HSC 2014 MX2 Marathon (archive) (4 Viewers)

Fade1233 · Sep 27, 2014

Re: HSC 2014 4U Marathon

fizzy_cyst said:
any takers?

jrahs 2013?

faisalabdul16 · Sep 30, 2014

Re: HSC 2014 4U Marathon

Nice question

glittergal96 · Sep 30, 2014

Re: HSC 2014 4U Marathon

faisalabdul16 said:
Nice question

View attachment 30965

The triangle inequality tells us

$b\leq a+c \Rightarrow b-c \leq a \Rightarrow (b-c)^2 \leq a^2.$

Summing this final inequality cyclically, we obtain

$2(a^2+b^2+c^2)-2(ab+ac+bc)\leq a^2+b^2+c^2\Rightarrow (a+b+c)^2\leq 4(ab+ac+bc).$

dunjaaa · Sep 30, 2014

Re: HSC 2014 4U Marathon

On the topic of triangles, here's a nice question I made:

Screen shot 2014-09-30 at 12.31.43 PM.png

integral95 · Sep 30, 2014

Re: HSC 2014 4U Marathon

dunjaaa said:
On the topic of triangles, heres a nice question I made: View attachment 30967

Well clearly r is at least positive since you can't have negative sides

(am I missing something else?)

r>1 ??

for part ii

let the sides be a,ar,ar^2

since
the triangle is right-angled then the Pythagoras theorem applies

(assuming r >1, implying ar^2 is the longest side)

a^2r^4= a^2r^2a^2

$r^4-r^2-1 = 0 \Rightarrow r^2 = \frac{1+\sqrt(5)}{2}$ and then take the positive square root.

faisalabdul16 · Sep 30, 2014

Re: HSC 2014 4U Marathon

integral95 said:
Well clearly r is at least positive since you can't have negative sides (am I missing something else?)

r>1 ??

for part ii

let the sides be a,ar,ar^2

since
the triangle is right-angled then the Pythagoras theorem applies

(assuming r >1, implying ar^2 is the longest side)

a^2r^4= a^2r^2a^2

$r^4-r^2-1 = 0 \Rightarrow r^2 = \frac{1+\sqrt(5)}{2}$ and then take the positive square root.

Why can't r be a fraction?

integral95 · Sep 30, 2014

Re: HSC 2014 4U Marathon

faisalabdul16 said:
Why can't r be a fraction?

well it can haha

I just realised there are 2 sets of solutions

Another case is where a^2 is the longest side instead (where $0<r<1$ )

And you'd be solving $\\ \\ r^4+r^2-1 \Rightarrow r^2 = \frac{-1+\sqrt{5}}{2}$

faisalabdul16 · Sep 30, 2014

Re: HSC 2014 4U Marathon

integral95 said:
well it can haha

I just realised there are 2 sets of solutions

Another case is where a^2 is the longest side instead (where 0<r<1)

And you'd be solving $\\ \\ r^4+r^2-1 \Rightarrow r^2 = \frac{-1+\sqrt{5}}{2}$

Exactly, you get 2 r values. However question says value of r not values, so I am assuming we have done something wrong in the first part, unless dunjaa has missed out a s in the word value

dunjaaa · Sep 30, 2014

Re: HSC 2014 4U Marathon

You got the second part of the question right, but the first part is wrong. Clearly, r>0, but the sides must also obey a condition.

integral95 · Sep 30, 2014

Re: HSC 2014 4U Marathon

faisalabdul16 said:
Exactly, you get 2 r values. However question says value of r not values, so I am assuming we have done something wrong in the first part, unless dunjaa has missed out a s in the word value

value(s) heh

integral95 · Sep 30, 2014

Re: HSC 2014 4U Marathon

dunjaaa said:
You got the second part of the question right, but the first part is wrong. Clearly, r>0, but the sides must also obey a condition.

Oh great it's not the triangular inequality is it.....

so like $ar^2 \leq ar+a \Rightarrow 0<r\leq\frac{1+\sqrt{5}}{2}$

dunjaaa · Sep 30, 2014

Re: HSC 2014 4U Marathon

Indeed it is haha

faisalabdul16 · Sep 30, 2014

Re: HSC 2014 4U Marathon

dunjaaa said:
Indeed it is haha

Oh wow hahahah

Nice question!

dunjaaa · Sep 30, 2014

Re: HSC 2014 4U Marathon

You got the upper bound correct, but the lower bound is not 0

integral95 · Sep 30, 2014

Re: HSC 2014 4U Marathon

dunjaaa said:
You got the upper bound correct, but the lower bound is not 0

$a\leq ar+ar^2 \ AND \ ar^2\leq ar+a \Rightarrow \frac{-1+\sqrt{5}}{2}\leq r \leq \frac{1+\sqrt{5}}{2}$

dunjaaa · Sep 30, 2014

Re: HSC 2014 4U Marathon

That's correct, good job

. It's a cool result since it ties in with the golden ratio (Kepler Triangle)

faisalabdul16 · Oct 3, 2014

Re: HSC 2014 4U Marathon

$$If P(x)=x^{m}(b^{n}- c^{n})+ b^{m}(c^{n}-x^{n}) + c^{m}(x^{n}-b^{n})\\ \\ \text{where m and n are positive integers, show that} \\ \\ x^{2}-x(b+c)+bc \\ \\ \text{is a factor of P(x)}$

Kurosaki · Oct 3, 2014

Re: HSC 2014 4U Marathon

faisalabdul16 said:
$$If P(x)=x^{m}(b^{n}- c^{n})+ b^{m}(c^{n}-x^{n}) + c^{m}(x^{n}-b^{n})\\ \\ \text{where m and n are positive integers, show that} \\ \\ x^{2}-x(b+c)+bc \\ \\ \text{is a factor of P(x)}$

Do you just sub in x=b and x=c, show that P(b) and P(c) equals zero, meaning that (x-b) and (x-c) are both factors, implying that their product, the given quadratic, is a factor?

faisalabdul16 · Oct 3, 2014

Re: HSC 2014 4U Marathon

Kurosaki said:
Do you just sub in x=b and x=c, show that P(b) and P(c) equals zero, meaning that (x-b) and (x-c) are both factors, implying that their product, the given quadratic, is a factor?

Yeap exactly what i did

I just wanted to see if anyone was going to pick up on that, because the polynomial looks "scary" however easy question.

Sy123 · Oct 13, 2014

Re: HSC 2014 4U Marathon

$\\ $Given$ \ a,b,c \ $are real$ \\ $Show that at least one of the following equations has at least one real root$ \\ \\ x^2 + (a-b)x + (b-c) = 0 \\ x^2 + (b-c)x + (c-a) = 0 \\ x^2 + (c-a)x + (a-b) = 0$

HSC 2014 MX2 Marathon (archive) (4 Viewers)

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