## Coordinate integer puzzle

Today’s puzzle is a nice question from the 2019 Ritangle competition. Ritangle is an integral maths competition aimed at students who study mathematics Level (or equivalent). They have tons of questions from the last few years on the website, many of which look very interesting. This one is one that I came across ages ago in Chris Smith’s (@aap03102) maths newsletter and it’s been sitting in my screenshots folder on my phone since. Here it is:

It’s an interesting question, and one I put a bit of thought into. First I considered what the midpoint was:

The mean of the x coordinates and the mean of the y coordinates.

Then I thought, what sums could I have for the coordinates?

Because all the coordinates, including the midpoint, had to be integers I could discount the odd sums:

Then I considered how I could get these sums. And what half of them would be:

Discounting any possibilities that would lead to repeating the same number in the sum and the midpoint it left limited possibilities. And when looking at which pairs involved all of the numbers 1-6 and nothing else it only left two possibilities:

Once I knew what the coordinates would be I could look at what their differences would be. Due to the nature of Pythagoras’s Theorem, it wouldn’t matter whether the coordinates were the x or y, or even which were paired, as they’d give the same length of line, so there was two possible lengths. 2rt2 and 4rt2. So the longest possible length is 4rt2.

I found this a nice puzzle to tackle, I suspect there are other ways to go about it, and I’d love to hear how you approached it, especially if you did it differently.

## Proving sums and differences

Today I want to look at a proof puzzle that I got from a previous addition of Chris Smith’s (@aap03102) brilliant maths newsletter. The puzzle is this:

It’s one I saw a few weeks ago, had a think about, got stuck and promptly forgot all about. But today I found the screenshot on my camera roll and thought I’d another go.

I tried a few things out that became fruitless endeavours, then I considers different ways to express the integers. As we were thinking I thought about modulo 100, what are the remainders we could have, as we are dealing with sums and differences we need to get the remainders to equal 0 for the sum or difference to be divisible.

Using mod 100 we would have 100 possible remainders. I only had 52 integers though, I I needed more thought. I considered this:

Using negatives I could express b as:

So I could get the magnitude of b to be one of 51 options. That meant that if I had 52 integers, they could be expressed this way and at least 2 of them would have the same value.

If the b values had the same sign, then their difference would be 0 hence the difference of the integers would be divisible by 100. Likewise if the signs were different then their sum would be divisible by 100.

I then considered how else I could write that.

I could express the integers as 100a + b (argh, why did I use a gain but for something different. Grrr.)

Then it follows the same way that at least 2 must have the same magnitude of b either the sum or the difference is divisible by 100. As shown here with the example b = 49:

A nice puzzle that I enjoyed exploring. I think there are probably nice proofs, certainly ones laid out better. I’d love to see yours.

## Consecutive integer puzzle

Today’s puzzle comes via Chris Smith’s (@aap03102) newsletter and was bet by Guney Mentes:

Chris’s newsletter is a lot of fun, full of great maths facts and great puzzles, if you aren’t a subscriber you definitely should be.

Anyway, this puzzle looked quite nice so I thought I’d have a quick go.

I decided I needed to think about it algebraically, rather than jump in an start guessing. So I considered a list of 8 consecutive integers starting with n.

*n, n+1, n+2, n+3, n+4, n+5, n+6, n+7*

This means the sum of the blue squares would be:

*3n + 3*

The yellow:

*2n + 7*

And the pink:

*3n + 18*

Equating the blue and yellow gives:

*3n + 3 = 2n + 7*

*n = 4*

That gives blue sum = 15, yellow sum = 15 and pink sum = 30.

I thought it interesting that the blue and yellow sums were equal to the sum of the first and last term. Then I realised that there was a good reason for this. The blue and yellow sums are each equal to a quarter of the total sum. The series is arithmetic with 8 terms. The sum of a series is equal to half the number of terms multiplied by the sum of the first and the last. It occurred to me then that this might be a good puzzle to use with a post 16 class to allow a discussion about summing series.

## Pythagorean Triples

So, pythagorean triples. For those who don’t know, pythagorean triples are right angled triangles with integer (whole number) side lengths. The most famous pythagorean triple is the 3, 4, 5 triangle.

These triangles are great for maths teachers, as they can be used to make the arithmetic easier when introducing higher mathematical concepts. They are often favoured by exam boards for the same reason. Recently I have been working on the Edexcel M3 module and it seems that *EVERY* triangle they use is a 3,4,5 or a multiple there of. Noticing this fact has help my students and I save lots of time, we lost 20 minutes to this by not spotting it:

Yesterday we were revising FP1 and one of my students was finding the modulus of -7 + 24i, as she was typing the required sum into her calculator I said “25” to the general amazement of the class. They were intrigued as to how I did it so quickly, I explained I knew that the 7,24,25 triangle was a pythagorean triple and this led onto a fascinating discussion.

They thought it was an impressive trick, and I told them that yes, it was, and another good trick (courtesy of @icecolbeveridge) was knowing that the angles if a 3,4,5 triangle are roughly 53.13 and 36.87 degrees!

We discussed this further, and they asked how many triples I knew “off the top of my head”. I explained that I could write them forever if we weren’t limited to primitive triples, because I could just add 3,4 and 5 to the respective sides. This lead to a nice discussion about primitive triples. Which are those where the side lengths are co-prime (have no common factors). And then I wrote the primitive ones I could remember:

*3,4,5
5,12,13
7,24,25
9,40,41*

This prompted the question “is it some sort of sequence? I can see the smallest side is the sequence of odd numbers.” I was impressed how quickly they had spotted than, but I then remembered the *8,15,17* triangle which has an even numbered smallest side. I also told them that I have a vague recollection that there is a way to use one primitive triple to generate another but I can’t remember it. (A quick Google search didn’t help, so perhaps I imagined it. If you know it, I’d love you to tell me!)

I then discussed Euclid’s formula for generating triples and then moved on to my favourite way:

*Take two unit fractions with denominators one apart, add them. The answer in its simplest form will be a fraction in which the numerator and the denominator form the short legs of a pythagorean triple.*

I’m not sure where I first heard this method, but I do love it and I often use it when working out questions to set my classes. I explained that this method shows there are infinitely many primitive triples, as it will generate infinitely many, but that I didn’t think it would generate them all. (Again, I don’t know if it will or not, I just imagine it won’t. Do send me a link to a proof either way if you have it!)

The class, interested and enthused as ever, wanted to see proof that this always worked. They had tried lots of numbers and they had all worked, but they knew this didn’t prove it. I then said we would prove it together.

I asked how we might start, had the word induction thrown at me, then one said,

*“could we start with 1/m + 1/(m+2), then square the top and bottom of the answer and see if it’s a square?*

I shouted “excellent” and so we did:

*Obviously this isn’t the whiteboard… I forgot to photograph it so have recreated it on paper.*

The start, adding the fractions:

Then squaring the numerator and denominator, manipulating the expression and factorising:

And finally, the all important statement:

Do you have a preferred method of generating triples? Or do you know any interesting facts about them? I’d love to hear them.

## An interesting discussion

Yesterday I was teaching perimeter of compound shapes with my year 9s. After they had been solving problems based on shapes made up solely from squares and rectangles one if the pupils asked “Sir, will the perimeter always be even?” I thought this was a great point for discussion so I opened it up for the class.

They decided that for the type of shape we were looking at the perimeter would indeed always be even, as you had to cover every distance twice, meaning 2 was always a factor of the perimeter. I was impressed with their reasoning, but a little disappointed that one had to prompt non-integer side lengths. When I did they quickly dismissed this with “of course we knew that sir, but be were only considering whole numbers!”

I then asked them to consider is this applied to all shapes. They quickly concluded that triangles could have odd perimeters with integer side lengths, and circles, then they extrapolated to any shape with an odd number if sides. They concluded quickly that regular polygons with an even number of sides couldn’t have an odd perimeter if it had integer side lengths. Irregular polygons provided a more difficult challenge, and I left them to ponder it over the weekend!

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