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

Sy123 · Jul 25, 2015

Re: HSC 2015 4U Marathon - Advanced Level

$\\ a,b,c \ $are sides of a triangle, show that$ \\\\ a^2(b+c-a) + b^2(a+c -b) + c^2(a+b-c) \leq 3abc$

dan964 · Jul 25, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Note sure if correct:
<removed> fixed version 2 posts below
added RTP line.

InteGrand · Jul 25, 2015

Re: HSC 2015 4U Marathon - Advanced Level

dan964 said:
Note sure if correct:
View attachment 32239
added RTP line.

Your LHS in the 'RTP' line is not always less than 2.5. E.g. in an equilateral triangle, all angles are 60º, and cos(60º) = ½, so your LHS is then 2•(½ + ½ + ½) = 3.

dan964 · Jul 25, 2015

Re: HSC 2015 4U Marathon - Advanced Level

InteGrand said:
Your LHS in the 'RTP' line is not always less than 2.5. E.g. in an equilateral triangle, all angles are 60º, and cos(60º) = ½, so your LHS is then 2•(½ + ½ + ½) = 3.

I realised that and was trying to look for which line was wrong.
It has been found.It was a dumb error in failing to carry a constant of 2 through on some terms.
<removed>
fixed version is below

InteGrand · Jul 25, 2015

Re: HSC 2015 4U Marathon - Advanced Level

dan964 said:
I realised that and was trying to look for which line was wrong.
It has been found.It was a dumb error in failing to carry a constant of 2 through on some terms.
View attachment 32240

It still says LHS ≤ 2.5 on the second last line (this could just be a typo, I haven't read through the proof, just skimmed it).

Sy123 · Jul 26, 2015

Re: HSC 2015 4U Marathon - Advanced Level

dan964 said:
I realised that and was trying to look for which line was wrong.
It has been found.It was a dumb error in failing to carry a constant of 2 through on some terms.
View attachment 32240

nvm

dan964 · Jul 26, 2015

Re: HSC 2015 4U Marathon - Advanced Level

all errors fixed now:

Sy123 · Jul 26, 2015

Re: HSC 2015 4U Marathon - Advanced Level

The proof looks solid, well done

Here was my approach:

$\\ $Let$ \ x = b+c-a, y = a+c - b, z = a+b-c$

$\\ $Since$ \ a,b,c \ $are sides of a triangle$ \ x,y,z \ $are all non-negative$$

$\\ c = \frac{x+y}{2} , b = \frac{x+z}{2} , a = \frac{y+z}{2}$

$\\ $Therefore$ \\\\ 3abc - (a^2(b+c-a) + b^2(a+c-b) + c^2(a+b-c)) \\\\ = \frac{3}{8} (x+y)(x+z)(y+z) - \frac{1}{4} (x(y+z)^2 + y(x+z)^2 + z(x+y)^2)$

$\\ = \frac{1}{8} \left( x(y^2+z^2) + y(x^2+z^2) + z(x^2+y^2) - 6xyz \right) \\\\ = \frac{1}{8} ( x(y^2+z^2-2yz) + y(x^2+z^2-2xz) + z(x^2+y^2-2xy) ) \\\\ = \frac{1}{8} (x(y-z)^2 + y(x-z)^2 + z(x-y)^2) \geq 0$

$\\ \therefore \ 3(x+y)(x+z)(y+z) \geq 2 (x(y+z)^2 + y(x+z)^2 + z(x+y)^2) \\ \\ \Rightarrow \ 3abc \geq a^2(b+c-a) + b^2(a+c-b) + c^2(a+b-c)$

InteGrand · Jul 26, 2015

Re: HSC 2015 4U Marathon - Advanced Level

dan964 said:
Can I post a question:
Needs fixing just wait...

$\cos{P}, \cos{Q}, \cos{R} $ are roots to the equation $ x^{3}+bx^{2}+cx+d.$ Show the area of the triangle with angles, P, Q and R is $ \frac{1}{4} \sqrt{c}$

You may assume or derive a suitable formula for the area of a triangle in terms of the sides or angles.

Knowing the angles of a triangle isn't enough to determine a unique area, since the triangles could have different sizes for given angles (they would all be similar triangles).

shervos · Jul 26, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Show that 4m^2+17n^2 and 4n^2+17m^2 cannot be both be perfect squares, where n, and m are positive integers

dan964 · Jul 26, 2015

Re: HSC 2015 4U Marathon - Advanced Level

shervos said:
Show that 4m^2+17n^2 and 4n^2+17m^2 cannot be both be perfect squares, where n, and m are positive integers

I'll rewrite it in latex for you:
$Show that $ 4m^{2}+17n^{2}$ and $4n^{2}+17m^{2} $ cannot be both be perfect squares, where n, and m are positive integers$

simpleetal · Jul 27, 2015

Re: HSC 2015 4U Marathon - Advanced Level

shervos said:
Show that 4m^2+17n^2 and 4n^2+17m^2 cannot be both be perfect squares, where n, and m are positive integers

by letting each of the expressions equal to u^2 and v^2 respectively, we can deduce that m and n must both be odd for u and v to be integers. Hence, we only consider the case where m and n areboth odd integers. But this is a contradiction as for each expression, upon division by 8, we get a remainder of 5 on the lhs(all odd squares have remainder 1 when divided by 8) while on the rhs we have remainder 1 (since u and v are both odd if m, n both odd)

glittergal96 · Jul 27, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Finishing off the cubic question (finally!):

It remains to show that for a polynomial with three real roots, the inequalities:

1. |bd-c| < 1-d^2

and

2. |b+d| < |1+c|

imply that all three roots live in the open unit interval.

As discussed in my previous posts, inequality 2 implies that p(1) > 0 and p(-1) < 0, which means there is at least one real root in the unit interval.

If all three roots DON'T live in the unit interval then either two of them are at least 1 or two of them are at most -1. (By considering the limiting behaviour of p at +inf and -inf.)
Let us assume we are in one of these cases for the sake of contradiction.

For inequality 1 to hold, we must of course have d^2 < 1. (*)

We also have that the polynomial q(1)= 1-d^2 - (c-bd) >= 1-d^2 - |c-bd| > 0. (q was defined earlier as the polynomial with roots the pairwise products of the roots of p).

This means that either zero or two of the quantities

$\alpha\beta,\alpha\gamma,\beta\gamma$

exceed 1, with the others being less than 1. (**)

As at least two of these roots have the same sign, this observation implies that all three roots must have the same sign.

So we are left with the possibilities:

$0 < \alpha < 1 \leq \beta,\gamma$

$\beta,\gamma \leq -1 < \alpha <0.$

In the first of these, we have $\alpha\beta < 1 < \alpha\gamma < \beta\gamma$ from (**).

This means we can multiply $1< \alpha\gamma \leq \alpha\gamma\cdot \beta =d$

which contradicts (**).

Similarly, in the latter of the two possibilities, we have:

$\alpha\beta < 1 <\alpha \gamma < \beta\gamma$

So $|d|=-\beta\cdot \alpha\gamma >1$

again in contradiction with (**).

This completes the proof.

glittergal96 · Jul 27, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Find all functions $f:\mathbb{Q}\rightarrow \mathbb{Q}$ such that

$f(xy)=f(x)f(y)-f(x+y)+1$

for all rational $x,y$ .

(Note that $\mathbb{Q}$ is the set of rational numbers.)

InteGrand · Jul 27, 2015

Re: HSC 2015 4U Marathon - Advanced Level

This is more like an Olympiad Marathon than a Q16-worthy HSC 4U Marathon haha.

VBN2470 · Jul 27, 2015

Re: HSC 2015 4U Marathon - Advanced Level

glittergal96 said:
Find all functions $f:\mathbb{Q}\rightarrow \mathbb{Q}$ such that

$f(xy)=f(x)f(y)-f(x+y)+1$

for all rational $x,y$ .

(Note that $\mathbb{Q}$ is the set of rational numbers.)

Set $x=y=0$ to get $f(0)=f(0)^2-f(0)+1 \Rightarrow (f(0)-1)^2=0 \Rightarrow f(0)=1$ . Not sure how to go from there..

I can see that the function defined by $f:\mathbb{Q}\rightarrow \mathbb{Q}$ such that $f:=f(x)=1$ is a solution, but not sure how to show it (let alone other possible solutions).

InteGrand · Jul 27, 2015

Re: HSC 2015 4U Marathon - Advanced Level

$Plugging in $y$ = $-x$ and rearranging gives $f(y)f(-y)=f(-y^2)$, though I don't know whether this will help much and I haven't worked on the problem too much. Haven't tried too much plugging in yet (which is usually what is required for these functional equations Q's).$

VBN2470 · Jul 27, 2015

Re: HSC 2015 4U Marathon - Advanced Level

InteGrand said:
$Plugging in $y$ = $x$ and rearranging gives $f(y)f(-y)=f(-y^2)$, though I haven't worked on the problem too much.$

I tried that as well as plugging in $y=-x$ , but got an equation which I'm not sure how to manipulate..

Not too experienced/familiar with these functional equations.

VBN2470 · Jul 27, 2015

Re: HSC 2015 4U Marathon - Advanced Level

InteGrand said:
$$f(xy)=f(x)f(y)-f(x+y)+1$, set $y=-x$ and use the discovery that $f(0)=1$ to cancel 1's on the R.H.S.$

Yep I got that for $y=-x$ , I think you meant $y=-x$ but instead you wrote $y=x$ which should instead give $f(x^2)=f(x)^2-f(2x)+1$ .. (3 posts ago)

simpleetal · Jul 28, 2015

Re: HSC 2015 4U Marathon - Advanced Level

I'll do this in two separate posts, because otherwise the equation is too long
$f(xy)=f(x)f(y)-f(x+y)+1.$ It's already been established that $f(0)=1, $and letting x=-1, and y=1 (to somehow try make use of f(0)), we get that$ f(-1)=f(-1)f(1),$ so we consider two cases:f(-1)=0 or f(1)=1. $f(1)=1,$ then, by letting y=1, we get $f(x)=f(1)-f(x+1)+1,$ and thus$ f(x+1)=1, $or$ f(x)=1.\\$ If$ f(-1)=0, $then by letting y=-1, (to try get an expression involving f(-1)) we get that $f(-x)=f(x)f(-1)-f(x-1)+1. $Thus,$ f(-x)=1-f(x-1).$Call this equation A. Subbing this into original equation, we get that $f(-xy)=(1-f(x-1))f(y)-f(-x+y)+1\rightarrow 1-f(xy-1)=(1-f(x-1))f(y)-f(-x+y)+1.$ The ones cancel off, so we get$ (1-f(x-1))f(y)-f(-x+y)+f(xy-1)=0.$ Letting x=2, and y=1, we get that$ f(1)-f(1)^2-f(-1)+f(1),$ and f(-1)=0, so (2-f(1))(f(1))=0. \\So f(1)=$2 or 0$.$

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