A pythagorean problem

Came across another nice puzzle from Mr Gordon (@MrGordonMaths):

First thing I noticed was that it was a pythagorean triple. My initial thought was that there might be a solution involving circle theorems, but then I realised that as an area was given this might be the best route.

As angle QPS IS 90, then the area of triangle QPS is 24 (6×8/2). That means a perpendicular from P to QS must be 4.8 (as QS is 10 and would be the base on this instance.

The area PRS is given as 8 so the area of triangle QPR must be 16 (24 – 8).

This means that 4.8x/2 must be 16 (where x is the length QR. So x = 32/4.8 = 6 ⅔ so length RS is 3⅓.

At this point I realised that I’d actually gone a slightly longer than necessary way. As 4.8 is also the perp height of PRS when RS is the base I could have used that triangle. 4.8y/2 = 8 where y is length RS, so y = 16/4.8 = 3⅓.

I then realized QPR and PRS were both triangles with the same perpendicular height, and that as the area for each triangle is bh/2 then the ratios of their areas would be the same aa their bases. So as its 16:8 (or 2:1) all I really had to do was split 10 in the ratio 2:1 to get 6⅔:3⅓, and pick the smaller one as the length PR.

A lovely problem with a nice solution.

Envelope puzzle and neat little square

I’ve been looking through my saved puzzles again and I found this nice little one in the maths newsletter from Chris Smith (@aap03102):

It’s a nice little question that took me some thinking about.

First I considered the half squares with hypotenuse 2. As these are isoceles RATs, that means their side length is rt2 so each has an area of 1.

Then I thought about the half square with hypotenuse 3. Again it’s an isoceles RAT so pythagoras’s theorem gives us a sidelength of 3/rt2. So an area of 9/4.

These 3 half squares add to 17/4. The area of the rectangle is 6 so the part not covered by the half squares must be 7/4.

When thinking about the overlap, we need to consider what that means. I assume it means the bit that goes over the others, so in this case 9/4 (area of the halfsquare) – 7/4 (area of the gal). Which is 1/2 cm².

I think this is an amazing little question that I cant wait to try out in class. It also got me thinking about the square with diagonal of 2 and area of 2. It’s an interesting square really, and the only one where we see this. Consider a square, side length, x. The diagonal is (2x²)^½, the area is x². If we equate the. And square both sides we have 2x² = x⁴ , or x⁴ – 2x² = 0, so x²(x² – 2) = 0. This generates a solution when x = 0 (which is trivial and discountable as no square can exist without side length. We also get positive and negative square roots if 2. But we can ignore the negative as lengths in this case are scalar, so we have one answer. It’s a neat little square.

1-9 math walk

Today I want to look at another puzzle I found on math walks (from Traci Jackson @traciteacher):

I love these 1-9 puzzles, and thought I’d have have a crack.

First I considered the 9, with the 1 gone already that means that the 9 must share a line with the 2 and the 3 to make 14.

That means that the 4 shares 1 line with a 2 and one with a 3. That means the 4 is one lines 4,2,8 and 4,3,7.

I considered these lines:

If I put the 7 on the left, I’d need the 6 at the intersection of the green lines. That would also mean that the 2 was above it, but I’d need another 6 on that line which doesn’t work.

So the 8 must be on the left and if we follow through we get:

I enjoyed this puzzle, if you have any cool 1-9 puzzles do send them on.

A surprising fraction?

The other day I saw this tweet from John Rowe (@MrJohnRowe):

I looked at the picture and decided the answer was probably a half. And thought about it a bit more. The 4 small circles in the top left should be the same as the white quarter circle below, and the 4 quarter circles in the top right should fit over the white circle below them. This was interesting to me, and I thought I’d look at the algebra.

Looking at the top left corner, the circles involved there have the smallest radius, so we will call that radius r. That means each circle has a radius of pi r².

Below it we have a square the same size and a quarter circle radius 4r, so the white area is (16pi r²)/4 or 4pi r² this is the same as the pink area in the square above. We can see from this that half of the left rectangle is pink. Or we can continue with our algebra. The area of each square is 16r² (its sidelength is 4r) so the pink but here is 16r² – 4pi r², so if we add this to the bit above it we have 16r² shaded (and 16r² white).

The top right has 4 shaded quarter circles, each with radius 2r, so the total shaded area is 4pi r². Below it the white circle has the same radius so same area 4pi r², again that makes the shaded area 16r² – 4pi r² so the shaded area in the right rectangle 16r², and in the big square 32r². The total area of the big square if 64r² (radius is 8r) so the shaded fraction is a half. Which is nice.

Now I know I started by saying I thought it looked like a half, which is what I did think at the time I first saw it. But I still think it’s a surprise fraction. I’ve done a lot of geometry puzzles. And many have included shapes like this, so when I look at this that knowledge helps me see. Most students would not have done anywhere near the amount of puzzles I have so wont have that foresight, and I think it would be a very surprising result that could open many up to the wow factor. I also think that it might be a good starting point for some rich class discussions.

I think it’s a great visual and a lovely answer.

Step perimeters and infinite coasts

A few weeks ago I came across this picture on twitter:

I looked at it and thought “13cm, that’s a great question to use for perimeter.” I like questions like this, where you can show that if you invest the sides to the opposite sides of little rectangles the perimeter stays the same. I’ve seen them referred to as “step perimeter” and “cross perimeter” problems. Usually it’s the cross version I use, or a rectangle with a smaller rectangle cut from a corner. I liked this question as it had many different cuts and I feel would be great to open up a discussion on class.

I then thought about the problem a lot more, you could make many cuts. You could make them at irregular intervals like this, or you coukd make them at regular ones. You can look at it many ways. I started thinking about using more and more regular intervals and how it would start to look like a straight line. I then started to consider that. Intuitively it feels like as the intervals get infinitesimally small the limit should be the straight line. But that’s not what happens.

Bizarrely after I’d been thinking on this for a long time I was scrolling through twitter and saw this:

Someone else had obviously been having similar thoughts to me.

I started thinking about it again. I changed it slightly in my head, I changed it to a 5cm by 12 cm rectangle. This was so i could visualise the difference better.

Obviously due to Pythagoras’s Theorem the diagonal of a 5×12 rectangle is 13. But the distance we get by travelling in a series of steps that are parallel to the sides is the semi perimeter- in this case 17. That’s 4cm difference.

What’s going on here? It feels counterintuitive and almost paradoxical. What is going on is that no matter how small the steps are, it’s still longer to take them than it is to cover the distance diagonally. And the sum of all these differences will always be 4 for a 5×12 rectangle.

When you draw them so small it looks diagonal, it’s not actually diagonal. Its still steps. But you cant see them due to the width of the line and the inaccuracies in drawing and human vision. If you zoom in then the steps are their.

It made me think about the coastline paradox. Which states that you can never accurately measure a coastline as it is fractal in nature. The smaller the unit of measurement you use the longer the coastline will be. Meaning that the length of the coastline grows to infinity as your unit of measurement shrinks towards zero, giving coastlines infinite length of you use infinitesimal measurements.

This kinda feels the same as the case if the steps that create the illusion of a straight line. In reality you can’t physically measure an infinitesimal length and you will find a length for the coastline. And in reality you won’t be able to draw the steps that are that small. But it’s a nice interesting result. One I’ve enjoyed thinking about so far, and one I expect I’ll be thinking about some more

Tilings and areas

My daughter and I had another play around with pattern blocks. Firstly we played around and made some patterns. She made this one that was pretty cool:

We talked about tiling the plane and how shapes tesselate. Looking at which shapes fit together. Then I asked her if she could make a repeating pattern.

She came up with this one. Which wasn’t exactly what I meant but cool non the less.

Then she made this one that was more what I had meant. At this point we discussed which colour had more shapes and which took up more of the area.

We had similar discussions about these two tilings. We discussed how the red and yellow had the same amount if area in the red yellow and green one even though the yellow had twice as many squares. She showed this by making a hexagon put if the trapeziums.

She said the green, blue and purple one looked 3d.

I agree. I mentioned briefly that it was to do with the angles if the lines and that you can get dotted paper to help draw 3d which has dots at these angles. We talked briefly about rotational and reflective symmetry too.

She then made a hexagon:

We talked about how much bigger it was. She said it looked about 4 times bigger. We then discussed what this mean, and looked at the areas. Counting triangles.

I showed her that we could do it without counting triangles. We then looked at the side lengths of the hexagons and discussed how and why this scale factor was different to the area one. I think this photo of the hexagons is an excellent visual to use when looking at similarity in secondary school. Normally I just talk about squares and rectangles but can see an excellent set of visuals using these shapes.

We then started to look at fitting shapes together round points and on a line. And we found that if you put the thin blue rhombuses together on a line you can get some cool patterns:

We didn’t get into angles that much, but I can certainly see this could be a great entry point to those discussions in future. I can also see that as well as similarity there can be further discussions around area and perimeter that build from using these shapes and I hope to explore this more in future sessions.

This is the tenth post in a series looking at the use of manipulatives in maths teaching. You can see the others here.

Exploring the link between addition and multiplication

Today’s Cuisenaire rod session was quite interesting. After aying and looking at some stuff that was similar to previous posts my daughter came up with this sequence:

(Again, please forgive the ordering the table is quite small).

She decided that she wanted to add how much each sequence was worth:

She started with tallys as she had used before, then asked if there was a quicker way. I got her to think about what was going on and she decided she could use multiplying :

After she did the one with 3s we had a discussion about the = symbol and what it meant and why it was wrong to use it the way she had initially.

When she did the 5 one she said “is that the wrong way round”, which led to a nice discussion on the commutative law.

After she had done a few she realised she could miss out a step:

When she did the one with the 8 I said she didn’t need to +1 on a different line and explained why, but she said she wanted to keep doing it to show it was separate. We then discussed the order of operations.

I think this task is an excellent way of seeing why multiplication would take precedence over addition when we come to looking at the order of operations.

I think this task and discussion were a good way to embed the link between repeated addition and multiplication, and to lay foundations for algebraic reasoning when it comes to collecting like terms. I can see that for older students it would also be a great way to show and think about the position to term relationship in a sequence.

This is the 9th post in a series about the use of manipulatives in teaching mathematics. The others can be viewed from here.

Returning numbers

I’ve recently discovered the website “Math Walks“, by Traci Jackson (@traciteacher). The site is full of pictures taken on walks that have happened during quarantine where Traci has used chalk on a paving stone to create a maths prompt to aid discussion. Sometimes they are puzzles, sometimes they are sequences, and sometimes another maths picture and they always seem to get me thinking about maths. This one is one I’ve been thinking about today:

It’s a great visual, and I really like it on so many levels. Obviously the task if ti find a solution that when you follow the path round you get your original answer. It struck me as interesting that this can be accessed on a number of different levels.

I could give this to my daughter and she would be able to complete trial and error and eventually find solutions to each of these, but theres much more too it than that.

I thought about how I would tackle this problem and decided I would use algebra:

I quite liked this as a forming and solving equations exercise, I think it’s accessible but bit too easy. Many students may struggle with the notation around the forming and many may get confused with the order they need to do things in.

I considered how one might challenge a student who does just guess, and I feel that asking them to prove whether their answer is the only one or not would be a good follow up question in this case, I think it would be unlikely that many using trial and error would get both answers for the one which includes a square.

I then considered if there was a trick to generating these puzzles, presumably you start with your answer then you can make sure you always get back there.

I think these ones are lovely, and I hope to use them at some point when we get back. I’d love to hear your thoughts on them. How would you approach them? How would you generate them?

Circles on a line

I saw this lovely question from Mr Gordon (@MrGordonMaths) the other day:

I looked at it and even though it said it was GCSE maths only it didn’t look at all obvious how to find an answer. It did look interesting though, I wondered how my y11s would get on with it. I thought I’d give it a try:

As always I drew a sketch:

I was looking at straight line shapes I could draw and realised the trapezium was the better option in this case:

From here it was a bit of Pythagoras’s theorem:

Which gave me all I needed for the final answer, which is 1:4. (Obviously I discounted the trival R=0 as it doesn’t make sense in the context of the question).

A nice little puzzle that I can see could be rather taxing for students despite using only concepts they will have learned in KS3 and 4. It’s the sort of question that can really help with problem solving ability and is one i will definitely try on my year 11s when we are back fully.

I’d love to see how you solved this one, especially if you took a different approach.

A great 1-9 puzzle

This number puzzle was one I really enjoyed and it came from @1to9puzzle :

When I looked at it I did think about setting up 8 equations with 8 unknowns and solving them as one big system. But then I figures there was probably a better way.

I looked at the sums and decided that the one summing to 10 would be a good place to start. That means I need 2 numbers that sum to 6. Which gives 1 and 5 (as 4 is already taken and we can only have 1 of each number). I knew the 2 on this diagonal needed to be 1 and 5 but wasnt sure which way round they were yet.

Then I wrote some number bonds to different numbers down. I considered the middle horizontal row. It needed to sum to 12. 9 and 3, 8 and 4, 7 and 5, 6 and 6. It couldn’t be 6 and 6 as I was only allowed 1 if each. I knew the 4 and 5 were already taken on the diagonal and in the centre so this line had to be 9 and 3.

The 5 and 3 couldn’t be on the right side together as if they were on the right that would leave me needing 10 to make 18. The 5 and 9 on this side would mean I needed 4 to make 18, so I needed the 1 in the bottom right. If I then had the 3 above it I’d need to add 14 but only had 1 more square so that meant I needed to put the 9 in the middle right. From there it was a case of simple addition and subtraction to fill in the rest:

I really enjoyed this one. Would love to hear how you did it, and do send me any others like this you find.