HSC 2016 MX1 Marathon (archive) (1 Viewer)

davidgoes4wce · Jul 4, 2016

Re: HSC 2016 3U Marathon

$$ I wasn't 100 \% sure why they use the reasoning 'corresponding angles' for part (b)$

$I was going to go along something of the form:$

$\angle \ BAC=\angle \ DCA $ (alternate angles), therefore AB $ \parallel DC$

Drongoski · Jul 4, 2016

Re: HSC 2016 3U Marathon

davidgoes4wce said:
$$ I wasn't 100 \% sure why they use the reasoning 'corresponding angles' for part (b)$

$I was going to go along something of the form:$

$\angle \ BAC=\angle \ DCA $ (alternate angles), therefore AB $ \parallel DC$

I think there is an unfortunate ambiguity in the use of the word "corresponding" in this case. In this usage, corresponding refers to our day-to-day usage of the term; not the "corresponding" as in "alternate" and "co-interior" which all 3 are technical terms. So the use of "corresponding" in the given explanation leads to this confusion.

leehuan · Jul 4, 2016

Re: HSC 2016 3U Marathon

davidgoes4wce said:
$$ I wasn't 100 \% sure why they use the reasoning 'corresponding angles' for part (b)$

$I was going to go along something of the form:$

$\angle \ BAC=\angle \ DCA $ (alternate angles), therefore AB $ \parallel DC$
image]

They are alternate angles to me.

davidgoes4wce · Jul 4, 2016

Re: HSC 2016 3U Marathon

The question is from Cambridge, Cambridge doesn't usually make many mistakes either so I just wanted to check.

leehuan · Jul 4, 2016

Re: HSC 2016 3U Marathon

Oh, reading what Drongoski said.

Proper reasoning is this

1. Therefore angle this = angle that (corresponding angles on congruent triangles)
2. Therefore || (pair of alternate angles equal)

Yeah thing with me is I always say "matching" angles in cong triangles cause I hate using the word corresponding (in geometry) outside of the context of parallelism

davidgoes4wce · Jul 4, 2016

Re: HSC 2016 3U Marathon

Would my reason be good enough as well?

InteGrand · Jul 4, 2016

Re: HSC 2016 3U Marathon

davidgoes4wce said:
Would my reason be good enough as well?

You'd need to give the reason as related to the congruent triangles part. If you said alternate angles as the reason for the equal angles, then that means you're assuming the lines are parallel, which is what we're trying to prove.

HeroicPandas · Jul 7, 2016

Re: HSC 2016 3U Marathon

mini8658 said:
Show by mathematical induction that [(a+b)/2]^2 (less than or equal to, I don't know how to type it with my keyboard) (A^n+B^n)/2.

Question asked to prove that A^(n+1)-A^nB+B^(n+1)-B^nA=(a-b)^2(A^n-1+A^n-2+...+B^n-1) and deduce A^(n+1)+B^(n+1) (greater than or equal to) A^nB+B^nA, given that A>0, B>0 and n is a positive integer beforehand, which I did get, and I think I'm supposed to use them in the induction, but I'm not sure how...

Assuming a is A and b is B, could there possibly be a typo in the question?

Suppose a = 0.1, b = 0.2 and n = 3. Then

$\left(\frac{a+b}{2} \right )^2 = \left(\frac{0.1+0.2}{2} \right )^2= 0.0225$

$\frac{a^n+b^n}{2} = \frac{0.1^3 + 0.2^2}{2} = 0.0205$

Therefore, statement is not true for all n.

Here are my guesses of possible typos from working on this question:

$\left(\frac{a+b}{2} \right )^{\color{red}{n}} \leq \frac{a^n+b^n}{2}$

OR

$a+b \geq 2$

EDIT: typo on denominator

InteGrand · Jul 7, 2016

Re: HSC 2016 3U Marathon

$\noindent I'm guessing the Q. is asking to prove the inequality $\frac{a^n + b^n}{2} \geq \left(\frac{a+b}{2}\right)^n$, for $a,b \geq 0$, $n$ a positive integer. This is a special case of the \textsl{power means inequality}, linked below.$

https://en.wikipedia.org/wiki/Generalized_mean

HeroicPandas · Jul 7, 2016

Re: HSC 2016 3U Marathon

InteGrand said:
$\noindent I'm guessing the Q. is asking to prove the inequality $\frac{a^n + b^n}{2} \geq \left(\frac{a+b}{2}\right)^n$, for $a,b \geq 0$, $n$ a positive integer. This is a special case of the \textsl{power means inequality}, linked below.$

https://en.wikipedia.org/wiki/Generalized_mean

Yeah! I made a little typo on the denominator

leehuan · Jul 7, 2016

Re: HSC 2016 3U Marathon

mini8658 said:
Drawing attention to this question, which was posted a few days ago and still has no replies. I checked the times, and this was posted shortly before another user posted a reply to a previous question so perhaps it was lost in the replies? But anyway, I'd really like this question answered thanks ^_^

Can you please retype what you're trying to prove using this? http://s1.daumcdn.net/editor/fp/service_nc/pencil/Pencil_chromestore.html

Copy and paste the entire syntax onto this forum.

InteGrand · Jul 7, 2016

Re: HSC 2016 3U Marathon

leehuan said:
Can you please retype what you're trying to prove using this? http://s1.daumcdn.net/editor/fp/service_nc/pencil/Pencil_chromestore.html

Copy and paste the entire syntax onto this forum.

It should just be what HeroicPandas and I wrote above.

leehuan · Jul 7, 2016

Re: HSC 2016 3U Marathon

InteGrand said:
It should just be what HeroicPandas and I wrote above.

Alright if you're sure then...

...well I am missing the obvious again for the inductive step:

$\text{Assumed: }\frac{a^k+b^k}{2} \ge \left( \frac{a+b}{2} \right)^k\\ \text{RTP: }\frac{a^{k+1}+b^{k+1}}{2} \ge \left( \frac{a+b}{2} \right)^{k+1}$

$\because ab \le \frac{a^2+b^2}{2}$

$\begin{align*}RHS&= \left( \frac{a+b}{2} \right) \left( \frac{a+b}{2} \right)^k\ \\ &\le \left( \frac{a+b}{2} \right)\left(\frac{a^k+b^k}{2}\right) \qquad (*) \\ &= \frac{a^{k+1}+b^{k+1}}{4} + \frac{a^kb+b^ka}{4} \\ &= \frac{a^{k+1}+b^{k+1}}{4} + \frac{ab(a^{k-1}+b^{k-1})}{4} \\ &\le \frac{a^{k+1}+b^{k+1}}{4} + \frac{(a^2+b^2)(a^{k-1}+b^{k-1})}{8}\\ &= \frac{3(a^{k+1}+b^{k+1})}{8} + \frac{ab(a^{k-2}+b^{k-2})}{8} \\ &\le \frac{7(a^{k+1}+b^{k+1})}{16} + \frac{ab(a^{k-3}+b^{k-3})}{16} \\ &\le \frac{15(a^{k+1}+b^{k+1})}{32} + \frac{ab(a^{k-4}+b^{k-4})}{32} \\ &\vdots \\ &\le \frac{(2^{?-1}-1)(a^{k+1}+b^{k+1}}{2^?} + \frac{ab(a^0+b^0)}{2^?}\end{align*}$

Paradoxica · Jul 7, 2016

Re: HSC 2016 3U Marathon

leehuan said:
Alright if you're sure then...

...well I am missing the obvious again for the inductive step:

$\text{Assumed: }\frac{a^k+b^k}{2} \ge \left( \frac{a+b}{2} \right)^k\\ \text{RTP: }\frac{a^{k+1}+b^{k+1}}{2} \ge \left( \frac{a+b}{2} \right)^{k+1}$

$\because ab \le \frac{a^2+b^2}{2}$

$\begin{align*}RHS&= \left( \frac{a+b}{2} \right) \left( \frac{a+b}{2} \right)^k\ \\ &\le \left( \frac{a+b}{2} \right)\left(\frac{a^k+b^k}{2}\right) \qquad (*) \\ &= \frac{a^{k+1}+b^{k+1}}{4} + \frac{a^kb+b^ka}{4} \\ &= \frac{a^{k+1}+b^{k+1}}{4} + \frac{ab(a^{k-1}+b^{k-1})}{4} \\ &\le \frac{a^{k+1}+b^{k+1}}{4} + \frac{(a^2+b^2)(a^{k-1}+b^{k-1})}{8}\\ &= \frac{3(a^{k+1}+b^{k+1})}{8} + \frac{ab(a^{k-2}+b^{k-2})}{8} \\ &\le \frac{7(a^{k+1}+b^{k+1})}{16} + \frac{ab(a^{k-3}+b^{k-3})}{16} \\ &\le \frac{15(a^{k+1}+b^{k+1})}{32} + \frac{ab(a^{k-4}+b^{k-4})}{32} \\ &\vdots \end{align*}$

Don't iterate inequalities. Find other inequalities to use in the inductive step.

HeroicPandas · Jul 8, 2016

Re: HSC 2016 3U Marathon

Here is a hint. Highlight to see. Also in LaTeX code to try to prevent immediate comprehension in-case you accidentally highlighted or ctrl+A
___
\frac{a^{k+1}+b^{k+1}}{2} \geq \frac{a^kb + b^ka}{2} = \boxed{?}\times \left( \frac{a^k + b^k}{2} \right ) + \boxed{?} \times \left(\frac{a^k + b^k}{2} \right ) - \boxed{?}
___

Paradoxica · Jul 8, 2016

Re: HSC 2016 3U Marathon

leehuan said:
Alright if you're sure then...

...well I am missing the obvious again for the inductive step:

$\text{Assumed: }\frac{a^k+b^k}{2} \ge \left( \frac{a+b}{2} \right)^k\\ \text{RTP: }\frac{a^{k+1}+b^{k+1}}{2} \ge \left( \frac{a+b}{2} \right)^{k+1}$

$\because ab \le \frac{a^2+b^2}{2}$

$\begin{align*}RHS&= \left( \frac{a+b}{2} \right) \left( \frac{a+b}{2} \right)^k\ \\ &\le \left( \frac{a+b}{2} \right)\left(\frac{a^k+b^k}{2}\right) \qquad (*) \\ &= \frac{a^{k+1}+b^{k+1}}{4} + \frac{a^kb+b^ka}{4} \\ &= \frac{a^{k+1}+b^{k+1}}{4} + \frac{ab(a^{k-1}+b^{k-1})}{4} \\ &\le \frac{a^{k+1}+b^{k+1}}{4} + \frac{(a^2+b^2)(a^{k-1}+b^{k-1})}{8}\\ &= \frac{3(a^{k+1}+b^{k+1})}{8} + \frac{ab(a^{k-2}+b^{k-2})}{8} \\ &\le \frac{7(a^{k+1}+b^{k+1})}{16} + \frac{ab(a^{k-3}+b^{k-3})}{16} \\ &\le \frac{15(a^{k+1}+b^{k+1})}{32} + \frac{ab(a^{k-4}+b^{k-4})}{32} \\ &\vdots \\ &\le \frac{(2^{?-1}-1)(a^{k+1}+b^{k+1}}{2^?} + \frac{ab(a^0+b^0)}{2^?}\end{align*}$

Hint: Use the fact that (aⁿ - bⁿ)(a-b) is non-negative for all real a and b.

leehuan · Jul 8, 2016

Re: HSC 2016 3U Marathon

Paradoxica said:
Hint: Use the fact that (aⁿ - bⁿ)(a-b) is non-negative for all real a and b.

Would not have thought about that at all. Just by substituting this result into line 3 of the proof the result falls out immediately.

$\begin{align*}(a-b)(a^k-b^k)&\ge 0\\ a^{k+1}+b^{k+1}& \ge ab^k+a^kb\end{align*}$

I was trying to think about why a^kb+ab^k≤a^k+1+b^k+1. Guess I've never seen that before to use it.

Paradoxica · Jul 8, 2016

Re: HSC 2016 3U Marathon

leehuan said:
Would not have thought about that at all. Just by substituting this result into line 3 of the proof the result falls out immediately.

$\begin{align*}(a-b)(a^k-b^k)&\ge 0\\ a^{k+1}+b^{k+1}& \ge ab^k+a^kb\end{align*}$

I was trying to think about why a^kb+ab^k≤a^k+1+b^k+1. Guess I've never seen that before to use it.

Well if you can't prove inequality as it holds, subtracting is always good for HSC questions, because usually it is possible to factorise and partition the expression into strictly positive terms.

davidgoes4wce · Jul 12, 2016

Re: HSC 2016 3U Marathon

Find the equation of the parabola whose axis is parallel to the y-axis, vertex is (2,-1) and has a tangent with equation y=2x-7.

I have no answer to compare against . This question (can't remember off the top of my head) but was used in the Trials at one of the schools in NSW.

leehuan · Jul 12, 2016

Re: HSC 2016 3U Marathon

The parabola takes the form (x-2)^2=4a(y+1)

A wolfram alpha approach: Sub y=2x-7 in and simplify the expression

Take the quadratic discriminant, set it equal to 0 and take your value for a

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