Sharing Problem #2 from the European Girls’s Mathematical Olympiad with my younger son

Saw this problem from the 2019 EGMO last week and though it would be a great one to share with kids:

My older son was working on something else tonight, so I talked through the problem with my younger son (he’s in 7th grade). The aim of the project tonight was not to solve the problem, but just to have him play around with a few simple cases and see if he could take a guess at the answer.

I started by just sharing the problem and making sure he understood the ideas and the constraints:

Having looked at the 2×2 case in the prior video, we moved on to the 4×4 case in this video. He had some pretty interesting ideas about how to check if we’d found the maximum number of dominoes:

Now that we’d satisfied ourselves with the 4×4 case, we moved to the 6×6 case. This case is a little harder, but still accessible to kids. Here’s his first attempt at a solution – the trouble is that we weren’t sure if it used the maximum number of dominoes:

It took a bit more experimenting to see that we’d used the most dominoes we could in the last video, and you can see him starting to understand some of the patterns in the problem as he experiments.

By the end of this video he guessed that the maximum number for the 8×8 grid would be 10.

The final challenge was an 8×8 grid. In the first attempt at filling in the board with dominoes we kept getting stuck – but his thoughts about the problem are very interesting:

With one more try through the 8×8 board we were able to fit 10 – yay!

This is a great problem to share with kids. Again, even if they can’t get all the way to understanding the general solution, there are plenty of things they can play with and understand – and also tons of ways to approach the problem!

Sharring problem A1 from the 2017 Putnam with kids

We had a snow day today and I finally got around to sharing a neat problem from the 2017 Putnam Exam with the boys.

When I first saw the problem I thought it would be absolutely terrific to share with kids:

I started off the project today by having them read the problem and spending a little bit of time playing around:

After the initial conversation the boys, I triehd to start getting a bit more precise. The first sequence of numbers they knew was in the set was 2, 7, 12, 17, . . . .

They were not sure if 4 was in the set or not. My first challenge problem to them was to show that if 4 was in the set, then 3 would be in the set.

My next challenge question was whether or not 1 would be in the set.

Now we moved on to one of the number theory aspects of the problem – is 5 in the set?

During this conversation my younger son noticed that we had found a number that was 1 mod 5.

Finally, we talked through how you could find 6 from the number my son noticed in the last video.

I’m really happy with how this project went. This problem is not one (obviously) that I would expect the kids to be able to solve on their own, but most of the steps necessary to solve the problem are accessible to kids. It was really neat to hear their ideas along the way.

Talking through two problems from the 2005 AMC 10

I really enjoy using old AMC problems to talk about math with the boys.

These two problems gave my older son some trouble today:

Tonight I had a chance to talk through these problems with them.

Here’s the probability problem:

Here’s the GCD problem:

A fun Fibonacci number surprise with a 1, 2, Sqrt[5] right triangle

My son had a little trouble with this problem from the 2009 AMC 10 B yesterday:

I should him how to solve the problem a few different ways – including by folding!

One of the ways I talked about was finding a rough answer by approximating $\sqrt{5}$ by the fraction 9/4. We found this number by looking at a calculator and seeing that $\sqrt{5} \approx 2.23607.$

As I thought about the problem more last night, I realized that 9/4 is part of the continued fraction approximation for $\sqrt{5}$. The first couple of approximations that you find using continued fractions are:

2, 9/4, 38/17, and 161/72.

If I approximate the hypotenuse of the original right triangle with those numbers, I get the following approximations for the length of BD, which are all ratios of consecutive Fibonacci numbers:

If you are familiar with continued fractions and especially the continued fraction approximations for the golden ratio, the emergence of the Fibonacci numbers probably isn’t a huge surprise. I missed it the first time, though, and think that students might really enjoy seeing this little Fibonacci surprise.

An AMC 10 problem with some neat lessons about a 15-75-90 right triangle

This problem (#22 from the 2014 AMC 10a) gave my son some trouble this morning:

We ended up having a nice talk about the problem this morning. To see if the ideas really sunk in, I asked him to talk through the solution tonight, and he did a nice job:

After we finished, I wanted to go back to the 2014 AMC 10 and just happened to notice that google was also showing that Art of Problem Solving had a video about the problem. So, I thought it would be fun to watch Richard Rusczyk’s solution. Turned out to be a lucky decision since his solution was totally different than the one we found:

It was neat to see this second solution – I learned a lot about 15-75-90 triangles today!

Working through a challenging AMC 10 problem

My son was working on a few old AMC 10 problems yesterday and problem 17 from the 2016 AMC 10a gave him some trouble:

I thought this would be a nice problem to go through with him. We started by talking through the problem to make sure that he understood it:

In the last video he had the idea to check the cases with 10 and 15 balls in the bucket, so we went through those cases:

Now we tried to figure out what was happening. He was having some difficulty seeing the pattern, so I spent this video trying to help him see the pattern. The trouble for me was that the pattern was 0, 1, 2, . . ., so it was hard to find a good hint.

Finally he worked through the algebraic expression he found in the last video:

This isn’t one of the “wow, this is a great problem” AMC problems, but I still like it. To solve it you have to bring in a few different ideas, and combining those different ideas is what seemed to give my son some trouble. Hopefully going through this problem was valuable for him.

Finding Cos(72)

My older son is learning trig out of Art of Problem Solving’s Precalculus book this year. Yesterday he was working on the “sum to product” section, which derives rules for expressions like Cos(x) + Cos(y). It reminded me of one of my all time favorite math contest problems:

Today I thought I would show him my solution to that problem. What we go through probably isn’t the best or easiest solution, but I think it is an instructive solution for someone learning trig.

We started by talking about the problem and how some of the ideas he was currently learning could help solve it:

At the end of the last video we’d found a nice equation that we derived from the original problem:

$\cos(36^o) - \cos(72^o) = 2 \cos(36^o) * \cos(72^o)$

Now we used the double angle formula to simplify even more and find a cubic equation satisfied by Cos(36):

Now we tried to find the solutions to the cubic equation we found in the last video. This part gave my son a bit of trouble, but he eventually got there.

Now we were almost home! We just had to compute the value of Cos(72) and we’d be able to solve the problem. That involved one last application of the double angle formula:

I think solving this problem from scratch would be far too difficult for just about any kid just learning trig. But, the fun thing about this problem is that the ideas needed to solve the problem are all within reach using elementary trig identities. So, I think that working through the solution to this problem is a nice exercise for kids.

Struggling through an AMC 8 problem

My younger son has been practicing for the AMC 8. This week we’ll be going over a few problems here and there that give him trouble. The problem from the practice test today was #16 from the 2016 AMC 8:

This problem really gave him some trouble – as you’ll see from his 5 min struggle below:

I was caught a bit by surprise over the difficulty he was having. It wouldn’t surprise me if the mistakes he was making were quite common mistakes for a problem like this, but I was stuck on what to do. So, I decided to show him one path that leads to the solution to the problem:

So, having shown him one way to solve the problem, I challenged him to find a different solution. Initially he struggled, but then he did something pretty clever:

I definitely struggle to see a good way forward when a problem is giving one of my kids as much trouble as this one was. Hopefully my son was able to see some of the important ideas in the problem after we talked through it in the second video. I really do like the solution he came up with in the third video, though, especially since it is more geometric and less reliant on calculation.

An introductory stars and bars problem

Yesterday a counting problem from my son’s math team homework gave him a little trouble. The problem went something like this:

There are 5 different types of fasteners and you need to buy 10 total. If you need to buy at least one of each, how many different ways can you do it?

First we talked about the problem and got their initial thoughts. Then I introduced the stars and bars counting idea:

Next I tried to go through a few more examples by changing the numbers a little. The main ideas seemed a little confusing to the boys and I’d hoped a few extra examples would help. Unfortunately things weren’t going so well.

The last example in the prior video confused my younger son, so I moved on to the next video to talk about that example in more detail. By the end of this example I hoped that the general idea had sunk in, but there was still a little confusion.

So, we talked through the problem a few more times. Now the ideas seemed to be sinking in. IF you have N groups of objects (in the original problem 5 fasteners) and you have to pick M total objects (in the original problem we were trying to pick 5 fasteners) you can represent the problem with M stars and N – 1 bars. So the total number of different ways to make the selections are (M + N – 1) “choose” M or, alternately, (M + N – 1) choose (N – 1).

Not all of our projects go super well. Here my mistake was thinking that I could introduce an advanced concept and the boys would immediately understand it. I feel like the ideas here are definitely within their grasp and will probably spend a bit more time this weekend covering the concept. Hopefully a few more examples will do the trick.

Working through two old contest problems

I’ve been sort of on pause doing new math with the boys for the last couple of weeks. I want them to find their stride with the new school year before seeing what additional enrichment math we can do at home.

So, while on pause they’ve just been working through problems from old AMC tests in the morning. When they finish we talk through some of the problems that gave them trouble. Both problems were pretty interesting lessons (for them and me, I mean) today.

Here’s what my older son had to say about problem 18 from the 2013 AMC 10b. It was fascinating to me how he counted the numbers in this problem.

and here’s what my younger son had to say about problem #17 from the 1985 AMC 8 (which then was the American Junior High School Math Exam). It was fascinating to me to see both how he played with the averages and how he found his arithmetic mistake.

I love using these old AMC problems to keep the kids engaged with math. It is always fun to see what sorts of ideas give them problems and just as fun to see their problem solving strategies.