HSC 2014 MX2 Marathon (archive) (2 Viewers)

Chlee1998 · Nov 28, 2014

Re: HSC 2014 4U Marathon

let O be the center of a circle with diameter AB. C is another distinct point on the circle and D is the point on AB such that CBD is 90degrees. M is the point such that BMO is 90 degrees and lies on BC. BD is three times OM. find angle ABC

SilentWaters · Nov 28, 2014

Re: HSC 2014 4U Marathon

Chlee1998 said:
let O be the center of a circle with diameter AB. C is another distinct point on the circle and D is the point on AB such that CBD is 90degrees. M is the point such that BMO is 90 degrees and lies on BC. BD is three times OM. find angle ABC

I take it you mean right angle CDB rather than CBD.

Let $OM = l$ and $BD = 3l$
Applying Pythagoras we obtain the following:

$\\BM = \sqrt{r^2 -l^2}$
$OD = 3l-r$ so considering $\Delta CDO, CD = \sqrt{6lr-9l^2}$
but considering $\Delta CDB$ , and noting that $BC = 2BM$ , (any radius perpendicular to a chord bisects that chord) an equivalent expression results: $\sqrt{4r^2 - 13l^2}$ . So by squaring both expressions, rearranging and simplifying we get $2r^2-3lr-2l^2 = 0$ . Factorising and discarding the negative value, $r=2l$ . Looking at $\Delta OMB$ , the required angle is $sin^{-1}\left(\frac{l}{r}\right) = sin^{-1}\left(\frac{1}{2}\right) = 30^{\circ}$ .

turntaker · Nov 28, 2014

Re: HSC 2014 4U Marathon

can someone make a 2015 thread. This is old

Sy123 · Nov 29, 2014

Re: HSC 2014 4U Marathon

turntaker said:
can someone make a 2015 thread. This is old

Tell the mods to change the name, better to use the same thread so people have access to older questions

Like in the other marathon

Jordie97 · Nov 29, 2014

Re: HSC 2014 4U Marathon

Hey i am assuming that everyone in here has completed their HSC now? if that is the case can you please fill in my survey? It is for my CAFS IRP and unfortunately I am struggling to get POST HSC students to take the time and fill it in. I would be 100% grateful if u could also send this link to a bunch of your friends whom also have completed the HSC (anyone from 2010+) would be greatly appreciated!!

The link is https://www.surveymonkey.com/s/59TSCVV i need about 15 more POST hsc students to fill it in!

Axio · Nov 29, 2014

Re: HSC 2014 4U Marathon

Jordie97 said:
Hey i am assuming that everyone in here has completed their HSC now? if that is the case can you please fill in my survey? It is for my CAFS IRP and unfortunately I am struggling to get POST HSC students to take the time and fill it in. I would be 100% grateful if u could also send this link to a bunch of your friends whom also have completed the HSC (anyone from 2010+) would be greatly appreciated!!

The link is https://www.surveymonkey.com/s/59TSCVV i need about 15 more POST hsc students to fill it in!

FrankXie · Dec 3, 2014

Re: HSC 2014 4U Marathon

$$ Given that $ a\sqrt{1-b^2}+b\sqrt{1-a^2}=1, $ prove $ a^2+b^2=1$

(i) by using a trigonometric substitution (ii) without using trigonometry.

(not an HSC-like question lol)

Axio · Dec 4, 2014

Re: HSC 2014 4U Marathon

FrankXie said:
$$ Given that $ a\sqrt{1-b^2}+b\sqrt{1-a^2}=1, $ prove $ a^2+b^2=1$

(i) by using a trigonometric substitution (ii) without using trigonometry.

(not an HSC-like question lol)

(i) let a=sinx, b=cosx

sinx * sqroot(1-1+sin^2 x) +cosx * sqroot(1-1+cos^2 x)
=sin^2 x + cos^2 x =1

Therefore, a^2 + b^2 = sin^2 x + cos^2 x =1.

FrankXie · Dec 4, 2014

Re: HSC 2014 4U Marathon

Axio said:
(i) let a=sinx, b=cosx

This 'let...' is illegal, coz if you already have this, don't you really need to prove a^2+b^2=1? it is just pythagoras

ymcaec · Dec 4, 2014

Re: HSC 2014 4U Marathon

Axio said:
(i) let a=sinx, b=cosx

sinx * sqroot(1-1+sin^2 x) +cosx * sqroot(1-1+cos^2 x)
=sin^2 x + cos^2 x =1

Therefore, a^2 + b^2 = sin^2 x + cos^2 x =1.

(ii)
$a\sqrt{1-b^2}=1-b\sqrt{1-a^2}\\a^2(1-b^2)=1-2b\sqrt(1-a^2)+b^2(1-a^2)\\2b\sqrt{1-a^2}=[1-(a^2-b^2)]^2\\4b^2(1-a^2)=1-2(a^2-b^2)+(a^2-b^2)^2\\0=1-2(a^2+b^2)+(a^2+b^2)^2\\a^2+b^2=\frac{2\pm \sqrt{2^2-4}}{2}=1$

FrankXie · Dec 4, 2014

Re: HSC 2014 4U Marathon

well done

Sy123 · Dec 5, 2014

Re: HSC 2014 4U Marathon

$\\ $Let$ \ F_n = \frac{1}{1+ \frac{1}{1 + \frac{1}{\dots}}} \ $where there are$ \ n \ $fractions$ \\ \\ $So, $ F_1 = \frac{1}{1} \ , \ F_2 = \frac{1}{1+ \frac{1}{1}} = \frac{1}{2} \ $and so on$$

$$i) Show that$ \ F_{n+1} = \frac{1}{1+ F_n} \ $for$ \ n > 1$

$$ii) Prove that$ \ 0 < F_n < 1 \ $for all$ \ n > 1$

$\\ $iii) You may assume that$ \ F_n \ $converges as$ \ n \ $increases without bound$ \\ \\ $This being the case, find$ \ \lim_{n \to \infty} F_n$

Axio · Dec 5, 2014

Re: HSC 2014 4U Marathon

FrankXie said:
This 'let...' is illegal, coz if you already have this, don't you really need to prove a^2+b^2=1? it is just pythagoras

Ok. I wasn't that confident about how I did it anyway

FrankXie · Dec 5, 2014

Re: HSC 2014 4U Marathon

Axio said:
Ok. I wasn't that confident about how I did it anyway

what you can let is a=sin x, b=sin y (because |a|<=1, |b|<=1, you can always do this), and try to show x,y are complementary. then you will have the result

HSC 2014 MX2 Marathon (archive) (2 Viewers)

Chlee1998

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SilentWaters

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turntaker

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Sy123

This too shall pass

Jordie97

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Axio

=o

FrankXie

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Axio

=o

FrankXie

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ymcaec

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FrankXie

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Sy123

This too shall pass

Axio

=o

FrankXie

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