You are currently browsing the category archive for the ‘puzzles’ category.

A geometric proof for the Arithmetic and Geometric Mean

There is more than one way to define the mean of a number.  The arithmetic mean is the mean we learn at secondary school – for 2 numbers a and b it is:

(a + b) /2.

The geometric mean on the other hand is defined as:

(x1.x2.x3…xn)1/n

So for example with the numbers 1,2,3 the geometric mean is (1 x 2 x 3)1/3.

With 2 numbers, a and b, the geometric mean is (ab)1/2.

We can then use the above diagram to prove that (a + b) /2 ≥ (ab)1/2 for all a and b. Indeed this inequality holds more generally and it can be proved that the Arithmetic mean ≥ Geometric mean.

Step (1) We draw a triangle as above, with the line MQ a diameter, and therefore angle MNQ a right angle (from the circle theorems).  Let MP be the length a, and let PQ be the length b.

Step (2) We can find the length of the green line OR, because this is the radius of the circle.  Given that the length a+b was the diameter, then (a+b) /2 is the radius.

Step (3) We then attempt to find an equation for the length of the purple line PN.

We find MN using Pythagoras:  (MN)2 = a2 +x2

We find NQ using Pythagoras:  (NQ)2 = b2 +x2

Therefore the length MQ can also be found by Pythagoras:

(MQ)2 = (MN) + (NQ)2

(MQ) = a2 +x2 + b2 +x2

But MQ = a + b.  Therefore:

(a + b) = a2 +x2 + b2 +x2

a2+ b2 + 2ab = a2 +x2 + b2 +x2

2ab = x2 +x2

ab = x2

x = (ab)1/2

Therefore our green line represents the arithmetic mean of 2 numbers (a+b) /2 and our purple line represents the geometric mean of 2 numbers (ab)1/2. The green line will always be greater than the purple line (except when a = b which gives equality) therefore we have a geometrical proof of our inequality.

There is a more rigorous proof of the general case using induction you may wish to explore as well.

This is an example of how an investigation into area optimisation could progress.  The problem is this:

A farmer has 40m of fencing.  What is the maximum area he can enclose?

Case 1:  The rectangle:

Reflection – the rectangle turns out to be a square, with sides 10m by 10m.  Therefore the area enclosed is 100 metres squared.

Case 2:  The circle:

Reflection:  The area enclosed is greater than that of the square – this time we have around 127 metres squared enclosed.

Case 3: The isosceles triangle:

Reflection – our isosceles triangle turns out to be an equilateral triangle, and it only encloses an area of around 77 metres squared.

Case 4, the n sided regular polygon

Reflection:  Given that we found the cases for a 3 sided and 4 sided shape gave us the regular shapes, it made sense to look for the n-sided regular polygon case.  If we try to plot the graph of the area against n we can see that for n ≥3 the graph has no maximum but gets gets closer to an asymptote.  By looking at the limit of this area (using Wolfram Alpha) as n gets large we can see that the limiting case is the circle. This makes sense as regular polygons become closer to circles the more sides they have.

Proof of the limit using L’Hospital’s Rule

Here we can prove that the limit is indeed 400/pi by using L’Hospital’s rule.  We have to use it twice and also use a trig identity for sin(2x) – but pleasingly it agrees with Wolfram Alpha.

So, a simple example of how an investigation can develop – from a simple case, getting progressively more complex and finishing with some HL Calculus Option mathematics.

Sequence Investigation

This is a nice investigation idea from Nrich.  The above screen capture is from their Picture Story puzzle.  We have successive cubes – a 1x1x1 cube, a 2x2x2 cube etc.

The cubes are then rearranged to give the following shape.  The puzzle is then to use this information to discover a mathematical relationship.  This was my first attempt at this:

13 = 12
23 = (1+2)2 – 12
33 = (1+2+3)2 – (1+2)2
43 = (1+2+3+4)2 – (1+2+3)2

n3 = (1+2+3+4+…+n)2 – (1+2+3+…+ (n-1))2

This is not an especially attractive relationship – but nevertheless we have discovered a mathematical relationship using the geometrical figures above. Next let’s see why the RHS is the same as the LHS.

RHS:

(1+2+3+4+…+n)2 – (1+2+3+…+ (n-1))2

= ([1+2+3+4+…+ (n-1)] + n)2 – (1+2+3+…+ (n-1))2

= (1+2+3+…+ (n-1))2 + n2 + 2n(1+2+3+4+…+ (n-1)) – (1+2+3+…+ (n-1))2

= n2 + 2n(1+2+3+4+…+ (n-1))

next we notice that 1+2+3+4+…+ (n-1) is the sum of an arithmetic sequence first term 1, common difference 1 so we have:

1+2+3+4+…+ (n-1) = (n-1)/2 (1 + (n-1) )

1+2+3+4+…+ (n-1) = (n-1)/2 + (n-1)2/2

1+2+3+4+…+ (n-1) = (n-1)/2 + (n2 – 2n + 1)/2

Therefore:

2n(1+2+3+4+…+ (n-1)) = 2n ( (n-1)/2 + (n2 – 2n + 1)/2 )

2n(1+2+3+4+…+ (n-1)) = n2 -n + n3 – 2n2 + n

Therefore

n2 + 2n(1+2+3+4+…+ (n-1)) = n2 + n2 -n + n3 – 2n2 + n
n2 + 2n(1+2+3+4+…+ (n-1)) = n3

and we have shown that the RHS does indeed simplify to the LHS – as we would expect.

An alternative relationship

13 = 12
13+23 = (1+2)2
13+23+33  = (1+2+3)2
13+23+33+…n3  = (1+2+3+…+n)2

This looks a bit nicer – and this is a well known relationship between cubes and squares.  Could we prove this using induction?  Well we can show it’s true for n =1.  Then we can assume true for n=k:

13+23+33+…k3  = (1+2+3+…+k)2

Then we want to show true for n = k+1

ie.

13+23+33+… k3  + (k+1)3= (1+2+3+…+k + (k+1))2

LHS:

13+23+33+… k3  + (k+1)3

= (1+2+3+…+k)2 + (k+1)3

RHS:

(1+2+3+…+k + (k+1))2

= ([1+2+3+…+k] + (k+1) )2

= [1+2+3+…+k]2 + (k+1)2 + 2(k+1)[1+2+3+…+k]

= [1+2+3+…+k]+ (k+1)2 + 2(k+1)(k/2 (1+k))   (sum of a geometric formula)

= [1+2+3+…+k]+ (k+1)2 + 2(k+1)(k/2 (1+k))

= [1+2+3+…+k] + k3+ 3k2 + 3k + 1

= (1+2+3+…+k)2 + (k+1)3

Therefore we have shown that the LHS = RHS and using our induction steps have shown it’s true for all n.  (Write this more formally for a real proof question in IB!)

So there we go – a couple of different mathematical relationships derived from a simple geometric pattern – and been able to prove the second one (the first one would proceed in a similar manner).  This sort of free-style pattern investigation where you see what maths you can find in a pattern could make an interesting maths IA topic.

This is a great puzzle which the Guardian ran last week:

Fill in the equations below using any of the basic mathematical operations, +, –, x, ÷, and as many brackets as you like, so that they make arithmetical sense.

10 9 8 7 6 5 4 3 2 1 = 2017

There are many different ways of solving this – see if you can find the simplest way!

Scroll down to see some possible answers.

I had a play around with this and this is my effort:

(10+(9 x 8 x 7) -(6 + 5) ) x 4 + 3 + (2 x 1) = 2017

An even nicer answer was provided on the Guardian – which doesn’t even need brackets:

10 x 9 x 8 x 7 / 6 / 5  x 4  x  3 + 2 – 1 = 2017

Any other solutions?

The Watson Selection Task – a logical puzzle

The Watson Selection Task is a logical problem designed to show how bad we are at making logical decisions.  Watson first used it in 1968 – and found that only 10% of the population would get the correct answer.  Indeed around 65% of the population make the same error.  Here is the task:

The participants were given the following instructions:

Here is a rule: “every card that has a D on one side has a 3 on the other.” Your task is to select all those cards, but only those cards, which you would have to turn over in order to discover whether or not the rule has been violated.  Each card has a number on one side and a letter on the other.

Give yourself a couple of minutes to work out what you think the answer is – and then highlight the space below where the answer is written in white text.

The correct answer is to pick the D card and the 7 card

This result is normally quite unexpected – but it highlights one of the logical fallacies that we often fall into:

A implies B does not mean that B implies A

All cats have 4 legs (Cats = A, legs = B, A implies B)
All 4 legged animals are cats (B implies A)

We can see that here we would make a logical error if we concluded that all 4 legged animals were cats.

In the logic puzzle we need to turn over only 2 cards, D and 7.  This is surprising because most people will also say that you need to turn over card with a 3.  First we need to be clear about what we are trying to do:  We want to find evidence that the rule we are given is false.

If we turn over the D and find a number other than 3, we have evidence that the rule is false – therefore we need to turn over D.

If we turn over the 7 and find a D on the other side, we have evidence that the rule is false – therefore we need to turn over the 7.

But what about the 3?  If we turn over the 3 and find a D then we have no evidence that the rule is false (which is what we are looking for).  If we turn over the 3 and find another letter then this also gives us no evidence that the rule is false.  After all our rule says that all Ds have 3s on the other side, but it doesn’t say that all 3s have Ds on the other side.

Are mathematicians better at this puzzle than historians?

Given the importance of logical thought in mathematics, people have done studies to see if undergraduate students in maths perform better than humanities students on this task.  Here are the results:

You can see that there is a significant difference between the groups.  Maths students correctly guessed the answer D7 29% of the time, but only 8% of history students did.  The maths university lecturers performed best – getting the answer right 43% of the time.

Making different mistakes

You can also analyse the mistakes that students made- by only looking at the proportions of incorrect selections.  Here again are significant differences which show that the groups are thinking about the problem in different ways.  DK7 was chosen by around 1/5 of both maths students and lecturers, but by hardly any history students.

You can read about these results in much more depth in the following research paper Mathematicians and the Selection Task – where they also use Chi Squared testing for significance levels.

This is a nice example of using some maths to solve a puzzle from the mindyourdecisions youtube channel (screencaptures from the video).

How to Avoid The Troll: A Puzzle

In these situations it’s best to look at the extreme case first so you get some idea of the problem.  If you are feeling particularly pessimistic you could assume that the troll is always going to be there.  Therefore you would head to the top of the barrier each time.  This situation is represented below:

The Pessimistic Solution:

Another basic strategy would be the optimistic strategy.  Basically head in a straight line hoping that the troll is not there.  If it’s not, then the journey is only 2km.  If it is then you have to make a lengthy detour.  This situation is shown below:

The Optimistic Solution:

The expected value was worked out here by doing 0.5 x (2) + 0.5 x (2 + root 2) = 2.71.

The question is now, is there a better strategy than either of these?  An obvious possibility is heading for the point halfway along where the barrier might be.  This would make a triangle of base 1 and height 1/2.  This has a hypotenuse of root (5/4).  In the best case scenario we would then have a total distance of 2 x root (5/4).  In the worst case scenario we would have a total distance of root(5/4) + 1/2 + root 2.  We find the expected value by multiply both by 0.5 and adding.  This gives 2.63 (2 dp).  But can we do any better?  Yes – by using some algebra and then optimising to find a minimum.

The Optimisation Solution:

To minimise this function, we need to differentiate and find when the gradient is equal to zero, or draw a graph and look for the minimum.  Now, hopefully you can remember how to differentiate polynomials, so here I’ve used Wolfram Alpha to solve it for us.  Wolfram Alpha is incredibly powerful -and also very easy to use.  Here is what I entered:

and here is the output:

So, when we head for a point exactly 1/(2 root 2) up the potential barrier, we minimise the distance travelled to around 2.62 miles.

So, there we go, we have saved 0.21 miles from our most pessimistic model, and 0.01 miles from our best guess model of heading for the midpoint.  Not a huge difference – but nevertheless we’ll save ourselves a few seconds!

This is a good example of how an exploration could progress – once you get to the end you could then look at changing the question slightly, perhaps the troll is only 1/3 of the distance across?  Maybe the troll appears only 1/3 of the time?  Could you even generalise the results for when the troll is y distance away or appears z percent of the time?

This is a classic puzzle which is discussed in some more detail by the excellent Wired article.  The puzzle is best represented by the picture below.  We have a hunter who whilst in the jungle stumbles across a monkey on a tree branch.  However he knows that the monkey, being clever, will drop from the branch as soon as he hears the shot being fired.  The question is therefore, at what angle should the hunter aim so that he still hits the monkey?

(picture from the Wired article – originally from a UCLA physics textbook)

The surprising conclusion is that counter to what you would expect, you should actually still aim at the monkey on the branch – and in this way your bullet’s trajectory will still hit the monkey as it falls.  You can see a video of this experiment at the top of the page.

You can use tracking software (such as the free software tracker ) to show this working graphically.  Tracker provides a video demo with the falling monkey experiment:

As you can see from the still frame, we have the gun in the bottom left corner, lined up with the origin, the red trace representing the bullet and the blue trace representing the falling monkey.

We can then generate a graph to represent this data.  The red line is the height of the bullet with respect to time.  The faint blue line (with yellow dots) is the height of the monkey with respect to time.  We can see clearly that the red line can be modeled as a quadratic.  The blue line should in theory also be a quadratic (see below):

but in our model, the blue line is so flat as to be better modeled as a linear approximation – which is shown in pink.  Now we can use regression technology to find the equation of both of these lines, to show not only that they do intersect, but also the time of that intersection.

We have the linear approximation as y = -18.5t + 14.5
and the quadratic approximation as y = -56t2+39t +0.1

So the 2 graphs will indeed intersect when -18.5t + 14.6 = -56t2+39t +0.1

which will be around 0.45 seconds after the gun is fired.

(A more humane version, also from Wired – where we can throw the monkey a banana)

Newtonian Mathematics

The next question is can we prove this using some algebra?  Of course!  The key point is that the force of gravity will affect both the bullet and the falling monkey equally (it will not be affected by the different weights of the two – see the previous post here about throwing cannonballs from the Leaning Tower of Pisa).  So even thought the bullet deviates from the straight line path lined up in the gun sights, the distance the bullet deviates will be exactly the same distance that the monkey falls.  So they still collide.  Mathematically we have:

The vertical height of the bullet given by:

y = V0t – 0.5gt2

Where V0 is the initial vertical speed, t is the time, g is the gravitational force (9.8)

The vertical height of the monkey is given by:

y = h – 0.5gt2

where h is the initial vertical height of the monkey.

Therefore these will intersect when:

V0t – 0.5gt2 = h – 0.5gt2
V0t = h
V0/h = t

And for any given non-zero value of V0 we will have a t value – which represents the time of collision.

Well done – you have successfully shot the monkey!

If you like this you might also like:

Throwing cannonballs off the Leaning tower of Pisa – why weight doesn’t affect falling velocity

War Maths – how cannon operators used projectile motion to win wars

How to Win at Rock, Paper, Scissors

You might think that winning at rock, paper, scissors was purely a matter of chance – after all mathematically each outcome has the same probability. We can express the likelihood of winning in terms of a game theory grid:

It is clear that in theory you would expect to win, draw and lose with probability 1/3.  However you can actually exploit human psychology to give yourself a significant edge at this game.  Below is a report of a Chinese study on the psychology of game players:

Zhijian and co carried out their experiments with 360 students recruited from Zhejiang University and divided into 60 groups of six players. In each group, the players played 300 rounds of Rock-Paper-Scissors against each other with their actions carefully recorded.

As an incentive, the winners were paid in local currency in proportion to the number of their victories. To test how this incentive influenced the strategy, Zhijian and co varied the payout for different groups. If a loss is worth nothing and a tie worth 1, the winning payout varied from 1.1 to 100.

The results reveal a surprising pattern of behavior. On average, the players in all the groups chose each action about a third of the time, which is exactly as expected if their choices were random.

But a closer inspection of their behavior reveals something else. Zhijian and co say that players who win tend to stick with the same action while those who lose switch to the next action in a clockwise direction (where R → P → S is clockwise).

So, for example if person A chooses Rock and person B chooses Paper, then person B wins.  Human nature therefore seems to mean that person B is more likely to stick to a winning strategy and choose Paper again, whilst person A is more likely to copy that previous winning behaviour and also choose Paper.  A draw.

So you can exploit this by always moving anticlockwise i.e R → S → P.  To look at our example again, person A chooses Rock and person B chooses Paper, then person B wins. This time person A follows his previous pattern and still chooses Paper, but person B exploits this new knowledge to choose Rock.  Player B wins.

You can play against a Wolfram Alpha AI player here.  This program will track your win percentage, and will also adapt its behavior to exploit any non-random behavior that you exhibit.  Even though you may not be conscious of your biases, they probably will still be there – and the designers of this simulator are confident that the program will be beating you after about 50 games.  Have a go!

There are some additional tips for winning at rock paper scissors – if you are in a single game competition then choose paper.  This is because men are most likely to choose rock, and scissors are the least popular choice.  Also you should try some reverse psychology and announce what you will throw.  Most opponents will not believe you and modify their throw as a result.

Rock, Paper, Scissors, Lizard, Spock

You can of course make the game as complicated as you wish – the version above was popularised (though not invented by) The Big Bang Theory.  The grid below shows the possible outcomes for this game:

And of course, why stop there?  Below is a 15 throw version of the game

If you’ve honed your strategy then maybe you could compete in the a professional rock, paper, scissors tournament – here you can watch the final of a \$50,000 Las Vegas competition.

If you liked this post you might also like:

Game Theory and Tic Tac Toe – Tic Tac Toe has already been solved using Game Theory – this topic also brings in an introduction to Group Theory.

Does it Pay to be Nice? Game Theory and Evolution. How understanding mathematics helps us understand human behaviour

Can you find a sequence of consecutive integers that add up to 1000?

This puzzle is based on the excellent book A First Step to Mathematical Olympiad Problems – which is full of problems that could be extended to become exploration ideas.

Step 1 – arithmetic formula

Our first step is to write out what we want:

a + (a+1) + (a+2) + … (a +n) = 1000

next we notice that the LHS is an arithmetic series with first term a, last term a+n and n+1 terms.  Therefore we can use the sum of an arithmetic sequence formula:

Sn = 0.5n(u1 + un)

Sn = 0.5(n+1)(a + a+n) = 1000

Sn = (n+1)(2a+n) = 2000

Step 2 – logic

However, we currently have 2 unknowns, n and a, and only 1 equation – so we can’t solve this straight away.  However we do know that both a and n are integers – and n can be taken as positive.

The next step is to see that one of the brackets (n+1)(2a+n) must be odd and the other even (if  n is odd then 2a + n is odd.  If n is even then n+1 is odd).   Therefore we can look at the odd factors of 2000:

Step 3 – prime factorisation

Using prime factorisation: 2000 = 24 x 5³

Therefore any odd factors must solely come from the prime factor combinations of 5 – i.e 5, 25 and 125.

Step 4 – trial and error

So we now know that either (n+1) or (2a+n) must be 5, 25, 125.  And therefore the other bracket must be 400, 80 or 16 (as 5 x 400 = 2000 etc).  Next we can equate the (n+1) bracket to one of these 6 values, find the value of n and hence find a.  For example:

If one bracket is 5 then the other bracket is 400.

So if (n+1) = 5 and (2a+n) = 400 then n = 4 and a = 198.

This means that the sequence: 198+199+200+201+202 = 1000.

If (n+1) = 400 and (2a+n) = 5 then n = 399 and a = -197.

This means the sequence: -197 + -196+ -195 … + 201 + 202 = 1000.

We follow this same method for brackets 25, 80 and 125,16.  This gives the following other sequences:

28+29+30+…+51+52 = 1000

-54+-53+-52+…+69+70 = 1000

-27+-26+-25+…+51+52 = 1000

55+56+57+…+69+70 = 1000

So with a mixture of mathematical formulae, prime factorisation, logic and trial and error we have our solutions.  A good example of how mathematics is often solved in reality!

Tetrahedral Numbers – Stacking Cannonballs

This is one of those deceptively simple topics which actually contains a lot of mathematics – and it involves how spheres can be stacked, and how they can be stacked most efficiently.  Starting off with the basics we can explore the sequence:

1, 4, 10, 20, 35, 56….

These are the total number of cannons in a stack as the stack gets higher.  From the diagram we can see that this sequence is in fact a sum of the triangular numbers:

S1 = 1

S2 1+3

S3 1+3+6

S4 1+3+6+10

So we can sum the first n triangular numbers to get the general term of the tetrahedral numbers. Now, the general term of the triangular numbers is 0.5n2 + 0.5n therefore we can think of tetrahedral numbers as the summation:

$\bf \sum_{k=1}^{n}0.5k+0.5k^2 = \sum_{k=1}^{n}0.5k+\sum_{k=1}^{n}0.5k^2$

But we have known results for the 2 summations on the right hand side:

$\bf \sum_{k=1}^{n}0.5k =\frac{n(n+1)}{4}$

and

$\bf \huge \sum_{k=1}^{n}0.5k^2 = \frac{n(n+1)(2n+1)}{12}$

and when we add these two together (with a bit of algebraic manipulation!) we get:

$\bf S_n= \frac{n(n+1)(n+2)}{6}$

This is the general formula for the total number of cannonballs in a stack n rows high. We can notice that this is also the same as the binomial coefficient:

$\bf S_n={n+2\choose3}$

Therefore we also can find the tetrahedral numbers in Pascals’ triangle (4th diagonal column above).

The classic maths puzzle (called the cannonball problem), which asks which tetrahedral number is also a square number was proved in 1878. It turns out there are only 3 possible answers. The first square number (1) is also a tetrahedral number, as is the second square number (4), as is the 140th square number (19,600).

We can also look at something called the generating function of the sequence. This is a polynomial whose coefficients give the sequence terms. In this case the generating function is:

$\bf \frac{x}{(x-1)^4} = x + 4x^2 + 10x^3 + 20x^4 ...$

Having looked at some of the basic ideas behind the maths of stacking spheres we can look at a much more complicated mathematical problem. This is called Kepler’s Conjecture – and was posed 400 years ago. Kepler was a 17th century mathematician who in 1611 conjectured that there was no way to pack spheres to make better use of the given space than the stack above. The spheres pictured above fill about 74% of the given space. This was thought to be intuitively true – but unproven. It was chosen by Hilbert in the 18th century as one of his famous 23 unsolved problems. Despite much mathematical efforts it was only finally proved in 1998.

If you like this post you might also like:

The Poincare Conjecture – the search for a solution to one of mathematics greatest problems.

### Website Stats

• 5,492,773 views