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Proving Pythagoras Like Einstein?
There are many ways to prove Pythagoras’ theorem – Einstein reputedly used the sketch above to prove this using similar triangles. To keep in the spirit of discovery I also just took this diagram as a starting point and tried to prove this myself, (though Einstein’s version turns out to be a bit more elegant)!
Step 1: Finding some links between triangles
We can see that our large right angled triangle has sides a,b,c with angles alpha and beta. Hopefully it should also be clear that the two smaller right angled triangles will also have angles alpha and beta. Therefore our triangles will all be similar. It should also be clear that the area of the 2 small triangles will be the same as the area of the large triangle.
Step 2: Drawing a sketch to make things clearer:
It always helps to clarify the situation with some diagrams. So, let’s do that first.
Step 3: Making some equations
As the area of the 2 small triangles will be the same as the area of the large triangle this gives the following equation:
We also can make the following equation by considering that triangles 2 and 3 are similar
We can now substitute our previous result for x into this new equation (remember our goal is to have an equation just in terms of a,b,c so we want to eliminate x and y from our equations).
We can also make the following equation by considering that triangles 1 and 2 are similar:
And as before, our goal is to remove everything except a,b,c from these equations, so let’s make the substitution for y using our previous result:
And if by magic, Pythagoras’ theorem appears! Remember that the original a,b,c related to any right angled triangle with hypotenuse c, so we have proved that this equation must always be true for right angled triangles.
You can explore some other ways of proving Pythagoras here. Which is the most elegant?
Find the average distance between 2 points on a square
This is another excellent mathematical puzzle from the MindYourDecisions youtube channel. I like to try these without looking at the answer – and then to see how far I get. This one is pretty difficult (and the actual solution exceptionally difficult!) The problem is to take a square and randomly choose 2 points somewhere inside. If you calculate the distance between the 2 points, then do this trial approaching an infinite number of times what will the average distance be? Here is what I did.
Simplify the situation: 1×1 square
This is one of the most important strategies in tackling difficult maths problems. You simplify in order to gain an understanding of the underlying problem and possibly either develop strategies or notice patterns. So, I started with a unit square and only considered the vertices. We can then list all the possible lengths:
We can then find the average length by simply doing:
2×2 square
We can then follow the same method for a 2×2 square. This gives:
Which gives an average of:
Back to a 1×1 square
Now, we can imagine that we have a 1 x 1 square with dots at every 0.5. This is simply a scaled version of the 2×2 square, so we can divide our answer by 2 to give:
3×3 square
Following the same method we have:
This gives an average of:
and if we imagine a 1×1 square with dots at every 1/3. This is simply a scaled version of the 3×3 square, so we can divide our answer by 3 to give:
We can then investigate what happens as we consider more and more dots inside our 1×1 square. When we have considered an infinite number then we will have our average distance – so we are looking the limit to infinity. This suggests using a graph. First I calculated a few more terms in the sequence:
Then I plotted this on Desmos. The points looked like they fit either an exponential or a reciprocal function – both which have asymptotes, so I tried both. The reciprocal function fit with an R squared value of 1. This is a perfect fit so I will use that.
This was plotted using the regression line:
And we can find the equation of the horizontal asymptote by seeing what happens when x approaches infinity. This will give a/c. Using the values provided by Desmos’ regression I got 0.515004887. Because I have been using approximate answers throughout I’ll take this as 0.52 (2sf). Therefore I predict that the average distance between 2 points in a 1×1 square will be approximately 0.52. And more generally, the average distance in an n x n square will be 0.52(n). This is somewhat surprising as a result – it’s not obvious why it would be a little over half the distance from 0 to 1.
Brute forcing using Python
We can also write a quick code to approximate this answer using Python (This is a Monte Carlo method). I generate 4 random numbers to represent the 2 x-coordinates and 2-y coordinates of 2 random points. I then work out the distance between them and repeat this 10 million times, then calculate the average distance. This gives:
Checking with the actual answer
Now for the moment of truth – and we watch the video to find out how accurate this is. The correct answer is indeed 0.52 (2sf) – which is great – our method worked! The exact answer is given by:
Our graphical answer is not quite accurate enough to 3 sf – probably because we relied on rounded values to plot our regression line. Our Python method with 10 million trials was accurate to 4 sf. Just to keep my computer on its toes I also calculated this with 100 million trials. This gave 0.5214126210834646 (now accurate to 5 sf).
We can also find the percentage error when using our graphical method. This is only:
Overall this is a decent result! If you are feeling extremely brave you might want to look at the video to see how to do this using calculus.
Extension: The average distance between 2 points in a unit circle
I modified the Python code slightly to now calculate the average distance between 2 points in a unit circle. This code is:
which returns an answer of 0.9054134561871364. I then looked up what the exact answer is. For the unit circle it is 128/(45 pi). This is approximately 0.9054147874. We can see that our computer method was accurate to 5 sf here. Again, the actual mathematical proof is extremely difficult.
Reflection
This is a nice example of important skills and techniques useful in mathematics – simplification of a problem, noticing patterns, graphical methods, computational power and perseverance!
Essential resources for IB students:
questions/”>Paper 3 Resources
I’ve put together four comprehensive pdf guides to help students prepare for their exploration coursework and Paper 3 investigations. The exploration guides talk through the marking criteria, common student mistakes, excellent ideas for explorations, technology advice, modeling methods and a variety of statistical techniques with detailed explanations. I’ve also made 17 full investigation questions which are also excellent starting points for explorations. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.
If you are a teacher then please also visit my new site: intermathematics.com for over 2000+ pdf pages of resources for teaching IB maths!
Further investigation of the Mordell Equation
This post carries on from the previous post on the Mordell Equation – so make sure you read that one first – otherwise this may not make much sense. The man pictured above (cite: Wikipedia) is Louis Mordell who studied the equations we are looking at today (and which now bear his name).
In the previous post I looked at solutions to the difference between a cube and a square giving an answer of 2. This time I’ll try to generalise to the difference between a cube and a square giving an answer of k. I’ll start with the same method as from the previous post:
In the last 2 lines we outline the 2 possibilities, either b = 1 or b = -1. First let’s see what happens when b = 1:
This will only provide an integer solution for a if we have:
Which generates the following first few values for k when we run through m = 1, 2,3..:
k = 2, 11, 26, 47
We follow the same method for b = -1 and get the following:
Which generates the following first few values for k when we run through m = 1, 2,3…:
k = 4, 13, 28, 49
These are the values of k which we will be able to generate solutions to. Following the same method as in the previous post this generates the following solutions:
Let’s illustrate one of these results graphically. If we take the solutions for k = 13, which are (17,70) and (17,-70), these points should be on the curve x cubed – y squared = 13.
This is indeed the case. This graph also demonstrates how all solutions to these curves will have symmetrical solutions (e, f) and (e, -f).
We can run a quick computer program to show that this method does not find all the solutions for the given values of k, but it does ensure solutions will be found for the k values in these lists.
In the code solutions above, results are listed k, x, y, x cubed, y squared. We can see for example that in the case of k = 11 our method did not find the solution x = 3 and y = 4 (though we found x = 15 and y = 58). So, using this method we now have a way of finding some solutions for some values of k – we’ve not cracked the general case, but we have at least made a start!
Essential resources for IB students:
Revision Village has been put together to help IB students with topic revision both for during the course and for the end of Year 12 school exams and Year 13 final exams. I would strongly recommend students use this as a resource during the course (not just for final revision in Y13!) There are specific resources for HL and SL students for both Analysis and Applications.
There is a comprehensive Questionbank takes you to a breakdown of each main subject area (e.g. Algebra, Calculus etc) and then provides a large bank of graded questions. What I like about this is that you are given a difficulty rating, as well as a mark scheme and also a worked video tutorial. Really useful!
The Practice Exams section takes you to a large number of ready made quizzes, exams and predicted papers. These all have worked solutions and allow you to focus on specific topics or start general revision. This also has some excellent challenging questions for those students aiming for 6s and 7s.
Essential Resources for IB Teachers
If you are a teacher then please also visit my new site. This has been designed specifically for teachers of mathematics at international schools. The content now includes over 2000 pages of pdf content for the entire SL and HL Analysis syllabus and also the SL Applications syllabus. Some of the content includes:
- Original pdf worksheets (with full worked solutions) designed to cover all the syllabus topics. These make great homework sheets or in class worksheets – and are each designed to last between 40 minutes and 1 hour.
- Original Paper 3 investigations (with full worked solutions) to develop investigative techniques and support both the exploration and the Paper 3 examination.
- Over 150 pages of Coursework Guides to introduce students to the essentials behind getting an excellent mark on their exploration coursework.
- A large number of enrichment activities such as treasure hunts, quizzes, investigations, Desmos explorations, Python coding and more – to engage IB learners in the course.
There is also a lot more. I think this could save teachers 200+ hours of preparation time in delivering an IB maths course – so it should be well worth exploring!
Essential Resources for both IB teachers and IB students
1) Exploration Guides and Paper 3 Resources
I’ve put together a 168 page Super Exploration Guide to talk students and teachers through all aspects of producing an excellent coursework submission. Students always make the same mistakes when doing their coursework – get the inside track from an IB moderator! I have also made Paper 3 packs for HL Analysis and also Applications students to help prepare for their Paper 3 exams. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.
The Mordell Equation [Fermat’s proof]
Let’s have a look at a special case of the Mordell Equation, which looks at the difference between an integer cube and an integer square. In this case we want to find all the integers x,y such that the difference between the cube and the square gives 2. These sorts of problems are called Diophantine problems and have been studied by mathematicians for around 2000 years. We want to find integer solution to:
First we can rearrange and factorise, using the property of imaginary numbers.
Next we define alpha and beta such that:
For completeness we can say that alpha and beta are part of an algebraic number field:
Next we use an extension of the Coprime Power Trick, which ensures that the following 2 equations have solutions (if our original equation also has a solution). Therefore we define:
We can then substitute our definition for alpha into the first equation directly above and expand:
Next we equate real and imaginary coefficients to give:
This last equation therefore requires that either one of the following equations must be true:
If we take the case when b = 1 we get:
Therefore:
If we take the case when b = -1 we get
Therefore our solution set is (a,b): (1,1), (1,-1), (-1,1), (-1,-1. We substitute these possible answers into our definition for y to give the following:
We can then substitute these 2 values for y into the definition for x to get:
These therefore are the only solutions to our original equation. We can check they both work:
We can see this result illustrated graphically by plotting the graph:
and then seeing that we have our integer solutions (3,5) and (3,-5) as coordinate on this curve.
This curve also clearly illustrates why we have a symmetrical set of solutions, as our graph is symmetrical about the x axis.
This particular proof was first derived by Fermat (of Fermat’s Last Theorem fame) in the 1600s and is an elegant example of a proof in number theory. You can read more about the Mordell Equation in this paper (the proof above is based on that given in the paper, but there is a small mistake in factorization so that y = 7 and y = -7 is erroneously obtained).
Essential resources for IB students:
1) Exploration Guides and Paper 3 Resources
I’ve put together four comprehensive pdf guides to help students prepare for their exploration coursework and Paper 3 investigations. The exploration guides talk through the marking criteria, common student mistakes, excellent ideas for explorations, technology advice, modeling methods and a variety of statistical techniques with detailed explanations. I’ve also made 17 full investigation questions which are also excellent starting points for explorations. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.
Hollow Cubes investigation
Hollow cubes like the picture above [reference] are an extension of the hollow squares investigation done previously. This time we can imagine a 3 dimensional stack of soldiers, and so try to work out which numbers of soldiers can be arranged into hollow cubes.
Therefore what we need to find is what numbers can be formed from a3-b3
Python code
We can write some Python3 code to find this out (this can be run here):
for k in range(1,200):
for a in range(0, 100):
for b in range(0,100):
if a**3-b**3 == k :
print(k,a,b)
This gives the following: (the first number is the number of soldiers and the 2 subsequent numbers are the 2 cubes).
1 1 0
7 2 1
8 2 0
19 3 2
26 3 1
27 3 0
37 4 3
56 4 2
61 5 4
63 4 1
64 4 0
91 6 5
98 5 3
117 5 2
124 5 1
125 5 0
127 7 6
152 6 4
169 8 7
189 6 3
We could perhaps investigate any patterns in these numbers, or explore how we can predict when a hollow cube has more than one solution. I’ll investigate which numbers can be written as both a hollow square and also a hollow cube.
Hollow squares and hollow cubes
list1=[]
for a in range(2, 50):
for b in range(2,50):
if a**2-b**2 !=0:
if a**2-b**2 > 0:
list1.append(a**2-b**2)
list2=[]
for j in list1:
for c in range(2,50):
for d in range(2,50):
if c**3-d**3 == j:
list2.append(c**3-d**3)
print(list2)
This returns the following numbers which can all be written as both hollow squares and hollow cubes.
[56, 91, 19, 117, 189, 56, 208, 189, 217, 37, 279, 152, 117, 448, 513, 504, 448, 504, 387, 665, 504, 208, 875, 819, 936, 817, 61, 999, 988, 448, 728, 513, 189, 1216, 936, 784, 335, 469, 1323, 819, 1512, 1352, 1197, 992, 296, 152, 1519, 1512, 1197, 657, 1664, 1323, 1647, 1736, 1701, 1664, 936, 504, 2107, 1387, 1216, 1027, 91, 2015, 279, 2232]
Hollow squares, cubes and hypercubes
Taking this further, can we find any number which can be written as a hollow square, hollow cube and hollow hypercube (4 dimensional cube)? This would require our soldiers to be able to be stretch out into a 4th dimensional space – but let’s see if it’s theoretically possible.
Here’s the extra code to type:
list1=[]
for a in range(2, 200):
for b in range(2,200):
if a**2-b**2 !=0:
if a**2-b**2 > 0:
list1.append(a**2-b**2)
list2=[]
for j in list1:
for c in range(2,200):
for d in range(2,200):
if c**3-d**3 == j:
list2.append(c**3-d**3)
print(list2)
for k in list2:
for e in range(2,200):
for f in range(2,200):
if k == e**4-f**4:
print(k)
Very pleasingly this does indeed find some solutions:
9919: Which can be formed as either 1002-92 or 223-93 or 104-34.
14625: Which can be formed as either 1212-42 or 253-103 or 114-24.
Given that these took some time to find, I think it’ll require a lot of computer power (or a better designed code) to find any number which is a hollow square, hollow cube, hollow hypercube and hollow 5-dimensional cube, but I would expect that there is a number out there that satisfies all criteria. Maybe you can find it?
Waging war with maths: Hollow squares
The picture above [US National Archives, Wikipedia] shows an example of the hollow square infantry formation which was used in wars over several hundred years. The idea was to have an outer square of men, with an inner empty square. This then allowed the men in the formation to be tightly packed, facing the enemy in all 4 directions, whilst the hollow centre allowed the men flexibility to rotate (and also was a place to hold supplies). It was one of the infantry formations of choice against charging cavalry.
So, the question is, what groupings of men can be arranged into a hollow square? This is a current Nrich investigation, so I thought I’d do a mini-investigation on this.
We can rethink this question as asking which numbers can be written as the difference between 2 squares. For example in the following diagram (from the Nrich task Hollow Squares)
We can see that the hollow square formation contains a larger square of 20 by 20 and a smaller hollow square of 8 by 8. Therefore the number of men in this formation is:
202-82 = 336.
The first question we might ask therefore is how many numbers from 1-100 can be written as the difference between 2 squares? These will all be potential formations for our army.
I wrote a quick code on Python to find all these combinations. I included 0 as a square number (though this no longer creates a hollow square, rather just a square!). You can copy this and run it in a Python editor like Repl.it.
for k in range(1,50):
for a in range(0, 100):
for b in range(0,100):
if a**2-b**2 == k :
print(k,a,b)
This returned the following results:
1 1 0
3 2 1
4 2 0
5 3 2
7 4 3
8 3 1
9 3 0
9 5 4
11 6 5
12 4 2
13 7 6
15 4 1
15 8 7
16 4 0
16 5 3
17 9 8
19 10 9
20 6 4
21 5 2
21 11 10
23 12 11
24 5 1
24 7 5
25 5 0
25 13 12
27 6 3
27 14 13
28 8 6
29 15 14
31 16 15
32 6 2
32 9 7
33 7 4
33 17 16
35 6 1
35 18 17
36 6 0
36 10 8
37 19 18
39 8 5
39 20 19
40 7 3
40 11 9
41 21 20
43 22 21
44 12 10
45 7 2
45 9 6
45 23 22
47 24 23
48 7 1
48 8 4
48 13 11
49 7 0
49 25 24
Therefore we can see that the numbers with no solutions found are:
2,6,10,14,18,22,26,30,34,38,42,46,50
which are all clearly in the sequence 4n-2.
Thinking about this, we can see that this can be written as 2(2n-1) which is the product of an even number and an odd number. This means that all numbers in this sequence will require an odd factor in each of their factor pairs:
eg. 50 can be written as 10 (even) x 5 (odd) or 2 (even) x 25 (odd) etc.
But with a2-b2 = (a+b)(a-b), due to symmetries we will always end up with (a+b) and (a-b) being both even or both odd, so we can’t create a number with a factor pair of one odd and one even number. Therefore numbers in the sequence 4n-2 can’t be formed as the difference of 2 squares. There are some nicer (more formal) proofs of this here.
A battalion with 960 soldiers
Next we are asked to find how many different ways of arranging 960 soldiers in a hollow square. So let’s modify the code first:
for a in range(0, 1000):
for b in range(0,1000):
if a**2-b**2 == 960 :
print(a,b)
Which gives us the following solutions:
31 1
32 8
34 14
38 22
46 34
53 43
64 56
83 77
122 118
241 239
General patterns
We can notice that when the number of soldiers is 1,3,5,7,9,11 (2n-1) we can always find a solution with the pair n and n-1. For example, 21 can be written as 2n-1 with n = 11. Therefore we have 10 and 11 as our pair of squares. This works because 112-102 = (11+10)(11-10) returns the factor pair 21 and 1. In general it always returns the factor pair, 2n-1 and 1.
We can also notice that when the number of soldiers is 4,8,12,16,20 (4n) we can always find a solution with the pair n+1 and n-1. For example, 20 can be written as 4n with n = 5. Therefore we have 6 and 4 as our pair of squares. This works because 62-42 = (6+4)(6-4) returns the factor pair 10 and 2. In general it always returns the factor pair, 2n and 2.
And we have already shown that numbers 2,6,10,14,18,22 (4n-2) will have no solution. These 3 sequences account for all the natural numbers (as 2n-1 incorporates the 2 sequences 4n-3 and 4n-1).
So, we have found a method of always finding a hollow square formation (if one exists) as well as being able to use some computer code to find other possible solutions. There are lots of other avenues to explore here – could you find a method for finding all possible combinations for a given number of men? What happens when the hollow squares become rectangles?
If you are a teacher then please also visit my new site: intermathematics.com for over 2000+ pdf pages of resources for teaching IB maths!
Normal Numbers – and random number generators
Numberphile have a nice new video where Matt Parker discusses all different types of numbers – including “normal numbers”. Normal numbers are defined as irrational numbers for which the probability of choosing any given 1 digit number is the same, the probability of choosing any given 2 digit number is the same etc. For example in the normal number 0.12345678910111213141516… , if I choose any digit in the entire number at random P(1) = P(2) = P(3) = … P(9) = 1/10. Equally if I choose any 2 digit number at random I have P(10) = P(11) = P(12) = P(99) = 1/100.
It is incredibly hard to find normal numbers, but there is a formula to find some of them.
In base 10, we are restricted to choosing a value of c such that 10 and c are relatively prime (i.e share no common factors apart from 1). So if we choose c = 3 this gives:
We can now put this into Wolfram Alpha and see what number this gives us:
So we can put the first few digits into an online calculator to find the distributions
0.000333333444444444444448148148148148148148148148148148148148148149382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049827160493827160493827160479423863312 7572016460905349794238683127572016460905349794238683127572016460 9053497942386831275720164609053497942386831275720164609053497942
4: 61
1: 41
8: 40
3: 38
0: 36
2: 33
7: 33
9: 33
6: 32
5: 10
We can see that we are already seeing a reasonably similar distribution of single digits, though with 4 and 5 outliers. As the number progressed we would expect these distributions to even up (otherwise it would not be a normal number).
One of the potential uses of normal numbers is in random number generators – if you can use a normal number and specify a digit (or number of digits) at random then this should give an equal chance of returning each number.
To finish off this, let’s prove that the infinite series:
does indeed converge to a number (if it diverged then it could not be used to represent a real number). To do that we can use the ratio test (only worry about this bit if you have already studied the Calculus Option for HL!):
We can see that in the last limit 3 to the power n+1 will grow faster than 3 to the power n, therefore as n increases the limit will approach 0. Therefore by the ratio test the series converges to a real number.
Is pi normal?
Interestingly we don’t know if numbers like e, pi and ln(2) are normal or not. We can analyse large numbers of digits of pi – and it looks like it will be normal, but as yet there is no proof. Here are the distribution of the first 100,000 digits of pi:
1: 10137
6: 10028
3: 10026
5: 10026
7: 10025
0: 9999
8: 9978
4: 9971
2: 9908
9: 9902
Which we can see are all very close to the expected value of 10,000 (+/- around 1%).
So, next I copied the first 1 million digits of pi into a character frequency counter which gives the following:
5: 100359
3: 100230
4: 100230
9: 100106
2: 100026
8: 99985
0: 99959
7: 99800
1: 99758
6: 99548
This is even closer to the expected values of 100,000 with most with +/- 0.25 %.
Proving that pi is normal would be an important result in number theory – perhaps you could be the one to do it!
Essential Resources for IB Teachers
If you are a teacher then please also visit my new site. This has been designed specifically for teachers of mathematics at international schools. The content now includes over 2000 pages of pdf content for the entire SL and HL Analysis syllabus and also the SL Applications syllabus. Some of the content includes:
- Original pdf worksheets (with full worked solutions) designed to cover all the syllabus topics. These make great homework sheets or in class worksheets – and are each designed to last between 40 minutes and 1 hour.
- Original Paper 3 investigations (with full worked solutions) to develop investigative techniques and support both the exploration and the Paper 3 examination.
- Over 150 pages of Coursework Guides to introduce students to the essentials behind getting an excellent mark on their exploration coursework.
- A large number of enrichment activities such as treasure hunts, quizzes, investigations, Desmos explorations, Python coding and more – to engage IB learners in the course.
There is also a lot more. I think this could save teachers 200+ hours of preparation time in delivering an IB maths course – so it should be well worth exploring!
Essential Resources for both IB teachers and IB students
1) Exploration Guides and Paper 3 Resources
I’ve put together a 168 page Super Exploration Guide to talk students and teachers through all aspects of producing an excellent coursework submission. Students always make the same mistakes when doing their coursework – get the inside track from an IB moderator! I have also made Paper 3 packs for HL Analysis and also Applications students to help prepare for their Paper 3 exams. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.
If you are a teacher then please also visit my new site: intermathematics.com for over 2000+ pdf pages of resources for teaching IB maths!
Volume optimization of a cuboid
This is an extension of the Nrich task which is currently live – where students have to find the maximum volume of a cuboid formed by cutting squares of size x from each corner of a 20 x 20 piece of paper. I’m going to use an n x 10 rectangle and see what the optimum x value is when n tends to infinity.
First we can find the volume of the cuboid:
Next we want to find when the volume is a maximum, so differentiate and set this equal to 0.
Next we use the quadratic formula to find the roots of the quadratic, and then see what happens as n tends to infinity (i.e we want to see what the optimum x values are for our cuboid when n approaches infinity). We only take the negative solution of the + – quadratic solutions because this will be the only one that fits the initial problem.
Next we try and simplify the square root by taking out a factor of 16, and then we complete the square for the term inside the square root (this will be useful next!)
Next we make a u substitution. Note that this means that as n approaches infinity, u approaches 0.
Substituting this into the expression gives us:
We then manipulate the surd further to get it in the following form:
Now, the reason for all that manipulation becomes apparent – we can use the binomial expansion for the square root of 1 + u2 to get the following:
Therefore we have shown that as the value of n approaches infinity, the value of x that gives the optimum volume approaches 2.5cm.
So, even though we start with a pretty simple optimization task, it quickly develops into some quite complicated mathematics. We could obviously have plotted the term in n to see what its behavior was as n approaches infinity, but it’s nicer to prove it. So, let’s check our result graphically.
As we can see from the graph, with n plotted on the x axis and x plotted on the y axis we approach x = 2.5 as n approaches infinity – as required.
An m by n rectangle.
So, we can then extend this by considering an n by m rectangle, where m is fixed and then n tends to infinity. As before the question is what is the value of x which gives the maximum volume as n tends to infinity?
We do the same method. First we write the equation for the volume and put it into the quadratic formula.
Next we complete the square, and make the u substitution:
Next we simplify the surd, and then use the expansion for the square root of 1 + u2
This then gives the following answer:
So, we can see that for an n by m rectangle, as m is fixed and n tends to infinity, the value of x which gives the optimum volume tends to m/4. For example when we had a 10 by n rectangle (i.e m = 10) we had x = 2.5. When we have a 20 by n rectangle we would have x = 5 etc.
And we’ve finished! See what other things you can explore with this problem.
Essential Resources for IB Teachers
If you are a teacher then please also visit my new site. This has been designed specifically for teachers of mathematics at international schools. The content now includes over 2000 pages of pdf content for the entire SL and HL Analysis syllabus and also the SL Applications syllabus. Some of the content includes:
- Original pdf worksheets (with full worked solutions) designed to cover all the syllabus topics. These make great homework sheets or in class worksheets – and are each designed to last between 40 minutes and 1 hour.
- Original Paper 3 investigations (with full worked solutions) to develop investigative techniques and support both the exploration and the Paper 3 examination.
- Over 150 pages of Coursework Guides to introduce students to the essentials behind getting an excellent mark on their exploration coursework.
- A large number of enrichment activities such as treasure hunts, quizzes, investigations, Desmos explorations, Python coding and more – to engage IB learners in the course.
There is also a lot more. I think this could save teachers 200+ hours of preparation time in delivering an IB maths course – so it should be well worth exploring!
Essential Resources for both IB teachers and IB students
1) Exploration Guides and Paper 3 Resources
I’ve put together a 168 page Super Exploration Guide to talk students and teachers through all aspects of producing an excellent coursework submission. Students always make the same mistakes when doing their coursework – get the inside track from an IB moderator! I have also made Paper 3 packs for HL Analysis and also Applications students to help prepare for their Paper 3 exams. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.
If you are a teacher then please also visit my new site: intermathematics.com for over 2000+ pdf pages of resources for teaching IB maths!
Narcissistic Numbers
Narcissistic Numbers are defined as follows:
An n digit number is narcissistic if the sum of its digits to the nth power equal the original number.
For example with 2 digits, say I choose the number 36:
32 + 62 = 45. Therefore 36 is not a narcissistic number, as my answer is not 36.
For example with 3 digits, say I choose the number 124:
13 + 23 + 43 = 73. Therefore 124 is not a narcissistic number as my answer is not 124.
The question is how to find all the narcissistic numbers less than 1000, without checking 1000 different numbers? Let’s start with 1 digit numbers.
1 digit numbers
01 = 0
11 = 1
21 = 2 etc.
Therefore all numbers from 0-9 are narcissistic.
2 digit numbers
For 2 digit numbers in the form ab we need the following:
a2 + b2 = 10a + b.
Therefore
a2 – 10a + b2 – b = 0.
Next if we choose a = 1,2,3,4,5 we get the following simultaneous equations:
b2 – b -16 = 0
b2 – b -21 = 0
b2 – b -24 = 0
b2 – b -25 = 0
None of these factorise for integer solutions, therefore there are no 2 digit solutions from 11 to 59. Trying a = 6,7,8,9 we find that we get the same as the first four equations. This is because a and 10-a give equivalent solutions. In other words, when a = 1 we get the equation b2 – b -9 = 0 and when a = 9 we also get the equation b2 – b -9 = 0. This is because:
for:
a2 – 10a
if we substitute a = (10 – a) we get
(10 – a)2 – 10(10 – a) = a2 – 10a.
Therefore we prove that there are no 2 digit narcissistic numbers.
3 digit numbers
First we list the cube numbers:
13 = 1, 23 = 8, 33 = 27, 43 = 64, 53 = 125, 63 = 216, 73 = 343, 83 = 512, 93 = 729.
and then consider 3 digit numbers of the form 1bc first. We need:
13+ b3 + c3 = 100 + 10b + c.
If our first digit is 1, then b3 + c3 need to add up to give us a number in the one hundreds, therefore:
99 ≤ b3 + c3≤ 198.
We can then check the cube numbers and see that the only possible combinations for a and b are 0 5, 5 0, 1 5, 5 1, 2 5, 5 2, 3 5, 5 3, 4 4, 4 5, 5 4. We can check these (only have to use the calculator for half as the reversed numbers give equivalent answers) and find that for 153 we do indeed get a narcissistic number i.e:
13+ 53 + 33 = 153.
Next we consider 3 digit numbers of the form 2bc first. We need:
192 ≤ b3 + c3≤ 292
This gives the following possibilities for b and c: 6 0, 0 6, 6 1, 16, 2 6, 6 2, 6 3, 3 6, 6 4, 4 6.
None of these give narcissistic numbers.
Next we consider 3 digit numbers of the form 3bc first. We need:
273 ≤ b3 + c3≤373
This gives the following possibilities for b and c: 6 4, 4 6, 6 5, 5 6, 7 1, 1 7, 7 2, 2 7, 7 3, 3 7, 7 0, 0 7.
Checking these we find 2 more narcissistic numbers:
370 = 33+ 73 + 03
371= 33+ 73 + 13
Using the same method, we can find that the only possibilities for 4bc are: 5 6, 6 5, 7 1, 1 7, 7 2, 2 7, 7 3, 3 7, 7 4, 4 7, 7 0, 0 7. Checking these gives us 1 more narcissistic number:
407= 43+ 03 + 73
We can carry on with this method checking up to the form 9ab (it gets easier as there are less combinations possible, but will find no more narcissistic numbers. Therefore we have all the narcissistic numbers less than 1000:
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 153, 370, 371, 407.
Is there a limit to how many narcissistic numbers there are?
Surprisingly there is a limit – there are exactly 88 narcissistic numbers in base 10. To see why we can consider the following:
In 3 digits the biggest number we can choose is 999. That would give 93+ 93 + 93 (or 3(9)3). This needs to give a number in the hundreds (102) otherwise it would be impossible to achieve a narcissistic number. Therefore with an n digit number the largest number we can make is n(9)n and if we can’t make a number in the 10n-1, then a narcissistic number is not possible. If we can prove that the inequality:
n(9)n < 10n-1
is true for some values of n, then there will be an upper bound to the narcissistic numbers we can make. We could simply plot this directly, but let’s see if we can convince ourselves it’s true for some n without using graphical software first. Let’s see if we can find an equality:
n(9)n = 10n-1
First we take log base 10 of both sides
log n(9)n = n-1
log(n) + nlog(9) = n-1
n(log9 -1) + logn +1 = 0
Next we make the substitution logn = u and therefore 10u = n. This gives:
10u(log9 -1) + u + 1 = 0.
Now we can clearly see that 10u will grow much larger than u + 1, so any root must be for u is small. Let’s see, when u = 1 we get a positive number (as log9 -1 is a negative number close to 0), but when u = 2 we get a negative number. Therefore we have a root between u = 1 and u = 2. Given that we made the substitution logn = u, that means we have found the inequality n(9)n < 10n-1 will hold for n somewhere between 101 and 10 2.
Using Wolfram we can see that the equality is reached when u = 1.784, i.e when n = 101.784 or approx 60.8. Therefore we can see that when we have more than 60 digit numbers, it is no longer possible to make narcissistic numbers.
Essential Resources for IB Teachers
If you are a teacher then please also visit my new site. This has been designed specifically for teachers of mathematics at international schools. The content now includes over 2000 pages of pdf content for the entire SL and HL Analysis syllabus and also the SL Applications syllabus. Some of the content includes:
- Original pdf worksheets (with full worked solutions) designed to cover all the syllabus topics. These make great homework sheets or in class worksheets – and are each designed to last between 40 minutes and 1 hour.
- Original Paper 3 investigations (with full worked solutions) to develop investigative techniques and support both the exploration and the Paper 3 examination.
- Over 150 pages of Coursework Guides to introduce students to the essentials behind getting an excellent mark on their exploration coursework.
- A large number of enrichment activities such as treasure hunts, quizzes, investigations, Desmos explorations, Python coding and more – to engage IB learners in the course.
There is also a lot more. I think this could save teachers 200+ hours of preparation time in delivering an IB maths course – so it should be well worth exploring!
Essential Resources for both IB teachers and IB students
1) Exploration Guides and Paper 3 Resources
I’ve put together a 168 page Super Exploration Guide to talk students and teachers through all aspects of producing an excellent coursework submission. Students always make the same mistakes when doing their coursework – get the inside track from an IB moderator! I have also made Paper 3 packs for HL Analysis and also Applications students to help prepare for their Paper 3 exams. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.
If you are a teacher then please also visit my new site: intermathematics.com for over 2000+ pdf pages of resources for teaching IB maths!
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?
Essential resources for IB students:
Revision Village has been put together to help IB students with topic revision both for during the course and for the end of Year 12 school exams and Year 13 final exams. I would strongly recommend students use this as a resource during the course (not just for final revision in Y13!) There are specific resources for HL and SL students for both Analysis and Applications.
There is a comprehensive Questionbank takes you to a breakdown of each main subject area (e.g. Algebra, Calculus etc) and then provides a large bank of graded questions. What I like about this is that you are given a difficulty rating, as well as a mark scheme and also a worked video tutorial. Really useful!
The Practice Exams section takes you to a large number of ready made quizzes, exams and predicted papers. These all have worked solutions and allow you to focus on specific topics or start general revision. This also has some excellent challenging questions for those students aiming for 6s and 7s.
Essential Resources for IB Teachers
If you are a teacher then please also visit my new site. This has been designed specifically for teachers of mathematics at international schools. The content now includes over 2000 pages of pdf content for the entire SL and HL Analysis syllabus and also the SL Applications syllabus. Some of the content includes:
- Original pdf worksheets (with full worked solutions) designed to cover all the syllabus topics. These make great homework sheets or in class worksheets – and are each designed to last between 40 minutes and 1 hour.
- Original Paper 3 investigations (with full worked solutions) to develop investigative techniques and support both the exploration and the Paper 3 examination.
- Over 150 pages of Coursework Guides to introduce students to the essentials behind getting an excellent mark on their exploration coursework.
- A large number of enrichment activities such as treasure hunts, quizzes, investigations, Desmos explorations, Python coding and more – to engage IB learners in the course.
There is also a lot more. I think this could save teachers 200+ hours of preparation time in delivering an IB maths course – so it should be well worth exploring!
Essential Resources for both IB teachers and IB students
1) Exploration Guides and Paper 3 Resources
I’ve put together a 168 page Super Exploration Guide to talk students and teachers through all aspects of producing an excellent coursework submission. Students always make the same mistakes when doing their coursework – get the inside track from an IB moderator! I have also made Paper 3 packs for HL Analysis and also Applications students to help prepare for their Paper 3 exams. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.
IB Maths Resources from Intermathematics 
