HSC 2015 MX2 Marathon (archive) (1 Viewer)

Sy123 · Feb 19, 2015

Re: HSC 2015 4U Marathon

$\\ $Out of the set of shapes$ \ \{ $equilateral triangle$, \ $square$, \dots, \ $regular$ \ n$-gon$ \ , \dots \ $circle$ \} \\ \\ $All of which have identical perimeter, show that the circle has the largest area$$

dan964 · Feb 20, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
$i) =17+17i=(\sqrt{17}cis(tan^{-1}\frac{1}{4})...$

$ii) $ use $ arg(ab)=arg(a) +arg(b) $ rule $$

correct method, yes

(I) is fine. modulus is $17*\sqrt{2}$ , argument is $\frac{\pi}{4}$ which I know you got based by the '...'
(II) is fine as well. As mentioned above by others, modulus is not needed.

tip: except in probability, 'and' usually implies '+'

dan964 · Feb 20, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $Out of the set of shapes$ \ \{ $equilateral triangle$, \ $square$, \dots, \ $regular$ \ n$-gon$ \ , \dots \ $circle$ \} \\ \\ $All of which have identical perimeter, show that the circle has the largest area$$

Step 1: To produce a formula.
Lets say each had a perimeter of X. X>0. Therefore each side is X/n, n>2 n is an integer

(A) and (B) can be proven by inscribing each REGULAR polygon inside a circle

(A) Each side subtends an angle of 2pi/n at the centre, forming an isosceles triangle
(B) Producing perpendiculars at each side bisects each side, producing sides of X/(2n) and a centre angle of (pi/n)

Using trig. to find the perpendicular gives h=cot(pi/n)*X/2n

Each triangle is therefore has an area of X^2/4n^2*cot(pi/n)

There are n triangles: A = (X^2/4n)*cot(pi/n) is the area (*)

Step 2: Consider the limiting case

The limit of y/tan(y) as y approaches 0 is 1.

Let y be pi/n
cot(pi/n) *pi/n as pi/n approaches 0 is 1.

Therefore as n approaches infinity: cot(pi/n)*1/n equals 1/pi

Side Note: From (*)
Let n approach infinity (the limiting case, i.e. a circle) gives X^2/4pi (by substituting 2pi * r you get the formula)

Differentiate (*) to get
X^2/4n^3 * cosec^2(pi/n) - X^2/2n^3 * cot(pi/n)

X^2/4n^3 * (D+1)^2 where D is cot(pi/n)

which is positive for n>2.
Since the area is positive for n=3, and the dA/dn is positive, as n increases, so does A. So therefore when n is a max, A is a max. Therefore at n --> infinity, A approaches a maximum. (X^2/4pi)
etc.

Therefore circle has the largest area if PERIMETER is constant for all regular polygon.

braintic · Feb 20, 2015

Re: HSC 2015 4U Marathon

dan964 said:
tip: except in probability, 'and' usually implies '+'

I couldn't find what this was referring to. Would you explain what you mean in the non-probabilistic sense, perhaps with an example or two.

dan964 · Feb 20, 2015

Re: HSC 2015 4U Marathon

braintic said:
I couldn't find what this was referring to.

It was a general note, in response to an earlier usage (of which was straightforward)
I understood what he meant, but felt like writing a note.

braintic said:
Would you explain what you mean in the non-probabilistic sense, perhaps with an example or two.

3 and 4 give 7 not 12.
https://www.google.com.au/#q=3+and+4+equals

or the classic joke of 9 and 10 gives 21 (actually 19 but whatever)

Sy123 · Feb 21, 2015

Re: HSC 2015 4U Marathon

dan964 said:
Step 1: To produce a formula.
Lets say each had a perimeter of X. X>0. Therefore each side is X/n, n>2 n is an integer

(A) and (B) can be proven by inscribing each REGULAR polygon inside a circle

(A) Each side subtends an angle of 2pi/n at the centre, forming an isosceles triangle
(B) Producing perpendiculars at each side bisects each side, producing sides of X/(2n) and a centre angle of (pi/n)

Using trig. to find the perpendicular gives h=cot(pi/n)*X/2n

Each triangle is therefore has an area of X^2/4n^2*cot(pi/n)

There are n triangles: A = (X^2/4n)*cot(pi/n) is the area (*)

Step 2: Consider the limiting case

The limit of y/tan(y) as y approaches 0 is 1.

Let y be pi/n
cot(pi/n) *pi/n as pi/n approaches 0 is 1.

Therefore as n approaches infinity: cot(pi/n)*1/n equals 1/pi

Side Note: From (*)
Let n approach infinity (the limiting case, i.e. a circle) gives X^2/4pi (by substituting 2pi * r you get the formula)

Differentiate (*) to get
X^2/4n^3 * cosec^2(pi/n) - X^2/2n^3 * cot(pi/n)

X^2/4n^3 * (D+1)^2 where D is cot(pi/n)

which is positive for n>2.
Since the area is positive for n=3, and the dA/dn is positive, as n increases, so does A. So therefore when n is a max, A is a max. Therefore at n --> infinity, A approaches a maximum. (X^2/4pi)
etc.

Therefore circle has the largest area if PERIMETER is constant for all regular polygon.

Yes well done, for the other users who attempted this question (your attempt was deleted in a site crash yesterday), the key part I was looking for is what I bolded

Drsoccerball · Feb 21, 2015

Re: HSC 2015 4U Marathon

does circle geometry work for conics? I know it says circle but surely conics have some rules as well?

braintic · Feb 21, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
does circle geometry work for conics? I know it says circle but surely conics have some rules as well?

Those 'rules' .... they are all those things you've derived in doing all those Conics questions.

Drsoccerball · Feb 21, 2015

Re: HSC 2015 4U Marathon

braintic said:
Those 'rules' .... they are all those things you've derived in doing all those Conics questions.

Oh that makes sense ahahahahahahh can you write the formulas that are assumed knowledge ?

Sy123 · Feb 22, 2015

Re: HSC 2015 4U Marathon

$\\ $a) Show that for complex$ \ u,v \ $then$ \ |u+v|^2 = |u|^2 + |v|^2 + uv \left(\frac{\bar{u}}{u} + \frac{\bar{v}}{v} \right)$

$\\ $b) Let the points$ \ X_0, X_1, X_2, \dots, X_{n-1} \ $represent the roots of$ \ z^n = 1 \ $on the Argand plane. Let$ \ P \ $represent some complex number with modulus$ \ 1 \\\\ $Show that$ \ \sum_{k=0}^{n-1} PX_k^2 \ $is independent of the position of$ \ P$

------------------------------------------------

$\\ $($ PX_k \ $is the distance from point$ \ P \ $to the point$ \ X_k $)$$

Sy123 · Feb 23, 2015

Re: HSC 2015 4U Marathon

$\\ $Show that the curves$ \ x^2-y^2 = a^2 \ $and$ \ xy =c^2 \ $(for constants$ \ a,c $), intersect at right angles$$

Ekman · Feb 23, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $Show that the curves$ \ x^2-y^2 = a^2 \ $and$ \ xy =c^2 \ $(for constants$ \ a,c $), intersect at right angles$$

$$Let the point of intersection be: $ x_1,y_1$

$$ Differential of: $ x^2-y^2=a^2 $ is $ \frac{dy}{dx}=\frac{x}{y}$

$$ Differential of: $ xy=c^2 $ is $ \frac{dy}{dx}=\frac{-y}{x}$

$\therefore \frac{x}{y} \times \frac{-y}{x} = -1 $ Thus at the point of intersection where two tangents are drawn, they are perpendicular to each other$$

Ekman · Feb 23, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $a) Show that for complex$ \ u,v \ $then$ \ |u+v|^2 = |u|^2 + |v|^2 + uv \left(\frac{\bar{u}}{u} + \frac{\bar{v}}{v} \right)$

$\\ $b) Let the points$ \ X_0, X_1, X_2, \dots, X_{n-1} \ $represent the roots of$ \ z^n = 1 \ $on the Argand plane. Let$ \ P \ $represent some complex number with modulus$ \ 1 \\\\ $Show that$ \ \sum_{k=0}^{n-1} PX_k^2 \ $is independent of the position of$ \ P$

------------------------------------------------

$\\ $($ PX_k \ $is the distance from point$ \ P \ $to the point$ \ X_k $)$$

$$a) $ (u+v)(\bar{u} + \bar{v}) = |u+v|^2$

$(u+v)(\bar{u} + \bar{v}) = |u|^2+|v|^2 +u\bar{v} +\bar{u}v = |u|^2+|v|^2 + uv \left(\frac{\bar{u}}{u} + \frac{\bar{v}}{v} \right)$

$b) $ |x_k - P| = Px_k $ (This is the modulus of the vector of $ PX_k $ )$

$|x_k - P|^2 = (Px_k)^2 = |x_k|^2 + |-p|^2 -p\bar{x_k} -\bar{p}x_k $ (Using part a) )$$

$\sum_{k=0}^{n-1} (Px_k)^2 = \sum_{k=0}^{n-1} |x_k|^2 + |-p|^2 -p\bar{x_k} -\bar{p}x_k$

$|x_k| = 1 $ and $ |p| = 1$

$\therefore $ As sum up between 0 and n-1, there is n amount of terms, where $ -p\bar{x_k} -\bar{p}x_k $ would cancel out as the amount of solutions of $ z^n-1=0 $ above the x-axis is equal to the amount below the x-axis. $$

$\therefore \sum_{k=0}^{n-1} (Px_k)^2 = 2n$

Sy123 · Feb 23, 2015

Re: HSC 2015 4U Marathon

Ekman said:
$$a) $ (u+v)(\bar{u} + \bar{v}) = |u+v|^2$

$(u+v)(\bar{u} + \bar{v}) = |u|^2+|v|^2 +u\bar{v} +\bar{u}v = |u|^2+|v|^2 + uv \left(\frac{\bar{u}}{u} + \frac{\bar{v}}{v} \right)$

$b) $ |x_k - P| = Px_k $ (This is the modulus of the vector of $ PX_k $ )$

$|x_k - P|^2 = (Px_k)^2 = |x_k|^2 + |-p|^2 -p\bar{x_k} -\bar{p}x_k $ (Using part a) )$$

$\sum_{k=0}^{n-1} (Px_k)^2 = \sum_{k=0}^{n-1} |x_k|^2 + |-p|^2 -p\bar{x_k} -\bar{p}x_k$

$|x_k| = 1 $ and $ |p| = 1$

$\therefore $ As sum up between 0 and n-1, there is n amount of terms, where $ -p\bar{x_k} -\bar{p}x_k $ would cancel out as the amount of solutions of $ z^n-1=0 $ above the x-axis is equal to the amount below the x-axis. $$

$\therefore \sum_{k=0}^{n-1} (Px_k)^2 = 2n$

Yep that is correct, your explanation for why those extra terms cancel out is a little vague, but that's the general idea

Ekman · Feb 23, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
Yep that is correct, your explanation for why those extra terms cancel out is a little vague, but that's the general idea

I would of drew an Argand diagram to demonstrate the cancelling out process.

Axio · Feb 23, 2015

Re: HSC 2015 4U Marathon

$P(asec\theta ,btan\theta ) \ $lies on the hyperbola$ \ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 $ \ with foci (\pm ae,0)$
$$Show that$ \ PS=a(esec\theta -1) \ and \ PS'=a(esec\theta +1)$

Ekman · Feb 23, 2015

Re: HSC 2015 4U Marathon

Axio said:
$P(asec\theta ,btan\theta ) \ $lies on the hyperbola$ \ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 $ \ with foci (\pm ae,0)$
$$Show that$ \ PS=a(esec\theta -1) \ and \ PS'=a(esec\theta +1)$

$PS=ePM $ and $ PS'=ePM' $ (Focus-Directrix Definition)$$

$ePM = e(a\sec\theta - \frac{a}{e}) = a(e\sec\theta - 1)$

$$ Similarly $ ePM' = e(a\sec\theta + \frac{a}{e}) = a(e\sec\theta + 1)$

$\therefore PS= a(e\sec\theta - 1) $ and $ PS'=a(e\sec\theta + 1)$

Sy123 · Feb 23, 2015

Re: HSC 2015 4U Marathon

Ekman said:
I would of drew an Argand diagram to demonstrate the cancelling out process.

Right, a more rigorous way of showing it would be to do this:

$\\ \sum_{k=0}^{n-1} |p - x_k|^2 = \sum_{k=0}^{n-1} (|p|^2 + |x_k|^2 - (p\bar{x_k} + \bar{p} x_k)) = \sum_{k=0}^{n-1} 2 - \sum_{k=0}^{n-1}(p\bar{x_k} + \bar{p} x_k) = 2n - p\sum_{k=0}^{n-1}\bar{x_k} - \bar{p} \sum_{k=0}^{n-1} x_k = 2n - p \overline{\sum_{k=0}^{n-1} x_k} - \bar{p} \sum_{k=0}^{n-1} x_k \\ \\ \sum_{k=0}^{n-1} x_k = 0 \ $(sum of roots of$ \ x^n - 1 = 0 ) \\ \\ \therefore \ \sum_{k=0}^{n-1} PX_k^2 = 2n$

Sy123 · Feb 23, 2015

Re: HSC 2015 4U Marathon

$\\ $Show that the hyperbola$ \ x^2 - y^2 = a^2 \ $and$ \ xy = c^2 \ $intersect at right angles$$

Drsoccerball · Feb 23, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $Show that the hyperbola$ \ x^2 - y^2 = a^2 \ $and$ \ xy = c^2 \ $intersect at right angles$$

Equation 1
$2x-2yy'=0$

$y'= \frac{x}{y}$
Equation 2
$y+y'x =0$

$y'=\frac{-y}{x}$

$\frac{-y}{x} \times \frac{x}{y}=-1$

$\therefore $ intersects at right angles $$

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