HSC 2013-14 MX1 Marathon (archive) (1 Viewer)

Capt Rifle · Mar 30, 2013

Re: HSC 2013 3U Marathon Thread

Nws m8 said:
this IS 3u, I remember doing this in 3u, too easy for 4u m8

Thought so, my mate needed help with it and sent it to me lol.

Capt Rifle · Mar 30, 2013

Re: HSC 2013 3U Marathon Thread

Ima post a hard conics question on 4u marathon see who gets it lol

Sy123 · Mar 30, 2013

Re: HSC 2013 3U Marathon Thread

asianese said:
1.

I believe you this time :3

======================

$In the above diagram, PQ is tangent to the circle with centre O$

$By considering what happens as P approaches Q, prove that the the radius of a circle to its tangent is always perpendicular$

asianese · Mar 30, 2013

Re: HSC 2013 3U Marathon Thread

If you wanted some 'evidence' you split the numerator and write out a few terms. Mass cancellation and you're left with 1.

RealiseNothing · Mar 31, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
I believe you this time :3

======================

$In the above diagram, PQ is tangent to the circle with centre O$

$By considering what happens as P approaches Q, prove that the the radius of a circle to its tangent is always perpendicular$

As P approaches Q, the length PO approaches the length OQ. Also the length PQ approaches 0. Hence Pythagoras' Theorem holds:

$OQ^2 + PQ^2 = PO^2$

So either angle OPQ or OQP is 90. But since the angles of a triangle add to 180, the other angle must also be 90. Hence since angle OQP is a constant, it is always 90.

RealiseNothing · Mar 31, 2013

Re: HSC 2013 3U Marathon Thread

Prove that:

$cosAcosB\geq\sqrt{cos(A-B)cos(A+B)}$

Given $0 \leq B \leq A \leq \frac{\pi}{4}$

Sy123 · Mar 31, 2013

Re: HSC 2013 3U Marathon Thread

RealiseNothing said:
As P approaches Q, the length PO approaches the length OQ. Also the length PQ approaches 0. Hence Pythagoras' Theorem holds:

$OQ^2 + PQ^2 = PO^2$

So either angle OPQ or OQP is 90. But since the angles of a triangle add to 180, the other angle must also be 90. Hence since angle OQP is a constant, it is always 90.

Hmm what makes you think Pythagoras Theorem will hold?

RealiseNothing · Mar 31, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
Hmm what makes you think Pythagoras Theorem will hold?

1am in the morning me wasn't thinking straight.

Sy123 · Apr 1, 2013

Re: HSC 2013 3U Marathon Thread

$15 Balls of equal unit radius are bounded by a large triangle like a Billiard rack, as shown below$

$Find the area of empty space$

$Consider all calculations within 2 dimensions$

Sy123 · Apr 1, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
I believe you this time :3

======================

$In the above diagram, PQ is tangent to the circle with centre O$

$By considering what happens as P approaches Q, prove that the the radius of a circle to its tangent is always perpendicular$

As Q approaches P it is clear that the length QO is decreasing, and that the minimum distance is when Q is at P.

So therefore the shortest distance from O to Q is when Q coincides with P.

Now within HSC geometry, the shortest distance is always perpendicular distance.

Since OP is shortest distance, OP is perpendicular to QP

shongaponga · Apr 3, 2013

Re: HSC 2013 3U Marathon Thread

$\\ $Use Mathematical induction to show that:$ \\ \\ x^n $ + $ y(2x+y)^{n-1} $ is divisible by $ (x+y); $ where n, x and y are positive integers.$$

HeroicPandas · Apr 5, 2013

Re: HSC 2013 3U Marathon Thread

shongaponga said:
$\\ $Use Mathematical induction to show that:$ \\ \\ x^n $ + $ y(2x+y)^{n-1} $ is divisible by $ (x+y); $ where n, x and y are positive integers.$$

$$ Show true for n = 1, x + y(2x+y)^0 = x+y \\ $ \therefore $ divisble by (x+y) \\ \\ $ $Assume try for n = k, \\ i.e x^k + y(2x+y)^{k-1} = (x+y)M ($ $where M is a positive integer) \\ \\ RTP: n = k + 1, \\ i.e. x^{k+1} + y(2x+y)^k = (x+y)Q ($ $Where Q is a positive integer) \\ \\ LHS = x^{k+1} + y(2x+y)^k \\ =x . x^k + y(2x+y)^{k-1} .(2x+y) \\ = x.x^k + (2x+y)[(x+y)M - x^k] ($ $Utilising assumption) \\ = (2x+y)(x+y)M -x.x^k - y.x^k \\ = (x+y)[(2x+y)M - x^k] \\ = (x+y)Q ($ $ where Q is a positive integer) \\ \\ $ $Insert conclusion$

Problème (i'm not French)

$\int \frac{1 + cotx}{1 - cotx}.dx$

asianese · Apr 5, 2013

Re: HSC 2013 3U Marathon Thread

The direction of your implication is incorrect. It should go: It is true for n=k+1 since it is true for n=k (you ASSUMED this).

HeroicPandas · Apr 5, 2013

Re: HSC 2013 3U Marathon Thread

asianese said:
The direction of your implication is incorrect. It should go: It is true for n=k+1 since it is true for n=k (you ASSUMED this).

Thank you!

asianese · Apr 5, 2013

Re: HSC 2013 3U Marathon Thread

$\int \frac{1+\cot x}{1-\cot x}dx \\ =\int \frac{1+\frac{\cos x}{\sin x}}{1-\frac{\cos x}{\sin x}} dx\\ = \int \frac{\sin x+\cos x}{\sin x-\cos x}dx \\ = \ln | \sin x-\cos x | +C$

HeroicPandas · Apr 5, 2013

Re: HSC 2013 3U Marathon Thread

asianese said:
$\int \frac{1+\cot x}{1-\cot x}dx \\ =\int \frac{1+\frac{\cos x}{\sin x}}{1-\frac{\cos x}{\sin x}} dx\\ = \int \frac{\sin x+\cos x}{\sin x-\cos x}dx \\ = \ln | \sin x-\cos x | +C$

yes

Drongoski, ur one was correct aswell

RealiseNothing · Apr 5, 2013

Re: HSC 2013 3U Marathon Thread

asianese said:
The direction of your implication is incorrect. It should go: It is true for n=k+1 since it is true for n=k (you ASSUMED this).

That's still not correct. You have to say "It is true for n=k+1 if it is true for n=k".

asianese · Apr 5, 2013

Re: HSC 2013 3U Marathon Thread

RealiseNothing said:
That's still not correct. You have to say "It is true for n=k+1 if it is true for n=k".

Ah soz. I typed (you assumed this) so I did mean what you wrote

RealiseNothing · Apr 5, 2013

Re: HSC 2013 3U Marathon Thread

asianese said:
Ah soz. I typed (you assumed this) so I did mean what you wrote

I know what you meant

I'm just nit-picking.

asianese · Apr 5, 2013

Re: HSC 2013 3U Marathon Thread

RealiseNothing said:
I know what you meant

I'm just nit-picking.

I'm all for nit-picking, don't get me wrong hehe.

Someone do my question from way back. The one about intersecting tangents.

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