2012 Year 9 &10 Mathematics Marathon (2 Viewers)

Ealdoon · Jul 5, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

I'm trying to explain it as well but I cannot even use LaTeX haha.

EDIT: Yay it worked!

Ealdoon · Jul 5, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

My proofs and explanations are horrible but lets hope this is correct! Here it is:
Let's use a hexagon as an example and start from the bottom right corner. Draw 3 lines from the corner to show the diagonals you can get from that point, you will find that you get 3 diagonals. Lets move on to the next corner which is the middle right corner. You will find that you only get 2 diagonals. If you go to the last corner which is the top right corner, you will only get 1 diagonal. This proves that each diagonal connects to each corner except itself and its 2 neighbouring corners.

Let "n" be the number of corners (you can use sides as well though since the number of corners = the number of sides). Thus: $n - 3$

Proof:
A hexagon has 6 corners. Using the formula $n - 3$ , we get 3 which means that there are 3 diagonals from a corner, which is true. However this formula is only for 1 corner

If we multiply the number of sides/corners by the formula we just found, we will get the number of diagonals. We will also need to divide it by 2 since diagonals only connect 2 corners once which halves the shape. Thus the formula is: $\frac{n(n - 3)}{2}$

Proof:
If a polygon has 700 sides, how many diagonals are there?

$\frac{n(n - 3)}{2}$

$\frac{700(700 - 3)}{2}$

= 243950 diagonals

kazemagic · Jul 5, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

Ealdoon said:
My proofs and explanations are horrible but lets hope this is correct! Here it is:
Let's use a hexagon as an example and start from the bottom right corner. Draw 3 lines from the corner to show the diagonals you can get from that point, you will find that you get 3 diagonals. Lets move on to the next corner which is the middle right corner. You will find that you only get 2 diagonals. If you go to the last corner which is the top right corner, you will only get 1 diagonal. This proves that each diagonal connects to each corner except itself and its 2 neighbouring corners.

Let "n" be the number of corners (you can use sides as well though since the number of corners = the number of sides). Thus: $n - 3$

Proof:
A hexagon has 6 corners. Using the formula $n - 3$ , we get 3 which means that there are 3 diagonals from a corner, which is true. However this formula is only for 1 corner

If we multiply the number of sides/corners by the formula we just found, we will get the number of diagonals. We will also need to divide it by 2 since diagonals only connect 2 corners once which halves the shape. Thus the formula is: $\frac{n(n - 3)}{2}$

Proof:
If a polygon has 700 sides, how many diagonals are there?

$\frac{n(n - 3)}{2}$

$\frac{700(700 - 3)}{2}$

= 243950 diagonals

Great explanation! And this one by ymcaec!

hard to explain. i'll try my best

If n = number of vertices (or sides)

n - 3 = a vertex can form diagonals with all other vertices except for the two next to it and itself
then times n (for each of the vertices)
divide by 2 because there are 2 ends for each diagonal (so the diagonals are counted twice)

Ealdoon · Jul 5, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

kazemagic said:
Great explanation!

Thanks

Demento1 · Jul 5, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

Ealdoon said:
My proofs and explanations are horrible but lets hope this is correct! Here it is:
Let's use a hexagon as an example and start from the bottom right corner. Draw 3 lines from the corner to show the diagonals you can get from that point, you will find that you get 3 diagonals. Lets move on to the next corner which is the middle right corner. You will find that you only get 2 diagonals. If you go to the last corner which is the top right corner, you will only get 1 diagonal. This proves that each diagonal connects to each corner except itself and its 2 neighbouring corners.

Let "n" be the number of corners (you can use sides as well though since the number of corners = the number of sides). Thus: $n - 3$

Proof:
A hexagon has 6 corners. Using the formula $n - 3$ , we get 3 which means that there are 3 diagonals from a corner, which is true. However this formula is only for 1 corner

If we multiply the number of sides/corners by the formula we just found, we will get the number of diagonals. We will also need to divide it by 2 since diagonals only connect 2 corners once which halves the shape. Thus the formula is: $\frac{n(n - 3)}{2}$

Proof:
If a polygon has 700 sides, how many diagonals are there?

$\frac{n(n - 3)}{2}$

$\frac{700(700 - 3)}{2}$

= 243950 diagonals

$Pretty much this. All you really need to know and consider is that upon one vertex, it can connect with the 2 neighbouring vertices and extend to the opposite vertex. After that, you should remember that only half of the diagonals you multiply to total are considered, as you are not double counting half of the polygon.$

I should type like this ^.

Ealdoon · Jul 5, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

Demento1 said:
$Pretty much this. All you really need to know and consider is that upon one vertex, it can connect with the 2 neighbouring vertices and extend to the opposite vertex. After that, you should remember that only half of the diagonals you multiply to total are considered, as you are not double counting half of the polygon.$

I should type like this ^.

Haha! It looks good doesn't it?

iBibah · Jul 5, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

All spot on.

Simple when you think about it, each vertex can join to every other one bar itself and the two next to it hence (n-3). This can happen for every vertex, hence n(n-3). Then we must cancel double hence n(n-3)/2.

Timske · Jul 7, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=\textup{If} ~A^x = 3 \\ \textup{Find}~ A^{10x - x} - 5" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\textup{If} ~A^x = 3 \\ \textup{Find}~ A^{10x - x} - 5" title="\textup{If} ~A^x = 3 \\ \textup{Find}~ A^{10x - x} - 5" /></a>

Ealdoon · Jul 7, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

Timske said:
<a href="http://www.codecogs.com/eqnedit.php?latex=\textup{If} ~A^x = 3 \\ \textup{Find}~ A^{10x - x} - 5" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\textup{If} ~A^x = 3 \\ \textup{Find}~ A^{10x - x} - 5" title="\textup{If} ~A^x = 3 \\ \textup{Find}~ A^{10x - x} - 5" /></a>

I am going to give this a go!

twinklegal19 · Jul 7, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

^That looks right

Ealdoon · Jul 7, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

twinklegal19 said:
^That looks right

Thanks

Demento1 · Jul 7, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

Ealdoon said:
I am going to give this a go!

<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{A^{10x}}{A^{x}}- 5 = \frac{(A^{x})^{10}}{A^{x}}- 5" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{A^{10x}}{A^{x}}- 5 = \frac{(A^{x})^{10}}{A^{x}}- 5" title="\frac{A^{10x}}{A^{x}}- 5 = \frac{(A^{x})^{10}}{A^{x}}- 5" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{3^{10}}{3}- 5" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{3^{10}}{3}- 5" title="\frac{3^{10}}{3}- 5" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex== (3)^{9} - 5" target="_blank"><img src="http://latex.codecogs.com/gif.latex?= (3)^{9} - 5" title="= (3)^{9} - 5" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex=\therefore A^{10-x}- 5 = 19678" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\therefore A^{10-x}- 5 = 19678" title="\therefore A^{10-x}- 5 = 19678" /></a>

That's correct. This is my simplified format of solving it. Didn't even notice the Q until just then...

$A^{9x}=3^9$

$\therefore 3^9 - 5 = 19678$

Edit: You can prove this is possible through basic logarithms.

twinklegal19 · Jul 7, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

Let's have some financial maths/consumer arithmetic

Tom has been investing his money for 5 years. Initially his investment was earning interest at 9% per annum compounded monthly. From the start of the global financial crisis 18 months ago, his investment began to lose value at 15% per annum each month.

i. If Tom initially invested $P, what was the value of his portfolio immediately before the global financial crisis? Give your answer in unsimplified form in terms of P.
ii. Tom’s investment portfolio is now worth $10 000. What was the amount of his original investment?

Demento1 · Jul 7, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

twinklegal19 said:
Let's have some financial maths/consumer arithmetic

Tom has been investing his money for 5 years. Initially his investment was earning interest at 9% per annum compounded monthly. From the start of the global financial crisis 18 months ago, his investment began to lose value at 15% per annum each month.

i. If Tom initially invested $P, what was the value of his portfolio immediately before the global financial crisis? Give your answer in unsimplified form in terms of P.
ii. Tom’s investment portfolio is now worth $10 000. What was the amount of his original investment?

I think you may have discovered a slightly weaker area in my mathematics. But anyways, I can hopefully solve the Q that you throw at me.

i) Amount before the global financial crisis:

$= P(1+\frac{9}{1200})^{42}$

ii) I cannot bother typing all the working out, although the working out soon ended up as this with me.

$P = \frac{10000}{\left ( \frac{403}{400} \right )^{42} + \left ( \frac{79}{80} \right )^{18}}$

Therefore: P = $4616.73 (nearest cent)

RealiseNothing · Jul 7, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

twinklegal19 said:
Let's have some financial maths/consumer arithmetic

Tom has been investing his money for 5 years. Initially his investment was earning interest at 9% per annum compounded monthly. From the start of the global financial crisis 18 months ago, his investment began to lose value at 15% per annum each month.

i. If Tom initially invested $P, what was the value of his portfolio immediately before the global financial crisis? Give your answer in unsimplified form in terms of P.
ii. Tom’s investment portfolio is now worth $10 000. What was the amount of his original investment?

I hate you for posting this question.

twinklegal19 · Jul 7, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

Demento1 said:
I think you may have discovered a slightly weaker area in my mathematics. But anyways, I can hopefully solve the Q that you throw at me.

i) Amount before the global financial crisis:

$= P(1+\frac{9}{1200})^{42}$

ii) I cannot bother typing all the working out, although the working out soon ended up as this with me.

$P = \frac{10000}{\left ( \frac{403}{400} \right )^{42} + \left ( \frac{79}{80} \right )^{18}}$

Therefore: P = $4616.73 (nearest cent)

You got the first part right but not the second, try again

(dw, I got it wrong the first time last year as well...)

RealiseNothing said:
I hate you for posting this question.

<3

ymcaec · Jul 7, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

Demento1 said:
I think you may have discovered a slightly weaker area in my mathematics. But anyways, I can hopefully solve the Q that you throw at me.

i) Amount before the global financial crisis:

$= P(1+\frac{9}{1200})^{42}$

ii) I cannot bother typing all the working out, although the working out soon ended up as this with me.

$P = \frac{10000}{\left ( \frac{403}{400} \right )^{42} + \left ( \frac{79}{80} \right )^{18}}$

Therefore: P = $4616.73 (nearest cent)

for ii. i got 9163.05 instead

twinklegal19 · Jul 7, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

ymcaec said:
for ii. i got 9163.05 instead

correct! Show the lines of working out?

Demento1 · Jul 7, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

ymcaec said:
for ii. i got 9163.05 instead

That's correct.

ymcaec · Jul 7, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

twinklegal19 said:
correct! Show the lines of working out?

Instead of "plus", it should be a "times"
$P = \frac{10000}{\left ( \frac{403}{400} \right )^{42} \times \left ( \frac{79}{80} \right )^{18}} \\ = 9163.05$ (2dp)$$

2012 Year 9 &10 Mathematics Marathon (2 Viewers)

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