2012 Year 9 &10 Mathematics Marathon (1 Viewer)

twinklegal19 · Jul 5, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

ymcaec said:
$$i. $\frac{1}{n} - \frac{1}{n+1}\\= \frac{n+1-n}{n(n+1)}\\=\frac{1}{n(n+1)}\\\\\\$ii. If n = 1, $n(n+1) = 1\times2 \\$Hence $(\frac{1}{1} - \frac{1}{1+1}) +(\frac{1}{2} - \frac{1}{2+1}) +(\frac{1}{3} - \frac{1}{3+1}) + ... + (\frac{1}{100} - \frac{1}{100+1}) \\\frac{1}{1}+(\frac{1}{2} +\frac{1}{3} + ... + \frac{1}{100}) - (\frac{1}{2} + \frac{1}{3} + ... + \frac{1}{100}) - \frac{1}{101}\\=\1 - \frac{1}{101}\\ =\frac{100}{101}$

perfect

twinklegal19 · Jul 5, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

Another question

ymcaec, perhaps let the others have a go before posting your solution this time?

~~EDIT: I think part i meant angle TZY not TXY?~~

EDIT: nvm

russ3l · Jul 5, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

twinklegal19 said:
Another question

ymcaec, perhaps let the others have a go before posting your solution this time?

we we have a go but that doesn't mean we will get it right

iBibah · Jul 5, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

^Good question.

ymcaec · Jul 5, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

twinklegal19 said:
Another question

ymcaec, perhaps let the others have a go before posting your solution this time?

EDIT: I think part i meant angle TZY not TXY?

part 1 TXY is right

ymcaec · Jul 5, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

twinklegal19 said:
Another question

ymcaec, perhaps let the others have a go before posting your solution this time?

EDIT: I think part i meant angle TZY not TXY?

just checking : iii. 1000 cm^3 & d. 71.57 degrees (71 d 34 m) ?

Demento1 · Jul 5, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

Well well well, back from table tennis and lunch yet again. Is any year 10 other then me or ymcaec who is willing to solve the question above? I really cannot be bothered looking back at the Qs and answering them at this stage as I am in the middle of reading this strange manifesto about maths given from pwoh.

ymcaec · Jul 5, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

Demento1 said:
Well well well, back from table tennis and lunch yet again. Is any year 10 other then me or ymcaec who is willing to solve the question above? I really cannot be bothered looking back at the Qs and answering them at this stage as I am in the middle of reading this strange manifesto about maths given from pwoh.

ive done it but not posting the solution
russ3l is doing it

iBibah · Jul 5, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

twinklegal19 said:
Another question

ymcaec, perhaps let the others have a go before posting your solution this time?

EDIT: I think part i meant angle TZY not TXY?

Nah its supposed to be TXY, you have to show angle XTY is 90 and thats done by ***********

twinklegal19 · Jul 5, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

iBibah said:
Nah its supposed to be TXY, you have to show angle XTY is 90 and thats done by ***********

ah oops my bad

a year 11 failing this haha

twinklegal19 · Jul 5, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

ymcaec said:
just checking : iii. 1000 cm^3 & d. 71.57 degrees (71 d 34 m) ?

part iii is correct

however for iv, angle TMZ is 67 degrees 23 minutes

iBibah · Jul 5, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

ymcaec said:
just checking : iii. 1000 cm^3 & d. 71.57 degrees (71 d 34 m) ?

That angle is incorrect.

Hint:

Let SZ meet DC at E

What is SE? EZ? SZ (or TZ)?

Use that ^

ymcaec · Jul 5, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

iBibah said:
That angle is incorrect.

Hint:

Let SZ meet DC at E

What is SE? EZ? SZ (or TZ)?

Use that ^

thanks
i was looking at the wrong angle

iBibah · Jul 5, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

Here's a quick one: How many diagonals are there in a 700 sided polygon?

Demento1 · Jul 5, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

iBibah said:
Here's a quick one: How many diagonals are there in a 700 sided polygon?

My sort of question which I enjoy.

Number of diagonals = $\frac{700(700-3)}{2}$

= 243950 distinct diagonals.

iBibah · Jul 5, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

Demento1 said:
My sort of question which I enjoy.

Number of diagonals = $\frac{700(700-3)}{2}$

= 243950 distinct diagonals.

Yep, now explain how to get that formula.

twinklegal19 · Jul 5, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

iBibah said:
Yep, now explain how to get that formula.

^Yeah, that's always the hard part

(aka no rote learning allowed hehe)

russ3l · Jul 5, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

ugh rote learning is much easier

just like memorising essays

iBibah · Jul 5, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

twinklegal19 said:
^Yeah, that's always the hard part

(aka no rote learning allowed hehe)

*harder. Its not too hard, just logical thinking, combinatoric style

russ3l said:
ugh rote learning is much easier just like memorising essays

Shouldn't have said that......

ymcaec · Jul 5, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

iBibah said:
*harder. Its not too hard, just logical thinking, combinatoric style

hard to explain. i'll try my best

If n = number of vertices (or sides)
$\frac{n(n-3)}{2}$

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)

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