HSC 2015 MX2 Marathon (archive) (1 Viewer)

Drsoccerball · Aug 25, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $Let the number of ways to move up a staircase of$ \ n \ $steps, where one can only move up 1 or 2 steps at a time, be$ \ s_n$

$\\ $Find a recurrence formula for$ \ s_n \ $and prove its correctness$$

I think braintic posted this question on the new X2 Permutations marathon it was a fibonnaci sequence

Sy123 · Aug 25, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
I think braintic posted this question on the new X2 Permutations marathon it was a fibonnaci sequence

correct, but can you give the reason why? (unless it was already posted)

Drsoccerball · Aug 25, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
correct, but can you give the reason why? (unless it was already posted)

braintic said:
Let S(n) be the number of ways of climbing n steps.

The first move is either 1 step or 2 steps.
If it is 1 step, there are S(n-1) ways of climbing the remaining n-1 steps.
If it is 2 steps, there are S(n-2) ways of climbing the remaining n-2 steps.
So S(n) = S(n-1) + S(n-2), for n>2
This is the definition of the Fibonacci series.
The base cases S(1) and S(2) must be actually counted ... S(1) = 1, S(2) = 2.
These are the 2nd & 3rd terms of the Fibonacci series.

.

Ekman · Aug 25, 2015

Re: HSC 2015 4U Marathon

$$ Given that $ (p+2)(q+2)(r+2) = 64 $ (where p,q,r are positive integers), and assuming the inequality $ (a+b+c)\geq3(abc)^{\frac{1}{3}}$

$$ Prove: $ pqr \leq 8$

porcupinetree · Aug 25, 2015

Re: HSC 2015 4U Marathon

Ekman said:
$$ Given that $ (p+2)(q+2)(r+2) = 64 $ (where p,q,r are positive integers), and assuming the inequality $ (a+b+c)\geq3(abc)^{\frac{1}{3}}$

$$ Prove: $ pqr \leq 8$

https://imgur.com/9wpDIWb

Ekman · Aug 25, 2015

Re: HSC 2015 4U Marathon

porcupinetree said:
https://imgur.com/9wpDIWb

Just one thing, how do you know that: $6\geq 3(pqr)^{\frac{1}{3}}$

porcupinetree · Aug 25, 2015

Re: HSC 2015 4U Marathon

Ekman said:
Just one thing, how do you know that: $6\geq 3(pqr)^{\frac{1}{3}}$

Aha. Good pickup, I didn't notice that

Drsoccerball · Aug 25, 2015

Re: HSC 2015 4U Marathon

porcupinetree said:
Aha. Good pickup, I didn't notice that

$$ You have two expresions $ p+q+r \geq 6 $ and $ p+q+r \geq 3(pqr)^{1/3}$

$$ Subtracting second by 1 st by second you get 0 \geq 3(pqr)^{1/3} -6$

Continue from yours

Ekman · Aug 25, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
$$ You have two expresions $ p+q+r \geq 6 $ and $ p+q+r \geq 3(pqr)^{1/3}$

$$ Subtracting second by 1 st by second you get 0 \geq 3(pqr)^{1/3} -6$

Continue from yours

But cant you subtract the 1st expression by the 2nd one and get the reverse?

Drsoccerball · Aug 25, 2015

Re: HSC 2015 4U Marathon

Ekman said:
But cant you subtract the 1st equation by the 2nd one and get the reverse?

Thats what i was thinking aahah tried getting passed your security in picking up mistakes

Ekman · Aug 25, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
Thats what i was thinking aahah tried getting passed your security in picking up mistakes

I answered the question like porcupinetree did in the exam, but when I looked back over the proof, I realised I didn't prove that $6\geq3(pqr)^{\frac{1}{3}}$

Zlatman · Aug 25, 2015

Re: HSC 2015 4U Marathon

Ekman said:
$$ Given that $ (p+2)(q+2)(r+2) = 64 $ (where p,q,r are positive integers), and assuming the inequality $ (a+b+c)\geq3(abc)^{\frac{1}{3}}$

$$ Prove: $ pqr \leq 8$

I'm not sure if there's much mathematical merit behind this, but if p, q and r are positive integers, wouldn't the only case for which $(p+2)(q+2)(r+2) = 64$ holds true be when p = q = r = 2 (pqr = 8)?

p, q or r cannot equal 1, since 3 isn't a factor of 64.

When any of them are an integer greater than 2, there's no possible way to make 64 from the 3 sets of brackets. If p = 3, 4, 5, then p + 2 = 5, 6 or 7, which aren't factors of 64. If p = 6 (p + 2 = 8), then (q+2)(r+2) = 8, which can't be true, since each set of brackets must be greater than or equal to 3. Any number greater than 6 will have the same result.

The only other time $(p+2)(q+2)(r+2) = 64$ holds true (with these conditions) is if we include 0 as a positive integer (which isn't right). If p = 0, then (q+2)(r+2) = 32, which can be made with q = 2 and r = 6, for example. In this case, pqr = 0 < 8.

I hope that makes some sense.

Ekman · Aug 25, 2015

Re: HSC 2015 4U Marathon

Zlatman said:
I'm not sure if there's much mathematical merit behind this, but if p, q and r are positive integers, wouldn't the only case for which $(p+2)(q+2)(r+2) = 64$ holds true be when p = q = r = 2 (pqr = 8)?

p, q or r cannot equal 1, since 3 isn't a factor of 64.

When any of them are an integer greater than 2, there's no possible way to make 64 from the 3 sets of brackets. If p = 3, 4, 5, then p + 2 = 5, 6 or 7, which aren't factors of 64. If p = 6 (p + 2 = 8), then (q+2)(r+2) = 8, which can't be true, since each set of brackets must be greater than or equal to 3. Any number greater than 6 will have the same result.

The only other time $(p+2)(q+2)(r+2) = 64$ holds true is if we include 0 as a positive integer (which isn't right). If p = 0, then (q+2)(r+2) = 32, which can be made with q = 2 and r = 6, for example. In this case, pqr = 0 < 8.

I hope that makes some sense.

I may have confused the conditions for p,q,r for another question. The one before it was also an inequalities question that required squeeze theorem so it was fun to prove. So I may have confused the conditions between those two questions

Zlatman · Aug 25, 2015

Re: HSC 2015 4U Marathon

Ekman said:
I may have confused the conditions for p,q,r for another question. The one before it was also an inequalities question that required squeeze theorem so it was fun to prove. So I may have confused the conditions between those two questions

I guess p, q and r could just be positive? ¯\_(ツ)_/¯

InteGrand · Aug 25, 2015

Re: HSC 2015 4U Marathon

$64=(p+2)(q+2)(r+2)$, $p,q,r>0$

$Note that by the AM-GM inequality for two variables, $p+2 \geq 2\sqrt{2p}=\sqrt{8}\sqrt{p}$, so similarly, $q+2\geq\sqrt{8}\sqrt{q}$ and $r+2\geq \sqrt{8}\sqrt{r}$.$

$\therefore 64=(p+2)(q+2)(r+2) \geq \sqrt{8}\sqrt{p}\cdot \sqrt{8}\sqrt{q}\cdot \sqrt{8}\sqrt{r}$

$\Rightarrow 64 \geq 8\sqrt{8}\sqrt{pqr}$

$\Rightarrow \frac{64}{8\sqrt{8}} \geq \sqrt{pqr}$

$$\Rightarrow \sqrt{8} \geq \sqrt{pqr}$ (as $\frac{64}{8\sqrt{8}}=\frac{8}{\sqrt{8}} =\sqrt{8}$)$

$$\Rightarrow 8 \geq pqr$ (squaring both sides, which preserves the inequality as both sides are positive). \blacksquare$

Ekman · Aug 25, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
$64=(p+2)(q+2)(r+2)$, $p,q,r>0$

$Note that by the AM-GM inequality for two variables, $p+2 \geq 2\sqrt{2p}=\sqrt{8}\sqrt{p}$, so similarly, $q+2\geq\sqrt{8}\sqrt{q}$ and $r+2\geq \sqrt{8}\sqrt{r}$.$

$\therefore 64=(p+2)(q+2)(r+2) \geq \sqrt{8}\sqrt{p}\cdot \sqrt{8}\sqrt{q}\cdot \sqrt{8}\sqrt{r}$

$\Rightarrow 64 \geq 8\sqrt{8}\sqrt{pqr}$

$\Rightarrow \frac{64}{8\sqrt{8}} \geq \sqrt{pqr}$

$$\Rightarrow \sqrt{8} \geq \sqrt{pqr}$ (as $\frac{64}{8\sqrt{8}}=\frac{8}{\sqrt{8}} =\sqrt{8}$)$

$$\Rightarrow 8 \geq pqr$ (squaring both sides, which preserves the inequality as both sides are positive). \blacksquare$

I guess that works, but it doesn't use the given inequality. How would we go about using the AM-GM inequality for 3 terms

InteGrand · Aug 25, 2015

Re: HSC 2015 4U Marathon

Ekman said:
I guess that works, but it doesn't use the given inequality. How would we go about using the AM-GM inequality for 3 terms

$You don't need to use the given one (at least in the way you worded the Q.). I haven't thought about how to do it using the 3-variable AM-GM inequality.$

InteGrand · Aug 25, 2015

Re: HSC 2015 4U Marathon

$Here's a way to do it using 3-variable AM-GM inequality.$

$64=(p+2)(q+2)(r+2)$

$$\Rightarrow 64=pqr+2(pq+qr+rp)+4(p+q+r)+8$ (expanding)$

$$\Rightarrow pqr= 64-2(pq+qr+rp)-4(p+q+r)-8$ (rearranging)$

$$\Rightarrow pqr= 56-2(pq+qr+rp)-4(p+q+r)$ (simplifying). \ \ (*)$

$Now, by the 3-variable AM-GM inequality, $pq+qr+rp\geq 3\sqrt[3]{p^2q^2 r^2}=3(pqr)^{\frac{2}{3}}$, and $p+q+r\geq 3 \sqrt[3]{pqr}$. Substituting these into (*), we have$

$pqr \leq 56-2 \times 3(pqr)^{\frac{2}{3}} - 4\times 3 \sqrt[3]{pqr}$

$\Rightarrow pqr \leq 56-6(pqr)^{\frac{2}{3}} - 12\sqrt[3]{pqr}$

$\Rightarrow pqr + 6(pqr)^{\frac{2}{3}} +12\sqrt[3]{pqr} \leq 56$

$$ \Rightarrow f(u):=u+ 6u^{\frac{2}{3}} +12\sqrt[3]{u} \leq 56$ (writing $u\equiv pqr$).$

$Since $f(u)$ is clearly monotonically increasing for positive $u$ and $f(8)=8+6\times 8^{\frac{2}{3}}+12\times \sqrt[3]{8}=56$, the above inequality implies that $u\leq 8$, i.e. $pqr\leq 8$, as required.$

Kaido · Aug 26, 2015

Re: HSC 2015 4U Marathon

Let p+2 = a , q+2 =b r+2=c

Therefore (p+2) + (q+2) + (r+2) > 3 cuberoot ((P+2)(Q+2)(r+2))
Simplifying: p+q+r > 6

Now p+q+r > 3 cuberoot (pqr)
Rearranging, pqr < (P+q+r/3)^3
therefore, pqr < (6/2)^3
pqr < 8

Does this work?

Edit: nvm, i see peeps have done this and i guess it doesnt work

Paradoxica · Aug 27, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $Prove the inequality$ \ x + x^3 - x^4 - x^6 < 1 \ $for all real$ \ x$

$(x^3-\frac{1}{2})^2=x^6-x^3+\frac{1}{4} \geqslant 0$
$(x^2-\frac{1}{2})^2=x^4-x^2+\frac{1}{4} \geqslant 0$
$(x-\frac{1}{2})^2=x^2-x+\frac{1}{4} \geqslant 0$
Adding these three inequalities yields:
$x^6+x^4+x^2-x^3-x^2-x+\frac{1}{4}+\frac{1}{4}+\frac{1}{4} > 0$
The strict inequality is true since all three inequalities are zero at different values of x.
Adding $\frac{1}{4}$ only furthers the strictness of the inequality, which leads to:
$x^6+x^4-x^3-x+1 > 0$
Which after re-arranging gives the desired inequality.
Now, my question is, how many marks would I get (compared to whatever would be given) in the HSC for doing it this way instead of calculus? (I have nothing against calculus, it's just that I will always try to solve a problem in the most elegant way I can (quickly) think of)

HSC 2015 MX2 Marathon (archive) (1 Viewer)

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This too shall pass

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not actually a porcupine

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not actually a porcupine

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be.

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