Tag contest math

A project inspired by an AMC 12 octagon problem

The problem pictured below from the 2003 AMC 12 gave my son some trouble:

Screen Shot 2016-12-26 at 9.58.08 AM.png

We talked through it together a few days ago, but I thought it would be fun to try to do an octagon-inspired math project today.

We started with the problem and then talked a bit about a 3d print we found on Thingiverse:

Next we took a look at a version of the 3d printed shape that we made from our Zometool set. You can’t make a regular octagon with a Zometool set, and the fact that our shape didn’t have a regular octagon led to a good discussion:

For the last part of the project we tried to find the volume of our truncated cube.

An AMC12 algebra problem that gave my son trouble

The problem below gave my son some trouble this morning:

Algebra.jpg

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 🙂

A challenge relating to a few problems giving my son trouble

I’ve seen some interesting ideas from Tracy Johnston Zager over the last week about the relationship between learning math and intuition. For example:

Although I’ve been traveling a bit for work this week the relationship between learning math and intuition has stayed in my head. Sometimes my thoughts have drifted to and old blog post about a problem from the European Girls’ Math Olympaid:

A Challenge / Plea to math folks

That post, in turn, was inspired by an old post by Tim Gowers where he “live blogged” his work while he solved a problem from the International Mathematics Olympiad.

It can be really hard for anyone to know what math intuition looks like because everyone sees polished solutions way more often than they see the actual process of doing math.

That’s part of the reason I make the “what a kid learning math can look like” posts – so everyone can see that the path kids (or anyone!) actually takes to the solution of a problem is hardly ever a straight line:

Our “what a kid learning math can look like” series

The other thing on my mind this week has been some old AMC 10 problems that have really given my older son some trouble. These are pretty challenging problems and require quite a bit of mathematical intuition to solve.

So, I’d like to make the same challenge with these problems that I made with the problem from the European Girls’ Math Olympiad – “live blog” yourself solving one of these problems. Post the though process rather than a perfect solution. Let people see *where* your mathematical intuition came into play.

problem-19

 

problem19

problem

Struggling through a challenging AMC 10 problem

This problem gave my son some difficulty yesterday – it is problem #19 from the 2011 AMC 10a

AMC10.jpg

Last night we talked through the problem. The talk took a while, but I was happy to have him slowly see the path to the solution. Here’s his initial look at the problem:

Next we looked at the equation y^2 - x^2 = 141. Solving this equation in integers is a nice lesson in factoring. Unfortunately by working a bit too quickly he goes down a wrong fork for a little bit.

In the last video we found that the original population of the town might have been 484, and it might have changed to 634 and then once more to 784. We had to check if 784 was a perfect square.

Finally, we needed to compute the approximate value (as a percent) of 300 / 484. The final step in this problem is a great exercise in estimating.

So, a really challenging problem, but also a great problem to learn from. We went through it one more time this morning just to make sure that some of the lessons had sunk in.

Why I love watching my kids learn math

Had a great night with the boys tonight. My older son was working on some old AMC 10 problems and we talked through one that stumped him for his movie:

It was #15 from the 2013 AMC 10a:

 

AMC10.jpg

Next I spent some time with my younger son. He’s been studying the basics of lines using Art of Problem Solving’s Introduction to Algebra book plus a little bit of Khan Academy (when I’ve been traveling for work). I asked him what he’d learned so far and loved his response. It was a great reminder of the joy of learning new math ideas for the first time:

Struggling with a challenging AMC 10 geometry problem

This problem from the 2016 AMC 10 a gave my son some trouble yesterday:

AMC Problem.jpg

We talked through it last night and it was interesting to see where his intuition was off:

After finding a solution in the last video that he knew was wrong (because it wasn’t one of the choices on the test) we looked back to see if there was alternate approach to the problem. This approach led us to find the actual solution and also the mistake in the first approach:

It is always fascinating to the though process on a challenging problem. Sometimes the thought process is so close to the right approach that the mistake is really tricky to spot.

A challenging counting problem from the 2011 AMC 8

My older son has been preparing for the AMC8 and this problem from 2011 gave him a little trouble:

 

2011

We talked through it this morning and he was still a little confused about why his original answer isn’t correct. The error is pretty subtle – especially for a test that gives you roughly 2 min per problem:

So, we talked for a bit more and he was able to find some numbers that he counted that did not fit the requirements of the problem:

Finally, when he got home from school tonight we revisited the problem and counted the number of solutions directly:

I like this problem a lot – it is a great one for helping you learn how to count carefully!

Po-Shen Loh’s coin problem part 2

Yesterday we began looking at a problem that Po-Shen Loh presented during his fantastic public lecture at the Museum of Math. Here’s his lecture:

and here’s the conversation we had yesterday about the problem in his talk:

Po-Shen Loh’s coin problem

It may help to look at the first part of our project before going through this one since we sort of dive in where we left off and don’t really explain the problem in great detail.

As a quick overview, though, yesterday’s problem was about a row of 6 boxes each containing one coin. You play a game in which the only rule is that you can take a coin out of one box and replace it with two coins in the box immediately to the right (so you cannot apply this move to the right most box). You can stop playing the game at any time and take all of the money in the boxes. The question is what is the most amount of money you could win by playing this game?

Today’s project is looking at a twist on that game (explained in the video below). That twist has a surprising impact on the value of the game. Finding the maximum value of the new game is a problem that is too difficult for kids, but playing this new game is a fantastic math activity for kids.

Here’s the introduction to the new game and some thoughts from the boys:

Next I had my older son play the new game to see what would happen. He just finished 6th grade hopefully this video shows how a 12 year old might think about this game:

After my older son’s walk through the game I had my younger son play. He just finished 4th grade, so hopefully this video shows how a 4th grader can think about this game (and also how a kid can learn a lot from seeing it played just once):

Finally, we talked for a bit about strategy for the game. Then I revealed the actual problem from the International Mathematics Olympiad. The boys were a little skeptical that you could ever find a number as big as the number from the problem (!!):

After we finished the boys played the game one more time off camera and found a value of 7*2^{27}. Not bad for their third try!

This is such a fun problem. I’m excited to see the math camp kids play around with this problem today and I also can’t wait to use a version of this project for 4th and 5th grade Family Math night next year.

I’m so happy to have seen Loh’s MoMath talk – I never would have thought that you could transform IMO problems into great project for kids!

Po-Shen Loh’s coin problem

Last year I saw this amazing presentation from Po-Shen Loh at the Museum of Math:

In the presentation Loh does something that seems close to impossible. He takes a difficult problem from the International Mathematics Olympiad (problem #5 from 2010) – a problem on which roughly 2/3 of the 2010 IMO participants received 0/7 points! – and turns it into a fantastic public lecture.

Tomorrow I’m going to give a talk at the East Coast Idea Math camp and I’m using problem in Loh’s lecture for one part of my talk. For our Family Math project today I used the problem in the first part of Loh’s talk with my kids.

Here’s how it went.

(1) First I introduced the problem and the boys gave their initial reaction:

(2) Having found a value of 63 cents in the last video next we tried to see if we could find some other values by moving the coins in different ways. We also had a nice discussion about how you could determine if 63 was the maximum value.

(3) Before talking more about why 63 is the maximum value, I wanted to have my younger son try an new approach to the game just to see if we’d ever find a value other than 63.

(4) Finally, we wrapped up by talking about one way you can see that the value of this game is 63. One lucky coincidence is that my younger son was learning about different bases this week. That coincidence helped him see the connection between this game and binary counting fairly quickly.

So, I love Loh’s presentation. It is so cool (and inspiring) to see him take a super challenging IMO problem and turn it into a public lecture. I won’t walk the kids all the way through the solution of the original problem tomorrow, but I will use Loh’s approach to create a big discussion about the 2nd part of the problem. Can’t wait to see what values the kids find for the 2nd game.

My respose to my challenge

Earlier today I wrote about problem #! from the 2016 European Girls’ Mathematical Olympiad:

A Challenge / Plea to Math Folks

I thought the problem was wonderful and “live blogs” of different people solving the problem would help people (kids especially) learn about different aspects of mathematical thinking.

Here’s the problem:

 

Screen Shot 2016-04-21 at 6.05.15 PM

And here’s a sketch of the process I went through to solve it. First, though, here’s what my scattered notes looked like by the time I got to the end:

Step 1: Total confusion after reading the problem. You just have a seemingly random set of numbers . . . how can you show that the max of one set of expressions is even related to the min of another set of expressions? No idea what to do.

Step 2:  Oh wait, I remember the arithmetic mean / geometric mean inequality – that’s what’s written down in the middle of my note paper.  You’ll see the scratched out inequality when I first wrote it down, because I remembered the inequality backwards.

This inequality comes from the simple idea that the quantity (a - b)^2   has to be greater than or equal to zero.  The idea is that any real numbers squared has to be greater than or equal to zero.  When you expend the algebraic expression you get  a^2 - 2ab + b^2 \geq 0 which simplifies to a^2 + b^2 \geq 2ab.

That’s what I’ve got written down in the middle of the page.   Feeling good at this point in the problem solving process!

Step 3:  Wait a minute, the inequality in the problem goes the other way.  It says less than or equal to not greater than or equal to.  What??  This problem seems to want me to prove the exact opposite of the arithmetic mean / geometric mean inequality.  But this inequality is so true that it has its own Wikipedia page:

The Arithmetic Mean / Geometric Mean inequality on Wikipedia

So proving the opposite is probably not going to be the right strategy!

Ummmm . . . . . Not feeling good anymore . . . .

Step 4:  Well . . .  x_{i}^2 + x_{x+1}^2 sort of looks like a geometric expression.  It relates to the distance between the point (x_i,x_{i+1}) and the point (0,0) in the plane – maybe there is a geometric angle to the problem.   What would the term 2x_j x_{j+1} mean in this setting?   It would be an area, or maybe twice the area of a rectangle, but how would I compare that to a length?   Hmmm . . .

No idea . . . .

Step 5:  Time for an example – you’ll see that I wrote down the numbers 1, 3, and 5 in the upper left hand corner of the paper and in the upper middle right of the paper I wrote down the values for the two sets of expressions in the problem:

 

The min of x_{i}^2 + x_{x+1}^2 was 10 and the max of 2x_j x_{j+1} was 30, so the inequality we are trying to prove seems to be true for the numbers 1, 3, 5.  Yay, I guess, but what now?

Step 6:  I started wondering about why the problem wanted you to use an odd number of numbers, so I tried to see if I could find a set of 4 numbers that didn’t work.  Maybe that would help me see something.  Just to keep the numbers as simple as possible I tried 1, 2, 1, 2 and found that the min / max inequality in the problem didn’t work.  x_{i}^2 + x_{i+1}^2 was always 5 and 2x_j x_{j+1} was always 4.

Interesting.

Step 7:  You’ll see that I’ve got a < b < c written down in the lower left hand part of my paper.   Maybe what was special about my 1, 3, 5 example is that the numbers were increasing.   When you have three increasing numbers like this a^2 + b^2 will be less than 2bc because 2bc is greater than 2b^2 and 2b^2 is greater than a^2 + b^2 since b is greater than a.  That’s what this part is saying:

That’s it for the notes in the lower left hand corner.  A nearly identical argument would show that three decreasing numbers would produce a solution to the problem, too.

By the way, once you find that one of the 2 x_{i} x_{i+1} expressions is greater than one of the x_{j}^2 + x_{j+1}^2 expressions, you’ve solved the problem because the max of the 2 x_{i} x_{i+1} expressions has to be greater than or equal to the one you’ve found and the min of the x_{j}^2 + x_{j+1}^2 expressions has to be less than or equal to the one you’ve found.  That means the max of the 2 x_{i} x_{i+1} terms will be larger than the min of the x_{j}^2 + x_{j+1}^2 terms.

So, yay, we can solve the problem if we have three increasing (or decreasing) numbers in  a row, but why does there have to be three numbers like this in a row?  It seems as if the numbers are chosen totally at random.  Hmmmmmmm . . . .

Step 8:  The triangle on the right hand side of my notes – it wasn’t just a doodle!

The numbers x_1 to x_n aren’t really arranged in a row.  The problem points out that the last comparison involves x_{n+1} which is equal to x_1.   The numbers are really going around in a circle – or a triangle as I’ve drawn.

As you move from one number to the next you will either increase or decrease (because if two are ever equal the inequality in the problem is an equality and the problem is solved immediately).  You are making n up or down movements as you move around the circle of numbers, but since n is odd you have an odd number of total steps around the circle, and  the number of times you move up in value as you step cannot be equal to the number times you move down in value.

Aha.

You can imagine going around the triangle and having 2 up comparisons and 1 down.  The two ups have to be next to each other because there are only three sides, so any arrangement of the two ups and one down has to have the two ups next to each other.  As soon as you have two ups in a row you know you have three increasing numbers.

The general case isn’t that much harder.  Assume that you have more ups than downs and you want to try to spread out the ups and downs as much as possible so that there are never two of the same type  in a row.  Put, for example,  an up in the first position, a down in the second, and etc.  The ups will always be in odd positions and the downs will be in even positions.  If you are really lucky and can spread them out as much as possible you’ll have an up in the last position (since that position is odd) with no ups or downs next to each other.

But WAIT, since the numbers are going around in a circle the last position is really next to the first position.  oops – even if you are lucky enough to be able spread the ups and downs out as much as possible you end up with two next to each other.  So . . . you are forced to have either two ups or two downs together (which ever type of comparison there are more of)  when you put an odd number of numbers around a circle!

That’s it – that’s the main idea in the problem!

I know this write up isn’t perfect and certainly not perfect enough for something that a contestant would need to write up during the contest, but I wasn’t going for perfection here.  I just wanted to communicate my thought process as I worked my way through the problem.

This was a fun problem to think through and I hope that some ideas about how people trained in mathematics think about problems like this one came through in my write up.