HSC 2013 MX2 Marathon (archive) (3 Viewers)

[Sy123]

Re: HSC 2013 4U Marathon

Your second line (angle BAO = angle CAP) assumes that A, O, T and P are collinear.

True dat, back to square one.

 

[twinklegal19]

Your second line (angle BAO = angle CAP) assumes that A, O, T and P are collinear.

dw sy pretty much everyone at the unsw maths seminar thing made the same mistake at least once (even the uni students running it lol)

 

[Sy123]

Re: HSC 2013 4U Marathon

dw sy pretty much everyone at the unsw maths seminar thing made the same mistake at least once (even the uni students running it lol)

I got it now relatively simple actually if you look at it what it means to prove something is collinear:

Let angle BOA = x , angle BOP = y

Now we need to prove x + y = 180
After showing various things like that OTP is collinear already (due to that property of intersecting circles and their centres)
And showing that various angles are 90 degrees, we then consider

Angle BAO = 90 -x
Angle CPA = 180 - y

Consider triangle APC

180 = 90 + 90 -x + 180 - y

therefore x + y = 180

Therefore A, O , T is collinear

O, P, T is already collinear

Therefore A, O, P, T is collinear




Attempting second one now

 

[twinklegal19]

I got it now relatively simple actually if you look at it what it means to prove something is collinear:

Let angle BOA = x , angle BOP = y

Now we need to prove x + y = 180
After showing various things like that OTP is collinear already (due to that property of intersecting circles and their centres)
And showing that various angles are 90 degrees, we then consider

Angle BAO = 90 -x
Angle CPA = 180 - y

Consider triangle APC

180 = 90 + 90 -x + 180 - y

therefore x + y = 180

Therefore A, O , T is collinear

O, P, T is already collinear

Therefore A, O, P, T is collinear




Attempting second one now

I don't think you can consider triangle CPA since you are assuming AOP is collinear in order to add all those angles up

 

[Sy123]

Re: HSC 2013 4U Marathon

I don't think you can consider triangle CPA since you are assuming AOP is collinear in order to add all those angles up

:/

Ok, how about this, we know that OPT is collinear, and we know that B and C are collinear.

So lets just extend these lines, they are not parallel hence they must meet at some point, lets call it A.

By the symmetry argument the same thing must apply to C D, that is if we extend the line CD, it will be OTP at the same point A due to it being symmetrical, and no the symmetry does not depend on the point A.

Hence A, O, P, T collinear???

 

[twinklegal19]

:/

Ok, how about this, we know that OPT is collinear, and we know that B and C are collinear.

So lets just extend these lines, they are not parallel hence they must meet at some point, lets call it A.

By the symmetry argument the same thing must apply to C D, that is if we extend the line CD, it will be OTP at the same point A due to it being symmetrical, and no the symmetry does not depend on the point A.

Hence A, O, P, T collinear???

I get what you mean but I'm not sure if that proof is rigorous enough lol, (perhaps others can clarify this)

Another way is to make a few constructions and do a little bit of angle chasing hehe

 

[Drongoski]

Re: HSC 2013 4U Marathon

I get what you mean but I'm not sure if that proof is rigorous enough lol, (perhaps others can clarify this)

..

 

[asianese]

Re: HSC 2013 4U Marathon

^ lol'ed

 

[twinklegal19]

..

Ahaha oops
It was 2:30am

 

[Drongoski]

Re: HSC 2013 4U Marathon

Ahaha oops
It was 2:30am

Don't worry. Wasn't trying to make you look bad.

 

[bleakarcher]

Re: HSC 2013 4U Marathon

This may help:

 

[Sy123]

Re: HSC 2013 4U Marathon

I got it in the end....











It can be shown that:











EDIT: Second one ->

First construct point M, which is the intersection with the small circle of AO, and point N, which is the extension of line AP to the larger circle
Consider triangles ABD and ACP, it can be shown that they are similar through AAA
Through that tangent secant theorem it can be shown that:

let AM=x

 

[Trebla]

Re: HSC 2013 4U Marathon

Suppose that f(x) is an even function and k is a constant. Show that



Hence evaluate

 

[twinklegal19]

I got it in the end....









therefore ECB = 180-y \ \ \ \ ($isosceles triangle and right angles$)

It can be shown that:











EDIT: Second one ->

First construct point M, which is the intersection with the small circle of AO, and point N, which is the extension of line AP to the larger circle
Consider triangles ABD and ACP, it can be shown that they are similar through AAA
Through that tangent secant theorem it can be shown that:

let AM=x

Not that I'm doubting your method or anything but can you explain this step a bit more please? I don't quite follow

Edit: why is angle CPT = angle EPT?

 

[asianese]

Re: HSC 2013 4U Marathon

Suppose that f(x) is an even function and k is a constant. Show that



Hence evaluate

Expand the integrand. Notice that the product of two even functions is even, and the product of an even and odd function of odd. Between the bounds of -a and a, the k*sinx*f(x) disappears whilst the cosx*f(x) doubles, giving the result.

I like the 2nd part! Do a nifty substitution and it ends in a reverse product rule .

 

[Sy123]

Re: HSC 2013 4U Marathon

Not that I'm doubting your method or anything but can you explain this step a bit more please? I don't quite follow

Yeah I made the wrong assumption that angle TPC = angle TPE
My method is invalid

 

[Immortalp00n]

Re: HSC 2013 4U Marathon

Practice Question: PROBABILITY:

I will start...

Q1. In how many ways can three groups of three numerals be chosen without replacement from the nine numerals 1,2,3,4,5,6,7,8,9 ?



Q2. In how many ways can a group of three letters be chosen without replacement from the nine letters of the world AUSTRALIA?



PRACTICE QUESTIONS: COMPLEX Numbers:

Q1. In an Argand diagram the vectors OP and OQ represent the complex numbers z and w repectively such that Triangle OPQ is equilateral.
What is an expression for z^2 +w^2.

.... more to come

thanks heaps bro
do u have any practise questions for 4unit? my half yearlies are soon man
god bless u so much my friend

 

[bleakarcher]

Re: HSC 2013 4U Marathon



Let angle BOF=x.
Now, angle AOD=x (triangles AOB and AOD are congruent)
Hence, angle BOD=2x
Hence, angle OBF=90-x
Hence, angle ABF=x (tangent is perpendicular to radius drawn to point of contact)
Now, triangle ABD is similar to triangle ACE
Hence, angle ACE=x since angle ACE corresponds to angle ABF
Hence, OPCG is a cyclic quadrilateral since angle FOB= angle BCG (i.e. exterior angle theorem holds)
Hence, angle BOG=180-x (opposite angles of a cyclic quad. are supplementary)
Hence, A, O, T and P are collinear.

Hope that's right.

 

[twinklegal19]

View attachment 27393

Let angle BOF=x.
Now, angle AOD=x (triangles AOB and AOD are congruent)
Hence, angle BOD=2x
Hence, angle OBF=90-x
Hence, angle ABF=x (tangent is perpendicular to radius drawn to point of contact)
Now, triangle ABD is similar to triangle ACE
Hence, angle ACE=x since angle ACE corresponds to angle ABF
Hence, OPCG is a cyclic quadrilateral since angle FOB= angle BCG (i.e. exterior angle theorem holds)
Hence, angle BOG=180-x (opposite angles of a cyclic quad. are supplementary)
Hence, A, O, T and P are collinear.

Hope that's right.

That theorem still requires FOG to be collinear and since F and G lie on the intervals AO and OP by construction then you're saying that AOP are collinear

oh wai-

 

[bleakarcher]

Re: HSC 2013 4U Marathon

That theorem still involves FOG to be collinear and since F and G lies on the intervals AO and OP by construction then you're saying that AOP is collinear

oh wai-

Damn, you're right lol.