Geometric/ Arithmetic Mean (1 Viewer)

pomsky · May 16, 2015

Hey guys!
Could someone explain what a GM and an AM is?
And also figure this question out haha.

$$The point M divides the internal AB externally in the ratio 1:$ \alpha$ in such a way that AB is the geometric mean of AM and FM. Find the exact value(s) of $\alpha$, and draw a diagram to show the results.$

Thanks : )

InteGrand · May 16, 2015

pomsky said:
Hey guys!
Could someone explain what a GM and an AM is?
And also figure this question out haha.

$$The point M divides the internal AB externally in the ratio 1:$ \alpha$ in such a way that AB is the geometric mean of AM and FM. Find the exact value(s) of $\alpha$, and draw a diagram to show the results.$

Thanks : )

$An arithmetic mean (AM) of $n$ numbers is the ``average'' of them that you're used to, i.e. $\frac{a_1 + a_2 + \cdots + a_n}{n}$. The geometric mean is similar, but instead of adding the numbers and dividing by the number of numbers, we multiply them and then take the $n$th root: $\mathrm{GM} = \sqrt[n]{a_1 a_2 \cdots a_n}$ (where each $a_i$ is non-negative).$

$(The $AM$ in your question refers to a line segment though.) For your question, what is $FM$? Is that a typo (meant to be $BM$)?$

pomsky · May 16, 2015

InteGrand said:
$An arithmetic mean (AM) of $n$ numbers is the ``average'' of them that you're used to, i.e. $\frac{a_1 + a_2 + \cdots + a_n}{n}$. The geometric mean is similar, but instead of adding the numbers and dividing by the number of numbers, we multiply them and then take the $n$th root: $\mathrm{GM} = \sqrt[n]{a_1 a_2 \cdots a_n}$ (where each $a_i$ is non-negative).$

$(The $AM$ in your question refers to a line segment though.) For your question, what is $FM$? Is that a typo (meant to be $BM$)?$

Yep! Sorry for the typo <> It is indeed meant to be BM.
Oh. I kind of understand where you would find AM questions, but not so much GM ones. How would you recognise one?

InteGrand · May 16, 2015

pomsky said:
Yep! Sorry for the typo <> It is indeed meant to be BM.
Oh. I kind of understand where you would find AM questions, but not so much GM ones. How would you recognise one?

GM often comes up in HSC 4U inequality Q's (AM-GM inequality).

pomsky · May 16, 2015

InteGrand said:
GM often comes up in HSC 4U inequality Q's (AM-GM inequality).

Haha I don't 4u

For the above Q, what does it mean by AB being the geometric mean of AM and BM? How would I do this question?

InteGrand · May 16, 2015

pomsky said:
Haha I don't 4u
For the above Q, what does it mean by AB being the geometric mean of AM and BM? How would I do this question?

They mean that the length of the interval AB is the geometric mean of the lengths of AM and BM, i.e. $AB = \sqrt{AM\times BM}$ .

Here's how to do the Q:

$$Draw $AB$ as a horizontal line interval with $A$ to the left of $B$. The fact that $M$ divides $AB$ externally in the ratio $1:\alpha$ means that $\frac{AM}{BM}=\frac{1}{\alpha}\Rightarrow \alpha = \frac{BM}{AM}$ (note $\alpha > 0$). We are also given that $AB = \sqrt{AM\times BM}$. There are two cases.$ .

$\underline{\text{CASE 1}: \textit{M}\text{ lies to the left of }\textit{A}}$

$In this case, it is clear from a sketch that $BM=AB + AM$. We can let $AM$ have unit length, then it is clear from a sketch that $BM=\alpha$ (also because $\alpha = \frac{BM}{AM}$) and so $AB=\alpha - 1$ (so $\alpha$ must be greater than 1, as $AB > 0$). Then $AB = \sqrt{AM\times BM}\Rightarrow \alpha -1 = \sqrt{1\times \alpha}\Rightarrow (\alpha -1)^2=\alpha$.$

$This quadratic equation in $\alpha$ can be solved to give $\frac{3\pm \sqrt{5}}{2}$. Since $\alpha > 1$, the only possible $\alpha$ for this case is $\alpha = \frac{3+\sqrt{5}}{2}$.$

$\underline{\text{CASE 2}: \textit{M}\text{ lies to the right of }\textit{B}}$

$In this case, it is clear from a sketch that $BM=AM-AB$. We can let $BM$ have unit length ($BM=1$), then since $\alpha = \frac{BM}{AM},$ we have $AM=\frac{BM}{\alpha}=\frac{1}{\alpha}$ and so $AB=\frac{1}{\alpha} - 1$ (so $\frac{1}{\alpha}$ must be greater than 1 $\Leftrightarrow \alpha < 1$, as $AB > 0$). Then $AB = \sqrt{AM\times BM}\Rightarrow \frac{1}{\alpha} -1 = \sqrt{1\times \frac{1}{\alpha}}$. This is the same equation as we had before, except replacing $\alpha$ with $\frac{1}{\alpha}$. So the solutions here are $\frac{1}{\alpha}= \frac{3\pm\sqrt{5}}{2}\Rightarrow \alpha = \frac{2}{3\pm\sqrt{5}}$. We can see that the only solution satisfying $\alpha < 1$ is $\alpha = \frac{2}{3+\sqrt{5}}$.$

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Geometric/ Arithmetic Mean (1 Viewer)

pomsky

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InteGrand

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pomsky

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InteGrand

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pomsky

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InteGrand

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