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# How would you use this puzzle in the classroom?

Twitter is a great source of puzzles and problems. I saw this problem recently posted by James Pearce.

What is nice about this problem is that there are many different ways to approach the solution. Maybe you would like to give it a go and then consider the replies to James’ Twitter post. https://twitter.com/MathsPadJames/status/1067485103474003975

When I complete these problems I always try to think how I can make best use of it in the classroom. To just present this problem as it is may, I think, has the potential of one student finding the answer relatively quickly, shouting out the answer and declaring task finished. Others may have a similar reaction to this reply on Twitter

## So what might I do?

First of all, maybe just give the information, not the question, in words.

#### A triangle ABC has sides AB = 6cm, AC = 10cm and BC = 8cm. A line is drawn from B to M where the point M is the mid-point of AC.

Can students draw the diagram? Does having to draw the diagram help students think through the problem and consider all of the information given?

The next step, once each group of students is happy with their diagram, is to ask them to write down as much as they can about what they can see. If they need to do some calculations; that’s fine. Hopefully students may find some areas, realise that the area of ABM and BCM are equal as they have the same length base and the same perpendicular heights, find some relationships between angles and hopefully, at least one group can share the fact that ABC is a right angled triangle so I know that all groups have this piece of information. It is interesting to see whether any group come up with what the length of BM is at this stage. If they do the challenge is to see whether they can calculate BM another way. Assuming no group has found the length of BM, the challenge is now set.

I would ask students to decide what strategy they are going to try. Depending upon the students some may consider Pythagoras, Cosine rule, Sine Rule but one or two groups I may suggest to look back at the circle theorems we have studied to see whether that may help.

I would aim to have more than one method of finding the solution. I initially used cosine rule, then used sine rule, before eventually realizing that AC must be the diameter of a circle with point B lying on the circumference of that circle. This makes M the centre of the circle and AM, BM and CM are all radii of length 5cm.