## A little circle problem

I’ve just seen this post from Colin Beveridge (@icecolbeveridge) answering this question:

Naturally I had a go at it before reading Colin’s solution. When I read his I found a lovely concise solution that we slightly different to mine, so I thought I’d share mine.

I started by just drawing a right angled triangle from the centre of the circle like so.

I seem to have cut off the denominator of 6 on the angle there. I know that the hypotenuse is 12-r (where r is radius) and the side opposite the known angle is r.

This means I can use the sine ratio of pi/6 to get

r/(12-r) = 1/2

Which leads to:

2r = 12 – r

Then

3r = 12

r = 4

Which is the same as Colin got.

I’ve seen questions like this on A level papers before and I know they often throw students, so I make sure I explore lots of geometry based problems and puzzles to combat this. I’d be interested to know which way you would approach this. Colin and I used a very similar approach, just differing in the point at which we introduced the 12. Which way did you do it?

## Circles puzzle

Here’s a lovely puzzle I saw on Brilliant.org this week:

It’s a nice little workout. I did it entirely in my head and that is my challenge to you. Do it, go on. Do it now….

Scroll down for my answer….

Have you done it? You better have…..

I looked at this picture and my frat thought was that the blue and gold areas are congruent. Thus the entire picture has an area of 70. There are 4 overlaps, each has an area of 5, so the total area of 5 circles is 90. Leaving each circle having an area of 18.

This is a nice mental work out and I feel it could build proprtional reasoning skills in my students. I am hoping to try it on some next week.

Did you manage the puzzle? Did you do it a different way?

*This post was cross posted to better questions *here*.*

## A puzzle with possibilities

Brilliant’s Facebook page is a fantastic source of brain teasers, they post a nice stream of questions that can provide a mental work out and that I feel can be utilised well to build problem solving amongst our students.

Today’s puzzle was this:

It’s a nice little question. But when I use it in class I will only use the graphic, as I feel the description gives away too much of the answer. Without the description students will need to deduce that the green area is a quarter of a circle radius 80 (so area 1600pi) with the blue semicircle radius 40 (so area 800pi) removed, leaving a green area of 800pi.

I find the fact that the area of the blue semi circle is equal to the green area is quite nice, and in feel that with a slight rephrasing the question could really make use of this relationship. Perhaps the other blue section could be removed or coloured differently and the question instead of finding the area could be find the ratio of blue area to green area.

Another option, one I may try with my further maths class on Friday, could be to remove the other blue section and remove the side length and ask them to prove that the areas are always equal, this would provide a great bit of practice at algebraic proof.

*Can you think of any further questions that could arise from this? I’d love to hear them!*

*This post was cross-posted to the blog Betterqs here.*

## Concentric Circles Area Puzzle

This morning I saw this post from Ed Southall (@solvemymaths):

And thought, that looks an interesting puzzle. I’ll have a little go. I think you should too, before reading any further…

Ok, so this is how I approached it. First I drew a sketch:

I assigned the arbitrary variables r and x to the radii of the larger and smaller circles respectively and used the fact that tangents are perpendicular to right angles, and the symmetry of isosceles triangles, to construct two right angled triangles.

I wrote an expression for the required area in r and x. Used Pythagoras’s Theorem to find an expression for x in terms or r, subbed it in and got the lovely answer of 25pi.

*An interesting little puzzle, did you solve it the same way? I’d love to hear alternative solutions.*

## Area Puzzle – Squares and Circles

This morning I came across this puzzle from Ed Southall (@solvemymaths):

And it seemed interesting. Looking at it is seems that A is the midpoint of the arc, so equidistant from b and the lower left corner of the square. This would mean that an isosceles triangle exists in the semi-circle, which in turn implies the square has side length 4rt2 (from Pythagoras’s Theorem, as the side of the square is a diameter).

This means the square has area 32cm^2. The semi-circles have radius 2rt2 and as such have area 4pi. That leaves just the white circle. The diagonal of the square is a diameter of this, and as the side lengths are 4rt2 the diagonal must be 8 (again via Pythagoras’s Theorem). This means the circle must have a radius of 4 and hence an area of 16pi.

So the sum of the orange areas must equal the square add the 4 Semi-Circles subtract the circle add the square.

Or:

2(32) + 4(4pi) – 16pi

Which simplifies to 64cm^2.

A lovely solution, and one which shows us that the areas of the orange crescents equal the area of the square.

Towards the end of the working I realised I could have used another property of Pythagoras’s Theorem, namely that the sum of the areas of semi-circles on the two shorter sides equals the area of the semi-circle on the hypotenuse. By splitting the square into two right angled triangles I could have reasoned that the 4 smaller semi-circles provide the same area as the large circle. Which means we would get the total area to be the area of the square, add the area of the circle (from the semi-circles) subtract the area of the circle (from the circle) add the area of the square. Which again simplifies to twice the area if the square. A much more elegant solution.

## Find the radius puzzle

Yesterday I came across this photo on my phone:

It’s a question that I found on a wall in one of the classrooms at school when I started in September and thought, “that’s a nice puzzle”, then promptly forgot about. When I found it, I thought “let’s explore this one.”

I sketched it out and had a look, my brain made its usually first stop at trigonometry.

After spending a while deriving (accidentally) that cos x = adj/hyp for right angled triangles using the sine rule, various trigonometric identities and Pythagoras’s Theorem I decided there must be a better way. I sketched the problem on a coordinate grid so the centre of the circle was at (r,0) when r is the radius (ie so the y axis is a tangent) and looked at the equation:

As I know the dimensions of the rectangle, I can deduce that the point (2,r-1) is on the circumference, sub these values in and solve for r.

When I have r= 1 or 5 I can discount 1 as I know r>2, thus the radius is 5 cm . I was surprised by this answer, as it felt like the answer should be 4, not sure why.

Bizarrely, as I was working through this Jo Morgan (@mathsjem) posted this which included this similar puzzle:

So I used the same method to solve that:

I was happy with the solution, but I had a real feeling that I was missing something obvious that would lead to a much more concise solution, so I sketched again, this time I dropped a perpendicular from the point on the circumference which meets the rectangle:

*A right angled triangle, with side lengths r-2, r-1, r. It’s only the most famous RAT of all, a 3,4,5 triangle!*

I followed the algebra:

A much more concise method. This method made me think Chris Smith’s (@aap03102) puzzle (the one Jo shared) would be better to use in class, as it’s much more likely that students will know a 3,4,5 triangle than a 20,21,29 triangle!

## Circles and Triangles

Regular readers will know that I love a good puzzle. I love all maths problems, but ones which make me think and get me stuck a bit are by far my favourite. The other day Ed Southall (@solvemymaths) shared this little beauty that did just that:

I thought “Circles and a 3 4 5 triangle – what an awesome puzzle”, I reached for a pen an paper and drew out the puzzle.

I was at a bit of a loss to start with. I did some pythag to work some things out:

Eliminated y and did some algebra:

Wrote out what I knew:

And drew a diagram that didn’t help much:

I then added some additional lines to my original diagram:

Which made me see what I needed to do!

I redrew the important bits (using the knowledge that radii meet tangents at 90 degrees and that the line was 3.2 away from c but the center of the large circle was 2.5 away):

Then considered the left bit first:

Used Pythagoras’s theorem:

Then solved for x:

Then briefly git annoyed at myself because I’d already used x for something else.

I did the same with the other side to find the final radius.

A lovely puzzle using mainly Pythagoras’s theorem, circle theorems and algebra so one that is, in theory at least, accessible to GCSE students.

I hope you enjoyed this one as much as I did!

## Share this via:

## Like this: