This investigation is based on the ancient Greek Apollonius Problem: Can you draw a circle that just touches three given circles? How many can you draw?
We tried this by hand and then on Geogebra following these instructions.
The maximum number of circles possible was unsolved for more than 2000 years, but we solved it! For the solutions and more see the links below:
Thanks to Barasa in Kenya for telling us about this one …
Challenge 1
Take 10 cards from a deck – A,2,3, … 10. Ace represents 1.
Arrange the cards in a pile face down so that when you show the first card it is the Ace. Remove this card. Then count to two putting the first card on the bottom of the pile. The second card should be the two. Show it and remove it. Then count to three, each time you say a number you should be putting a card on the bottom and then when you get to 3 you show the 3 and remove it from the pile.
Arrange the cards so that this works for every number from A to 10. By the time you get to 9 you should only have two cards left in your hand! Cycle through them counting 1, 2, 3, … 9 and show the 9. And then you just have 10 left and the trick is over.
Challenge 2
Try to arrange the cards to perform the same trick except you don’t remove the card when you show it. The card is just placed at the bottom of the pile and you start counting the next number.
You should find out that this is impossible for 10 cards. Is there a number of cards that it would work for?
We tried a dice puzzle today set a while back by Matt Parker.
You only have the tool above – three dice inside a cube.
But you want to play a game that requires throwing a single dice and then a game that requires throwing two dice and adding their score together. Is there a way that you can use the sum of these three dice to simulate both of these scenarios?
Task 1
First work out the distribution of numbers you can score with three dice
Can you map this in a nice way to the distribution for 1 dice?
The Irish logarithm is an algorithm invented by Percy Ludgate in 1909 for multiplying single-digit numbers. The idea was to program a computer to do these calculations. The algorithm uses two tables to perform the multiplication. With just these tables you can calculate products up to 9×9 simply by adding two numbers together.
Here are blank versions of the two tables below (and as a pdf).
Our challenge was to fill in these tables. Here is the start of an attempt that would not work …
1 x 2 would be 1+2 so the correct answer of 2 needs to be in the box called 3.
1×3 would be 1+3 so the correct answer of 3 needs to be in the box called 4.
2×3 would be 2+3 so the correct answer of 6 needs to be in the box called 5.
So far so good ….
But … 2×2 would be 2+2 so the correct answer of 4 needs to go in the box called 4. But the box called 4 already has a 3 in it, so our initial choice of numbers is a bad one.
Can you do better?
It is very difficult to come up with the solution from scratch (but do try!). Here are a few numbers already filled in, if you would like a starting point (and as a pdf).
“The 100 prisoners problem is a mathematicalproblem in probability theory and combinatorics. In this problem, 100 numbered prisoners must find their own numbers in one of 100 drawers in order to survive. The rules state that each prisoner may open only 50 drawers and cannot communicate with other prisoners after the first prisoner enters to look in the drawers. At first glance, the situation appears hopeless, but a clever strategy offers the prisoners a realistic chance of survival.
Anna Gál and Peter Bro Miltersen first proposed the problem in 2003.”
In maths club we tried this problem with just 10 prisoners. You can read about the strategy here.
We analysed the percentage chance of winning by looking at the different possible cycles for 10 prisoners.
There are twelve identical looking islanders and a seesaw. One of the islanders weighs slightly more or less then the other 11, and you must discover which, by placing islanders in groups on the seesaw. However only three measurements are allowed.