## 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.

## Circles and an octagon

Here’s an interesting puzzle that came via Diego Rattaggi (@diegorattaggi) and involves circles and octagon.

I started as always with a diagram:

I labelled some sides up, then changed my mind and changed labels as I was thinking about taking a coordinate geometry approach and didn’t want to have used x and y. As it happened I didnt use that approach anyway. While looking at the sketch I realised that the triangle AOC was a right angled isoceles. Due to an error a few weeks back I wanted to double check I wasn’t making an assumption here so did some work to justify this was the case:

I was using some similar reasoning to this hexagon puzzle, I could justify that to had to be an isoceles, and that extending the lines gave an isoceles, I could just that the vetex was definitely on diameter I’d drawn and was equidistant from both circles in x and y, but felt that wasn’t enough, and if it wasn’t definitely the centre there could be multiple solutions, then I saw a different version in my screenshot:

Once I had this information it was fairly straightforward using Pythagoras’s Theorem:

At this point I realised the ratio if the radii squared was the same as the area so that’s all I needed:

I got to the end and realised I had my fraction upside down so I flipped it over.

This was an interesting puzzle, and I think I will need to think further on the case where the centre being the exact of the right triangle wasn’t specified. I might need to look on geogebra.

## 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.

## Circles in a semi circle

Came across this puzzle from Le Bécachel Sébastien @le_becachel:

I loved the look of it so I thought I’d give it a try.

First I drew a sketch, then redraw it withoutthe 3 circles:

Considered a triangle:

Went down a rabbit hole of trigonometry identities:

Figured that this was probably not the best route to go on so sketches it again:

Realised that due to the nature of circles the tangents cut it into 3 congruent sectors so the angle must be 60 degrees.

Used the fact that tangents meet radii at 90 and the symmetry of each sector to create a right angled triangle that allowed me to see that the radius if the small circle was a third the radius if the large circle. Thus the area of the small circle was one ninth the radius of the whole large circle.

We had 3 of the small circles so that’s a third of the large circle or 2/3 of the semi circle.

This was a lovely problem and I think this solution is really nice, just wish I’d not fallen down the rabbit hole on my way to finding it!

How did you solve it?

## A hexagon and some interior lines

Today’s puzzle comes from Eylem Gercek Boss (@_eylem_99) and it’s a nice quick one that I loved, and includes a hexagon, which an awful.lot of puzzles I find at the moment seem to do!:

Initially there wasn’t an obvious solution to me so I sketched it put and labelled a load of things.

Then started writing what I knew:

I had 3 parallel lines equally spaced, so I had 2 similar triangles. I knew the diagonal was double the side length. I had enough to form an eqution:

2x = (1/2)x + 12

3x = 24

x = 8

From here I could easily work out the area:

A nice little problem that got me thinking about problem solving. I didn’t see a solution immediately, although perhaps I should have, so I just started jotting down what I knew until I saw a way forward. This is a key still that students need, just being able to consider what they do know an look at what that means in the context of the question. I think this question would be really good to use with students to model that thought process.

Do you have a different solution? I’d love to hear it.

## Two circles and a trapezium

This puzzle is one that really got me thinking about a number of things, and had me stuck for a little while. I found it on twitter, where it was shared by Elyem Gercek Boss (@_eylem_99) . Here it is:

I started, as always, with a sketch:

I feel that I should mention here that the radius OF wasnt added until much later. In fact adding this line was my breakthrough moment to be honest. I had been struggling for a bit when I thought to add it in. But more of that later. The first thing I did was this:

I saw I come express AC in terms of the 2 radii, I then moved on from that and looked at the similar triangles I had and came up with x = ((r1)^2)/r2

I then worked out I had another similar triangle:

And decided this must be the route to the answer. I did various things:

Including this that was a load of work to get a result I already knew.

After some dead ends and obsessing over the similar triangles I looked back to my original triangle and that’s when I drew the line:

I realised that as I already had an equation in r1, r2 and x and a right angled triangle featuring these on the sides this would be a route to go down.

I realised I had a a quadratic:

At this point I could express the square of one radius over the square of the other, and as area is directly proportional to the square of the radius this was all I needed:

A really nice problem that had me thinking lots. I’m certainly there will be more concise solutions. If you have one, I’d love to see it! Please let me know in the comments or on social media

## 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?

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