Movies and maths

Another maths blog that I follow, Mathspig, has made a series of posts debunking ten popular movie clichés. I won't go into too much detail here, as I'll just be repeating the posts themselves, so here they are:

10 Movie Clichés Debunked with Maths (introductory post)

1. You can duck bullets
2. You can out run a fireball
3. You can out run machine gun fire
4. If you are involved in a car chase, hijacking, explosion, volcanic eruption or alien invasion, you will not go into shock
5. You can jump out of an airplane [sic] that is about to crash and catch the baddie who took the last parachute
6. If the pilot dies the control tower can talk you down
7. You can safely jump from a burning skyscraper / bridge / aircraft into water
8. You can survive any amount of beating in a fight
9. A single match or candle will light a room the size of a football stadium
10. You can out run a lava flow

Number 5 reminded me of this (if you get bored you might want to skip to around 6 minutes to get what I'm talking about...):

Hare Lift - Watch more Videos at Vodpod.

What is zero to the power of zero?

To answer this, we need to think about what powers are.

What are powers?

A power is a number written small, slightly above and to the right of something, like this: 10³. In this case, it means 10 times 10 times 10 - or "ten multiplied by itself three times". The same way, x⁴ = x times x times x times x.

Multiplying powers

What happens if you have to multiply 10² by 10³? You could do it by saying "10 squared is 100 and 10 cubed is 1,000, and if you times them together you get 100,000," just by counting the noughts. What you're really doing, even if you don't realise it, is using the multiplying powers rule.

10² x 10³ = (10 x 10) x (10 x 10 x 10) = 10 x 10 x 10 x 10 x 10 = 10⁵

When you multiply powers together, you add the numbers above* and leave the bottom number alone. x¹² times x⁹⁷ = x¹⁰⁹. Easy, eh?

Dividing powers

It is a well-established fact that maths was invented by the same pedantically-minded two-year-old that invented the German language. That's what makes it easy. If multiplying powers together makes you do an addition sum, it makes Logical Sense that if you divide powers, you have to do a take away. That's exactly what happens. Try it if you don't believe me:

10⁵ ÷ 10³ = 10².

Powers of 0

How can you multiply something by itself no times? Well, you can't, really. But with powers, you can. You can turn it into a division calculation:

10⁰ = 10¹ ÷ 10¹ = 10 ÷ 10 = 1.

This is true... as long as the number underneath - 10, in this case - is not exactly zero.

Why not?

The reason why not is, 0 to the power of anything** is 0. Zero multiplied by anything, even itself, is zero. So we have a bit of a conflict: zero to the power of anything is 0 and anything to the power of 0 is 1.

Now, when maths sees a conflict, it has several options. One is to decide arbitrarily that one rule is more important than the other. Another is to figure out which is more important. A third is to throw up its hands and say "It just doesn't work," which is precisely what it does in this case.

0⁰ is undefined, just the same as 0÷0. You can't do it, it breaks the calculator. There is no answer that makes sense. It's an excellent question, and a really disappointing answer.

* As long as the number below is the same
** Anything positive, strictly

The new Odeon Premiére Reward Card: is it worth it?

The Odeon cinema chain has recently brought out its own reward card, but there's a catch: you have to pay for it. They've made it even more confusing by introducing three different levels of payment, each accompanied by a different number of 'free' rewards points.

The question is, is it worth forking out for a card? And if it is, which level do I go for?

I used the power of maths to work out an answer to both questions, and posted it at my regular blog, here.

Averages (for a list of numbers)

There are three kinds of average you need to know about for now: the MEAN, the MEDIAN and the MODE. They aren't the only ones, but they're the important ones for GCSE.

Let's say you have a list of numbers: 20, 19, 18, 18, 17, 13, 11, 10 and 8.

The MODE is the MOST COMMON number - here, it's 18. You can remember it by making an elaborate MO! sound, but be careful not to sound like a cow. If you do French, you can remember that la mode means 'fashion' - the mode is the most fashionable number in the list.

The MEDIAN is the MIDDLE number. It's 17 for this list - you get there by counting in from both ends until you reach the middle. Sometimes you get two middle numbers (when the length of the list is even). If the middle numbers are the same, that's your median; otherwise, the median is the number halfway between them (so, if you had 17 and 13 as your middle numbers, the median would be 15). You can remember that median sounds like medium, which means middling; or, if you speak American, you'll know that the middle of a highway (what we call the central reservation) is called the median strip because it's in the middle.

The MEAN is the nasty one. It's the meanest thing they can ask, hence the name. What you have to do is add all of the numbers up and divide by how many there are. The numbers here add to 116, which we have to divide by 9 to get 14.

What do 'nth' and 'nth term' mean in maths?

When you're describing how to get somewhere you might tell someone to 'go up the 5th street on the left'. This just means go past the first four, and go up the next one. When choosing an item of clothing from a rail, you might say 'the 7th one along'. This just means count along 7 and pick that one.

If I wanted to simplify that description of how to find the 5th road, or the 7th item of clothing (or even the 18th bottle on the wall, the 48th marble in the bag or the 239th name on a list), I might say something like this:

"To find the nth object, just count along the objects until you get to n."

This might look more complicated at first, but think about what it means: In maths, we can use a letter to:

1. represent numbers we don't know yet, or
2. work as place holder for something that works with any number.

In the example above, it's the second case that works for us. We could swap any number for n, and the statement would still make sense:

"To find the 5th object, just count along the objects until you get to 5."
"To find the 8th object, just count along the objects until you get to 8."
"To find the 846th object, just count along the objects until you get to 846.

What's the point?
In the example above, it may seem a little pointless, but remember that this is an over-simplified example. A slightly more complicated example is:

Imagine that a shop sells sea shells. The price of the shells is three for £1. The shop keeper could say, then, that:

"The number of sea shells I give a customer is three times the number of pounds the customer gives me."

Using our notation, we can write down the number of shells that he'd have to give his customer with a bit less effort. If the customer gives £n, then he gets 3 x n shells in return. If you can remember that 'n' stands for the number of pounds the customer gives, then remembering (and writing) "3 x n" takes a lot less effort than remembering "multiply the number of pounds by 3".


What about 'nth term'?
The 'nth term' is to do with sequences. A sequence is basically a list of numbers that follows some kind of rule. If you'd like to know more about sequences, please leave a comment and ask!

Take the following sequence as an example:

5, 8, 11, 14, 17, 20, 23, 26, 29... and so on.

This sequence goes on forever, and writing the whole thing down would take us just as long. We could just say "the sequence starts at 5, then you add three on each time." Or you could write it in the following way:

3n + 2

Then, if you wanted to find the 1st number in the list, you'd just say that n = 1 (3 x 1 + 2 = 5: it works!). If you wanted to find the 2nd, then n = 2 (3 x 2 + 2 = 8: that works too!).
The beauty of writing it down in this way is that you could also skip straight to finding the 48th number in the list: n would be replaced by 48 (3 x 48 + 2 = 146).
Writing down a sequence in this way is called writing down the 'nth term rule'. That just means that if we know which number term* we want to find (for example, the 8th, the 17th or the 4,243rd), we just swap n for that number




* A 'term' is just a posh word for each number in the list. For example, the first 'term' in the sequence we're looking at here is 5. The 7th 'term' in this sequence is 23: it's just the 7th number along when you write the whole list out.

What is a quadrillion?

The word 'quadrillion' comes from the same family of number names as 'ten', 'hundred', 'million' and 'googol'. It's just a name for a particular number.

But which number?

Another way of saying 'one quadrillion' would be 'one thousand million million'. What's that in digits? Let's work it out:

We know that 'one thousand' is written like this: 1,000 (a one with three zeroes, or 10³),

and 'one million' is written like this: 1,000,000 (a one with six zeroes or 10⁶).

So 'one thousand million' is just the 'million' bit with the 'one' replaced with 'one thousand', like this: 1,000,000,000.

So 'one thousand million million' should just be the 'million' bit with the 'one' replaced with 'one thousand million', like this: 1,000,000,000,000,000.

So 'one quadrillion' is 1,000,000,000,000,000 (a one with fifteen zeroes, or 10¹⁵).


Facts* about the word 'quadrillion'

• The SI prefix for a quadrillion is 'peta-'. This means that a 'petabyte' is one quadrillion bytes, or a 'petametre' would be a quadrillion metres.
• 'Quadrillion' can also refer to the number 'one million million million million', or 1,000,000,000,000,000,000,000,000, or 10²⁴, in the 'long scale' system of number naming. This definition is, however, falling out of use in the English language, which is why I have demonstrated the first definition in more detail (which uses the 'short scale' number naming conventions).
• The SI prefix for the long scale definition of a quadrillion is 'yotta-'. In the short scale system, 'yotta-' is the prefix for a 'septillion'.

*Whether they're interesting or not is entirely down to personal preference and circumstances.

Geometry question

Here's a nice geometry question a student recently asked me about:

The surface area of a cube is 150 sq. cm. What is the length of its diagonal in cm?

The one measurement that's always useful when we're looking at cubes is the length of the side. How do we figure that out here?

Well, there are six sides to a cube, they're all the same size and they're all squares. So each of them has an area of 150÷6 = 25 square centimetres. If the area of a square is 25, the length of the side is √25 = 5cm.

Now comes the tricky bit. When I'm doing 3-d geometry problems, I find it helps to think about a rectangular room with four rectangular walls. In this case, we're trying to find the distance from one corner on the floor to the opposite corner at the top of the wall, and I imagine a bit of string running from one to the other.

Straight away, I can see a right-angled triangle - one side runs diagonally across the floor and the other up the wall, and the string makes the hypotenuse.

We know how big the wall is - it's 5cm, like all of the other sides of the cube. However, we don't know how far it is across the floor. To find that, we need to consult with everyone's favourite bean-eating ancient Greek mathematician, Pythagoras.

The line across the floor is also the hypotenuse of a right-angled triangle - again, its shorter sides are both 5cm long. Pythagoras says the hypotenuse squared is equal to one of the short sides squared plus the other short side squared: H² = 5² + 5² = 50. So the line across the floor is √50 cm long - a smidge over 7cm.

Now we have to work out the longer diagonal. The line on the floor is now one of the shorter sides, along with the wall. So D² = (√50)² + 5² = 50 + 25 = 75. The main diagonal is √75 cm - around 8.66cm.

Bonus question: why is this the same as 5√3?

In which year will I be twice your age?

This question, of course, depends on who you are, and who "I" is*, but we can work out a general case using algebra. First, though, I'll use myself and my friend John as:


An example
If we know the year of birth of any particular person we can work out their age by subtracting their birth year from the current year.

• TeaKay was born in 1982 and it's now 2009, so my age is      2009 - 1982 = 27      years old**.
• John was born in 1955, and it's still 2009, so his age is      2009 - 1955 = 54      years old.

Now, you can see already that John will become twice my age this year (2009), but how could we work it out if it wasn't so easy?

How would we work out how old each person is in any given year? Well, we'd use the same method- subtract their birth year from the year you're looking at. In general, we could say that:

• TeaKay is/was/will be      y - 1982      years old in the year y.
• John is/was/will be      y - 1955      years old in the year y.

To answer the question we want to find what year a must be for John's age to be twice TeaKay's age. So we can write:     John's age = 2 x TeaKay's age

Algebraically, that would look like this:

     y - 1955     =     2 x (y - 1982)

Now all we have to do is solve to find a value for a:

     y - 1955     =     2y - 3964               Expand the brackets

     y + 2009    =     2y                          Use the inverse to get the numbers on the same side of the '=' sign

            2009   =       y                          Use the inverse to get the letters on the same side of the '=' sign

So now we have proven that John will be twice as old as TeaKay in 2009.


A general case
We can go a step further and develop a formula for finding out when anyone will be twice as old as anyone else:

Call your people Person A and Person B. Person A is always the oldest of the two. Now call a the year that Person A was born in, and b the year that Person B was born in. In the year y:

• Person A will be      y - a      years old.
• Person B will be      y - b      years old.

So Person A will be twice as old as Person B when:

     y - a      =      2(y - b)

So now we can solve to find a value for y, given any years a and b:

     y - a      =      2y - 2b

  y - a + 2b =      2y

      -a + 2b =       y

And neaten up:

                y = 2b - a

So, in English, this formula says "to find the year in which Person A will be twice as old as Person B, double Person B's birth year and then subtract Person A's birth year."

Which is why mathematicians like algebra so much!

*Difficult to parse, I know, but grammatically sound in the sense that I mean it...
** Of course this assumes that your birthday has already happened- I won't actually be 27 until October 10th, but lets simplify in order to get the idea across.

The formula for the mode and the median of grouped data

Any student at GCSE level in the UK needs to know about the three averages- the mean, median and mode. Of these, the mode is the simplest to find- it is the most occurring data point. The median is not much more difficult- it's the middle data point. Even at the higher end of the GCSE spectrum, when you need to find averages from tables in which the data has been grouped, finding the mode is easy: just look for the group with the largest associated frequency. Finding the median is, again, not much more involved: work out in which group the middle value falls. Or is it as simple as that?

Don't get me wrong- the above paragraph describes what you, as a higher level GCSE student, need to understand in order to find the mode and median for a grouped frequency table, and if you're feeling confused you probably shouldn't read any further. But strictly speaking, you haven't found the mode or the median. You've found the modal class and median class of the data. That is, the class (or group) in which the median and mode lie.


So how do I find the mode and median?
Let me just stress again that for the purposes of your GCSE, what you've been told to do to find the modal and median classes is exactly what you need to do. This post is purely for those who want to stretch themselves a little further than the bounds of their GCSE course.

It must be noted that, such is the nature of grouped frequency tables, it is not possible to calculate a definite average for the data. A value calculated for the mean, median or mode of grouped data must be referred to as an estimate.

The mode for grouped data
You can calculate the mode for a grouped frequency table by using the following formula:






Where:

• L is the lower class boundary of the modal class.
• f_m is the frequency associated with the modal class.
• f₁ is the frequency of the class before to the modal class.
• f2 is the frequency of the class after the modal class.
• h is the difference between the upper and lower bounds of the modal class.

The median for grouped data
You can calculate the median for a grouped frequency table using the following formula:

Where:

• L is the lower class boundary of the median class.
• f is the frequency associated with the median class.
• n is the total number of observations (i.e. the total of the Frequency column).
• c is the cumulative frequency up to the class before the median class.
• h is the difference between the upper and lower bounds of the median class.

Where do these formulae come from?

For the mode, imagine plotting the frequency of each group with its midpoint, and then joining the points up with a  smooth curve. With most distributions you would see a single peak to this curve. The formula calculates an estimate for the point on the x-axis that is directly below this peak.

For the median, imagine plotting a cumulative frequency graph of the data in your table. To find an estimate of the median, you would find half of your total frequency on the cumulative frequency axis, draw a line horizontally to your hand-drawn c.f. curve, and then drop vertically to the x-axis. Whatever this value is would be your estimate for the median. The formula just does this for you.

Finding the nth term of a sequence (the easy kind)

Everyone dreads the nth term questions, but once you get the hang of them, they're not actually all that hard.

A typical question might be: A sequence of numbers starts 2, 5, 8, 11... . What is the next number in the sequence? Find an expression for the nth term of the sequence.

The method

The next number is 14, obviously - you just add three each time. The difference between the terms is 3 - a term is simply a number in the sequence.

To find the nth term of the sequence, we look for a different sequence that goes up in threes - a good one would be the three times table.

A sidetrack that explains things

The nth term of the three times table is 3n. What does that even mean? It means that if you pick a number - say, 74 - I can tell you that the 74th number in the three times table is three times 74, or 222.

In that case, you picked 74 to be n and I replaced the n in the equation with 74. If I wanted the 100th term, n would be 100.

But the three times table isn't the same as this sequence!

Quite right. The sequence goes 2, 5, 8, 11... and the three times table goes 3, 6, 9, 12.... To get from the three times table to the sequence, we have to take away one. So the expression for the nth term of that series is 3n - 1.

Can you give me another example?

Why, of course! Let's look at -4, 3, 10, 17.... Here, the difference between terms is 7, so we're going to start from the seven times table - which has the expression 7n.

To get from 7, 14, 21... to -4, 3, 10..., we need to take away 11 - so the expression for the nth term of the sequence -4, 3, 10... is 7n - 11.

Derren Brown and Lottery Magic

So, Derren Brown. Guessing lottery numbers. Correctly*. What's the probability of him getting everything right by blind luck?

Let's say Derren's less-talented brother, Gordon, also played the lotto on Wednesday and - by a sheer fluke - also guessed that 2, 11, 23, 28, 35 and 39 would be drawn.

What's probability of the first ball drawn being on Gordon's ticket?

There are six balls on the ticket, and 49 in the machine - Sapphire - they're using for the draw. The probability of any particular ball coming up is one out of 49, and we have six that would be good news - so the probability of the first ball being on the ticket is 6/49.

And the second ball?

Now we only have five balls left in the machine that we want to see. There is also one fewer ball in the machine, so the probability of Gordon liking the second ball is 5/48.

How about the rest of them?

The rest work the same way. Each time there is one fewer ball that helps us to win, so the top goes down by one; each time there is also one fewer ball in the machine, so the bottom goes down by one each time as well. The third ball would be 4/47, the fourth 3/46, the fifth 2/45 and the final ball 1/44.

Now what do we do with those?

We just multiply them all together. We get 720 / 10,068,347,520, which works out to be 1 / 13,983,816 - just a little better than the one in fourteen million you might have heard mentioned.

How do you know it was a trick?

Easy. Derren Brown is a magician. Also, if he knew the numbers before the draw, he'd have shown them before the draw. And probably bought a ticket.

* Of course it's a trick. The deep maths he's talking about? Nuh-uh.

Factorising quadratics

What is a quadratic?

You need to be able to recognise these. One of the tell-tale signs is that your teacher will ask 'what kind of expression/equation is this?'. The other is that it has an x² in it* - the word 'quadratic' comes from the Latin for square.

What do you do with them?

Most often, you have to factorise them (put into brackets) or to solve them (figure out what x has to be). In this post, we're not going to talk about solving them using the quadratic formula, which you use when they ask for answers to a number of significant figures or decimal places. Instead, we'll look at putting them into brackets.

What's the method?

1. Take the number that doesn't have any xs attached to it, including any minus sign


2. List the factor pairs of this number (remember to think about positive and negative options), and what the numbers add up to


3. Find a pair that adds up to the middle number


4. Rewrite the x term of the expression with the x split up into the numbers you've just found (see below for an example)


5. You should now have an expression with four parts to it. Find a common factor for the first two and a common factor for the second two - bracket them off


6. Tidy up the brackets


7. If it's an equation, figure out what x has to be to make the brackets equal 0

Let's start with an example.

Say we have to solve x² -5x + 4 = 0**.

Step 1: We want the number on its own, which is +4
Step 2: What are the factor pairs of +4? We could have:

Factors	Times to make	Add to make
+1 and +4	+4	+5
+2 and +2	+4	+4
-1 and -4	+4	-5
-2 and -2	+4	+4

Step 3: We want them to add to the middle number - in this equation, it's -5. So the pair we want is -1 and -4.
Step 4: We can rewrite -5x as (-1x + -4x): that makes our equation x² -1x -4x + 4
Step 5: The first two terms have an x in common, so we can write them as x ( x-1 ). The second two terms have a -4*** we can take out, so we can write them as -4 (x-1). Hey, look! The bracket is the same for both of them!
Step 6: That means we can write the whole equation as (x-4)(x-1) = 0.
Step 7: It's an equation (there's an equals sign), so we think about what could make each bracket 0. If x=4, the first bracket is 0 and 0 multiplied by anything is 0 so the equation works. If x=1, the second bracket is 0, so the equation also works. The two answers are x=4 and x=1.

What if there's a number in front of the x²?

Excellent question, I was hoping you'd ask that. The method is almost identical, except that instead of just taking the number on its own, you multiply it by the number of x²s there are. For instance, if you had 4x² + 5x -6=0, instead of taking -6 as your number, you'd take 4 times -6 = -24.

Let's do that.

-24 has a lot of factors.

Factors	Times to make	Add to make
+1 and -24	-24	-23
+2 and -12	-24	-10
+3 and -8	-24	-5
+4 and -6	-24	-2
-1 and +24	-24	+23
-2 and +12	-24	+10
-3 and +8	-24	+5
-4 and +6	-24	+2

We want -3 and +8 because they add up to +5: the middle number. We can split up the 5x into -3x + 8x to get:

4x² - 3x + 8x - 6 = 0

If we take an x out of the first pair and a 2 out of the second, we get:

x(4x - 3) + 2(4x-3) = (x+2)(4x-3) = 0

And we're done! x = -2 or x = 3/4.

* Strictly, the x² has to be the highest power - if there's an x³ in the equation, it's a cubic, x⁴ is a quartic - you don't need to know those, you just need to know that they're not quadratics.

** CHORUS: "We have to solve x² - 5x + 4 = 0!"

*** We could take out a 4, sure. However, we really want the bracket afterwards to be the same - generally, we want to take out the minus sign if there is one).

What is a google / googol?

The famous web search engine Google took its name from the maths word 'googol'. But what does it mean? Well, just like words such as 'four,' 'pi,' 'six thousand' and 'one hundred and thirty billion,' 'one googol' is a number. Specifically, the word 'googol' belongs in the same family as hundred, thousand, million, billion and so on: it is a power of ten. What's a power of 10? That's a bigger question for another post (please ask us to get going on it if you really want to know!), but a quick recap: Powers tell you how many times to multiply a number by itself (for example, 5³ means 5 x 5 x 5). A power of 10 is just the number ten 'raised to a power' (for example, 10⁴ is read as "ten to the power of 4," and means 10 x 10 x 10 x 10, which is ten thousand (10,000)). So what's a googol? A googol is

• "ten to the power of one hundred," or
• 10¹⁰⁰, which means:

10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10

• Which equals:

10,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000 That's quite big.

Perpendicular Bisectors

What is a perpendicular bisector?
Look at the words by themselves, first of all:

• Perpendicular: This means 'at right-angles to.' For example, walls are usually perpendicular to the floor, and table legs are usually perpendicular to the table top.
• Bisector: This means 'something that cuts exactly in half.' For example, if you snap a KitKat finger and the two pieces you have left over are exactly the same size, you have bisected it.

When you're asked to construct a perpendicular bisector, you'll be expected to draw a line that cuts another line exactly in half. Sometimes you'll be given two points and asked to find where something can go if it needs to be the same distance between them. Just pretend that the two points are opposite ends of a line* and use the same method as outlined below.


Tools
Wait! Before you start, you'll need the following tools:

• Pencil (diagrams must be drawn in pencil if you're working towards an exam, and your pencil needs to be well sharpened);
• A ruler or other straight edge;
• A pair of compasses**;
• A rubber (eraser) in case you make mistakes, or to rub out construction lines.
  

How to construct a perpendicular bisector
I'll be assuming that you've already got a line to bisect. If not, now's the time to draw one. Draw any line you like, but if you need a suggestion, go for one that's about 10cm long and draw it in the middle of your page to give yourself plenty of space. Now follow these steps:

Step 1.
Open up your compasses. It doesn't matter how wide, as long as it's at least half the distance between the two ends of your line.

Step 2.
Place the point of your compasses on one end of the line that you're trying to bisect; it doesn't matter which end. Now draw a circle with your compasses in the usual way.

Step 3.
DON'T CHANGE THE SIZE OF YOUR COMPASSES! <--- This bit is really, really, really important. It won't work if you close your compasses or open it further, and you'll have to start again.
Place the point of your compasses on the other end of your line, and draw a circle in the same way.

• If you've done it correctly, you should have two circles that overlap and are exactly the same size. If they're not the same size, or they don't overlap, you'll need to start again and see what you've missed out from these instructions!

Step 4.
Use your ruler to help you join up the two places where the circles cross over with a straight line. Extend your line as far as you can on either side of the diagram

• You've drawn your perpendicular bisector! The next bit is just about tidying up.

Step 5.
You might want to use your rubber to get rid of the circles and make it easier to see the result. Be warned though, that many exam questions will ask you to leave your construction lines visible - in this case, your circles are your construction lines and you should not rub them out!

Now take a look at the perpendicular bisectors example question to see a 'real life' example.


More information
When you become more confident at drawing this type of construction, it'll become clearer that it's not necessary to draw the whole of each circle in order to find the perpendicular bisector. With a bit of practise you'll get used to how much of the circle (or what size arc) you need to draw to get things done.

Other uses
The method for drawing a perpendicular bisector can also be used to:

• Find the midpoint of a line
• Find the midpoint between two points (draw a line between the two points, then find the midpoint of that)
• Find the locus of points equidistant from two given points
  

* Strictly speaking, a line segment.
** That's one set of the pointy type of compass, not two of the magnetic type. If you'd like help setting up your compasses, please comment and we'll sort out a post for it!

Perpendicular Bisectors example question

After a long and bloody war, the empires of Euclidia and Right Anglia have reached a tentative peace treaty over the disputed Bearing Strait. The diplomats from either side have agreed that the only fair thing to do is to divide the Strait down the middle - any point that's closer to the Euclidian airport will belong to Euclidia, and anything closer to the printing press at the Right Anglian port of High Potter will henceforth be Right Anglian territory. We got two e-mails - one from the cartographers on each side - asking almost an almost identical question: we don't trust the other guys to divide the Strait up fairly on the map, so we want to make sure they draw the dividing line exactly right - otherwise there's likely to be another long and bloody war over control of the Parallelo pipeline that runs under the Straits.

We sent the same reply to both sets of map-makers:

Dear Sirs, For this difficult and probably futile task, you're going to need a pair of compasses and a straight edge. Being cartographers, you'll probably have those handy. In your pencil cases there, look. No, don't poke each other with the pointy bits. For heaven's sake. Now, put the pointy bit in one of the cities and stretch the compass out well over halfway to the other, and draw a big circle. Without changing how wide the compass is, do the same thing from the other city. You should have two points where the circles cross. Those two points are going to be on the new border. Yes, I know it's cool. You can put the compasses back in their protective wrapping now. The last thing to do is to join them up using a ruler - extend it way beyond the crossing points and bingo! There's your new border. Assuming you haven't cheated, you should both end up with the same line.

Peace, The Maths Questions Bloggers.

What's the probability of being dealt a blackjack?

What is a blackjack? Blackjack (or pontoon, twenty-one or – if you're French - vingt-et-un), is a card game. If you've never heard of it, you can find more info and rules here. The important rules as far as this question goes are:

• Aces can be worth either 1 or 11 points (as the player prefers).
• Picture cards (Jack, Queen and King) are worth 10 points.
• All other cards are worth their number value.
• The goal is to get as close as you can to a score of 21 with the cards you're dealt without going over.
• Each player is dealt two cards to start the game.

To be 'dealt a blackjack' means that the first two cards that you're given add up to 21, and you're done without needing any more cards. If you're dealt this hand, it's impossible for you to lose (although you could tie). So what's the probability of receiving this unbeatable hand? To answer this question, for simplicity's sake we'll assume that the cards are coming from a full, standard pack. And we'll look at two equivalent ways of doing it: with logic and with a tree diagram. The way with logic: how can you get 21 with only two cards? This is fairly simple – the only way you can get a score of 21 using just two cards is to get an ace (11 points) and either a 10 or a picture card (worth 10 points each). Obviously, it doesn't matter which order you draw these in, because 10 + 11 = 11 + 10. We're only interested in events when we get a card that's worth either 10 (10, J, Q, K) or 11 (Ace), so we can ignore everything else. So there are only two possible ways we can get 21:

1. We're dealt an 11, and then a 10. (11 + 10 = 21)
2. We're dealt a 10, and then an 11. (10 + 11 = 21)

Option 1:

• First event (getting an 11 card): There are 4 aces in a pack, and we've got a full pack (52 cards), so the probability of getting one of the aces first is 4 out of 52 (4/52).
• Second event (getting a 10 card): There are 16 cards worth 10 in a pack (4 of each of 10, J, Q, K), but we've already got one of the cards from the back, so the probability of getting a 10 now is 16 out of 51 (16/51).

We want event 1 AND event 2 to happen, so we have to multiply the probabilities: 4/52 x 16/51 = 64/2652. Option 2:

• First event (getting a 10 card): There are 16 cards worth 10, as before, but this time we've got a full pack, so the probability is 16/52.
• Second event (getting an 11 card): There are 4 aces, as before, but this time we've got rid of one card, so the probability is 4/52.

Again, we want event 1 AND event 2 to happen: 16/52 x 4/51 = 64/2652* Now, we want either option 1 OR option 2 to happen (because either way will give us a blackjack). There's an OR there, so we've got to add the two probabilities together: 64/2652 + 64/2652 = 128/2652. The probability bit's finished. All that's left is to tidy up, and make the numbers easier to read:

• Simplify the fraction: 128/2652 = 32/663

The probability tree method We start off by thinking about what can happen with the first card – it can be an ace (which is great), a face card (which is good) or something else (which is useless, as far as blackjacks go). We draw a branch for each of these possibilities starting on the left – upwards to A for ace, left to F for face card, and downwards to O for other. There are four aces in a pack of 52, so the probability on the A branch is 4/52. There are sixteen face cards, so the F branch is 16/52; and thirty-two other cards, so the O branch has a probability of 32/52. Now, if we started with an ace, if we're only interested in a blackjack, the next card is either a card worth 10 (yay!) or something else (boo!). We draw two branches from the A: F for a face card or ten, and A or O for everything else. Sixteen out of the 51 cards left in the pack are worth ten, so the branch to the F in the top right has a probability of 16/51 - one way of getting a blackjack has a probability of (4/52) x (16/51) = (64/2652), which cancels down to 16/663. If instead we had a face card first, we want an ace next - there are four of those left out of 51, so the probability of that branch (to the A in the middle on the right) is (16/52) x (4/51) = (64/2652) again, or 16/663. Adding the two probabilities together gives 32/663 – the same number as the other way. What on earth is 32/663? That fraction doesn't really tell us much – it's a very strange number. Let's try turning it into a decimal or a percentage:

• 32 / 663 = 0.04826546... (about 0.048, or about 4.8%).

Another way of looking at it is to turn it upside-down to find the odds: 663 ÷ 32 is (aptly) about 21 – so if you played 21 games you'd only expect to get one blackjack. Something to think about How many games would you have to play so that the probability of getting at least one blackjack was greater than 0.5?

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