HSC 2015 Maths Marathon (archive) (1 Viewer)

InteGrand · Oct 14, 2015

Re: HSC 2015 2U Marathon

Mr_Kap said:
so does seating Alice not matter to the probability in this case because it is a circle, and circles can start from anywhere?

Basically yeah; this is one of the things you learn in 3U perms and combs (symmetry of circular arrangements). So at the start, all seats are essentially the same, because we can rotate the circle containing just one person onto any other arrangement with just one person. To account for this rotational symmetry, the first person's position can just be fixed. This is why in perms and combs we would say the number of ways to sit 1 person at the table containing 6 seats is just 1; we wouldn't say 6, because these 6 "ways" are all really just the same (rotations of one another) (the chairs are assumed identical for these problems).

A way to think about it is, wherever Alice sits, the chairs needed are those Seats 1 and 2 that are two seats away from her chosen seat. So it doesn't matter where she sits, because we'll be doing the same 1/5 • 1/4 thing regardless.

InteGrand · Oct 14, 2015

Re: HSC 2015 2U Marathon

If you wanted, you could say there's a 1/6 chance of Alice sitting in a given seat, and for each seat, get the 1/10 result for that seat, and then do this for the six seats that Alice could go to, and you'd end up with the same answer as you'd do 6•(1/6 • 1/10) = 1/10 (i.e. taking into account the probability of Alice sitting in a specific seat just gets cancelled out by then having to account for the number of seats she can sit in, so we just can fix Alice's position).

Mr_Kap · Oct 14, 2015

Re: HSC 2015 2U Marathon

Thx for that Integrand. Appreciate it.

rand_althor · Oct 15, 2015

Re: HSC 2015 2U Marathon

Let ABC be a triangle such that AB = 3, BC = 4, AC = 5. Let X be a point in the triangle. Find the minimal possible value of AX² + BX² + CX².

Paradoxica · Oct 16, 2015

Re: HSC 2015 2U Marathon

rand_althor said:
Let ABC be a triangle such that AB = 3, BC = 4, AC = 5. Let X be a point in the triangle. Find the minimal possible value of AX² + BX² + CX².

$$Although I cannot prove it, the answer appears to be exactly $28-7\sqrt{2}$

InteGrand · Oct 16, 2015

Re: HSC 2015 2U Marathon

rand_althor said:
Let ABC be a triangle such that AB = 3, BC = 4, AC = 5. Let X be a point in the triangle. Find the minimal possible value of AX² + BX² + CX².

$Place the triangle (clearly a right-angled one) onto the Cartesian plane, with $A=(0,3)$, $B=(0,0)$, $C=(4,0)$. Let $X=(x,y)$. We can search for the global minimum of the objective function and see whether it is in the triangle. Note that $X$ lies within the triangle if and only if $y\leq 3 - \frac{3}{4}x$ and $y\geq 0,0\leq x \leq 4$.$

$We have $\mathcal{D}:= AX^2 + BX^2 + CX^2 = x^2 + (y-3)^2 + x^2 + y^2 + (x-4)^2 + y^2 = 2x^2 + (x-4)^2 + 2y^2 + (y-3)^2$. Let $f(x,y)=2x^2 + (x-4)^2 + 2y^2 + (y-3)^2$, so $\mathcal{D}=f(x,y)$. It is clear that $f$ is convex so the stationary point we find is a global minimum.$

$To find this minimum $(x^*,y^*)$, set $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ equal to 0:$

$$\frac{\partial f}{\partial x} = 4x + 2(x-4)=0 $ if and only if $x=x^* = \frac{4}{3}$.$

$$\frac{\partial f}{\partial y} = 4y + 2(y-3) =0 $ if and only if $y=y^* = 1$.$

$We check whether $(x^*,y^*)$ lies inside the triangle; clearly $3-\frac{3}{4}x^* = 3-\frac{3}{4}\times \frac{4}{3} = 3-1=2 \geq y^*$, and also, $0\leq x^* \leq 4$ and $0\leq y^*$. So the point is in the triangle.$

$Therefore, $\mathcal{D}_{\text{min.}}=f(x^*,y^*)=2\left(\frac{4}{3}\right)^2 +(1-3)^2+ 2\times 1^2 +\left(\frac{4}{3}-4 \right)^2=\frac{50}{3}$.$

$(Edit: realised it's more HSC accessible by just expanding the expression for $f(x,y)$, completing squares, and then reading off the minimum since squares are non-negative (not necessarily faster since more tedious). (Still need to confirm that it's inside the triangle.))$

Mr_Kap · Oct 16, 2015

Re: HSC 2015 2U Marathon

InteGrand said:
$(Edit: realised it's faster / HSC accessible by just expanding the expression for $f(x,y)$, completing squares, and then reading off the minimum since squares are non-negative. (Still need to confirm that it's inside the triangle.))$

Can u write working for this way please?

InteGrand · Oct 16, 2015

Re: HSC 2015 2U Marathon

Mr_Kap said:
Can u write working for this way please?

$\mathcal{D}=f(x,y) = 2x^2 + (x-4)^2 + 2y^2 + (y-3)^2$

$$\Rightarrow f(x,y) = 2x^2 + x^2 -8x + 16 + 2y^2 + y^2 -6y + 9$ (expanding)$

$$ = 3x^2 -8x + 3y^2 -6y + 25$ (collecting like terms)$

$$ = 3\left(x^2 -\frac{8}{3}x\right) + 3\left(y^2 -2y\right) + 25$ (factoring out the leading coefficient of the degree 2 terms to begin completing the square)$

$= 3\left(x^2 -\frac{8}{3}x + \left(\frac{4}{3} \right)^2 - \left(\frac{4}{3} \right)^2\right) + 3\left(y^2 -2y + 1 -1\right) + 25$

$= 3\left( x - \frac{4}{3} \right)^2 -3\left(\frac{4}{3}\right)^2 + 3(y-1)^2 -3+ 25$

$$ = 3\left( x - \frac{4}{3} \right)^2 + 3(y-1)^2+\frac{50}{3}$ (collecting the constants)$

$$\geq \frac{50}{3}$ with equality if and only if $x= \frac{4}{3} $ and $y=1$, since $\left( x - \frac{4}{3} \right)^2 \geq 0$ (equality iff $x=\frac{4}{3}$) and $(y-1)^2\geq 0$ (equality iff $y=1$) (`iff' is a real term and means `if and only if').$

$So the minimum value is $\frac{50}{3}$ and occurs precisely when $X=\left(\frac{4}{3},1 \right)$, which we can confirm is inside the triangle.$

InteGrand · Oct 16, 2015

Re: HSC 2015 2U Marathon

The concept of optimisation of multivariate functions isn't inside the 2U course as far as I know though, so I don't think this question would appear, although it is doable with 2U knowledge. If it did appear, you'd be guided through the steps.

Trebla · Oct 16, 2015

HSC 2015 2U Marathon

InteGrand said:
The concept of optimisation of multivariate functions isn't inside the 2U course as far as I know though, so I don't think this question would appear, although it is doable with 2U knowledge. If it did appear, you'd be guided through the steps.

You can have multivariate optimisations provided that there are enough constraint conditions that can convert it into a univariate optimisation.

InteGrand · Oct 16, 2015

Re: HSC 2015 2U Marathon

Trebla said:
You can have multivariate optimisations provided that there are enough constraint conditions that can convert it into a univariate optimisation.

Yeah I know; by multivariate I meant where the variables stay independent.

Paradoxica · Oct 16, 2015

Re: HSC 2015 2U Marathon

...Wow... I completed the square and then decided to give up. lol, not sure what happened to me then and there.

Flop21 · Oct 16, 2015

Re: HSC 2015 2U Marathon

gimme some questions

rand_althor · Oct 16, 2015

Re: HSC 2015 2U Marathon

$$Given that $f(x)+2f(8-x)=x^2$ for all real $x$, find $f(2)$.$

laters · Oct 16, 2015

Re: HSC 2015 2U Marathon

rand_althor said:
$$Given that $f(x)+2f(8-x)=x^2$ for all real $x$, find $f(2)$.$

f(2)+2f(6)=4
f(6)+2f(2)=36

Solving simultaneously will give f(2)=68/3.

InteGrand · Oct 16, 2015

Re: HSC 2015 2U Marathon

$And in general, we can find $f(x)$ by replacing $x$ with $8-x$ in the original equation to obtain $f(8-x)+2f(x)=(8-x)^2$, and then solve with the original equation. One finds that $f(x)=\frac{x^2 -32x+128}{3}$ for all real $x$.$

rand_althor · Oct 16, 2015

Re: HSC 2015 2U Marathon

$$1. Differentiate $\log_e{\left (\frac{\sqrt{x}}{2x+1} \right )}$. \\\indent 2. Calculate the volume of solid generated when the curve $y=\frac{1}{2} \left (e^{x}+e^{-x} \right )$ is rotated about the $x$-axis between $x=-3$ and $x=3$. \\\indent 3. Prove $\int^{e^2}_e ~\frac{1+\ln{x}}{x\ln{x}}~\mathrm{d}x=1+\ln{2}$.$

milkytea99 · Oct 17, 2015

Re: HSC 2015 2U Marathon

Do you think this year's 2uxpaper will be hard again?

photastic · Oct 17, 2015

Re: HSC 2015 2U Marathon

milkytea99 said:
Do you think this year's 2uxpaper will be hard again?

Most likely considering most exams have been harder on an odd year.

milkytea99 · Oct 17, 2015

Re: HSC 2015 2U Marathon

photastic said:
Most likely considering most exams have been harder on an odd year.

Omfg....

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