7. Special segments

Lesson

Finding the **mean**, or **average**, of numbers is a very useful idea. Taking the average of two numbers is something you have probably done many times before, both within mathematics and in everyday life. But there is more than one way to take an average, and to find out why we might want to take a "different" average in certain contexts, we should start by revisiting the "regular" average.

Sam and Alex have just finished high school, and they are traveling to Hawaii together to celebrate. They are allowed one bag of luggage each.

If they want to split their luggage equally between the two bags, how much should Alex give to Sam?

You might be able to figure this out already, but let's take our time and write this problem down in mathematical language. If the amount of luggage we want to transfer between them is $x$`x` kg, then we need to solve this equation:

$x+15=37-x$`x`+15=37−`x`

We then solve this equation to find $x=11$`x`=11 kg. Remember that this is the amount of luggage we need to move, so to find the resulting weight of each bag we add this value to Sam's luggage ($11+15=26$11+15=26 kg), and to check, we also subtract this weight from Alex's luggage to see if it's the same ($37-11=26$37−11=26 kg). In summary, we move $11$11 kg from Alex's luggage to Sam's, and then they will both have $26$26 kg.

Now let's generalize this approach - what if Sam has $A$`A` kg of luggage and Alex has $B$`B` kg, for any numbers $A$`A` and $B$`B`? What value of $x$`x` solves $x+A=B-x$`x`+`A`=`B`−`x`, and what will the value of $x+A$`x`+`A` be?

Going through the same process we used above tells us that the amount we need to move from $B$`B` to $A$`A` is $x=\frac{B-A}{2}$`x`=`B`−`A`2, and each bag then has

$x+A=\frac{B-A}{2}+A$`x`+`A`=`B`−`A`2+`A`$=$=$\frac{A+B}{2}$`A`+`B`2 kg.

This is the arithmetic mean, and it coincides with our current understanding of an "average" - add the numbers together, and divide by two! Using this approach, we end up with two bags of $\frac{A+B}{2}$`A`+`B`2 kg weighing the same as one bag of $A$`A` kg and one bag of $B$`B` kg.

We can also think about this generalized problem in terms of a number line: the length that sits halfway between $A$`A` and $B$`B` is half the total length:

A bakery is known around the world for its chocolate fudge slabs. People buy them for parties and other catered events.

They bake the fudge in baking tins that measure $25$25 cm by $81$81 cm.

One day, the oven breaks and needs to be replaced. The new oven is slightly smaller than the old one - it cannot fit trays larger than $70$70 cm in any dimension! This means their existing tins can't be used. The bakery wants to keep selling the same amount of fudge at the same price, but they need the tins to be a different size. They say they want to try out a square tin instead - what should the side lengths of these new tins be?

For their first attempt, they take the **arithmetic mean** of the original dimensions - $\frac{25+81}{2}=53$25+812=53 cm - and make square tins of this size. But these are too big - the fudge mixture ends up being too shallow, and the entire batch is overcooked and ruined.

So they go back and try to figure it out a different way. They start by calculating the area of the base of the original tin, multiplying the sides together: $25\times81=2025$25×81=2025 cm^{2}.

They then take the square root: $\sqrt{2025}=45$√2025=45 cm.

They check to make sure that square tins with each side length $45$45 cm will make exactly the same amount of fudge as a tin with one side $25$25 cm and one side $81$81 cm:

$45\times45=2025$45×45=2025 cm^{2}

They make square tins of this new side length, and the test batch is cooked to perfection. After only a short delay the fudge is back on the menu, and they have averted the crisis.

In Exploration 1 we found that two bags of $26$26 kg was "the same" as a bag of $15$15 kg and a bag of $37$37 kg. In Exploration 2 we found that a tin with two sides of $45$45 cm was "the same" as a tin with $25$25 cm on one side and $81$81 cm on the other. What it means to be "the same" depended on context - in Exploration 1, we found the **arithmetic mean**, and in Exploration 2, we did something a little different.

In general, we say that the geometric mean of two numbers is the number we get when we multiply them together and take the square root.

The geometric mean

The geometric mean of $A$`A` and $B$`B` is $\sqrt{AB}$√`A``B`.

In this diagram, the two triangles are similar:

What is the length $Y$`Y` of the common side?

Note that the common side is the hypotenuse of the smaller triangle, though it is a short side of the larger triangle. The fact that the two triangles are similar means that the ratios of their common sides are equal:

$\frac{64}{Y}=\frac{Y}{16}$64`Y`=`Y`16

This equation is very similar to the equation we had for the arithmetic mean, but we are using division rather than subtraction to express the relationship. Rearranging, we find $Y^2=64\times16$`Y`2=64×16, or $Y=\sqrt{64\times16}$`Y`=√64×16$=$=$32$32.

This time, instead of adding the numbers together and halving them, we **multiplied** the numbers together and found their **square root**.

Now that we have met both these means, it is time to explore this applet. It illustrates the arithmetic and the geometric mean of two numbers, each between $0$0 and $1$1:

Notice in particular that the geometric mean is **never larger** than the arithmetic mean, and this is true for all numbers, not just numbers between $0$0 and $1$1. The geometric mean is useful in contexts where we want to find an average of quantities that are usually multiplied together.

Let's revisit the arithmetic mean first.

Lindsey decides at the last minute that he wants to travel with Alex and Sam to Hawaii. They have $30$30 kg of luggage they want to bring with them. How should the three travelers split their luggage now?

Before: Two groups of $\frac{A+B}{2}$`A`+`B`2 is the same as one lot of $A$`A` and one lot of $B$`B`.

After: Three groups of $\frac{A+B+C}{3}$`A`+`B`+`C`3 is the same as one lot of each of $A$`A`, $B$`B`, and $C$`C`.

So now they should split $\frac{15+37+30}{3}=26$15+37+303=26 kg between all three of them.

The generalized arithmetic mean

The arithmetic mean of $n$`n` numbers $A_1$`A`1, $A_2$`A`2, $\ldots$…, $A_n$`A``n` is:

$\frac{A_1+A_2+\ldots+A_n}{n}$`A`1+`A`2+…+`A``n``n`

Now let's generalize the geometric mean in the same way.

A very special customer wants to receive the same amount of fudge as the famous classic fudge slab, but they want a **cube** of fudge. If the original slabs were $5$5 cm high, what is the dimensions of a fudge cube of the same volume?

Before: Multiply $A$`A` and $B$`B` together, and find the **square **root.

After: Multiply $A$`A`, $B$`B`, and $C$`C` together, and find the **cube **root.

So the bakery should source a tin that is a cube, with each side $\sqrt[3]{25\times81\times5}$^{3}√25×81×5$\approx$≈$21.6$21.6 cm long.

The generalized geometric mean

The geometric mean of $n$`n` numbers $A_1$`A`1, $A_2$`A`2, $\ldots$…, $A_n$`A``n` is the $n$`n`^{th} root of their product:

$\sqrt[n]{A_1A_2\ldots A_n}$^{n}√`A`1`A`2…`A``n`

What is the geometric mean of $6$6, $10$10, $60$60, $75$75, and $90$90?

Find the exact value of $h$`h` in the following triangle.

A proposal for a Martian colony includes a design for hemispherical domes. Each dome has a radius of $54$54 meters, and is supported by four vertical pillars placed $27$27 meters from the edge of the dome.

How tall must the pillars be to reach the dome? Give your answer as a simplified radical.