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Science communication is important in today's technologically advanced society. A good part of the adult community is not science savvy and lacks the background to make sense of rapidly changing technology. My blog attempts to help by publishing articles of general interest in an easy to read and understand format without using mathematics. You can contact me at ektalks@yahoo.co.uk

Saturday, 29 August 2026

What Has Probability to do With Scrabble, Dice, and Extreme Climate Events? An Introduction to Probabilities for the Non-specialist.

 The blog is written with the express aim to encourage everybody, particularly students, to carry out fun experiments with dice to learn how probabilities work.

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My wife and I regularly play scrabble and over the years jotter books were piling up in the filing system.  It occurred to me that  scrabble scores from our games - 359 over four years - might be a good set of data (collected unintentionally) that one can analyse to see if any identifiable patterns are present.  I thought that since in every move in scrabble chance is an overriding factor (sitting with six vowels can be so frustrating), the final scores must reflect this (the role of chance) in some way. 

I added up both scores and with a bin size of 20 points, plotted the distribution and fitted a Gaussian Probability Function to obtain the following result (slide 1)

The standard error in the mean (defined as standard deviation divided by the square root of the number of points) is 10.8, and if we repeat this exercise many times, then 2 out of 3 times, the mean score will be in the range 559 to 581  (more on this later).

This was all good fun  - I was not expecting such a good fit to a random distribution that the Gaussian represents.  Obviously repeating this exercise many times is not very practical,  and it seemed sensible to consider throwing dice as a much better option to understand outcomes in random events - so that is what I did next. 

Why a good understanding of probability is so important?    Probability measures the  likelihood of an event happening on a scale of 0 to 1. Weather forecast, risk assessment in finance and business, health test results are some examples of how we depend on probability outcome of events.  Probabilities govern every aspect of our lives - after all the most successful scientific theory (quantum mechanics) tells us that what happens in nature is governed by interacting microscopic particles that do not follow exact predictable paths but follow those defined by chance and probability. It is built into our lives and there is nothing you or I can do about it. The only option we have is to educate ourselves.

Humans are hopelessly poor in their understanding of probabilities: Daniel Kahneman's book Thinking, Fast and Slow is my favourite read.  It tells us about the various cognitive biases and mental shortcuts that human intuition depends on, making us fundamentally unequipped for statistical (probabilistic) thinking. In making decisions, we give much weight to cognitive biases like the confirmation bias (stick to what we already believe), the availability heuristic (give much importance to what readily comes to mind), gambler's fallacy (past random events influence future outcomes) etc.  The result is poor, flawed decision making.  

Collectively, we are unappreciative of the role that mathematics plays in our daily lives - many surveys find that people do not understand how mathematics governs our lives and there is little incentive to educate ourselves that way.  

In the following, I shall follow the example of my experiments with throwing dice - an excellent way to understand probabilities and outcome of random events.  Then I discuss how probabilities are related to disorder in a system, and why extreme climate events appear to be happening more frequently.

My Experiments with Six Dice:  I found this to be an ideal way to learn about probability outcome in random events.  A die is a cube with each of the six sides marked with a different digit from 1 to 6.  Throwing a die results in its top face displaying the number printed on it.  The number is totally random, and the chance of the number being any one from 1 to 6 is equally likely.

I took 6 such dice, put them in a cylindrical box, shook them well and emptied the box on to a plane surface. The exposed faces had numbers which were noted and the throw repeated a total of 100 times.  

We note that each throw is completely independent of the previous throws, and the 100 events provide an excellent statistically random dataset.  Obviously, one would expect that each of the numbers 1 to 6 will appear 100 times in the final count as there is equal chance of each number appearing in a throw.  But, remember that in random events the standard deviation is square root of N ( = ✓N) - that is (explained later) 2 out of 3 times (68% chance), the result will be within the range 士 ✓N. I found the following for the 100 throws:

              Number       Frequency 

                   1                          107 

                         2                          102 

                         3                            98

                         4                          104

                         5                            93 

                         6                            96  

The results are between 90 and 110 and show that the sample is statistically sound.

In the next step, I summed the six numbers displayed in each throw.  One expects the sum to lie in the range from 6 (all dice display number 1) to 36 (all dice display number 6).  In order to increase the frequency of occurrence, I binned them by adding two adjacent bins and obtained 14 data points.  These are plotted in the next slide along with a Gaussian Distribution fit to the data. Slide 2:

The slide shows a reasonable fit to the data but it is not as good as in the case of the scrabble data (Slide 1).  This represents the smaller sample size of 100 dice throws as compared to 359 for the scrabble data.  The bigger the sample size, the better approximation to a gaussian distribution would be expected.

Interestingly, the dice experiment also predicts several other probabilities that we can check in the data.  

For example, one could check the probability of all numbers being different - how many times in the 100 throws, did the six dice display all six numbers. For a totally random set, the prediction is 1.54% or 1.54 times in 100 throws.  In the data, I had 2 occasions when all six numbers were different - a really good result.

In fact, there are several ways that we can look at the data to work out probabilities of different combinations happening.  I show some of the combinations with their predicted probabilities and the actual observed events. For ease of understanding, I shall label the six die faces as A, B, C, D, E and F.  The results are shown in Slide 3.


Making Sense of the Results:  Now, we have sufficient information to analyse and understand the behaviour of systems where we observe an outcome determined by the internal ordering of its components.  The outcome is something we can observe/measure and is called a macrostate. The macrostate is formed in the different ways the internal components of system combine - we can not observe these internal states (they are called the microstates of the system).

A very good example is that of an assembly of gas molecules in a container.  The large number of gas molecules move randomly and collide with each other and with the container walls to produce the observable properties like the pressure and temperature of the gas.  Pressure and temperature are the macrostates of the gas - we can observe them.  The positions and velocities of the molecules in the container are the microstates of the system. Gas molecules change their position and velocities millions of times every second and the microstates of the system change accordingly.  Many different microstates can produce the same observable macrostate.

 Back to the Experiment with Dice:  We are in a good position to understand how the number of microstates in a system determine the likelihood of different macrostates manifesting, and provide a definition of the probability of realising a particular outcome. 

Remember that microstates are specific arrangement of components of the system and every individual microstate is equally likely.  The system fluctuates between microstates constantly.  Macrostates are general observable states.

One Die:  In one die, the six faces are numbers 1 to 6.  The throw of the die results in one of the numbers facing upwards.  This may be any number from 1 to 6.  

The macrostate of the die is the number showing on the top face and there are six possible macrostates of a die.  The microstates of the die are the six numbers on its sides - any one of them has the same likelihood of being on top.  Hence the probability of throwing any number is 1/6.  

Two Dice:  In two dice, the six faces are numbers 1 to 6. When two 6-sided dice are rolled, there are 6X6 = 36 possible microstates -> 11,12,13,...; 21,22,23,...; 31,32,33,...;  41,42,43...; 51,52,53... and 61,62,63.... 

Any of the 36 combinations has equal probability to appear on the top faces and hence its probability is 1/36.

If we wish to calculate the likelihood of a double appearing then there are six different ways (11, 22, 33, 44, 55 and 66) this may happen.  Hence there is 6 in 36 or 1 in 6 (16.7%) chance of a double appearing (probabilities are often quotes as %). If we want a double six (a specific double) then the chance to obtain a double six is 1/36 or 2.78%.

A larger number of microstates for a combination increases the likelihood of that combination appearing.  For example, if we wish to obtain a sum equal to 8 for the two up-facing sides, then one needs to work out the different numbers or microstates that can add to 8.  In the 2 dice case, the possibilities are   2,6;  3,5; 4,4; 5,3 and 6,2  -  five microstates in all.  The probability of sum 8 appearing is 5/36 = 13.9%.  

Can you work out the probabilities of other sums appearing? Remember the possible range is from 2 to 12. combination 7 is the most likely at 16.67% and 4 or 10 have a probability of 8.33%.  

It is easy to verify these predictions by throwing two dice and repeating the process at least 100 times to accumulate enough statistics - the more trials the closer the results will be to the predicted probabilities.

As a final point, if we throw six normal dice, then the possible combinations (number of microstates) is 6^6 = 46,656   --> a very large number and each of them is equally probably in a throw of 6 such dice.


The Gaussian Distribution Curve: Also know as the Normal Distribution or the Bell-shaped Curve.  We are now ready to discuss as to why the frequency, with which a particular combination (macrostate) appears, seems to follow the bell-shaped curve depicted in Slides 1 and 2. In random events (the sum in a scrabble game or throwing dice), the final result depends on how the microstates have organised themselves - and that is indeed a random process independent of what had appeared in the past.  Each microstate is equally likely and what happens in a single throw is a random selection of one of the possibilities.  When the number of throws is increased substantially, the combinations with greater likelihood (greater number of microstates) happen more often and the frequency distribution starts to show a bell-shaped curve (known as the Gaussian Distribution in mathematics).  I present Slide 2 again in the following to make this point.

In the throw of 6 dice, the sum of over 30 is highly improbable as several of the dice will have to throw multiple sixes and fives; and there are only a relatively small number of combinations - for example to get a score of 34, one need to have combinations one four+5 sixes or two fives+four sixes (a total of 6 + 15 = 21 microstates) and an overall probability of 21/46656 or 0.045%.  

However, a throw for the sum to be around 20 has 32 combinations with 4221 different ways (microstates) to roll a 20 with an overall probability of 4221/46656 or 9.05%, and hence shows up more often.  

(please go back to the section on the throw of 2 dice to see how the sum of 7 was more probable than the sum of 4 or 10).

Slide 3 presents a breakdown for the probabilities of several combinations when 6 dice are thrown repeatedly.  The observed probabilities do not always agree exactly with the predictions but the agreement overall is very good indeed.  If we can extend the number of throws to 500 from 100 used for slides 2 and 3, then the agreement is expected to get much better.

The six dice measurements take a couple of hours and the analysis is another two hours of counting.  I highly recommend this exercise as it gives a great feel for probabilities happening in real life. 

Returning back to the Gaussian Distribution Curve, the question one might ask is what a bell-shaped curve to do with measurement of random events.  Most random events concentrate around a mean that is the most probable outcome - it has the highest number of microstates.  To obtain a result near either extreme will require a few microstates to concentrate at the extremes at the same time (see above example of 6 sided dice rolling).

In fact, a fundamental rule in statistics (The Central Limit Theorem) states that when you add many independent variables together, their total sum forms a bell-shaped Gaussian curve.  I now discuss the main features of the Gaussian Curve.

The Gaussian Distribution Curve: 



Gaussian distribution is an excellent way of describing and understanding random events.  If enough measurements are made, the values of a random variable follows the bell-shaped Gaussian curve.  We can quantitatively understand the process under study.  Importantly, it tells us what the most probable value of the mean is and how the probabilities of other values diminish as we look for measured values away from the mean. In particulae 99.6% of probable outcomes are with 3 standard deviation of the mean value. However, there is a finite probability of a measurement falling at 4 sigma or even at 5 or 6 sigma away from the mean albeit with vanishingly small probability - note that this value is not zero and unexpected events do happen in life.  
For example, the mean height of young boys in OECD countries follow a Gaussian distribution of mean 176 cm with a standard deviation sigma of 7.1 cm. 

 Brandon Marshall from Suffolk measured 224 cm that is more than 6 standard deviation away from the mean!  

I give another example of extreme weather events happening more frequently due to global warming - a consequence of climate change.

An Example of Extreme Weather Events:  The weather is a great example of random events and we all talk about, particularly in the UK, of how unpredictable the weather is.  What is noteworthy is that in recent years, there have been many extreme weather events relating to hot days, flooding, very heavy rain falls, droughts, wild fires etc.  Once in a decade events are happening every year and it is possible to understand why such a shift in frequency of extreme events might be expected from global warming.  Let us talk about daily maximum temperatures that have been observed with much higher frequency. The following is a qualitative description of what happens in a warming world.

In a stable climate, the distribution of  maximum temperatures at a place on a particular day of the year will follow a Gaussian distribution curve with a defined mean and a standard deviation.  Temperatures that are say 3 sigma away from the mean (on either side of the maximum) will be much less likely and could happen once or twice a century.

However, if the distribution is shifted towards higher temperatures by half the standard deviation then temperatures that were 3 sigma away from the mean are only 2.5 sigma away on the hotter side and 3.5 sigma away on the colder side.  The probability of extremely hot days increases significantly and even mildly hotter days become much more probable - see slide below.  On the other side extremely cold days become even less probable - exactly what we have been experiencing in the UK and globally.

https://ektalks.blogspot.com/2021/12/slides-part-1-relating-to-climate.html 

End Note:  The blog has introduced some basic concepts relating to probabilities.  Probabilities rule our lives - making a decision is rarely definitive and one always has a choice.  The way humans make choices is somewhat dodgy as cognitive biases come in strongly.
For truly random events, probabilities follow a well defined prescription.  For a large enough sample, Gaussian or Normal distribution effectively describes what we observe and helps to make predictions of likelihood of certain trends happening.  
The experiments with dice provide an excellent way of learning about probabilities.  The concepts of macro- and micro- states help us in making sense why probabilities behave the way the do.  


Wednesday, 19 August 2026

The Circle of Apollonius - Another way of looking at Circles

 

We learn at school that...

A circle is a plane figure bounded by one curved line such that all straight lines, drawn from a certain point within it to the bounding line, are equal. The bounding line is called its circumference and the point, its centre.      — Euclid, Book 1, Elements.

Circles have no edges, vertices or corners - they are perfectly smooth.  Among all closed shapes with the same perimeter, the circle encloses the greatest possible area (the isoperimetric theorem). Circular shapes frequently appear in nature, art, engineering, architecture etc. 

A circle has no beginning and no end- it symbolizes perfection and eternity. (see also)

This is how I had always viewed a circle  -  a two dimensional round shape where every point on its perimeter is the same distance from the centre.  That is until, by chance, I met Apollonius in my local library. I have been walking the corridors of science departments of universities all over the world for the past 66 years and still did not know about Apollonian circles. I wish to set this right.

Slide 0:


The Circle of Apollonius:  is the set of all points (P) where the distance (PB) to one fixed point (B) divided by the distance (PA) to a second point (A) equals a constant positive number k.

For k ≠ 1, the set of points (locus of P) forms a true circle surrounding the point with the shorter of the lengths PA and PB.  The centre of the circle lies on the line passing through A and B.  

For k= 1, the set of points (locus of P) forms a straight line that is the perpendicular bisector of the line segment between A and B.

First, I would discuss some important characteristics of  Apollonian circles.  Then we shall look at the analysis to show that the locus of points P is indeed a circle.

Slide 1:  



Points to note:

  • Points A and B are the foci of the circle, and lengths PA and PB are called the focal radii
  • The line connecting the two foci intersects the circle at two points (C and D) with CD equal to the diameter of the circle (= 2R). 
  • The focal radii AP and BP form a right angle at P ( APB = 90 degrees).  (more on this later)
  • For k = 1, AP = BP.  The locus of P is a straight line that is the perpendicular bisector of the line AB. This is demonstrated in the slides below.
  Slide 2:
Slide 3:

Family of Apolonian Circles:  Apollonius actually defined a second set of circles that pass through the two points A and B. This set of circles cross the first set (described above) orthogonally, viz.
, the tangent lines drawn to each circle at the intersection point are perpendicular to each other.  
This is shown in the next slide.
Slide 4:  


A Proof of Apollonian Circles:  In the following, I show that points P as described above (slide 1) indeed describe a circle.  There are some elegant proofs available in the literature, but I feel these require a good background in mathematics.  Here, I shall describe a simpler method - surprisingly I have not come across this method in publications accessible to me.

Given a line segment between two points A and B, it is always possible to construct a circle that intersects the line passing through A and B at two points C and D.  Points C and D are such that BC and BD are k times AC and AD respectively.  This is explained in the next three slides.

 Slide 5: 
Slide 6:

Slide 7:
Slide 8:
The algebra for k <1 follow exactly the same steps but point B is swapped with point A.  The Apollonian circle encloses point B for k <1.  Also see Slide 2.
 
A Geometrical Proof of the Appolonian Circles:
In the following, I discuss quite a neat proof that is based on geometry - in particular on the angle bisector theorem.  This theorem is not well known and  I shall first discuss it in slides 9,10 & 11.  It is possible to jump to slide 12 without loss of continuity - though, the angle bisector theorem gives a good taste of how geometry works.  
Slide 9:


Slide 10:
 




Slide 11:


Having established the angle bisector theorem, we shall use it to define the Apollonian circles.

Slide 12:



Slide 13:


Slide 14:


Appendix


End Note:  The mathematical prowess of the ancient Greek mathematicians was amazing - particularly in Geometry.  I had never imagined that the simplest shape of a circle can be so exciting to analyse.  I found working on the Apollonian circles feature so satisfying - not only because it provides an alternate way of defining a circle but also the analysis around  the topic only involves the use of simple algebra and geometry - just goes to show that the maths we learn at the high school level is already very powerful and may be used to provide deep insights in to how things work around us. 

Apollonian circles are very useful and for those who wish to learn more about them, I shall refer to some nice descriptions available (1, 2, 3).

Friday, 24 July 2026

Monkey, Coconuts and the Three Sailors Problem - an Easier, More Tractable Version - but with all the Intrigue and Excitement

 Monkey and the coconut puzzle is an ancient puzzle.  Wiki has a detailed article on the puzzle with several different solutions and a list of references - but also note what Wiki says:

 "The problem is notorious for it confounding difficulty  ....  has become a staple in recreational mathematical collections"

There are two versions of the puzzle that are widely discussed.  Solving either version algebraically or through trial & error quickly leads to endless cycles and logical dead ends - it can be a frustrating exercise.

In this feature, I present a simpler, manageable version of the puzzle that is more tractable but still retains all the infuriating aspects of the original puzzle.  The change I have made is to reduce the number of sailors to three from five.

The puzzle (version1) is now stated as in the following:

Three sailors and a monkey, shipwrecked on a desert island, spent the first day gathering coconuts.  During the night, one sailor woke up and decided to take his share.  He divided the coconuts into three equal piles but had one coconut left over that he gave to the monkey.  He hid his pile and put the rest all back together.  Soon after, the second sailor woke up and did the same thing; he also had one coconut left over that he gave to the monkey.  The third sailor woke up next and did the same thing, also giving the extra coconut to the monkey.  In the morning, the sailors divided the coconuts that were left into three equal shares with one coconut left over for the monkey.  They all knew that a good number of coconuts were missing, but each sailor was guilty as the others and they kept quiet.  How many coconuts were present in the beginning?

In the alternate statement of the puzzle (version2), the division in the morning results in the coconuts divided exactly among the three sailors with no coconut left for the monkey.

In the following, we shall solve the puzzle using algebra and then try some ingenious methods that may be used.  The three sailor problem is more realistic in terms of the number of coconuts collected in a day - the five sailor version requires several thousand coconuts at the start.

A Numerical Approach

Slide 1: 


Slide 2: 

Slide 3: 

Slide 4: 


From the above discussion, it follows that since (-2,-1) is a solution, the original number of coconuts were -2+81 or 79.  
The morning division provides -1+8 or 7 coconuts to each sailor. 
It also follows that any additions of 81 to N would also be a solution; therefore, 160, 241, 322 ... are also possibilities, but the minimum number of coconuts would be 79.
An alternate way to solve eq.9 is described in the next slide 
Slide 5: 


Let us now look at version2 of the puzzle in which the number of coconuts in the morning divides equally between the sailors, and none is left for the monkey.  Version2 is generally quoted as a much more difficult variation of the puzzle - we look at it in the following:

Version2_ Coconuts divide equally in the morning with none left over

In fact, our analysis of version1 is applicable up to the point when the third sailor had his turn to divide the coconuts - the analysis from equations 1 to 5 is totally valid but equation 6 is now different as there is no leftover coconut for the monkey in the morning division.  This is discussed in slide 6.

Slide 6:

Slide 7:

A word of caution about the use of the above method (used in slide 7) to solve the linear Diophantine equation.  We are only looking for integral solutions and must check that both the starting number of coconuts and the final distribution provide integral numbers.  If not, then we might have to use more sophisticated methods to solve the problem.  From my attempts, it seems that all Version1 problems give acceptable solutions.  In version2, problems with odd number of sailors give good solutions but the method fails for even number of sailors - although the equations obtained are correct, the simple method of solving the equation fails to provide integral solutions for an even number of sailors.    

Summary of the three Sailor Puzzle:
 The following slide lists the distribution of coconuts in both versions of the three sailor monkey and coconuts puzzle:

Slide 8:

Some Observations from the Above Discussion: 

Let us call the number of sailors as n.

We notice for n = 3, version2 with no coconuts left over in the morning division could be solved as easily as version1 where there was a coconut left for the monkey.  

Slide 9:


Negative Coconuts (An ingenious solution for version1)
The next two slides describe an interesting method of solving the problem:


The monkey and the coconut problem is an interesting puzzle to work on.  In the literature, we can find some rather sophisticated analyses of the problem but these are largely intractable for the general audience.  Here, I have given a detailed solution of the problem, which is simple to follow and, hopefully, enjoyable.   

Appendix:  For completeness, I give a calculation of the division of the coconuts for 5 sailors.  Both versions 1 and 2 of the puzzle with 5 sailors are discussed in slides 5.1 to 5.4

Slide 5.1:

Slide 5.2:
Slide 5.3:
Slide 5.4:

Sunday, 12 July 2026

The Iso-perimetric Theorem - a Delightful Play with the Famous Martin Gardner's Twelve Matchstick Puzzle

 Martin Gardner (1914 - 2010) published his 12 matchstick puzzle in the November 1957 Issue of Scientific American (SciAm)  The puzzle was republished by SciAm on 2nd May 2026.  The puzzle (slightly rephrased by me) states:

Slide 1:

Slide 2:
The puzzle has been very popular, but it's solution has a twist that can be quite frustrating until you find it. The solution is available in Appendix 1 (If you want to attempt it before seeing the solution then try to start by first forming a 3,4,5 Pythagorean triangle of area 6 square units).  

The 12 matchstick puzzle raises several questions - for example, 

(a)  What is the maximum and minimum area of polygons that may be formed with the 12 matchsticks?
(b)  Are there other solutions to the problem with polygon area equal to 4 sq units?
(c) Can one form polygons of area 3 sq units?
(d) Can one form polygons of area 2 sq units?
etc.
 I shall address these questions in the following but also see reference.

Maximum Area:  We are in good territory here - as all 12 matchsticks have the same length, the 12-sided polygon formed is a regular polygon.  The iso-perimetric theorem states that  for n-sided polygons (n-gons) with the same perimeter, the regular polygon (all sides of the same length) has the largest area. 
The 12 sided polygon formed is called a dodecagon or 12-gon.  It follows from the  reference that a regular 12-gon approximating a circle will have the maximum enclosed area.  This is calculated in the next slide:
Slide 3: 

Minimum Area: There is very little guidance available here - except by trying different configurations.  The slide shows an example that, in principle, gives an area approaching zero if matchsticks have zero width.
Slide 4:

Constructing 12-gons with various enclosed areas:  Slide 4 provides a means of varying the enclosed area by adjusting the value of d (separation between the two rows of the matchsticks).  As pointed out in the slide, by making d equal to the length of a matchstick, the area enclosed is equal to 4 sq units - this solves the puzzle that Martin Gardner had asked.

In the following, we shall expand the scope of the puzzle and find solutions for enclosed area to be an integral number from 2 to 9 sq units. Generally, there is more than one way to achieve a particular value of the enclosed area.  The idea I have followed is to start with a simple geometric shape and explore the way to modify the shape to achieve the desired enclosed area.

1. Rectangular Shapes:  There are three possible rectangles that one may form with 12 matchsticks:  of sides (5,1), (4,2) and (3,3). We look at them individually: 
Rectangle with sides 5 and 1 units: This is a simple rectangle with enclosed area equal to 5 sq units.  This is shown in the figure below:
 Slide 5:


Rectangle with sides 4 and 2 units: This is a rectangle with enclosed area equal to 8 sq units. However, by moving sticks as shown in slides 6 and 7, it is possible to obtain enclosed area of 8, 7, 6 and 5 sq units.  
Slide 5: 


Rectangle with sides 3: This is a square with enclosed area equal to 9 sq units. However, by moving sticks as shown in slide 7, it is possible to obtain enclosed area of 8, 7, 6 and 5 sq units. By using equilateral triangles as in slide 5, we can also obtain a 12-gon of enclosed area equal to 4 sq units.
Slide 7:


2. Triangular Shapes:  The 12-gon may be arranged as a Pythagorean triangle with side lengths of 3, 4 and 5 units. The area enclosed in the right-angled triangle is 6 sq units. By moving sticks, It is possible to construct 12-gons of area 5, 4, 3 and 2 sq units.  The cases for 5, 3 and 2 sq units are shown in slides 8 to 11.  
Slide 8:
Slide 9:
Slide 10:
Slide 11:

Final Word:
  The 12-gon matchstick problem is great for spending time in thinking of various geometrical structures in 2-dimensions - I think it is a wonderful learning exercise for school children and also for adults.  
I have only given examples of some simple structures, but it is possible to construct more complicated looking designs - especially with the use of equilateral triangles to add and subtract areas to a basic polygon. I have desisted from adding several slides that are quite complex.  I am, however, not able to construct 12-gons with enclosed areas of 10 and 11 square units.  I believe that one will have to start with a dodecagon shape and work from there.  Any suggestions will be very welcome.

Thanks for reading - please pass on the link of this blog to others who you feel might enjoy the mental exercise!

Appendix 1