HSC 2015 MX1 Marathon (archive) (3 Viewers)

Drsoccerball · Sep 21, 2015

Re: HSC 2015 3U Marathon

$(x+1)^n +2$

$2+\binom{n}{0} +\binom{n}{1}x +...+\binom{n}{n}x^n$

$$ Let $ x=6, 7^n+2 = \binom{n}{0} +2 +\binom{n}{1}6+...+\binom{n}{n}6^n$

$3( 1+ \binom{n}{1}2 +...+\binom{n}{n}3^{n-1} 2^n$

$\therefore $ always divisible by 3 for all positive integers n $$

Drsoccerball · Sep 21, 2015

Re: HSC 2015 3U Marathon

Im sick and tired of latex not working on this site

Ambility · Sep 21, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
Im sick and tired of latex not working on this site

It's not the site's fault, it's codecogs'.

davidgoes4wce · Sep 22, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
Im sick and tired of latex not working on this site

Do a photobucket job like I do.

rand_althor · Sep 22, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce · Sep 23, 2015

Re: HSC 2015 3U Marathon

How harsh are markers if you don't put in the specific units required for measurements like area or volume of a curve? (eg units, units^2 and units ^3)?

dan964 · Sep 23, 2015

Re: HSC 2015 3U Marathon

If the question gives things in cm, yes you would probably lose a mark.

davidgoes4wce · Sep 23, 2015

Re: HSC 2015 3U Marathon

Someone give me a kick start to this. This is the last question from Sydney Boys High School in Surry Hills. I've been thinking about the distance OM. My thinking is OM is in the centre of the triangle? I've tried using half the distance between CM, one-third the distance of the height from CM (presuming its the centre of a triangle)

math.tutorvista.com/geometry/centroid.html

rand_althor · Sep 23, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
Someone give me a kick start to this. This is the last question from Sydney Boys High School in Surry Hills. I've been thinking about the distance OM. My thinking is OM is in the centre of the triangle? I've tried using half the distance between CM, one-third the distance of the height from CM (presuming its the centre of a triangle)

math.tutorvista.com/geometry/centroid.html

$$We can find $\cos{\theta}$ using the cosine rule, but to do that we need the lengths of all three sides in $\Delta VMC$. We know that $VC=\sqrt{5}a$. \\ \\$\Delta VAB$ is an equilateral triangle as $VA=VB=AB=2a$. Therefore $\angle{VBM}=60^{\circ}$. Since $VM$ bisects $AB$, $VM \perp AB$. Therefore $\angle{VMB}=90^{\circ}$. Now we know that $\Delta VMB$ is right-angled so we can use Pythagoras' Theorem to find the length of $VM$. \\ \\ In $\Delta CMB$, we have $CB=3a$, $BM=a$. Since $CM$ bisects $AB$, $CM \perp AB$. Now we know that $\Delta CMB$ is right-angled so we can use Pythagoras' Theorem to find the length of $CM$. Now that you know the lengths of all three sides in $\Delta VMC$ ($VM=\sqrt{3}a,VC=\sqrt{5}a,MC=2\sqrt{2}a$), you can use the cosine rule ($a^2=b^2+c^2-2bc\cos{A}$) to find $\cos{\theta}$.$

mrpotatoed · Sep 23, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
Someone give me a kick start to this. This is the last question from Sydney Boys High School in Surry Hills. I've been thinking about the distance OM. My thinking is OM is in the centre of the triangle? I've tried using half the distance between CM, one-third the distance of the height from CM (presuming its the centre of a triangle)

math.tutorvista.com/geometry/centroid.html

That brought back some bad memories bro... took me 20 mins just to get the first part

davidgoes4wce · Sep 24, 2015

Re: HSC 2015 3U Marathon

mrpotatoed said:
That brought back some bad memories bro... took me 20 mins just to get the first part

I am assuming you study there?

mrpotatoed · Sep 24, 2015

Re: HSC 2015 3U Marathon

yes

VBN2470 · Sep 24, 2015

Re: HSC 2015 3U Marathon

NEW QUESTION:

Prove that if $m$ and $n$ are positive integers, then $(mn)!\geq{m^{n}(n!)^{m}}$ .

leehuan · Sep 25, 2015

Re: HSC 2015 3U Marathon

Woah... is that actually suitable for 3U? I can't work my way around multidimensional induction

davidgoes4wce · Sep 26, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
$$We can find $\cos{\theta}$ using the cosine rule, but to do that we need the lengths of all three sides in $\Delta VMC$. We know that $VC=\sqrt{5}a$. \\ \\$\Delta VAB$ is an equilateral triangle as $VA=VB=AB=2a$. Therefore $\angle{VBM}=60^{\circ}$. Since $VM$ bisects $AB$, $VM \perp AB$. Therefore $\angle{VMB}=90^{\circ}$. Now we know that $\Delta VMB$ is right-angled so we can use Pythagoras' Theorem to find the length of $VM$. \\ \\ In $\Delta CMB$, we have $CB=3a$, $BM=a$. Since $CM$ bisects $AB$, $CM \perp AB$. Now we know that $\Delta CMB$ is right-angled so we can use Pythagoras' Theorem to find the length of $CM$. Now that you know the lengths of all three sides in $\Delta VMC$ ($VM=\sqrt{3}a,VC=\sqrt{5}a,MC=2\sqrt{2}a$), you can use the cosine rule ($a^2=b^2+c^2-2bc\cos{A}$) to find $\cos{\theta}$.$

Thanks. When I first saw the question , I didn't see the cosine rule coming into play. I was using SOH/CAH/TOA throughout so that opened up a new perspective for me.

davidgoes4wce · Sep 26, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
Someone give me a kick start to this. This is the last question from Sydney Boys High School in Surry Hills. I've been thinking about the distance OM. My thinking is OM is in the centre of the triangle? I've tried using half the distance between CM, one-third the distance of the height from CM (presuming its the centre of a triangle)

math.tutorvista.com/geometry/centroid.html

With part (v) to this question I was wondering if you could use related rates in order to solve for the maximum? (Or calculus maximum/minimum values) I need a kick start to this question.

psyc1011 · Sep 26, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
With part (v) to this question I was wondering if you could use related rates in order to solve for the maximum? (Or calculus maximum/minimum values) I need a kick start to this question.

Maximise LHS by maximising RHS and how do you maximise RHS? (hint: fraction)

Ding dong, calculus problem without needing to go nuts on differentiating that junk mindlessly.

davidgoes4wce · Sep 26, 2015

Re: HSC 2015 3U Marathon

Someone give me a gee up for this question.

I don't want to look at my formulae book or solutions. I want someone on here to tell me the right formula to use.(without giving the answer) I learn more from discussing with people.

checkm8y · Sep 26, 2015

Re: HSC 2015 3U Marathon

Use the angle at the centre is twice the angle at the circumference

InteGrand · Sep 27, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
How would you get the "tan's"

$Let the parabola be $y=Ax(x-2d)$ ($A<0$, $d>0$). Clearly, the $x$-value of the apex of the triangle (which is also the vertex of the parabola) is $d$. We can find this point (parabola's vertex) easily by substituting $x=d$ into $y=Ax(x-2d)$, so the vertex is $V\left(d,-Ad^2 \right)$. So $\tan \alpha = \frac{y_V}{x_V}=\frac{-Ad^2}{d}=-Ad$. Also, for the parabola, $y^\prime = 2Ax - 2Ad$. So at the origin, the slope of the tangent, which is $\tan \theta$, is equal to $2A\cdot 0 - 2Ad=-2Ad$. So $\tan \theta = -2Ad=2\tan \alpha$, as required.$

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