HSC 2016 MX2 Marathon ADVANCED (archive) (1 Viewer)

RealiseNothing · Nov 28, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Paradoxica said:
The Cauchy-Schwarz Inequality is not something that can be cited in the exam.

Pretty easy to prove it though.

Drsoccerball · Nov 28, 2015

Re: HSC 2016 4U Marathon - Advanced Level

RealiseNothing said:
Pretty easy to prove it though.

Now that you changed your photo you no longer feel like a bad ass...

omegadot · Nov 28, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Paradoxica said:
The Cauchy-Schwarz Inequality is not something that can be cited in the exam.

lita1000 said:
uhhhh Cauchy Schwarz isn't needed to solve that question mate.

RealiseNothing said:
Pretty easy to prove it though.

Would anyone care to offer up a solution for the benefit of others?

omegadot · Nov 28, 2015

Re: HSC 2016 4U Marathon - Advanced Level

$$\noindent \textbf{Next Question}$

$$\noindent Let$\\\begin{align*}f(x) &= a_0 x^m + a_1 x^{m - 1} + \cdots + a_m\end{align*}\\$and$\\\begin{align*}g(x) &= b_0 x^n + b_1 x^{n - 1} + \cdots + b_n\end{align*}\\$where $\deg f \leqslant \deg g - 1$ and $a_0,b_0 \neq 0.$

$$\noindent If $g(x)$ has $n$ distinct roots $\alpha_1,\alpha_2,\ldots,\alpha_n$ such that the decomposition of $\frac{f(x)}{g(x)}$ into partial fractions is given by$

$\frac{f(x)}{g(x)} = \frac{A_1}{x - \alpha_1} + \frac{A_2}{x - \alpha_2} + \cdots + \frac{A_n}{x - \alpha_n},$

$$\noindent show that $\\A_1 + A_2 + \cdots + A_n = \begin{cases}0 & \text{if } \deg f < \deg g - 1\\[1ex]\displaystyle{\frac{a_0}{b_0}} & \text{if } \deg f = \deg g - 1\end{cases}$

Paradoxica · Nov 29, 2015

Re: HSC 2016 4U Marathon - Advanced Level

lita1000 said:
uhhhh Cauchy Schwarz isn't needed to solve that question mate.

If you have a solution faster than Cauchy Schwarz then I concede.

Until then

$\large INELEGANT!!!$

RealiseNothing · Nov 29, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Drsoccerball said:
Now that you changed your photo you no longer feel like a bad ass...

Girlfriend did it, cbf changing back.

glittergal96 · Nov 30, 2015

Re: HSC 2016 4U Marathon - Advanced Level

For the polynomial question without assuming C-S:

$P(x)P(x^{-1})=\sum_{k=0}^n \sum_{j=0}^n a_ka_jx^{k-j}=\sum_{k=0}^n a_k^2 +\sum_{k<j}a_ka_j(x^{j-k}+x^{k-j})\\ \\ \geq \sum_{k=0}^n a_k^2 + 2\sum_{k<j} a_ka_j = \left(\sum_{k=0}^n a_k\right)^2=P(1)^2.$

Also, I wouldn't consider C-S an "elegant" solution to the inequality as posed to a high school student given that the inequality in this question in the first place is just a special case of C-S itself, and C-S is outside assumed knowledge.

That's kind of like citing AM-GM in n variables in a proof of AM-GM in 3 variables. Sure your proof will then be short, but it is only artificially short, as if you fleshed it out into a full path from MX2 assumed knowledge to the claimed statement, you are first walking past your target statement and then doubling back.

glittergal96 · Nov 30, 2015

Re: HSC 2016 4U Marathon - Advanced Level

omegadot said:
$$\noindent \textbf{Next Question}$

$$\noindent Let$\\\begin{align*}f(x) &= a_0 x^m + a_1 x^{m - 1} + \cdots + a_m\end{align*}\\$and$\\\begin{align*}g(x) &= b_0 x^n + b_1 x^{n - 1} + \cdots + b_n\end{align*}\\$where $\deg f \leqslant \deg g - 1$ and $a_0,b_0 \neq 0.$

$$\noindent If $g(x)$ has $n$ distinct roots $\alpha_1,\alpha_2,\ldots,\alpha_n$ such that the decomposition of $\frac{f(x)}{g(x)}$ into partial fractions is given by$

$\frac{f(x)}{g(x)} = \frac{A_1}{x - \alpha_1} + \frac{A_2}{x - \alpha_2} + \cdots + \frac{A_n}{x - \alpha_n},$

$$\noindent show that $\\A_1 + A_2 + \cdots + A_n = \begin{cases}0 & \text{if } \deg f < \deg g - 1\\[1ex]\displaystyle{\frac{a_0}{b_0}} & \text{if } \deg f = \deg g - 1\end{cases}$

From the definition of g, we have

$g(x)=b_0\prod_{i=1}^n (x-\alpha_i).$

So upon multiplying both sides of the equation

$\frac{f(x)}{g(x)}=\sum_{j=1}^n A_j(x-\alpha_j)^{-1}$

by $\prod_{i=1}^n (x-\alpha_i)$ , we are left with

$\frac{f(x)}{b_0}=\sum_{j=1}^n A_j \prod_{i\neq j} (x-\alpha_i).$

The claim then follows by equating the coefficients of degree n-1 on both sides.

Paradoxica · Dec 5, 2015

Re: HSC 2016 4U Marathon - Advanced Level

$\noindent Let $FGH$ be a triangle with circumcircle $A$ and incircle $B$, the latter in tangential contact at $J$ on $GH$. Let $C$ be a circle tangent to $FG$, $FH$ and $A$, and let $D$ be the point where $C$ and $A$ are tangential, as shown here:$

$$Prove $\angle FGH = \angle GDJ$

lita1000 · Dec 22, 2015

Re: HSC 2016 4U Marathon - Advanced Level

A difficult question which one of my students asked me the other day:

$$Consider the concentric ellipses$\ \frac{x^2}{a^2}+\frac{y^2}{b^2} $and$ \frac{x^2}{c^2}+\frac{y^2}{d^2}, $ where $ b<a<c $ and $ b<d<c. $The tangent at $E(x_0,y_0)$ intersects with the outer ellipse at $ P(x_1,y_1)$ and $ Q(x_1,y_1).$ By considering the expressions of $x_1x_2 $ and $ y_1y_2,$ and given that the tangents at P and Q are always perpendicular regardless of the choice of E on the inner ellipse, prove that $c^4=a^2(c^2+d^2)$

lita1000 · Dec 22, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Let's see if one of the BOS members can crack it

Carrotsticks · Dec 22, 2015

Re: HSC 2016 4U Marathon - Advanced Level

lita1000 said:
Let's see if one of the BOS members can crack it

$\\ $Pick two points $ P $ and $ Q $ such that $PQ$ is vertical and the tangents are perpendicular. The chord $ PQ $ must be a vertical tangent to the internal ellipse. By symmetry, if the tangents are to be perpendicular, then the tangent at say $ P$ in the first quadrant must be equal to $ -1. $ It is not too difficult to show that this occurs at $ \left ( \frac{c^2}{\sqrt{c^2+d^2}}, \frac{d^2}{\sqrt{c^2+d^2}} \right ). $ But the $x$ coordinate of this point must be equal to $a$, so $a=\frac{c^2}{\sqrt{c^2+d^2}} $ and the result follows.$$

InteGrand · Dec 22, 2015

Re: HSC 2016 4U Marathon - Advanced Level

lita1000 said:
A difficult question which one of my students asked me the other day:

$$Consider the concentric ellipses$\ \frac{x^2}{a^2}+\frac{y^2}{b^2} $and$ \frac{x^2}{c^2}+\frac{y^2}{d^2}, $ where $ b<a<c $ and $ b<d<c. $The tangent at $E(x_0,y_0)$ intersects with the outer ellipse at $ P(x_1,y_1)$ and $ Q(x_1,y_1).$ By considering the expressions of $x_1x_2 $ and $ y_1y_2,$ and given that the tangents at P and Q are always perpendicular regardless of the choice of E on the inner ellipse, prove that $c^4=a^2(c^2+d^2)$

$Tangent at $P$ has equation $\frac{x_1x}{c^2}+\frac{y_1 y}{d^2}=1$, which has slope $m_P = -\frac{\frac{x_1}{c^2}}{\frac{y_1}{d^2}}=-\frac{x_1 d^2}{c^2 y_1}$ for $y_1\neq 0$. Similarly, the slope of the tangent at $Q$ is $m_Q = -\frac{x_2 d^2}{c^2 y_2}$ for $y_2 \neq 0$. Since the tangents are perpendicular, we have $m_P m_Q = -1 \Rightarrow \frac{x_1 d^2}{c^2 y_1}\cdot \frac{x_2 d^2}{c^2 y_2}=-1 \Rightarrow \frac{x_1 x_2}{y_1 y_2}=-\frac{c^4}{d^4},y_1,y_2 \neq 0$.$

$As this happens for any choice of $(x_0,y_0)$ on the inner ellipse such that neither of $y_1,y_2$ is 0, choose $(x_0,y_0)=(a,0)$ so that the tangent is vertical. It is then easy to show that the two intersection points are $\left(a,\pm d \sqrt{1-\frac{a^2}{c^2}} \right)$ by solving simultaneously $x=a$ with the equation of the outer ellipse.$

$So $x_1 x_2 = a^2 $ and $y_1 y_2=-d^2\left(1-\frac{a^2}{c^2} \right)$. So $ \frac{x_1 x_2}{y_1 y_2}= -\frac{c^4}{d^4} \Rightarrow \frac{a^2}{-d^2 \left(1-\frac{a^2}{c^2} \right) } =-\frac{c^4}{d^4}$. Rearranging this yields the result.$

Edit: done above!

KX · Jan 3, 2016

Re: HSC 2016 4U Marathon - Advanced Level

Okay got sent here by a user called Integrand, can someone post a good question please?

Paradoxica · Jan 3, 2016

Re: HSC 2016 4U Marathon - Advanced Level

KX said:
Okay got sent here by a user called Integrand, can someone post a good question please?

I'll requote the unsolved problems from the previous marathons.

InteGrand said:
$\noindent Let $ABC$ be a triangle with a point $D$ on $AB$, $E$ on $BC$ and $F$ on $AC$. It is given that $AD=BE=CF$ and that $\triangle DEF$ is equilateral.$

$Show that $\triangle ABC$ is equilateral.$

lita1000 said:
$\\ $Show that for $a,b,c >0,\\ (1-a)^2+(1-b)^2+(1-c)^2\\ \geq \frac{(1-a^2)(1-b^2)c^2}{(ab+c)^2}+\frac{(1-c^2)(1-b^2)a^2}{(bc+a)^2}+\frac{(1-a^2)(1-c^2)b^2}{(ac+b)^2}$

HeroicPandas said:
$$\noindent$ a,b,c$, are sides of a triangle satisfying $a^2+b^2+c^2=3$.\\Prove $a^3b+b^3c+c^3a \geq ab+bc+ca$

InteGrand · Jan 3, 2016

Re: HSC 2016 4U Marathon - Advanced Level

Paradoxica said:
I'll requote the unsolved problems from the previous marathons.

I do remember some of these had solutions posted up (e.g. definitely remember simpleetal posting up his solution to his problem).

InteGrand · Jan 3, 2016

Re: HSC 2016 4U Marathon - Advanced Level

Paradoxica said:
I'll requote the unsolved problems from the previous marathons.

$Also, not sure about dan964's Q. If $f(nx)$ and $g(nx)$ are identical degree $n$ polynomials, they are equal for more than $n$ input values, so they are identical polynomials. Then if $f(x)^2 + g(x)^2=1$ (by this I assume he meant identically 1), we have $f(x)^2 + f(x)^2 \equiv 1\Rightarrow f(x) \equiv \pm \frac{1}{\sqrt{2}}$. This doesn't have anything to do with $n\geq 5$.$

Paradoxica · Jan 3, 2016

Re: HSC 2016 4U Marathon - Advanced Level

InteGrand said:
$Also, not sure about dan964's Q. If $f(nx)$ and $g(nx)$ are identical degree $n$ polynomials, they are equal for more than $n$ input values, so they are identical polynomials. Then if $f(x)^2 + g(x)^2=1$ (by this I assume he meant identically 1), we have $f(x)^2 + f(x)^2 \equiv 1\Rightarrow f(x) \equiv \pm \frac{1}{\sqrt{2}}$. This doesn't have anything to do with $n\geq 5$.$

Yeah, I couldn't really wrap my mind around that. Someone needs to summon him and acquire a clarification of the question, or perhaps the corrected version.

Also, there's my circle geo problem up there, which is beyond the scope of the Ext. 2 course.

Librah · Jan 3, 2016

Re: HSC 2016 4U Marathon - Advanced Level

Prove there can be no positive integers a,b,c for n>2 that can satisfy the equation a^n+b^n=c^n. Popular question in many past papers.

Paradoxica · Jan 3, 2016

Re: HSC 2016 4U Marathon - Advanced Level

Librah said:
Prove there can be no positive integers a,b,c for n>2 that can satisfy the equation a^n+b^n=c^n. Popular question in many past papers.

.

Pierre de Fermat said:
Cuius rei demonstrationem mirabilem sane detexi hanc marginis exiguitas non caperet.

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