/2) \sqrt{xy + 3}

and then further to:

\sqrt{ (x - 1)^2 + 4} + (1/2) \sqrt{4 - (x - 1)^2}

Now we are just down to a fairly straightforward calculus problem, and I’ll let Mathematica do the heavy lifting since the algebra isn’t that interesting:


We can see visually that the minimum occurs at x = 1 from the plot, and the plot of the derivative further confirms that there is only one critical point. The value of the last expression at x = 1 is indeed 3 as we were hoping.

So, Jensen’s inequality, a bit of calculus, and a nice glass of scotch shows that the original inequality is indeed true.

Thanks to Alexander Bogomolny for the problem, and to Nassim Taleb for his solution that got me thinking about the problem.


I received a note from Alexander Bogomoly over night. He spotted an error in the calculation:


and I thought my kids having trouble sleeping and waking me up at 5:00 am today was a bad start to the day!

But it seems that I’ve gotten very lucky – both learning from my carelessness in applying Jensens inequality and that the path forward from Bogomolny’s correction is easier than the path I actually took.

Starting here – we wish to show that:

(1/2) \sqrt{x^2 + 3} + (1/2) \sqrt{y^2 + 3} + (1/2) \sqrt{xy + 3} \geq 3

The correction shows that the expression on the left hand side is \geq than

\sqrt{ (\frac{x + y}{2})^2 + 3} + (1/2) \sqrt{xy + 3}

but since x + y = 2, the first piece of this expression is equal to 2 and the 2nd expression simplifies as before. So we are left with

2 + (1/2) \sqrt{4 - (x - 1)^2}

and this expression has a maximum of 3 at x = 1.

That means that the expression we were trying to show to be greater than 3 is indeed greater than 3, and the expression in the original tweet is greater than 6.

I’m grateful to Alexander Bogomolny for spotting the error in my original argument.

A hand waving approach to a problem posted by Cut the Knot

Saw this tweet yesterday:

It was a fun problem to think about and the two solutions on the site use the Stolz-Cesaro Lemma, which is basically l’Hospital’s rule for sums.

Through the various Christmas preparations yesterday I was wondering if there was a simple way to see why the limit exists in the first place. What follows below isn’t a rigorous proof (or even close to one!) but instead how I convinced myself that the limit probably does exist.

Since seeing Tim Gowers “live blog” his solution to an old IMO problem, I’ve been interested in occasionally sharing the solution process rather than polished solutions to problems. Two examples of problems I’ve used for that idea are below:

A Challenge / Plea to math folks

A challenge relating to a few problems giving my son trouble

So, for the problem at hand, here’s my “hand waving” approach to convincing myself that the limit even existed:

We know that:

\lim_{x\to\infty} E_n = \lim_{x\to\infty} (1 + \frac{1}{n})^n = e

and that

\lim_{x\to\infty} H_n = \lim_{x\to\infty} (1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{n}) \approx \ln{n} + \gamma

So I’ll approximate the difference we are considering like this for large n:

(E_{n+1})^{H_{n+1}} - (E_n)^{H_n}

\approx e^{H_{n+1}} - e^{H_n}

\approx (e^{H_n})(e^{\frac{1}{n+1}} - 1)

\approx (e^{\ln{n} + \gamma}) (\frac{1}{n+1} + \frac{1}{2(x+1)^2} + \ldots )

\approx (e^{\gamma}) * n * (\frac{1}{n+1} + \frac{1}{2(n+1)^2} + \ldots )

\approx e^{\gamma}

The fact that this hand waving approach arrives at the “right” answer is just a coincidence as I’m playing pretty fast and loose with limit rules. But, at least I now have some indication that this strange (and lovely!) \infty - \infty limit might actually exist.

Just for fun here’s what the expression looks like for n up to 1,000,000:


Definitely a fun little problem to noodle over 🙂

An AMC12 algebra problem that gave my son trouble

The problem below gave my son some trouble this morning:


When he got home from school we talked about it in more detail and it seemed to make more sense for him than it did this morning. The problem is a nice introductory algebra / quadratic problem:

Next I showed him a similar solution, but where “x” represented a different number:

Finally – just for a completely different way of looking at the problem – I wanted to show him a way that we could use the choices to help us find the solution. This is sort of cheating, but he was very confused by the problem this morning and I wanted to show him a way to get a little un-stuck when you are stuck.

Also, we got interrupted by the guy servicing our furnace – so sorry the video jumps in the middle 🙂

Examples of kids and problem solving

Both kids gave nice examples of the problem solving process in the two videos we did last night, so I wanted to highlight those videos with a short blog post.

First up was my younger son. He’s learning algebra this year and has a really nice way of thinking and talking through problems. I love how deliberate he is and how he discovers his own mistakes. The problem that he’s working on here is to find 3 solutions to the equation 3A – 5B = 9.

Next up was my older son. The problem he’s working on is an old Mathcounts problem, and it is pretty challenging:

What fraction of the first 100 triangular numbers are divisible by 7?

His work is a nice example of, for lack of a better phrase, the discovery process. Initially he does not see how to solve the problem, but I love his path to the solution.

After he finished I showed him two other approaches to solving the problem, just to help him see how a few other ideas in math can connect to this problem:

I wanted to share these examples to show that problem solving in math isn’t all about speed. A slow, deliberate process is a great way to get to the solution of a problem.

What a kid learning math can look like – struggling with a geometry problem

The problem my son wanted to work on this morning seemed fun to me:

A right triangle has hypotenuse 8 and area 8, what is the perimeter of this triangle?

The combination of algebra and geometry required to solve this problem gave my son some trouble. The first three videos below show his thought process while working through the problem. The last video is a recap of the solution he found.

(1) The first part is a struggle to find any path that leads to the solution

(2) The second part shows the struggle to find how to use the two algebraic identities that we have to help solve the problem:

(3) The third part is the solution to the problem:

(4) The last part is a recap of the solution. I was hoping that going through it one more time would help him understand a few of the ideas a little better.

Summer math talks

I’ll be giving two talks at math camps this summer. The first is at the east coast Idea math camp at the beginning of July and the second is at a math camp at Williams college in the middle of July.

I’m super excited to be able to have the opportunity to give these talks and can’t wait for the chance to interact with the students at the two camps.

The topics I’ll cover – not surprisingly – will come from some of the projects I’ve done with the boys this year. The talk at Williams is about 45 minutes shorter than the Idea math talk, so one or two of the topics below will get cut out.

Here’s are the ideas I’d like to cover:

(1) Larry Guth’s “No Rectangles” Problem

Larry Guth’s “No Rectangles” problem

I covered the 3×3 and 4×4 cases with the 2nd and 3rd graders at my younger son’s school as part of Family Math night and the kids loved the problem. With high school students I’d like to try to explore some of the larger cases and also discuss why this is a difficult problem for computers to solve.

Patrick Honner also showed me this related problem which I’ll leave as a challenge for the students 🙂

(2) Ann-Marie Ison’s Math Art

Our projects with Ann-Marie Ison’s art

I’m still waiting to hear what sort of projection capability I’ll have at the two events, but oh do I hope I have the ability to share this program with the students:

The explorations you can do with this simple modular arithmetic idea are incredible.


(3) A problem from Po-Shen Loh’s MoMath talk

What I love about Loh’s talk is that he takes an extremely difficult problem – one from the 2010 International Mathematics Olympiad – and turns it into a talk that is accessible to the public.

His approach is so accessible that I talked through the first part of the problem with my 4th grade son:

I’m very excited to hear the different guesses that the students have for the answer to Loh’s two questions.

(4) Bjorn Poonen’s N-Dimensional Sphere problem


Here’s the problem and our project on the problem:

A strange problem I overheard Bjorn Poonen discussing

Bjorn Poonen’s sphere problem

I’m guessing that not all of the kids will have seen geometry beyond 3 dimensions, so this problem will take a little bit of setting up.  Luckily the only complicated bit of math that they need to understand it is the Pythagorean theorem and I’m guessing that all of them will know that theorem.

I was blown away by the answers to Poonen’s questions when I finally worked through them.  This was also one of the most enjoyable projects that I’ve done with the boys this year.

I hope I have enough time to show the students the fun relationship between \pi and e hiding in this problem, too:

A fun surprise in Bjorn Poonen’s n-dimensional sphere problem


Can’t wait to talk about these problems with the kids!

Michael Fenton’s counting problem

[I hope this write up makes sense – I wrote it while my kids were watching Lord of the Rings “Two Towers” – so there was a bit of a distraction in the background]

Saw this problem posed on Twitter yesterday:

I like Fenton’s request to share solutions, and actually made a similar request previously with a problem from the Euorpean Girls’ Math Olympiad:

A Challenge / Plea to Math Folks

Having made a similar request, I’m sort of obligated to respond to this one!

There are a couple of different ways to approach Fenton’s problem, and I’d guess that the counting technique of “inclusion / exclusion” would probably be a common choice of how to attack the problem.

However, since the numbers were small enough I thought it would be fun to see if I could count the possibilities directly.

The main ideas I used to approach the problem were these:

(1) I’m going to try to count the arrangements with no songs by the artists next to each other. If I can find the probability of finding such an arrangement then 1 – that probability is the probability of having at least one pair.

(2) Next, I’ll ignore the different songs by different artists and just look for arrangements of the 8 numbers 1,1,2,2,3,3,4,4 with no numbers that are the same next to each other. If I can count these sets, the the arrangements of the number of songs with no songs by the same artist next to each other is just 2^4 times the number of these sets.

(2) For any arrangement of the 8 numbers – say 1,4,3,2,3,4,1,2 – a unique permutation of the numbers 1,2,3,4 will change the arrangement to one in which we encounter the numbers in increasing order. In my example, simply switching 2 and 4 changes the example to 1,2,3,4,3,2,1,4.

For a second example. Consider 3,2,4,3,2,1,4,1. If I map 1 to 3, 2 to 2, 3 to 4, and 4 to 1 I have the new sequence – 1,2,3,1,2,4,3,4 – in which we, again, encounter the numbers in the order 1,2,3,4. So, using this idea, if I can count the number of arrangements with no pairs in which the numbers appear in order, all I need to do is multiply by 4! to get the total number of arrangements of similar patterns.

So, let’s count:

(A) Arrangements of the form 1,2,3,4,_,_,_,_

There are 3 choices for the 5th numbers, and after that choice the remaining 3 numbers can be arranged in any of the 3! ways. So, there are 3*3! arrangements of this form, or 18.

(B) Arrangements of the form 1,2,3,1,_,_,_,_

There are two cases to consider:

Case 1: the 5th number is 4. There are then 2 choices for the 6th number (2 or 3) and 2 ways to arrange the remaining 2 numbers ( 4,2 or 2,4, for example, if the first choice was 3). Thus there are 2*2 = 4 arrangements of this form.

Case 2: The 5th number is 2 or 3. So, there are 2 choices for the 5th number. 4 must be the 6th number otherwise the arrangement would end 4,4 which we don’t want. The final two numbers are then forced. Thus, there are 2 arrangements of this form.

So, a total of 6 arrangements begin 1,2,3,1

(C) Arrangements of the form 1,2,3,2,_,_,_,_

This case is essentially the same as case (B) – so we get 6 more arrangements.

We’ve now covered all of the cases of the form 1,2,3,_,_,_,_,_ and the only other cases begin with 1,2,1 since the numbers must appear in order.

(D) Arrangements of the form 1,2,1,2,_,_,_,_

There is only 1 – which sort of surprised me (!) but the remaining slots are forced to be 3,4,3,4 by the increasing condition and the condition that no numbers appear next to the same number.

(E) Arrangements of the form 1,2,1,3,_,_,_,_

As with (B) above, we have two cases:

Case 1: The 5th number is 2.

Here the arrangement is forced to finish 4,3,4. So, there is only 1 arrangement in this case.

Case 2: The 5th number is 4.

Here we have 2 choices for the 6th number – 2 or 3. For either choice the remaining two numbers can be arranged in any order (4,2 or 2,4 if the choice for the 6th number was 3, for example). Thus there are 4 cases here.

So, part (E) has 5 total cases.

Putting it all together.

So, that’s all the cases, and we end up with 18 + 4 + 2 + 4 + 2 + 1 + 1 + 4 = 36 total arrangements in which (i) the numbers appear in increasing order, and (ii) there are no pairs.

As mentioned at the beginning, when we distinguish between the two 1’s, the two 2’s, the two 3’s, and the two 4’s, the number of arrangements is multiplied by 16. Also, when we no longer require the numbers to appear in order we multiply the number of arrangements by 24.

So, the total number of arrangements with no pairs is 36*16*24. The total number of arrangements is 8!, so the probably of having no pairs is:

(36 * 16 * 24) / 8! = (36 * 16 ) / ( 8 * 7 * 6 * 5) = 12 / 35.

Thus, the probability of having at least one pair is 23 / 35.