Lycee Francais SAMI Maths Club

Taxicab geometry is a form of geometry, where the distance between two points A and B is not the length of the line segment AB but the sum of the shortest horizontal and vertical distances between the two points. Example:

The first challenge is to try to find what a midpoint would be in taxicab geometry. Here is an example to help:

Since the distance between A to the midpoint is the same as the distance between the point B and the midpoint, the midpoint is at the same distance from A and B. Can you spot any more midpoints, if there are any? Can you pick two different points that do not have a midpoint?

The second task is to find what a perpendicular bisector looks like on taxicab geometry.

The third task is to try to draw a circle in the taxicab geometry.

Next, try to draw an equilateral triangle and a rhombus.

If you can do all that, try to draw other geometrical shapes you know on taxicab geometry and post them in the comments.

For more info have a look here.

This week we explored a brilliant puzzle on MathPickle.

The link above will send you a series of slides explaining the puzzle, and here is a the first page of trampoline puzzles to try.

A cryptarithm is a type of mathematical puzzle in which the digits in numerical calculations are replaced by letters of the alphabet.

Alphametics are cryptarithms in which letters form meaningful words, often in meaningful phrases. There are only a few simple rules for these puzzles:

1. The same letter always stands for the same digit, and the same digit is always represented by the same letter. So if a P stands for a 2, every P in the puzzle is a 2. If a 3 is represented by a K, every 3 in the puzzle will be a K.

2.The digit zero is not allowed to appear as the left-most digit in any of the numbers in the puzzle. For example, if the word FOOD represents an addend in a puzzle, the F may not be a zero.

3.Most alphametics have unique solutions (there is only one possible answer).

One of our favourite websites is brilliant.org, so today we spent time on it solving different puzzles. You have to register an account, but then you have access to great mathematics and other science questions.

Here is one of the puzzles we worked on:

As it was Pi Day on Tuesday (14th of March is 3.14 in American style date!) we thought we would try and estimate the value of Pi by simulating throwing darts randomly at a square dartboard like below and seeing how many land inside the circle.

One way to do this is to generate a random point by typing the following in the input bar.

(RandomUniform[-1, 1], RandomUniform[-1, 1])

This will generate a point with x coordinate inbetween -1 and 1 and y coordinate inbetween -1 and 1.

Then right click on the point and select “Trace On”.

In the View menu there is a command called “Recompute All Objects”. The shortcut for this is to hold down Ctrl and press R.

You should see your random point going to different places in the screen. You can do this say 80 times.

Alternatively, you can create 80 points on a spreadsheet with the Random Uniform command and then “Create List of Points”. See here for a working applet with this method.

Either way you can now estimate the value of Pi!!!

Count the number of points inside the circle
Divide this by 80 to get the proportion of points that are inside the circle.
Area of the circle is π * 1² = π
Area of square is 2*2=4
So this proportion should be π/4.
So if we multiply our proportion by 4 we get an estimate for π.

The more points you use the better the estimate will be.

Good luck, and happy Pi Day!

Think of an positive integer.

If it is equal to 1, stop.

If it is even, divide it by 2.

If it is odd, multiply it by 3 and add 1.

With the new number you get, repeat the instructions above.

If you picked the number 6, your sequence would look like this:

3, 10, 5, 16, 8, 4, 2, 1

Did your sequence end up at 1? Mathematicians have guessed (hypothesised) that all starting choices will eventually end up at 1, but no one has been able to prove this. If you can find just one example that disproves this theory then you could claim a prize of 120 million Japenese Yen!  Here is a fun example of a counterexample that took a long time to find.

Try a few starting numbers by hand. Then you could try and use the power of a spreadsheet such as Geogebra, or by writing some code on Python and let the computer do the hard work for you!

Open up Geogebra.

You will need to use the Mod  command:

Mod[ <Dividend Number>, <Divisor Number> ]

to check if the number is even.

and the If command:

If[ <Condition>, <Then>, <Else> ]

to choose what to do if it is even and what to do if it is odd.

Put together, here is the formula you should put into cell A2, once you have put a starting number in A1.

=If[Mod[A1, 2] == 0, A1 / 2, A1*3 + 1]

Then just hover your mouse pointer in the bottom right corner and click the left button and hold down and drag down lots of cells. You should see the sequence appear.

If you want to be super clever you could try and combine two If statements so that if the cell was equal to 1 it would stop calculating and just say “STOP”.

You could also try this activity on Python, click on the menu bar on the top left to make it full screen.

Good luck!

We started off by playing the card game Dobble. If you haven’t seen it before, the game consists of a set of cards like the one above with 8 symbols on each card. You compete in a small group of people to be the first to spot a common symbol with your topmost card and a card in the middle.

The question is, what is the maximum number of cards you could have in the Dobble pack so that there is always exactly one identical symbol between any two given cards, and this identical symbol is not the same for all the cards (that would be a boring game!).

This is a hard question to answer straight away, try first to create a “Dobble” set of cards with just 2 symbols per card.

Then try and create a set with 3 symbols per card.

Any patterns you spot should help you to answer the hard question!

Here is a nice blog on the maths behind Dobble.

Cribbage

We played a simplified version of Cribbage, and looked at some of the interesting maths behind it.

Rules

Game for up to 8 players. Each person is dealt six cards. Players choose four cards to keep. Then one card is turned up in the centre of the table and counts as part of each player’s hand. Ace is considered the low card, and king high. The scoring is as follows:

Fifteens. Each card is assigned a value. Ace through 10 are the face value of the card, and jack, queen and king have value 10. Each combination that totals fifteen is awarded 2 points.

Pairs. Each pair of cards, ace through king, is awarded 2 points.

Runs. Each run of three or more cards is awarded the number of points equal to the length of the run – a run of three is worth 3 points, a run of four, 4 points, and a run of five, 5 points. In this instance runs are not counted in multiple ways. For example, A ,2, 3, 4, 5 is not counted as one run of five, two runs of four and three runs of three, but only as a single run of five.

The person with the highest score after four rounds is the winner.

Example hand (scores 16 points)

Questions to think about:

What do you think is the minimum and maximum scores possible?

Can you find the maximum hand?

Are there any impossible numbers inbetween?

Are some points totals more common than others?  How could we know for sure?

See this page for some of the answers to these questions.

Here are the full version rules of Cribbage

Find all the different quadrilaterals you can make by joining four dots on a 9-dot grid.

Different in this case means not congruent – i.e. none of your quadrilaterals should be able to be formed by rotating, translating or reflecting one of your other quadrilaterals.

Here is a couple to get you started:

How many can you find?

Some of these puzzles are taken from a nice book by Alex Bellos called Can you solve my problems? Alex Bellos does a fun puzzle blog on the Guardian every two weeks.

The Four Stacks

Start with eight coins in a row, and create four stacks of two coins in four moves. A move consists of moving one coin to the left or right by hopping over two coins and landing on the third one along. You can hop over single coins or stacks.

Tait’s Teaser

The aim of the puzzle is to start with two different types of coin (or heads and tails of the same coin) in the arrangement above and get to the arrangement below in as few ‘moves’ as possible.

A move consists of moving two adjacent coins at the same time. You can move them anywhere in the same line, but you can’t switch the two coins around as you do so.

The answer is not five!

Frogs and Toads

Place six coins as shown above. The white ones represent toads and the grey ones are frogs. Frogs and toads can only move by hopping over one other frog or toad to an empty space. Toads can move to the right, and frogs to the left. Can you rearrange them into the position below?

See here for an interactive version of this puzzle created by NRich which allows you to change the number of frogs and toads.

Star Puzzle

Draw a star and try to place 9 coins on any of the black vertices. You can place coins by starting from a vertex which is empty, moving in a straight line and counting 1,2,3. Number 1 is the vertex you start on, number 2 may or may not have a coin on it, and number 3 is where you place your coin.

Two possible opening moves are shown below

Here is an attempt that has gone wrong! 7 coins have been placed but there is no way to place any more, because you must start on an empty vertex.

Here is a great little applet coded by Etienne Royer-Gray to play with: