HSC 2014 MX2 Marathon (archive) (1 Viewer)

Chlee1998 · Nov 10, 2014

Re: HSC 2014 4U Marathon

chlee1998 said:
here's a nice q (well in the syllabus carrot and sean lol)

abc is a triangle with ab = 360, bc = 240 and ac = 180. The internal and external bisectors of cab meet bc and bc produced at p and q respectively. Find the radius of the circle which passess through a, p and q.

also, do not use a calculator under no cricumstances

Chlee1998 · Nov 10, 2014

Re: HSC 2014 4U Marathon

RealiseNothing said:
The problem wasn't so much it wasn't in the syllabus, but more so that the maths was really flawed.

haha alright

FrankXie · Nov 10, 2014

Re: HSC 2014 4U Marathon

Chlee1998 said:
Here's a nice Q (well in the syllabus carrot and Sean lol)

ABC is a triangle with AB = 360, BC = 240 and AC = 180. The internal and external bisectors of CAB meet BC and BC produced at P and q respectively. Find the radius of the circle which passess through A, P and Q.

Use the bisector angle theorem, i.e.,
$AB:AC=BP:CP, AB:AC=BQ:CQ$
So
$CP=240\times\frac13=80, CQ=240$
Therefore, the radius is 160 because triangle APQ is right angled at A and PQ is the diameter.

Sy123 · Nov 14, 2014

Re: HSC 2014 4U Marathon

$\\ $Let$ \ ABC \ $be a triangle with point$ \ X \ $on side$ \ BC $. If$ \ AB = c, AC = b, AX = p, BX = m, $and$ \ XC = n, \ $then, prove$ \ ap^2 +amn = b^2m+c^2n.$

FrankXie · Nov 14, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
$\\ $Let$ \ ABC \ $be a triangle with point$ \ X \ $on side$ \ BC $. If$ \ AB = c, AC = b, AX = p, BX = m, $and$ \ XC = n, \ $then, prove$ \ ap^2 +amn = b^2m+c^2n.$

Apply the Cosine rule once in each of triangles ABX and ACX, then multiply the first equation by n second by m and add up.

Axio · Nov 19, 2014

Re: HSC 2014 4U Marathon

z satisfies |z-i|=Imz+1. Sketch/determine the locus of P(x,y) representing z. If A is the point (0,1) on the Argand diagram and B is the y-intercept of the tangent of the locus at P on the Argand diagram, show that angleAPB=angleABP.

integral95 · Nov 19, 2014

Re: HSC 2014 4U Marathon

Axio said:
z satisfies |z-i|=Im(z+1). Sketch/determine the locus of P(x,y) representing z. If A is the point (0,1) on the Argand diagram and B is the y-intercept of the tangent of the locus at P on the Argand diagram, show that angleAPB=angleABP.

After fudging the algebra you get $\\ y = \frac{x^2+1}{2}$ Then you realised that A is actually the focus of the parabola and B lies on the directrix, so therefore AB = PA from the focus directrix definition of the parabola. There fore the angles are equal as they are base angles of isos triangle

(yeah I'm actually not too sure about the directrix part)

integral95 · Nov 19, 2014

Re: HSC 2014 4U Marathon

Axio said:
z satisfies |z-i|=Im(z+1). Sketch/determine the locus of P(x,y) representing z. If A is the point (0,1) on the Argand diagram and B is the y-intercept of the tangent of the locus at P on the Argand diagram, show that angleAPB=angleABP.

integral95 said:
After fudging the algebra you get $\\ y = \frac{x^2+1}{2}$ Then you realised that A is actually the focus of the parabola and B lies on the directrix, so therefore AB = PA from the focus directrix definition of the parabola. There fore the angles are equal as they are base angles of isos triangle

(yeah I'm actually not too sure about the directrix part)

Yeah sorry that's not right at all

I'll let P be point (Xo,Yo)

$$Tangent of P is$ \ \ y - y_0 = x_0(x-x_0) \\ $Sub x = 0 and B is point (0,y_0-(x_0)^2))$ \\ \therefore\ AB = 1-(y_0-(x_0)^2) \\ \\ $Note that$ \ \ y_0 = \frac{(x_0)^2+1}{2} \Rightarrow (x_0)^2 = 2y_0 -1 \\ \\ AB = y_0 \\ \\ $Using distance formula for AP and the substitution$ \\ \\ AP = y_0 \\ \therefore AP = AB$

So APB is an isosceles triangle so therefore the angles are equal

Axio · Nov 19, 2014

Re: HSC 2014 4U Marathon

^ I removed the brackets from Im to make it easier to do.

Axio · Nov 22, 2014

Re: HSC 2014 4U Marathon

By using the binomial theorem on cos5theta and solving the equation 32x^5 -40x^3 +10x -1=0, find the exact form of $cos\frac{\pi}{15}cos\frac{7\pi}{15}cos\frac{13\pi}{15}cos\frac{19\pi}{15}$ .

Sy123 · Nov 22, 2014

Re: HSC 2014 4U Marathon

$\\ $Let$ \ C_n(\cos \theta) = \cos n \theta \ $for some polynomial$ \ C_n \ $defined on$ \ -1 \leq x \leq 1$

$\\ $i) Show that$ \\ \\ C_{n}(x) = \begin{cases} 1, \ n = 0 \\ x, \ n= 1 \\ 2xC_{n-1}(x) - C_{n-2}(x), \ n \geq 2 \end{cases}$

$\\ $ii) Using De Moivere's Theorem (or otherwise), show that$ \\\\ \sum_{n=0}^{\infty} C_n(x) x^n = 1 \ $for$ \ -1 < x < 1$

dan964 · Nov 22, 2014

Re: HSC 2014 4U Marathon

Second part, does it still work for x=0?
Is it supposed to be 0<|x|<1

I must be missing something because I keep getting Cn=1 when n=0?

Sy123 · Nov 22, 2014

Re: HSC 2014 4U Marathon

dan964 said:
Second part, does it still work for x=0?
Is it supposed to be 0<|x|<1

C_0(x) is supposed to be 1, my bad, editing now

And yes it should work for x = 0, (see when n=0)

Axio · Nov 22, 2014

Re: HSC 2014 4U Marathon

i)

$cosn\theta =cos^{n}-\binom{n}{2}sin^{2}\theta cos^{n-2}\theta +\binom{n}{4}sin^{4}\theta cos^{n-4}\theta...\\ =cos^{n}-\binom{n}{2}(cos^{n-2}\theta-cos^{n}\theta) +\binom{n}{4}(cos^{n-4}\theta-2cos^{n-2}\theta+cos^{n}\theta )...\\\Rightarrow x^{n}-\binom{n}{2}(x^{n-2}-x^{n})+\binom{n}{4}(x^{n-4}-2x^{n-2}+x^{n})...\\when, n=0, x^0=1\\ when,n=1, x^1=x\\\\2xC_{n-1}(x)-C_{n-2}(x)\\=2x(x^{n-1}-\binom{n-1}{2}(x^{n-3}-x^{n-1})+\binom{n-1}{4}(x^{n-5}-2x^{n-3}+x^{n-1})...)-x^{n-2}-\binom{n-2}{2}(x^{n-4}-x^{n-2})+\binom{n-2}{4}(x^{n-6}-2x^{n-4}+x^{n-2})...\\=C_{n}(x)$ (Don't know how to simplify last bit)

dan964 · Nov 22, 2014

Re: HSC 2014 4U Marathon

Axio said:
By using the binomial theorem on cos5theta and solving the equation 32x^5 -40x^3 +10x -1=0, find the exact form of $cos\frac{\pi}{15}cos\frac{7\pi}{15}cos\frac{13\pi}{15}cos\frac{19\pi}{15}$ .

Sy123 · Nov 22, 2014

Re: HSC 2014 4U Marathon

Axio said:
i)

$cosn\theta =cos^{n}-\binom{n}{2}sin^{2}\theta cos^{n-2}\theta +\binom{n}{4}sin^{4}\theta cos^{n-4}\theta...\\ =cos^{n}-\binom{n}{2}(cos^{n-2}\theta-cos^{n}\theta) +\binom{n}{4}(cos^{n-4}\theta-2cos^{n-2}\theta+cos^{n}\theta )...\\\Rightarrow x^{n}-\binom{n}{2}(x^{n-2}-x^{n})+\binom{n}{4}(x^{n-4}-2x^{n-2}+x^{n})...\\when, n=0, x^0=1\\ when,n=1, x^1=x\\\\2xC_{n-1}(x)-C_{n-2}(x)\\=2x(x^{n-1}-\binom{n-1}{2}(x^{n-3}-x^{n-1})+\binom{n-1}{4}(x^{n-5}-2x^{n-3}+x^{n-1})...)-x^{n-2}-\binom{n-2}{2}(x^{n-4}-x^{n-2})+\binom{n-2}{4}(x^{n-6}-2x^{n-4}+x^{n-2})...\\=C_{n}(x)$ (Don't know how to simplify last bit)

That looks a little hand-wavy tbh (unless by "last bit" you mean those lines in your proof for part (i))

There is an easier way to do it though

Axio · Nov 22, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
That looks a little hand-wavy tbh (unless by "last bit" you mean those lines in your proof for part (i))

There is an easier way to do it though

Meant this. Feel free to share the easier method

.

Sy123 · Nov 22, 2014

Re: HSC 2014 4U Marathon

Axio said:
Meant this. Feel free to share the easier method .

HINT: Try to prove the relationships for C_n(cos theta) instead of C_n(x)

FrankXie · Nov 22, 2014

Re: HSC 2014 4U Marathon

The easier way for part (i):

$\cos n\theta+\cos (n-2)\theta=2\cos\theta\cos (n-1)\theta$

Sy123 · Nov 22, 2014

Re: HSC 2014 4U Marathon

FrankXie said:
The easier way for part (i):

$\cos n\theta+\cos (n-2)\theta=2\cos\theta\cos (n-1)\theta$

Yep pretty much what I was getting at

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