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This is the Transum Newsletter for the month of May 2026. It begins, as usual, with the puzzle of the month.
King Arthur is organising a "Knight's Quiz Night" for five of his most enthusiastic knights. He has set eleven chairs, labelled A to K, at his round table. Knight 1 sits in chair A but the other knights are wondering where they should sit according to the King's rules.
Place Knights 2, 3, 4 and 5 in four of the remaining chairs so that all of the following rules are obeyed:
Where should the five knights sit?
If you get an answer, I'd love to hear how you solved the puzzle (or your students solved it). If you want an extension, consider 6 knights and 13 chairs. Fire off an email with your thoughts to gro.musnarT@rettelsweN
While you think about that, here are some of the key resources added to the Transum website during the last month.
Hidden Quadratics comes in three levels of practice where students get to spot the quadratic hiding inside exponential, trigonometric, and fraction equations. The final challenge is a real-life word problem that requires them to set up the equation themselves. The whole thing is self‑marking, of course, with optional hints, so it’s perfect for independent revision or a quick consolidation lesson.
With the addition of just one button, Snake Sort has taken on a whole new dimension. A hidden snake can now disappear from the collection, turning a straightforward sorting task into a simple but engaging game of deduction.
The activity is all about arranging snakes with different coloured sections into a logical order. As learners sort systematically, they begin to see how a good strategy helps them spot duplicates, notice patterns and realise when one possibility is missing. Hunting for the hidden snake gives real purpose to the sorting, because the better the strategy, the easier it is to spot the gap.
Don's Expressions is a new algebra exercise that was originally created by the late Don Steward. It's all about getting students to read line diagrams and turn them into algebraic expressions. Each question shows a set of horizontal bars stacked on top of each other, lined up against dashed guide lines. The top bars are labelled with numbers or with expressions like x, 2x, x+4 and so on. Underneath are some mystery bars labelled a, b and c. The job is to work out the length of each mystery bar in terms of x and numbers, just by looking at which dashed lines it stretches between and adding or subtracting the pieces above.
Pyramidenstumpf. What a great word, and it was the subject of the first email I had received from Allan for a couple of months. Allan teaches in Germany and often lets me know if something in the curriculum seems unfamiliar to someone educated in the English system. I too had never seen this formula before for the frustum of a pyramid, so after convincing myself that it is equivalent to the ‘subtract the volume of the virtual chopped off smaller pyramid from the volume of the virtual big pyramid’ method I present it to you here.
$$ \frac{h}{3}(a_1^2 +a_1a_2 + a_2^2) $$
There's also the formula for the frustum of a cone given in the German textbook.
Netflix contacted me last month seeking permission to use Equatero on a South Korean reality game show called The Devil's Plan. In the programme contestants play both collaborative and competitive strategy games in order to win a cash prize. I don’t know what episode it will be on, so if you spot Transum on the telly please let me know!
One phrase I have been thinking about recently is “realistically possible”. It sounds meaningful, but I wonder how many people could actually turn it into a number on a probability scale. That thought stayed with me, so I was interested to hear an episode of BBC Radio 4’s More or Less looking at the language people use when talking about chance and uncertainty.
In the programme, researcher Adam Kucharski explores the rather slippery words often used in the media, such as “likely”, “possible” and similar expressions. He talks about a quiz he created to find out how different people interpret these phrases, and the results show just how varied those interpretations can be. What sounds clear to one person may suggest a very different probability to someone else. It is a good reminder that words about chance are not always as precise as they first appear.
If this is a topic that interests you, you might like to explore the three related activities on the Transum website with your classes: Probability Words, Probability Washing Line and Likelihood.
Good luck to all those taking exams in May. There may still be time for some last-minute exam question practice and reading my favourite Exam Tips.
In last month's Newsletter I wrote about what may be the fastest survey in history. It takes about 15 seconds and simply involves clicking on a few bubbles. It is to let me know which age groups and curricula you teach so that I can focus my efforts on creating new resources that will be useful to Transum subscribers. The data is coming in and already I see some interesting comparisons. If you are a subscriber and have not yet given your 15 seconds you will find it on your Members Only page, on the left-hand side near the bottom of the Your Details section.
Finally the answer to last month's puzzle, which was:
Quazmere Quadrant has straight edges of length 24 metres. Inside the quadrant, an orange orchard is shaped like a semicircle, with its diameter lying along one of the quadrant's straight edges and of the same length.
What would be the maximum possible radius of the Green Glade, which is also a semicircle with its diameter along the other straight edge of the quadrant, if it touches, but does not overlap, the orange orchard?
The answer is 8 metres. Please take a moment to glance at the wonderful variety of solution methods below shared by Transum users around the world. I really enjoyed reading about how the problem can be solved in different ways.
That's all for now, good luck with this month's puzzle and remember that opportunity is not a lengthy visitor, so seize the day!
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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Chris, Scotland
Wednesday, April 1, 2026
"Here's a wee Desmos interactive I bashed together this morning with a slider that allows you to alter the size of the radius of the Glade:
[Click the image above to go to the interactive version.] "
Mala, New Zealand
Wednesday, April 1, 2026
"If we draw a line segment from the centre of the green glade, through the common tangent, to the centre of the orange orchard, we get a right-angled triangle with the right angle at the centre of the circle (whose quadrant is shown).
Let the radius of the green glade be \(x\) metres and let \(y\) be the length of the remainder of the radius of the quadrant, which means \(y=24-x\)
Using Pythagoras’ theorem, \((x+y)^2 + 12^2 = (x +12)^2\)
After simplifying this equation and substituting \(y= 24-x\) into it, we get \(x= 8\) metres. This is the maximum possible radius of the Green Glade. "
Leonard, United States
Friday, April 3, 2026
"Can you confirm that the radius of the green semicircle is 8?
I connected the semicircle centers, assigned “x” to the green radius, called upon Pythagoras, then solved.
If the triangle formed is actually 3-4-5, that seems like something of a miracle to me. I will continue to investigate further.
$$ \begin{align} (12+x)^2 &= 144 + (24-x)^2 \\ 144 + 24x + x^2 &= 720 - 48x + x^2 \\ 72x &= 576 \\ x &= 8 \end{align} $$"
Wil, England
Friday, April 3, 2026
"If the smaller radius is \(r\), we can form a right-angled triangle of sides \((r+12)\) (hypotenuse; both the radius 12 and the radius \(r\) going to the point of tangency are at right-angles to the tangent, so it's a straight line), \((24 - r)\), 12. Use Pythagoras in this triangle and it works out that \(r = 8\).
Mind you, the fact that it's such a simple number suggests to me that there is an 'obvious' method. "
Rick, United States
Saturday, April 4, 2026
"Since the orange orchard semicircle consumes one side of the quadrant, the diameter is 24 meters, and its radius is 12 meters. Draw a line from the center of each semicircle. This will form a right triangle with side 1 equal to the radius of the orange orchard, which is 12 meters. Let \(r\) represent the radius of the green grove. Then, side 2 is the length of the side of the quadrant minus the radius of the green grove \((24-r)\). The hypotenuse is the sum of the two radiuses \((12+r)\). Using the Pythagorean Theorem:
\(12^2+(24-r)^2=(12+r)^2\)
Expand each squared term.
\(144+576-48r+r^2=144+24r+r^2\)
Subtract \(144\) and \(r^2\) from both sides and add \(48r\) to both sides.
\(576 = 72r\)
Divide both sides by 72.
\(r = 8\)
Final answer:
The radius of the green grove is 8 meters"
Allan, Germany
Tuesday, April 7, 2026
"Here is the answer from my Dad. I doubt he is your oldest honorary friend of Transum but he must be pretty close at 81 years old.
"
Juliet Edworthy, St Albans School, UK
Friday, April 24, 2026
"We set this puzzle during Maths Club this week and asked our participants to write up their ideas and solutions. They were ably supported by Daniel M (yr 11). I am attaching their work, which we are all quite proud of.
Solution 1 :: Solution 2 :: Solution 3
Thank you for setting such interesting problems. "