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HSC 2015 MX2 Marathon ADVANCED (archive) (1 Viewer)

Thread starter Trebla
Start date Dec 6, 2014

Status: Not open for further replies.

seanieg89

Well-Known Member

#201

Re: HSC 2015 4U Marathon - Advanced Level

$A subset $X$ of the plane is said to be \emph{convex} if the line segment joining any two points in $A$ lies entirely within $A$.\\ Given that $X_1,X_2,...,X_n$ are convex sets in the plane ($n\geq 3$) such that the intersection of any three of these sets is nonempty, prove that \\$\bigcap_{j=1}^n X_j$ is nonempty.$

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abecina

Member

#202

Re: HSC 2015 4U Marathon - Advanced Level

$A function $f(x) = ax^2 +bx+c$ has real coefficients and satisfies $|f(x)|\leq 1$ for all $x\in[0,1]$. \\ Find the maximal value of $|a|+|b|+|c|$$

A

abecina

Member

#203

Re: HSC 2015 4U Marathon - Advanced Level

Hi seanieg89. I can get some sort of induction argument going, but was wondering if there is a proof by contradiction. It seems just right for one!?

seanieg89

Well-Known Member

#204

Re: HSC 2015 4U Marathon - Advanced Level

abecina said:
Hi seanieg89. I can get some sort of induction argument going, but was wondering if there is a proof by contradiction. It seems just right for one!?

I think my method was an induction, but might have involved contradiction in proving one of the steps.

Not sure about whether or not you can avoid induction.

S

simpleetal

Member

#205

Re: HSC 2015 4U Marathon - Advanced Level

seanieg89 said:
$A subset $X$ of the plane is said to be \emph{convex} if the line segment joining any two points in $A$ lies entirely within $A$.\\ Given that $X_1,X_2,...,X_n$ are convex sets in the plane ($n\geq 3$) such that the intersection of any three of these sets is nonempty, prove that \\$\bigcap_{j=1}^n X_j$ is nonempty.$

sorry but can you explain what the question is asking for? I don't know what that last symbol is supposed to mean.

S

simpleetal

Member

#206

Re: HSC 2015 4U Marathon - Advanced Level

abecina said:
$A function $f(x) = ax^2 +bx+c$ has real coefficients and satisfies $|f(x)|\leq 1$ for all $x\in[0,1]$. \\ Find the maximal value of $|a|+|b|+|c|$$

Assuming that the symbol means "for x between and including 0, and 1", I get an integer answer. Is it an integer answer? I'd like post my solution, but only if I'm sure that there's no sillies.

Trebla

Administrator

Administrator

#207

Re: HSC 2015 4U Marathon - Advanced Level

seanieg89 said:
$A subset $X$ of the plane is said to be \emph{convex} if the line segment joining any two points in $A$ lies entirely within $A$.\\ Given that $X_1,X_2,...,X_n$ are convex sets in the plane ($n\geq 3$) such that the intersection of any three of these sets is nonempty, prove that \\$\bigcap_{j=1}^n X_j$ is nonempty.$

I think this question is beyond the Ext2 course (the whole idea of convex sets and the intersections of such sets).

seanieg89

Well-Known Member

#208

Re: HSC 2015 4U Marathon - Advanced Level

simpleetal said:
sorry but can you explain what the question is asking for? I don't know what that last symbol is supposed to mean.

Do you understand the definition of a convex set in the first sentence?

Then the second sentence is just asking:

If we have a finite collection of convex sets in the plane such that any three sets from this collection share a common point, then show that there exists a point common to ALL of the sets in this collection.

Sy123

This too shall pass

#209

Re: HSC 2015 4U Marathon - Advanced Level

$\\ $Prove that if$ \ \alpha, \beta , \gamma\ $are roots of a monic cubic polynomial, then these roots are real if and only if$ \ (\alpha - \beta)^2(\beta - \gamma)^2(\gamma - \alpha)^2 > 0$

seanieg89

Well-Known Member

#210

Re: HSC 2015 4U Marathon - Advanced Level

Sy123 said:
$\\ $Prove that if$ \ \alpha, \beta , \gamma\ $are roots of a monic cubic polynomial, then these roots are real if and only if$ \ (\alpha - \beta)^2(\beta - \gamma)^2(\gamma - \alpha)^2 > 0$

Sy123 said:
$\\ $Prove that if$ \ \alpha, \beta , \gamma\ $are roots of a monic cubic polynomial, then these roots are real if and only if$ \ (\alpha - \beta)^2(\beta - \gamma)^2(\gamma - \alpha)^2 > 0$

This is only necessarily true if:
a) The polynomial has real coefficients. (Otherwise any three different complex numbers with the same imaginary part can be non-real roots of a polynomial with positive $\Delta$ .)
b) The polynomial has distinct roots. (If any roots have multiplicity > 1, then $\Delta=0$ . Still, all roots will be real in this case).

Under these assumptions, we now prove your claim.

The only if is trivial.

Now suppose such a cubic has not all roots real. As non-real roots come in conjugate pairs, we must have roots $\alpha,\beta,\overline{\beta}$ for some real $\alpha$ and some non-real $\beta$ .
So

$\Delta=(\alpha-\beta)^2(\alpha-\overline{\beta})^2(\beta-\overline{\beta})^2=|\alpha-\beta|^4(2i \textrm{Im}(\beta))^2 < 0.$

So any cubic polynomial with positive $\Delta$ must have all roots real and distinct.

Interesting point: The obvious generalisation of this quantity alone doesn't tell us whether the roots are all real if we are talking about higher degree polynomials! (Eg for a quartic, you can have two pairs of conjugate roots, and so two negative terms in $\Delta$ with positive product.) A potentially fun challenge problem is to construct a quantity in terms of a monic quartic's coefficients (maybe assume that the ^3 coefficient is zero to simplify the calculations. this can be arranged by a linear substitution anyway) that is positive if and only if it's roots are distinct reals.

Last edited: Apr 23, 2015

A

abecina

Member

#211

Re: HSC 2015 4U Marathon - Advanced Level

yep, integer

G

glittergal96

Active Member

#212

Re: HSC 2015 4U Marathon - Advanced Level

abecina said:
$A function $f(x) = ax^2 +bx+c$ has real coefficients and satisfies $|f(x)|\leq 1$ for all $x\in[0,1]$. \\ Find the maximal value of $|a|+|b|+|c|$$

Let $y_1=f(0),y_2=f(1/2),y_3=f(1).$

(Note that the evaluation at any three points uniquely determines a polynomial of degree <= 2, and that the conditions of the question imply that $|y_j|\leq 1$ ).

Solving the resulting simultaneous equations gives a,b,c in terms of the $y_j$ .

We get

$a=2y_1-4y_2+2y_3$
$b=-y_1+4y+2-3y_3$
$c=y_3$

Now the triangle inequality tells us

$|a|\leq 2|y_1-y_2|+2|y_3-y_2|\leq 2(|y_1|+|y_2|)+ 2(|y_3|+|y_2|)\leq 8$
$|b|\leq 3|y_2-y_3|+|y_2-y_1|\leq 8$

So $|a|+|b|+|c|\leq 17$ , and we can obtain equality with $y_1=1,y_2=-1,y_3=1$ , that is $a=8,b=-8,c=1$ for example.

G

glittergal96

Active Member

#213

Re: HSC 2015 4U Marathon - Advanced Level

seanieg89 said:
$A subset $X$ of the plane is said to be \emph{convex} if the line segment joining any two points in $A$ lies entirely within $A$.\\ Given that $X_1,X_2,...,X_n$ are convex sets in the plane ($n\geq 3$) such that the intersection of any three of these sets is nonempty, prove that \\$\bigcap_{j=1}^n X_j$ is nonempty.$

Okay, so let $S(n)$ be the claim for a given positive integer $n\geq 3$ .

Let's assume that we know $S(4)$ and $S(n)$ for some $n\geq 4$ .

Then we will show that $S(n+1)$ is true.

To see this, let $\mathcal{A}=\{X_j\}_{j=1}^{n+1}$ be such a collection of convex sets.

Then write $Y_j=X_j\cap X_{n+1}.$ As the intersection of two convex sets is convex (I can write a separate proof for this if its not obvious enough for me to assume), we have that $\mathcal{B}=\{Y_j\}_{j=1}^n$ is a collection of convex sets.

Moreover, for any distinct $j_1,j_2,j_3\in \{1,\ldots,n\}$ , we have $Y_{j_1}\cap Y_{j_2}\cap Y_{j_3}=X_{j_1}\cap X_{j_2}\cap X_{j_3}\cap X_{n+1}\neq \emptyset$
where the last inequality follows from $S(4)$ and our assumption that every triple of sets from $\mathcal{A}$ has a common element.

In other words we have shown that every triple of sets from $\mathcal{B}$ has a common point.

From $S(n)$ we conclude that $\cap_{j=1}^{n+1} X_j= \cap_{j=1}^n Y_j \neq \emptyset$ , which is precisely the statement $S(n+1)$ .

This reduces the problem to proving $S(4)$ , which can be tackled geometrically. I will do this in a separate post now to make the whole proof more readable.

G

glittergal96

Active Member

#214

Re: HSC 2015 4U Marathon - Advanced Level

Suppose $S(4)$ were not true.

Then given such convex sets $X_1,X_2,X_3,X_4$ , we could find distinct points in the plane $x_1,x_2,x_3,x_4$ such that $x_j$ is only in the three sets that AREN'T $X_j$ .

But any four points in the plane define a unique convex quadrilateral Q.

Without loss of generality suppose this quadrilateral has vertices $x_1,x_2,x_3,x_4$ in anti-clockwise order.

Then by convexity, the diagonal $x_1x_3$ lies in $X_2\cap X_4$ and the diagonal $x_2,x_4$ lies in $X_1\cap X_3$ .

As the diagonals of a convex quadrilateral intersect, that means the point of intersection of these diagonals must lie in every $X_j$ .

This completes the proof of $S(4)$ , and hence the whole problem.

C

Chlee1998

Member

#215

Re: HSC 2015 4U Marathon - Advanced Level

glittergal96 said:
Suppose $S(4)$ were not true.

Then given such convex sets $X_1,X_2,X_3,X_4$ , we could find distinct points in the plane $x_1,x_2,x_3,x_4$ such that $x_j$ is only in the three sets that AREN'T $X_j$ .

But any four points in the plane define a unique convex quadrilateral Q.

Without loss of generality suppose this quadrilateral has vertices $x_1,x_2,x_3,x_4$ in anti-clockwise order.

Then by convexity, the diagonal $x_1x_3$ lies in $X_2\cap X_4$ and the diagonal $x_2,x_4$ lies in $X_1\cap X_3$ .

As the diagonals of a convex quadrilateral intersect, that means the point of intersection of these diagonals must lie in every $X_j$ .

This completes the proof of $S(4)$ , and hence the whole problem.

Can you post a problem which you found to be quite tough from when you were year 12?

G

glittergal96

Active Member

#216

Re: HSC 2015 4U Marathon - Advanced Level

Chlee1998 said:
Can you post a problem which you found to be quite tough from when you were year 12?

I can't really remember off the top of my head. Try some of the problems from olympiads.

C

Chlee1998

Member

#217

Re: HSC 2015 4U Marathon - Advanced Level

Prove that (n+1)^(n+1) + (-n)^n is not divisble by 9 for any natural number n.

S

simpleetal

Member

#218

Re: HSC 2015 4U Marathon - Advanced Level

Chlee1998 said:
Prove that (n+1)^(n+1) + (-n)^n is not divisble by 9 for any natural number n.

$haven't seen this thread in a long time! $ $Okay, here's what I get for this q. note that all numbers are k mod 9, where k = 0--> 8. $ $so (n+1)^{(n+1)} +(-n)^n $ [text\]{is congruent to}$ (k+1)^{k+1} +(-k)^k mod 9 $where k = 0--> 8.$ $ $testing manually for k = 0 to 8 shows that the expression is not 0 mod 9 for any such k, and thus $ (n+1)^{(n+1)}$ $+ (-n)^n $is not divisble by 9 for any natural number n.$

Last edited: Jun 7, 2015

RealiseNothing

what is that?It is Cowpea

#219

Re: HSC 2015 4U Marathon - Advanced Level

Define a function by:

$f_p(k) = m$

Where $p$ is prime, $k$ is real, and $m$ is the highest integer power of $p$ that divides $k$ .

Note that $m$ doesn't have to be integer. i.e. $f_2(10^{\frac{1}{3}}) = f_2(2^{\frac{1}{3}} \times 5^{\frac{1}{3}}) = \frac{1}{3}$

You are given that the function defined has the property that:

$f_k(\frac{p}{q}) = f_k(p) - f_k(q)$

Use this function to prove that the square root of all non-square positive integers is irrational.

Last edited: Jun 11, 2015

I

InteGrand

Well-Known Member

#220

Re: HSC 2015 4U Marathon - Advanced Level

RealiseNothing said:
Define a function by:

$f_p(k) = m$

Where $p$ is prime, $k$ is a positive integer, and $m$ is the highest power of $p$ that divides $k$ .

Note that $m$ doesn't have to be integer. i.e. $f_2(10^{\frac{1}{3}}) = f_2(2^{\frac{1}{3}} \times 5^{\frac{1}{3}}) = \frac{1}{3}$

You are given that the function defined has t
he property that:

$f_k(\frac{p}{q}) = f_k(p) - f_k(q)$

Use this function to prove that the square root of all non-square positive integers is irrational.

How can we have the example of $10^{\frac{1}{3}}$ ? Didn't you say $k$ needed to be integer?

Last edited: Jun 11, 2015

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