## Interesting angle puzzle

Here is a nice little puzzle that has appeared from Ed Southall (@solvemymaths) I think it is a good little puzzle to get your brain going and one that should be usable in a secondary classroom as the maths is not particularly advanced. It would probably be a good one to get some students problem solving and I may give it to some students this week.

**My approach:**

I looked at this and assumed it is all regular. I labelled the three important points ABC and my first instinct was to draw a line from A to C to make a triangle. I decided not to do this, however, when I noticed that If I drew it to the point I have labelled B then I could get a nice isosceles trapezium:

From here it was just a case of using my knowledge of angles in quadrilaterals, other polygons, round a point etc to find the reflex angle required.

First I used knowledge of regular pentagons to see that angle AEF must be 162.

Then I used my knowledge of isosceles trapeziums and the knowledge that AEFB is an isosceles trapezium to work out that BAE and ABF are both 18.

Then I considered ABCD, again I know its an isosceles trapezium. I also know that ADGE is a square therefore i can work out that DCB and ABC both equal 72.

This means the reflex angle reuired must be 288.

*I’ve been looking at it further, and I’m not sure I can see any other ways that would work. But if you spot a different way then I would love to hear it.*

## 7 years on

Seven years ago today I started this blog. I had only been teaching a few years and had discovered Twitter and many teacher and math teacher blogs. I was enjoying reading the many blogs I’d found and had noticed some of them spoke of how they felt it had made them more reflective practitioners. In particular I remember Dave Gale’s (@reflectivemaths) blog Reflective Maths Teacher (I think he used a different host then though) had given me many ideas for the classroom and Dave had said his blog had helped him.

So there I was, lessons had finished o what was my daughters due date. There had been no signs of contractions and I had 30 minutes before a twilight CPD session, so I thought “why not?”. I opened this wordpress account and posted this short post reflecting on am extension task I had made up on the spot as two of my year 7 students had completed all the work I had planned for them before the lesson was up.

I didn’t write another post for about 5 months (I had a newborn baby and a full timetable), and not many people read any posts in that first year.

Since then my blogging has been up and down in its frequency. I’ve shared things I’ve done, shared thoughts on pedagogy, on education policy and maths in general. I have found blogging to be a good way to frame my thoughts on many things and start conversations on maths and on teaching with people I wouldn’t otherwise have communicated with. I feel I have learned a lot. It has made me a better teacher and a better mathematician.

I’m currently in a period of time where my blogging is more frequent again, and I hope to continue with that. I just thought I’d mark this little anniversary with a short post and say thanks to everyone who has read, commented, shared and discussed the thoughts I’ve posted here.

Onwards we go.

## The circumcircle and the law of sines

The other day I posted my solution to a lovely puzzle I’d seen that Ed Southall (@solvemymaths) had come up with, I assumed there would be better more concise solutions and ask for people to send me their solutions. I got a number of them, including this one from Xxxygorzao (@toodrunkforzero) :

At first one of the steps wasn’t obvious to me. It was the step that uses a / Sin A = 2r. I questioned Xxxygorzao who told me that it was a property of all triangles that the ratio of side / opposite angles is equal to the diameter of the circumcircle. I thought this was a rule I had forgotten, but after thinking about it I’m not entirely sure if it is or if it’s one I have never encountered before. Xxxygorzao offered to link me to the proof of this theorem, but I thought it would be nice to try and find one myself as it didn’t jump out as self-evident that it was true.

I started with a circle, and drew a triangle with 3 points on the circumference (n.b. I used a pair of compasses to draw the circle, but this was just because I’m rubbish at sketching circles, it is just for sketch purposes.):

I labelled the points A, B and C, and intend to use the notation A, B and C for the angles at each point and a, b and c for their opposite sides as is conventional in the law of sines. I then sketched chord bisectors and labelled the midpoints D, E and F. Then I labelled their intersection point, and hence the centre of the circle O.

I considered the triangle A, D, O. I know that angle ADO is a right angle and that AD is half of b, as it’s a perpendicular bisector. I also know that the hypotenuse is a radius.

I looked back at my circle and realised that by inscribed angle theorem AOC is equal to 2 x ABC, or 2 x B in this case.

AOC is an isosceles triangle made from 2 radii, and thus the perpendicular bisector of the base is also an angle bisector, giving angle AOD to be equal to B.

The rest then followed from right-angled trigonometry.

It’s a nice fact to know, and I found justifying it to myself this way was both fun and allowed be to gain an insight and an understanding into why this fact is. Obviously, I’ve only shown it for the ratio with B and b here, but the others can be shown in a similar way, and they don’t need to be as they follow from the sine rule.