HSC 2015 MX2 Marathon (archive) (1 Viewer)

asdfghjklqwerty · Sep 29, 2015

Re: HSC 2015 4U Marathon

Oooh ok if the border is root 12 then I got the answer via shells. The only issue I had was subbing in 3 as a border and I was getting root -3. Thanks guys. Answers 7pi btw

InteGrand · Sep 29, 2015

Re: HSC 2015 4U Marathon

seanieg89 said:
Haven't seen mechanics questions here for a while.

$Suppose a particle moves in simple harmonic motion with amplitude $A$ and angular frequency $\omega$.\\ \\ Find a function $F(x)$ such that $\int_a^b F(x)\, dx$ is equal to the proportion of time that the particle spends in the interval $a\leq x\leq b$ for any $a < b$, with both $a$ and $b$ smaller in absolute value than $A$.\\ \\ For an arbitrary interval $I=[a,b]$ inside $[-A,A]$, define $D(I)=\frac{1}{b-a}\int_a^b F(x)$. ($D(I)$ tells us how much time the particle spends in an interval relative to the size of the interval.) How large can $D(I)$ be?$

$I can't be bothered writing everything up now but we can show that we may take $F(x) = \frac{1}{\pi}\cdot \frac{1}{\sqrt{A^2 - x^2}}$. For the second part, it is clear that for a given interval length, the proportion of time spent in that interval would be greatest if it is closest to the endpoints of $[-A,A]$, since the particle is slowest there. So we should take say $b=A$ to make $D(I)$ as large as possible. Then we can show that $D(I)$ is equal to $\frac{1}{\pi(A-a)}\cdot \cos ^{-1}\left( \frac{a}{A}\right) \to \infty$ as $a\to A$, which can be shown using L'H$\hat{\text{o}}$pital's rule. I'm guessing there was a more elegant way to show it though.$

InteGrand · Sep 30, 2015

Re: HSC 2015 4U Marathon

$Let $\mathcal{E}$ be a given ellipse. Let $A_1 A_2$ be a given chord of $\mathcal{E}$, and let $B_1 B_2$ be another chord of $\mathcal{E}$, with $A_1 A_2$ parallel to $B_1 B_2$. Let $A$ and $B$ be the midpoint of $A_1 A_2$ and $B_1 B_2$ respectively. Show that the line segment $AB$ passes through the centre of $\mathcal{E}$.$

Drsoccerball · Sep 30, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
$Let $\mathcal{E}$ be a given ellipse. Let $A_1 A_2$ be a given chord of $\mathcal{E}$, and let $B_1 B_2$ be another chord of $\mathcal{E}$, with $A_1 A_2$ parallel to $B_1 B_2$. Let $A$ and $B$ be the midpoint of $A_1 A_2$ and $B_1 B_2$ respectively. Show that the line segment $AB$ passes through the centre of $\mathcal{E}$.$

I think you have to mention that both or none of the two chords pass through the origin.

InteGrand · Sep 30, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
I think you have to mention that both or none of the two chords pass through the origin.

$Oops, I mistyped the question slightly. Retyped below:$

$Let $\mathcal{E}$ be a given ellipse. Let $A_1 A_2$ be a given chord of $\mathcal{E}$, and let $B_1 B_2$ be another chord of $\mathcal{E}$, with $A_1 A_2$ parallel to $B_1 B_2$. Let $A$ and $B$ be the midpoint of $A_1 A_2$ and $B_1 B_2$ respectively. Similarly let $C_1 C_2$ and $D_1 D_2$ be parallel chords of $\mathcal{E}$, with midpoints $C$ and $D$ respectively. Show that the lines $AB$ and $CD$ intersect at the centre of $\mathcal{E}$.$

$(Alternatively, change the original wording so that it says ``show that the line $AB$ passes through the origin'' (rather than ``line segment'').)$

Ekman · Sep 30, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
$Oops, I mistyped the question slightly. Retyped below:$

$Let $\mathcal{E}$ be a given ellipse. Let $A_1 A_2$ be a given chord of $\mathcal{E}$, and let $B_1 B_2$ be another chord of $\mathcal{E}$, with $A_1 A_2$ parallel to $B_1 B_2$. Let $A$ and $B$ be the midpoint of $A_1 A_2$ and $B_1 B_2$ respectively. Similarly let $C_1 C_2$ and $D_1 D_2$ be parallel chords of $\mathcal{E}$, with midpoints $C$ and $D$ respectively. Show that the lines $AB$ and $CD$ intersect at the centre of $\mathcal{E}$.$

$(Alternatively, change the original wording so that it says ``show that the line $AB$ passes through the origin'' (rather than ``line segment'').)$

You also added 2 extra chords which is kinda redundant, because its the exact same proof, only different pronumerals

InteGrand · Sep 30, 2015

Re: HSC 2015 4U Marathon

Ekman said:
You also added 2 extra chords which is kinda redundant, because its the exact same proof, only different pronumerals

Yes. The original Q was meant to be with the intersection of the two lines joining the midpoints of two pairs of chords, but I forgot to add the second pair of chords in the original Q. But it turns out it doesn't matter. The only typo originally was basically "line segments" instead of "lines".

Ekman · Sep 30, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
$Oops, I mistyped the question slightly. Retyped below:$

$Let $\mathcal{E}$ be a given ellipse. Let $A_1 A_2$ be a given chord of $\mathcal{E}$, and let $B_1 B_2$ be another chord of $\mathcal{E}$, with $A_1 A_2$ parallel to $B_1 B_2$. Let $A$ and $B$ be the midpoint of $A_1 A_2$ and $B_1 B_2$ respectively. Similarly let $C_1 C_2$ and $D_1 D_2$ be parallel chords of $\mathcal{E}$, with midpoints $C$ and $D$ respectively. Show that the lines $AB$ and $CD$ intersect at the centre of $\mathcal{E}$.$

$(Alternatively, change the original wording so that it says ``show that the line $AB$ passes through the origin'' (rather than ``line segment'').)$

Judging from your comment after the question, it is safe to assume the centre of the ellipse is at the origin. It doesn't matter, it only makes it easier for me to type out the solution.

$\mathcal{E} = \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $ where a, b are real$

$$ Let chord $ A_{1} A_{2} $ have the equation: $ y=mx+c $ and chord $ B_{1} B_{2} $ be: $ y=mx+d$

$$Substitute the chord equation into the ellipse: $ x^2(b^2+a^2 m^2) + 2mca^2 x + a^2 c^2 - a^2 b^2 = 0$

$A: \left(\frac{-mca^2}{b^2 + a^2 m^2}, \frac{b^2 c}{b^2+ a^2 m^2} \right) $ (Dividing sum of roots by 2 as it is the midpoint, and substituting back into the equation) $$

$Replace c with d to find the similar coordinates of B$

$m_{OA} = \frac{b^2}{-m a^2} = m_{OB} = m_{AB} $ hence O, A and B are collinear as they have equal gradients $$

Repeat solution for points C and D

InteGrand · Sep 30, 2015

Re: HSC 2015 4U Marathon

Ekman said:
Judging from your comment after the question, it is safe to assume the centre of the ellipse is at the origin. It doesn't matter, it only makes it easier for me to type out the solution.

$\mathcal{E} = \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $ where a, b are real$

$$ Let chord $ A_{1} A_{2} $ have the equation: $ y=mx+c $ and chord $ B_{1} B_{2} $ be: $ y=mx+d$

$$Substitute the chord equation into the ellipse: $ x^2(b^2+a^2 m^2) + 2mca^2 x + a^2 c^2 - a^2 b^2 = 0$

$A: \left(\frac{-mca^2}{b^2 + a^2 m^2}, \frac{b^2 c}{b^2+ a^2 m^2} \right) $ (Dividing sum of roots by 2 as it is the midpoint, and substituting back into the equation) $$

$Replace c with d to find the similar coordinates of B$

$m_{OA} = \frac{-m a^2}{b^2} = m_{OB} = m_{AB} $ hence O, A and B are collinear as they have equal gradients $$

Repeat solution for points C and D

I think you accidentally gave the reciprocal of the slope at the end instead of the slope, but it doesn't matter much.

And yeah, we can define the origin to be the centre of the ellipse.

Ekman · Sep 30, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
I think you accidentally gave the reciprocal of the slope at the end instead of the slope, but it doesn't matter much.

And yeah, we can define the origin to be the centre of the ellipse.

Whoops my bad, but it shouldn't really matter. As long as you can prove the gradients are independent of the y intercepts of the chords, it should be all fine.

InteGrand · Sep 30, 2015

Re: HSC 2015 4U Marathon

Ekman said:
Whoops my bad, but it shouldn't really matter. As long as you can prove the gradients are independent of the y intercepts of the chords, it should be all fine.

Yeah, that's why I said it doesn't matter much.

Oh, and I realised I meant "centre", not "origin", in my rewording of the Q.

Ekman · Sep 30, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
Yeah, that's why I said it doesn't matter much.

Oh, and I realised I meant "centre", not "origin", in my rewording of the Q.

If thats the case then you have to replace the x^2 and y^2 with (x-h)^2 and (y-k)^2 and prove that (h, k), A and B are collinear which makes the working out a bit more messier

InteGrand · Sep 30, 2015

Re: HSC 2015 4U Marathon

Ekman said:
If thats the case then you have to replace the x^2 and y^2 with (x-h)^2 and (y-k)^2 and prove that (h, k), A and B are collinear which makes the working out a bit more messier

$Don't need to; given an arbitrary ellipse, we can define the origin of our plane to be at the ellipse's centre, the $x$-axis to be the along the major axis, and the $y$-axis to be along the minor axis.$

$(Also, a general ellipse in the plane need not be of the form $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$; this is just the general equation of an ellipse whose axes are parallel to the coordinate axes. A general ellipse, which can be rotated etc., has general equation $Ax^2 + Bxy + Cy^2+Dx + Ey + F = 0$, where $B^2 - 4AC<0$. In fact, depending on the sign of the quantity $B^2 - 4AC$, which is called the \textit{discriminant}, we obtain different types of conic sections. Further reading:$

https://en.wikipedia.org/wiki/Conic_section#Cartesian_coordinates $)$

lita1000 · Oct 1, 2015

Re: HSC 2015 4U Marathon

$$A tall building stands on level ground. The nozzle of a water sprinkler is positioned at a point P on the ground at a distance d from a wall of the building. Water sprays from the nozzle with speed V and the nozzle can be pointed in any direction from P. Suppose that $V= 2(gd)^{1/2}. $Show that the portion of the wall that can be sprayed with water is a parabolic segment with area$ (5/2)(d^2 * (15)^{1/2})$

Drsoccerball · Oct 1, 2015

Re: HSC 2015 4U Marathon

lita1000 said:
$$A tall building stands on level ground. The nozzle of a water sprinkler is positioned at a point P on the ground at a distance d from a wall of the building. Water sprays from the nozzle with speed V and the nozzle can be pointed in any direction from P. Suppose that $V= 2(gd)^{1/2}. $Show that the portion of the wall that can be sprayed with water is a parabolic segment with area$ (5/2)(d^2 * (15)^{1/2})$

Are you sure about this answer? Because the area under the curve is

$d^2(\frac{tan\theta}{2} - \frac{sec^2(\theta)}{24}$ The parabola's area is dependent on the angle.

InteGrand · Oct 1, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
Are you sure about this answer? Because the area under the curve is

$d^2(\frac{tan\theta}{2} - \frac{sec^2(\theta)}{24}$ The parabola's area is dependent on the angle.

This problem needs to be considered in 3D a little. You need to consider the maximum "sideways" angle you can turn the nozzle so that you still just hit the wall at the bottom of the wall, and for every sideways angle inbetween these extreme "sideways" angles, find the "upwards" angle that allows you to hit highest up the wall (this will be the point on the parabola in question).

This is a question from an old HSC (can't remember which year, but would be between 1960s and 1980s.)

lita1000 · Oct 1, 2015

Re: HSC 2015 4U Marathon

lita1000 said:
$$A tall building stands on level ground. The nozzle of a water sprinkler is positioned at a point P on the ground at a distance d from a wall of the building. Water sprays from the nozzle with speed V and the nozzle can be pointed in any direction from P. Suppose that $V= 2(gd)^{1/2}. $Show that the portion of the wall that can be sprayed with water is a parabolic segment with area$ (5/2)(d^2 * (15)^{1/2})$

Bump

InteGrand · Oct 1, 2015

Re: HSC 2015 4U Marathon

lita1000 said:
Bump

Was available two posts above your bump...lol.

Drsoccerball · Oct 1, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
Was available two posts above your bump...lol.

So was my solution right D: except i didnt find theta ?

InteGrand · Oct 1, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
So was my solution right D: except i didnt find theta ?

$No, you need to vary the angle of projection $\theta$, as well as the ``sideways'' angle ($\phi$ say).$

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