HSC 2013 MX2 Marathon (archive) (3 Viewers)

twinklegal19 · Jul 18, 2013

Re: HSC 2013 3U Marathon Thread

albertcamus said:
Nah as in isn't this the 3U thread?

nvm

haha oops, got mixed up between the threads

Sy123 · Jul 18, 2013

Re: HSC 2013 3U Marathon Thread

twinklegal19 said:
An interesting volumes question

It took me ages to visualise everything lol

Through drawing a diagram, we can see that the side of a square, s is given by: $s=2r-2y$

Hence area of one square is: 4(r-y)^2

Volume thus by integrating from r to -r, i.e. from the top to the bottom, we get 32/3 r^3

Hopefully that's right.

====

$\frac{z-i}{z+1} \ \ $is purely imaginary, find the locus of the point on the Argand plane representing$ \ \ z$

I'm still trying to do that polynomial one....

========

twinklegal19 · Jul 19, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
Through drawing a diagram, we can see that the side of a square, s is given by: $s=2r-2y$

Hence area of one square is: 4(r-y)^2

Volume thus by integrating from r to -r, i.e. from the top to the bottom, we get 32/3 r^3

not quite

beolun · Jul 19, 2013

Re: HSC 2013 4U Marathon

Find the exact value of sin18.

asianese · Jul 19, 2013

Re: HSC 2013 4U Marathon

$\frac{z-i}{z+1} \ \ $is purely imaginary, find the locus of the point on the Argand plane representing$ \ \ z$

Is this what you mean?

$$For $\alpha \in \mathbb{R},$

$\dfrac{z-i}{z+1}=\alpha i \Rightarrow z-i=\alpha i z+\alpha i \\ z(1-\alpha i) = \alpha i+i \Rightarrow z=i\dfrac{(\alpha+1)}{1-\alpha i}\cdot\dfrac{1+\alpha i}{1+\alpha i} \\ = i\dfrac{(\alpha+1)+i(\alpha^2+\alpha)}{1+\alpha^2} \\ = -\dfrac{\alpha^2+\alpha}{1+\alpha^2}+i\dfrac{\alpha+1}{1+\alpha^2}.$

Sy123 · Jul 19, 2013

Re: HSC 2013 4U Marathon

asianese said:
$\frac{z-i}{z+1} \ \ $is purely imaginary, find the locus of the point on the Argand plane representing$ \ \ z$

Is this what you mean?

$$For $\alpha \in \mathbb{R},$

$\dfrac{z-i}{z+1}=\alpha i \Rightarrow z-i=\alpha i z+\alpha i \\ z(1-\alpha i) = \alpha i+i \Rightarrow z=i\dfrac{(\alpha+1)}{1-\alpha i}\cdot\dfrac{1+\alpha i}{1+\alpha i} \\ = i\dfrac{(\alpha+1)+i(\alpha^2+\alpha)}{1+\alpha^2} \\ = -\dfrac{\alpha^2+\alpha}{1+\alpha^2}+i\dfrac{\alpha+1}{1+\alpha^2}.$

Not quite, that might be its locus, but we don't know what alpha is (i.e. Re(z-i)/(z+1) = 0 )

===

Also twinklegal, my explanation is wrong, the variable square is above the fixed square of units y, but still using similar triangles as we wake a triangle to the top of the thing, I still get the same answer :/

EDIT: NVM, we just take from O to a side of the square directly, then pythagoras theorem, to get sidelength 2sqrt(r^2-x^2), squaring this and integrating from -r to r we get 16r^3/3?

nightowl · Jul 19, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$\frac{z-i}{z+1} \ \ $is purely imaginary, find the locus of the point on the Argand plane representing$ \ \ z$

Is everyone allowed to answer these questions?

$\displaystyle\Re\left(\frac{x+(y-1)i}{x+1+yi}\right)=0\Rightarrow\Re((x+(y-1)i)(x+1-yi))=0\\ \Rightarrow x^2 +x+y^2 -y=0$

It's a circle (nearly all of one anyway

) and the radius and centre are easily found.

rolpsy · Jul 19, 2013

Re: HSC 2013 4U Marathon

rolpsy said:
$\\\text{Let } a, b, c \text{ be real numbers such that the roots of the cubic equation }\\[2pt]x^3 + ax^2 + bx + c = 0 \text{ are real}\\[5pt]\text{Prove these roots are bounded above by } \frac{2\sqrt{a^2 - 3b} - a}{3}$

Hint: Choose $\alpha$ as the largest root(s) and consider the discriminant of

$\frac{x^3 + ax^2 + bx + c}{x - \alpha}$

asianese · Jul 19, 2013

Omg loooooool I've forgotten so much of 4u haha. Sorry lol

twinklegal19 · Jul 19, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Also twinklegal, my explanation is wrong, the variable square is above the fixed square of units y, but still using similar triangles as we wake a triangle to the top of the thing, I still get the same answer :/

EDIT: NVM, we just take from O to a side of the square directly, then pythagoras theorem, to get sidelength 2sqrt(r^2-x^2), squaring this and integrating from -r to r we get 16r^3/3?

yep

Sy123 · Jul 19, 2013

Re: HSC 2013 4U Marathon

rolpsy said:
Hint: Choose $\alpha$ as the largest root(s) and consider the discriminant of

$\frac{x^3 + ax^2 + bx + c}{x - \alpha}$

Dividing using long division, we get that:

$\frac{x^3 + ax^2 + bx + c}{x - \alpha} = x^2+(a+\alpha)x + (b+\alpha(a+\alpha))$

Discriminant is greater than zero:

$(a+\alpha)^2 - 4(b + \alpha(a+\alpha)) > 0$

Expanding:

$a^2-2\alpha a - 3\alpha^2 -4b > 0$

$0> 3\alpha^2 +2a \alpha -a^2+4b$

Solving for alpha:

We find that the answer directly after as an upper bound

Sy123 · Jul 19, 2013

Re: HSC 2013 4U Marathon

$$For a triangle of side lengths$ \ \ a, b, c$

$$Show that the area of such a triangle is$ \ \ A = \sqrt{s(s-a)(s-b)(s-c)}$

$$Where$ \ \ s= \frac{a+b+c}{2}$

$Further show that the radius of a circle inscribed inside the triangle has a radius$

$\sqrt{\frac{(s-a)(s-b)(s-c)}{s}}$

study1234 · Jul 19, 2013

Re: HSC 2013 4U Marathon

Prove that four consecutive terms in a row of the Pascal triangle cannot form an AP.

Sy123 · Jul 20, 2013

Re: HSC 2013 4U Marathon

study1234 said:
Prove that four consecutive terms in a row of the Pascal triangle cannot form an AP.

The numbers in pascal's triangle can be represented by binomial coefficients, the nth row and kth spot number is given by: $\binom{n}{k}$

Thus, if there exists an AP in Pascal's triangle, for some $0\leq k \leq n$

$\binom{n}{k} , \binom{n}{k+1} , \binom{n}{k+2} , \binom{n}{k+3}$

Is an arithmetic progression, if it is an arithmetic progression then the difference between the terms must be equal:

$\binom{n}{k+1} - \binom{n}{k} = \frac{n!}{(n-k)! k!} - \frac{n!}{(n-k-1)! (k+1)!} = \frac{n!(n-2k-1)}{(n-k)! (k+1)!} = D(n, k)$

$D(n, k+1) = \frac{n!(n-2(k+1) - 1)}{(n-(k+1))!(k+2)!} = D(n,k) \frac{n-k}{k+2}$

$D(n, k+2) = \frac{n!(n-2(k+2)-1)}{(n-(k+2))!(k+3)!} = D(n,k) \frac{(n-k)(n-k-1)}{(k+2)(k+3)}$

$$If the sequence forms an arithmetic progression, then$ \ \ D(n,k) = D(n,k+1)$

$D(n,k) = D(n,k) \frac{n-k}{k+2} \Rightarrow k = \frac{n-2}{2}$

$If all 4 terms form the AP then$

$D(n,k)=D(n,k+1)=D(n,k+2)$

Take the second equality:

$D(n,k) \frac{n-k}{k+2} = D(n,k) \frac{(n-k)(n-k-1)}{(k+2)(k+3)} \Rightarrow k = \frac{n-4}{2}$

$There exists no$ \ k \ \ $such that$

$k=\frac{n-2}{2} = \frac{n-4}{2}$

$Thus an arithmetic progression cannot be established$

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

$The points$ \ \ A_1, A_2, \dots, A_n \ \ $are points on the Argand diagram that represent the roots of unity of the$ \ nth \ $degree$

$P \ $is the point on the argand diagram, such that its distance to the origin is constant at 1 unit$$

$$Prove$ \ \ \sum_{k=1}^n PA_k^2 = 2n$

$$Where$ \ \ PA_k \ \ $is the distance between$ \ P \ $and$ \ A_k$

seanieg89 · Jul 20, 2013

Re: HSC 2013 4U Marathon

$Let $(x_1,x_2,\ldots,x_n)$ be an increasing sequence of real numbers and let $(y_1,y_2,\ldots,y_n)$ be an arbitrary rearrangement thereof. \\Prove that the expression $\sum_{j=1}^n x_jy_j$\\ is maximised if the sequence $y$ is increasing, and minimised if $y$ is decreasing.$

Sy123 · Jul 20, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
$Let $(x_1,x_2,\ldots,x_n)$ be an increasing sequence of real numbers and let $(y_1,y_2,\ldots,y_n)$ be an arbitrary rearrangement thereof. \\Prove that the expression $\sum_{j=1}^n a_jb_j$\\ is maximised if the sequence $y$ is increasing, and minimised if $y$ is decreasing.$

Do you mean x and y instead of a and b inside the sum?

seanieg89 · Jul 20, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Do you mean x and y instead of a and b inside the sum?

yep, cheers.

Sy123 · Jul 20, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
$Let $(x_1,x_2,\ldots,x_n)$ be an increasing sequence of real numbers and let $(y_1,y_2,\ldots,y_n)$ be an arbitrary rearrangement thereof. \\Prove that the expression $\sum_{j=1}^n x_jy_j$\\ is maximised if the sequence $y$ is increasing, and minimised if $y$ is decreasing.$

$$Since$ \ \left{y_k \right} \ \ $is an arbitrary permutation of$ \ \left{x_k \right}$

$x_1^2+x_2^2+ \dots + x_n^2 = y_1^2+y_2^2+\dots+y_n^2$

$x_1^2+y_1^2 \geq 2x_1y_1$

$x_2^2+y_2^2 \geq 2x_2 y_2$

$x_3^2+y_3^2 \geq 2x_3y_3$

.
.
.
$x_n^2+y_n^2 \geq 2x_n y_n$

$\therefore 2(x_1^2+ \dots + x_n^2) \geq 2(x_1y_1 + \dots + x_ny_n )$

$x_1 x_1 + x_2 x_2 + \dots + x_n x_n \geq x_1 y_1 + \dots + x_n y_n$

$$The only permutation of$ \ \{x_k \} \ \ $that is increasing is$ \ (x_1, x_2, \dots , x_n)$

$$And$ \ (y_1, y_2, \dots, y_n) \ $is any permutation of$ \ \{x_k \}$

$Thus the specific permutation such that if$ \ \{ y_k \} = \{ x_k \} \ \ $then the inqequality, and the expression$ \ \ \sum_{j=1}^n x_j y_j \ \ $is maximised$

=======

I'll have a go at the minimisation later.

study1234 · Jul 20, 2013

Re: HSC 2013 4U Marathon

Show that, if the four points representing the complex numbers z1, z2, z3, z4 are concyclic, the fraction (z1-z2)(z3-z4)/(z3-z2)(z1-z4) must be real.

Sy123 · Jul 21, 2013

Re: HSC 2013 4U Marathon

study1234 said:
Show that, if the four points representing the complex numbers z1, z2, z3, z4 are concyclic, the fraction (z1-z2)(z3-z4)/(z3-z2)(z1-z4) must be real.

Let $P_k \ $represent$ \ z_k$

$\angle P_1 P_2 P_3 = \arg(z_1-z_2) - \arg(z_3-z_2)$

$\angle P_3P_4P_1 =\arg(z_3-z_4) - \arg(z_1-z_4)$

Add them side by side, opposite angles of a cyclic quadrilateral add to pi so, and use the identity:

$\arg(ab) = \arg(a) + \arg(b)$

$\arg \left(\frac{a}{b}\right) = \arg(a) - \arg(b)$

We get:

$\pi = \arg \left( \frac{(z_1-z_2)(z_3-z_4)}{(z_3-z_2)(z_1-z_4)} \right)$

Since the argument of the complex number $w=\frac{(z_1-z_2)(z_3-z_4)}{(z_3-z_2)(z_1-z_4)}$ is pi, that means $w$ is real.

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