Right, answers to yesterdays question!

So we started with a single square which had an area of 1. After the first iteration of the pattern, we had four new squares, each having an area of 1/3 X 1/3 = 1/9. On the next iteration we had 20 new boxes, each with an area of 1/9 X 1/9 = 1/81. Write down what you know so far:
Squares total
area sum
area total
1 4/9
1 56/81

As this shape is created using a pattern, there will be a pattern for the total area (that is a colloquial logic proof, but you use those principles if you want a ‘solid’ proof). All that we have to do is find the pattern that is forming from these areas, so I’m going to rewrite the area sums in a slightly different way for the 3rd iteration: 1 + (51-50)/91 + (52-51)/92. (I presume most people know 52=5X5 = 25).
The amount of boxes is also in a pattern, they are going up in powers of 5.
So, for the first iteration, the area = 1.
Second iteration = 1 + (51-50)/91.
Third iteration = second iteration + (52-51)/92
Fourth iteration = third iteration + (53-52)/93
So what will the sum be for any iteration?
Let’s try and find the sum for the nth iteration. As you will notice, the area of any iteration is equal to the area of all the previous ones + the area of the new one, meaning that the sum can be written as a pattern (which we’ve already shown) but this means that we can now use the summation symbol…
(∞,i=1) ∑ i
This symbol is an easier way of writing out the sum 1+2+3+4+….all the way to infinity (and beyond!). very simply, the lower limit (i=1) is the number you start with, ∞ is the upper limit (So you continue adding up to this number) and the pattern between each number is (in this case) 1.
If we had
(∞,i=1) ∑ (i ^2)
It would mean add up the numbers 1 + 4 +9 +16….all the way to infinity, whereas
(5,i=1) ∑( i^2)
Is equal to 1+22+32+42+52
So I hope you now understand summations.
Getting back to our problem: we are trying to find the sum of all the areas, we know that the area of the nth iteration is equal to the sum of all the previous iterations plus the new one, and we’re noticing that the 4th iteration is equal to the previous 3 plus (53-52)/93. Notice now that there are three powers, two of which are 3, the other is a 2. 3 is 4-1, 2 is 4-2. It is therefore possible to write this iteration as (54-1-54-2)94-1.
If you now replace the 4 with an n, you get the general form: (5n-1-5n-2)/9n-1.
Realising this, we can make the pattern for the area, the nth iteration =
So if we write it with a summation symbol:
1+(n,i=2) ∑((5^(n-1)-5^(n-2))/(9^(n-1))
Will give you the total area for whatever value of n you use.
But we can simplify it down. Looking at just the part in the summation symbol : (5n-1-5n-2)/9n-1 we can simplify it to 5n-1(1-5-1)/9n-1. 1-51 = 4/5, so =(4/5)*(5n-1)/(9n-1), so the summation can be written as
1+(4/5) (n,i=2)∑(5^(n-1)/9(n-1))
Note that this series is for the first iteration of the pattern (the first time the squares are added).
We could simplify it to
1+(4/5) (n,i=1) ∑(5^n/9^n)
(I appologise for the hideous format of the summations, i struggle with HTML….)
But the original question was what is the total area that this shape will make? To do this, we need to look at what the sum looks like at infinity, ie, we want to look at the area at the millionth iteration of the pattern (just an example, we actually want to use a number that is even bigger) so we want to find the answer to the equation
1+(4/5)(∞,i=1) ∑ (5^n/9^n)
So let’s have a closer look at the summation.
The first term is going to be 5/9, the second will be 25/81, 125/243…etc which is a geometric sequence.
A geometric sequence is a sequence where each term is a multiple of the term before, so 1,2,4,8,16,32,64….would be a geometric sequence. Adding up terms like these can be rather difficult without using a formula, so it’s rather…lucky?… that someone has found them ( I may at some stage do a blog on the history of the arithmetic and geometric sequences) and the formula for finding the sum is as follows: given that the ratio between the terms of a geometric series lies between -1 and 1, the sum of the terms to infinity can be expressed as a/(1-r), where a is the first term in the sequence and r is the ratio. If we had the series 1; 1/2 ; 1/4 ; 1/8 etc, a would be 1, and r would be 1/2 so the equation would be 1/(1/2)=2.
There is an interesting story about this particular series as well, but that will be another time….
Now, using the above formula for the equation given, we can solve just the summation as follows:
(5/9)/(1-(5/9)) = (5/9)/(4/9) =5/4
Substituting it in, we get 1 +(4/5)X(5/4)=2 …and this is the total area of the shape!


truthful guests of patterns

Let me introduce to some logic games/ideas…
There is an island with only two tribes on it. Tribe number 1 is the “truth tellers”, tribe number 2 is the “liars”. The truth tellers can only tell the truth, the liars can only lie. You’re visiting the island and come across three of the inhabitants. You ask the first what tribe he is from, but he responds in a voice so low and guttural you cannot make out what he said. You ask the second one what the first one said and he states “He said he is a truth teller”. the final inhabitant looks solemnly at the second, then, turning to you says “he lied to you”.
Which tribe was the third speaker from?
I suggest you think about this first before reading on, it’s quite a fun intellectual exercise to do, and requires nothing more than 5 mins of contemplation.
The trick with this question is to look first not at the third speaker, but one of the other two. Now, we don’t know what the first speaker said, so maybe we should see if we can find out what he said.
Imagine the first speaker was a truth teller. He would then tell us he is a Truth teller. But if he was a liar, he’d still tell us he was a truth teller! This tells us that the second speaker is a truth teller, which tells us that the last speaker is a liar.
A slightly more complicated version of this story: the truth tellers love to eat outsiders (ie: you) and the liars love to help outsiders get off the island safely. Both tribes respect the other, and so if you are with a liar, the truth tellers will not try and abduct you and if you are with a truth teller the liars will not try and save you. Inhabitants will never take an outsider, the outsider must always choose to go with them, regardless of the tribe. There are no differences between the tribes other than their honesty and what they do with outsiders. The only way to get off the island is with the liars, and if an outsider doesn’t choose a tribe s/he will starve to death…
So. Imagine you’re on the island and you run into 3 inhabitants. In as few questions as possible, how do you determine who you go with?
I like the idea of asking them questions about themselves (it’s more personal that way…and requires some well thought out questions). I’ll give them names for illustration: Andy, Bob and Carlise. Then you could ask Bob “if I asked Andy which tribe he was from, what would he say?”, asking Carlise the same question but about Bob, then asking Andy about Carlise. Whichever tribe they mention is the tribe they’re from. A simpler way is to ask them whether it’s raining or not!
There are many variations on these – I think there is a book on them, but I can’t seem to find it. Some of the variations involve different conditions for the inhabitants (they all tell the truth on certain days of the week) language issues, etc.
Now, as stated, these are logic games/puzzles. Some people may not realise it ties into mathematics – but it does! All these games can be broken down into symbols and letters and linked together, giving you a set of tools to solve the problems with. If you want to know more about mathematical logic, study it, look it up online, or ask me!
I would now like to go on a small rant if I may about something that bugs me about people who don’t know much about maths…why do people always expect that because you study maths you are going to be a whiz at mental arithmetic? There are people who are amazing at mental arithmetic, and there are others who quite enjoy working out the patterns of arithmetic ( indeed, I’m one of the latter. For example, take any three digit number where the 1st and 3rd digit differ by at least 2. Reverse the number. Subtract the smaller number from the bigger one. Take your new number, and reverse it. Add those two numbers together. And your answer is 1089. Always. And those patterns are very interesting to analyse as they make cool math tricks) but not every mathematician likes or is good at simple calculations – it’s why we have calculators, although it is definitely better for your brain if you spend some time calculating it yourself! In truth, expecting all mathematicians to be able to calculate quickly and easily in their head is like expecting an engineer to build a boeing 747 from a junk site, or a chef to make a wedding cake with ease, or an artist to take stunning photos, all musicians to be able to sing opera. Whilst there are some that can, it does not mean that all of them can.
Moving on 🙂
Using the logic we played with before, we can attack almost any problem and solve it. However, it is probably best to sort out a method of approach first!
First, find the simplest version of the problem.
See how it behaves.
Slowly complicate the problem toward the original.
look for patterns.
prove the patterns.
(if you read the blog about Fibonacci’s rabbits, it ties into the idea of simplifying the problem first).
there are many mini steps that can fit in there, and some steps that could be there but aren’t, and some that are in front of others that should be behind them….but this is just a general guide line.
So now for the next problem.
Mr and Mrs Senzigic were having a party. They invited 4 other couples around, and had a very good time. At about midnight, Mr Senzigic gathered everyone around and asked them all how many hands they had shaken. To his amazement, everyone had shaken a different amount of hands (even himself), ie, one had shaken 0, one 1, one 2…all the way to 8. Nobody had shook the same person twice, and nobody had shaken their spouses hand. How many hands did Mrs Senzigic shake?
On first look, this problem seems impossible. How, without knowing more detail, can you even begin to hope to solve to this??? The only thing you can deduce is that Mrs Senzigic shook the same number of hands as someone else.
So lets take it down to a simple state of affairs, with only two couples, the Hosts, and The Guests. HM will refer to Mr Host, HF will be Mrs Host, GM Mr Guest, GF Mrs Guest.
Let’s imagine that HM – being the nice man he is – shakes every hands except his wife and his own. This means that his wife has shook none, but both of the guests have shook one. if one of the guests shakes hands again, they will be on two shakes, but they can only get there if they shake HF, which would mean she has shaken two hands as well. This would cause a problem, and you would get a similar situation if you started by assuming HF had shaken everybody’s hand. The only option left the is that one of the Guests shakes two hands, so lets assume GF shakes both HM and HF hands, then they each have one, GF has two, and GM has none.
if you look at the same situation with 2 Guest couples, you find that Mr Guest 1 would shake 4 hands, his wife none, Mrs Guest 2 three hands, Mr Guest 2 one hand, and the Hosts would shake the 2 hands each.
The proof for this is very simple. Start with the last couple to arrive. They walk in, walk around the room greeting everybody, one of them shaking everybody’s hand (8 shakes), the other shaking none(0 shakes). The same for the couple before them, they would have walked in and one of them would have shaken everybody’s hand who was there (6 shakes for now) and their partner shaking none. When the couple arrived after them, the earlier couple would have both been shaken by the shaker in the new couple, meaning that the earlier couple would now have shook 7 and 1 hands respectively. The same pattern continues for the first two couples to arrive. Because of this, it ends up that every couple shakes a total of 8 hands. S0 8 and 0, 7 and 1, 6 and 2, 5 and 3…4 and 4?! but that would mean that there was a repeat in the numbers!!
and then you remember that that Mrs Senzigic shook the same amount of hands as one other person…therefore she must be in the couple of four and four, therefore Mrs Senziga shook four hands.
Hope you followed that, if not, let me know :).
Now for the final problem – and this was in an assignment for a fairly difficult paper that I’m not doing – yet.
Imagine a square, 1×1 units.

 Divide each of these sides into three new sides.
 On the middle piece of each sides, draw a new square with three new sides.

Repeat this pattern for each side.

What is the total area of this shape?
Have fun! Email me if you want the solution.