HSC 2014 MX2 Marathon ADVANCED (archive) (3 Viewers)

Sy123 · Jul 18, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Fade1233 said:
A hsc question? Remember doing this.

I doubt it would be, substitutions in inequalities is very rarely ask as far as I have experienced

Davo_01 · Jul 19, 2014

Re: HSC 2014 4U Marathon - Advanced Level

RealiseNothing said:
Prove that, for $m \geq n$

$\frac{(2m)!(m-n)!}{(m)!(2m-n)!} > 2^n$

Shouldnt it have equality with n=1
anyways here is my attempt:

$Suppose we have 2m balls, m of which are red and the remainder being blue$

$The probability of selecting n red balls without replacement is given by:$

$P_1=\dfrac{\binom{m}{n}}{\binom{2m}{n}}$

$While with replacement the probability is given by:$

$P_2=\dfrac{1}{2^n}$

$$Note that without replacement; the probability of selecting a red ball in successive selections become increasingly less while with replacement the probability will remain as$\; \dfrac{1}{2}$

$\therefore\; P_2 \geq P_1 \rightarrow \frac{(2m)!(m-n)!}{(m)!(2m-n)!} \geq 2^n$ .

RealiseNothing · Jul 19, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Davo_01 said:
Shouldnt it have equality with n=1
anyways here is my attempt:

$Suppose we have 2m balls, m of which are red and the remainder being blue$

$The probability of selecting n red balls without replacement is given by:$

$P_1=\dfrac{\binom{m}{n}}{\binom{2m}{n}}$

$While with replacement the probability is given by:$

$P_2=\dfrac{1}{2^n}$

$$Note that without replacement; the probability of selecting a red ball in successive selections become increasingly less while with replacement the probability will remain as$\; \dfrac{1}{2}$

$\therefore\; P_2 \geq P_1 \rightarrow \frac{(2m)!(m-n)!}{(m)!(2m-n)!} \geq 2^n$ .

Very nice. I had a similar method using probability but with a slightly different scenario.

Sy123 · Jul 20, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Here is my attempt (note more technical and less elegant)

$\\ $Note that the given expression can be simplified into the form$ \\ \ \binom{2m}{n} - 2^n\binom{m}{n} > 0 \\ \\ $This translates the problem into:$ \\ \\ $Prove that the co-efficient of$ \ x^n \ $in$ \ (1+x)^{2m} - (1+2x)^m \ $is positive$ \\ \\ $We begin by simplifying, but firstly, it is known that$ \\ \\ A^n - B^n = (A-B)(A^{n-1} + A^{n-2}B + \dots + AB^{n-2} + B^{n-1}) \\ \\ $So$ \ (1+x)^{2m} - (1+2x)^m = (1+2x+x^2)^m - (1+2x)^m \\ \\ = ((1+2x+x^2) - (1+2x)) \sum_{k=1}^m (1+2x+x^2)^{k} (1+2x)^{n-k} \\ = x^2\sum_{k=1}^m (1+2x+x^2)^{k} (1+2x)^{n-k} \\ \\ $Notice, the summand contains only positive operations, so it is impossible for the coefficient of$ \ x^n \ $in this sum to be negative, it is non zero as well since the$ \ x^n \ $term appears in the first sum term$ \\ \\ $So the coefficient of$ \ x^n \ $in$ \ (1+x)^{2m} - (1+2x)^m \ $is positive$ \\ \\ \Rightarrow \ \frac{(2m)! (m-n)!}{m! (2m-n)!} > 2^n \ \square$

Moonus · Jul 23, 2014

Re: HSC 2014 4U Marathon - Advanced Level

hey anto this is munt, tell me at school if u get this message

GoldyOrNugget · Jul 24, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Combinatorics

1) Count the number of possible subsets of an N-element set.

Let $A = \left\{a_1, a_2, \ldots, a_n\right\}$

2) Calculate the sum of all subset sums of A.
2) Calculate the sum of all subset products of A.

glittergal96 · Jul 25, 2014

Re: HSC 2014 4U Marathon - Advanced Level

GoldyOrNugget said:
Combinatorics

1) Count the number of possible subsets of an N-element set.

Let $A = \left\{a_1, a_2, \ldots, a_n\right\}$

2) Calculate the sum of all subset sums of A.
2) Calculate the sum of all subset products of A.

1.

$2^N.$

(Each element is either a subset member or not.)

2.

$2^{n-1}\sum_{k=1}^n a_k$

(Each ak is a member of exactly half of the subsets.)

3.

$\prod_{k=1}^n (1+a_k)$

(Don't really know how to explain this one, just expand it out and you get what you want with the convention that the empty product is 1.)

Fade1233 · Jul 25, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
Here is my attempt (note more technical and less elegant)

$\\ $Note that the given expression can be simplified into the form$ \\ \ \binom{2m}{n} - 2^n\binom{m}{n} > 0 \\ \\ $This translates the problem into:$ \\ \\ $Prove that the co-efficient of$ \ x^n \ $in$ \ (1+x)^{2m} - (1+2x)^m \ $is positive$ \\ \\ $We begin by simplifying, but firstly, it is known that$ \\ \\ A^n - B^n = (A-B)(A^{n-1} + A^{n-2}B + \dots + AB^{n-2} + B^{n-1}) \\ \\ $So$ \ (1+x)^{2m} - (1+2x)^m = (1+2x+x^2)^m - (1+2x)^m \\ \\ = ((1+2x+x^2) - (1+2x)) \sum_{k=1}^m (1+2x+x^2)^{k} (1+2x)^{n-k} \\ = x^2\sum_{k=1}^m (1+2x+x^2)^{k} (1+2x)^{n-k} \\ \\ $Notice, the summand contains only positive operations, so it is impossible for the coefficient of$ \ x^n \ $in this sum to be negative, it is non zero as well since the$ \ x^n \ $term appears in the first sum term$ \\ \\ $So the coefficient of$ \ x^n \ $in$ \ (1+x)^{2m} - (1+2x)^m \ $is positive$ \\ \\ \Rightarrow \ \frac{(2m)! (m-n)!}{m! (2m-n)!} > 2^n \ \square$

This hsc math?

Sy123 · Jul 25, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Fade1233 said:
This hsc math?

Yes
But if this was presented as a HSC question it would be split up into at the very least 3 parts

Fade1233 · Jul 25, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
Yes
But if this was presented as a HSC question it would be split up into at the very least 3 parts

I can't see questions like those in hsc though

Fade1233 · Jul 25, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
Yes
But if this was presented as a HSC question it would be split up into at the very least 3 parts

Ok that answers my question

Sy123 · Jul 25, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Not that hard compared to other problems in this thread but I think it belongs here more than in the other marathon:

$$Prove$ \ \binom{n}{k} \leq \frac{n^n}{k^k (n-k)^{(n-k)}}$

glittergal96 · Jul 26, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
Not that hard compared to other problems in this thread but I think it belongs here more than in the other marathon:

$$Prove$ \ \binom{n}{k} \leq \frac{n^n}{k^k (n-k)^{(n-k)}}$

$n^n=((n-k)+k)^n=(\sum_{j=0}^n \binom{n}{j}k^j(n-k)^{n-j} )\geq \binom{n}{k}k^k(n-k)^{n-k}.$

Sy123 · Jul 28, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Again not too hard:

$\\ $Given$ \ a,b,c > 0 \ $prove that$ \\ \\ \frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} \geq \frac{3}{2}$

glittergal96 · Jul 28, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
Again not too hard:

$\\ $Given$ \ a,b,c > 0 \ $prove that$ \\ \\ \frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} \geq \frac{3}{2}$

Let $x=a+b, y=a+c, z=b+c.$

Then $LHS=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} = \frac{1}{2} (\frac{x+y-z}{z}+\frac{x-y+z}{y}+\frac{-x+y+z}{x})\geq \frac{6-3}{2}=\frac{3}{2}$

using that the sum of a positive number and it's reciprocal is at least 2, which we know from $(\sqrt{t}-1/\sqrt{t})^2 \geq 0.$

Sy123 · Jul 28, 2014

Re: HSC 2014 4U Marathon - Advanced Level

glittergal96 said:
Let $x=a+b, y=a+c, z=b+c.$

Then $LHS=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} = \frac{1}{2} (\frac{x+y-z}{z}+\frac{x-y+z}{y}+\frac{-x+y+z}{x})\geq \frac{6-3}{2}=\frac{3}{2}$

using that the sum of a positive number and it's reciprocal is at least 2, which we know from $(\sqrt{t}-1/\sqrt{t})^2 \geq 0.$

Yep exactly my solution

Sy123 · Jul 29, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Here is a good one:

$\\ $The circumradius of a triangle is the radius of the circle that circumscribes the triangle$ \\ \\ $Let the cirumradius of a triangle with sides$ \ a,b,c \ $be$ \ R \\ \\ $Show that$ \\ \\ R \geq \frac{a^2+b^2}{2\sqrt{2a^2+2b^2-c^2}}$

glittergal96 · Jul 29, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
Here is a good one:

$\\ $The circumradius of a triangle is the radius of the circle that circumscribes the triangle$ \\ \\ $Let the cirumradius of a triangle with sides$ \ a,b,c \ $be$ \ R \\ \\ $Show that$ \\ \\ R \geq \frac{a^2+b^2}{2\sqrt{2a^2+2b^2-c^2}}$

$RHS=\frac{a^2+b^2}{2\sqrt{a^2+b^2+2ab\cos(C)}}\leq \frac{a^2+b^2}{2(a+b)} \leq \frac{2R(a+b)}{2(a+b)}=R$

with the last inequality coming from the fact that a and b are chord lengths in a circle with diameter 2R.

Sy123 · Jul 29, 2014

Re: HSC 2014 4U Marathon - Advanced Level

glittergal96 said:
$RHS=\frac{a^2+b^2}{2\sqrt{a^2+b^2+2ab\cos(C)}}\leq \frac{a^2+b^2}{2(a+b)} \leq \frac{2R(a+b)}{2(a+b)}=R$

with the last inequality coming from the fact that a and b are chord lengths in a circle with diameter 2R.

Wow much shorter than what I had in mind, using cosine rules and manipulations, and clever last inequality

Are you an olympiad student?

glittergal96 · Jul 29, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
Wow much shorter than what I had in mind, using cosine rules and manipulations, and clever last inequality

Are you an olympiad student?

Whoops, my first inequality is wrong

. I still get the feeling that there is a short solution from that first expression though, I will try again after netball.

No, I wish!

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