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I am currently surrounded! There is big hairy spider (bigger than my hand) outside the back door and a humongous (slight exaggeration) snake on the front drive. [This is not setting the scene for a maths puzzle; this is actually happening.] I think I am safer inside writing this newsletter so let’s begin with the puzzle of the month:

In triangle ABC the angle at A is the half of the mean of the angles at B and C. Find the size of the angle at A.

The answer is at the end of this newsletter.

I have just uploaded the new Transformations help video and am surprised that it ended up being so long (45 minutes). The video is to help pupils do the Transformations exercise which has seven levels so actually I would only expect anyone to watch the section of the video that helps with the level of the exercise they are doing so that’s roughly six minutes of video watching time per sitting. That’s not too bad.

An additional level has been added to the Pythagoras exercise which contains Dr Colin Foster’s favourite "Pythagoras" tasks. Colin is the Mathematical Association President Designate and gave a wonderful keynote address at the recent MA 150th Anniversary Conference.

Another conference speaker (@mrsouthernmaths) gave me permission to use his Coordinate Geometry Table on the website. The challenge is to find elegant, efficient ways to fill the more difficult-to-fill gaps in the table. I’m using it with Year 11 students already.

A new set of exercises on finding the Area and Perimeter of a Parallelogram has been added to the website with quite a challenging final question for the top level. Sometimes areas can be found without using formulas but by other problem solving techniques. Can you solve it?

Transum subscriber Ann Roberts requested a set of exercises on Polynomial Division and I decided that Level 5 should be a really difficult challenge revived from a book written in the nineteenth century. Ann has been doing a great job checking the answers of this marathon of a maths exercise. I learnt for the first time that traditional long division is not necessarily the best way to solve this sort of question and an alternate method is provided (video in the help tab) by Beth who was previously a student of Lucy, a former colleague of mine.

If Then Trigonometry is not a big resource but makes a nice addition to the Common Trigonometric Ratios exercises.

Make Them Right is a drag-and-drop activity in which pupils must sort the given lengths into Pythagorean triples. I used artistic licence to state that the diagrams were not drawn to scale because if they were the task would be too easy.

Groups of Four is a fun diversion based on the groups challenge in the TV programme Only Connect. The Transum version is based on mathematical concepts though it does get a little tricky when you get to Level 6. At the time of writing eight times as many people have claimed a trophy for Level 1 compared to Level 6.

Deconstructing Graphs asks pupils to fill in the tables of values by identifying the coordinates of key points on graphs. There are algebraic graphs, a scatter graph and a cumulative frequency graph.

I adapted the Areas Investigation on the website to produce Tridots, a challenge to find all the triangles that can be found by joining dots on a three by three grid of dots. I was reminded of this challenge by another great conference talk given by Colleen Young.

Whilst remaining in the house this morning (due to spider and snake without masks outside) I was interviewed for the Math Mindset Podcast. The interview took place over Zoom with me here in Thailand and host Maura Ollo in New Jersey USA. The time difference was almost twelve hours so I deduced we were on opposite sides of the world (ignoring latitude differences) and I remain so impressed with the power of technology today. I am old enough to remember the poor quality and high cost of long distance phone calls not so long ago.

During the interview I told the tale of the Missing Pound (which I skilfully converted to dollars) and included the solution I had heard from Danny Baker. You’ll need to play the audio at the bottom of the June 19th Starter page if you are not up to speed with this one!

Now the answer to the puzzle of the month. There are many ways to think of the situation but my algebraic solution, using the letters A, B and C to represent the size of the angles, is as follows:

The angles of a triangle sum to 180 so A + B + C = 180 [Let this be equation number 1]

The mean of the angles at B and C is (B + C) ÷ 2 so half of this is (B + C) ÷ 4 which is equal to A.

A = (B + C) ÷ 4 [Let this be equation number 2]

Multiply both sides of equation 2 by four to find a value for B + C.

B + C = 4A

Substitute this value of B + C into equation number 1

A + 4A = 180

5A = 180

A = 36

So the angle is A 36 degrees. The angles B and C could take any positive values that sum to 144.

That's all for now, look after yourself and, if you can, look after someone else too.

John

P.S. Try to avoid doing calculus when you are thirsty. You have heard the warning, don't drink and derive!

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