HSC 2016 MX2 Marathon ADVANCED (archive) (5 Viewers)

DatAtarLyfe · Jan 3, 2016

Re: HSC 2016 4U Marathon - Advanced Level

May i ask, are some of these questions simply harder problem solving question, that are considered "harder 4u" but in actuality, can be done if you sit there long enough and think about it, or does it simply require methodologies outside the syllabus? I'm not sure if i should even attempt these in case they need some special theorem and i'd be wasting my time trying to solve it.

leehuan · Jan 3, 2016

Re: HSC 2016 4U Marathon - Advanced Level

A few are out of the syllabus, that's why I leave this thread alone and plan to come back to it after I do uni maths.

DatAtarLyfe · Jan 3, 2016

Re: HSC 2016 4U Marathon - Advanced Level

Do these questions require higher level of maths, or just intellect?

Sy123 said:
$\\ $Show that$ \ y^2 = x^2 + x + 1 \ $has no integer solutions$$

InteGrand said:
$Let $ABC$ be a triangle with a point $D$ on $AB$, $E$ on $BC$ and $F$ on $AC$. It is given that $AD=BE=CF$ and that $\triangle DEF$ is equilateral.$

$Show that $\triangle ABC$ is equilateral.$

leehuan · Jan 3, 2016

Re: HSC 2016 4U Marathon - Advanced Level

DatAtarLyfe said:
Do these questions require higher level of maths, or just intellect?

InteGrand's question just requires intellect.

Idk about Sy's. But his questions in the 3U marathon last year were 100% intellect.

DatAtarLyfe · Jan 3, 2016

Re: HSC 2016 4U Marathon - Advanced Level

Sweet, will defs try it out tomorrow then

RealiseNothing · Jan 3, 2016

Re: HSC 2016 4U Marathon - Advanced Level

DatAtarLyfe said:
May i ask, are some of these questions simply harder problem solving question, that are considered "harder 4u" but in actuality, can be done if you sit there long enough and think about it, or does it simply require methodologies outside the syllabus? I'm not sure if i should even attempt these in case they need some special theorem and i'd be wasting my time trying to solve it.

I don't think uni maths is required for the questions in this thread tbh.

Uni maths has quite a different flavour to high school maths.

leehuan · Jan 3, 2016

Re: HSC 2016 4U Marathon - Advanced Level

RealiseNothing said:
I don't think uni maths is required for the questions in this thread tbh.

Uni maths has quite a different flavour to high school maths.

Too many of the questions, though not seeming impossible, play around with my mind a bit too much for me to bother attempting them

RealiseNothing · Jan 3, 2016

Re: HSC 2016 4U Marathon - Advanced Level

leehuan said:
Too many of the questions, though not seeming impossible, play around with my mind a bit too much for me to bother attempting them

Yep there's no denying they can be very hard, but still possible for a smart HSC student given they try for long enough.

glittergal96 · Jan 3, 2016

Re: HSC 2016 4U Marathon - Advanced Level

If x > 0, then x^2 < 1+x+x^2 < (1+x)^2
if x < -1 then (1+x)^2 < 1+x+x^2 < x^2.

So x=-1 or 0. (We cannot find a square strictly between two consecutive squares).

So the only integer solutions are
x=-1, y=+-1 and x=0,y=+-1.

Paradoxica · Jan 4, 2016

Re: HSC 2016 4U Marathon - Advanced Level

Sy123 said:
$\\ $Find all pairs of non-negative integers$ \ (x,y) \ $that satisfy$ \\ x^3y + x + y = xy + 2xy^2$

This is my first ever attempt at a diophantine equation, so I'm not certain it's 100% rigorous.

$\noindent From inspection, it is clear (0,0), (1,1) and (2,2) are all solutions to the equation. Divide both sides by $xy$ to obtain:$ \\ x^2 + \frac{1}{x} + \frac{1}{y} = 1+2y \\ $The RHS is an integer, so the LHS must be an integer. There are no possibilities for $x$ and $y$ other than (1,1) and (2,2), as any other pair of positive integers would give us a non-integer on the LHS. Therefore, the only solutions are the ones obtained by inspection.$

leehuan · Jan 4, 2016

Re: HSC 2016 4U Marathon - Advanced Level

Paradoxica said:
This is my first ever attempt at a diophantine equation, so I'm not certain it's 100% rigorous.

$\noindent From inspection, it is clear (0,0), (1,1) and (2,2) are all solutions to the equation. Divide both sides by $xy$ to obtain:$ \\ x^2 + \frac{1}{x} + \frac{1}{y} = 1+2y \\ $The RHS is an integer, so the LHS must be an integer. There are no possibilities for $x$ and $y$ other than (1,1) and (2,2), as any other pair of positive integers would give us a non-integer on the LHS. Therefore, the only solutions are the ones obtained by inspection.$

Can I ask how you managed to achieve the solution (2,2) by inspection? Doesn't seem so obvious.

InteGrand · Jan 4, 2016

Re: HSC 2016 4U Marathon - Advanced Level

leehuan said:
Can I ask how you managed to achieve the solution (2,2) by inspection? Doesn't seem so obvious.

$\noindent I don't know if he did it this way, but one way would be to look at the divided equation he has. He then noticed what are the only integer pairs that make the L.H.S. an integer. Then, just test these ones to see if they are solutions. This enables us to pick up these solutions even if we hadn't seen them before the division. The main thing to do before division would be to check the cases of 0, since we can't divide by 0.$

Paradoxica · Jan 4, 2016

Re: HSC 2016 4U Marathon - Advanced Level

leehuan said:
Can I ask how you managed to achieve the solution (2,2) by inspection? Doesn't seem so obvious.

Compare x^3 y and 2xy^2. Suppose they are homogenous, i.e. they take on a value that achieves similar algebraic degree. As 0 and 1 have already been found, The only possibility is 2. Then since 2x2 = 2+2, (2,2) is clearly a solution to the diophantine equation and we are done.

This is all a matter of intuition. Sorry if you don't possess the same weirdness of intuition I do.

JusticeTackle · Jan 4, 2016

Re: HSC 2016 4U Marathon - Advanced Level

Paradoxica said:
I'll requote the unsolved problems from the previous marathons.

Can 2015ers not spoil the answers for these? However, anyone who is sitting the HSC in the 2016 should feel free to answer these. Also, can someone clarify what realisenothing means when he says that m is the highest integer power but contradicts themselves in the next line when they say that m doesn't necessarily have to be an integer?

Also, the two questions posted by Sy have already been resolved by previous members ages ago, so I don't see why you included those? You should remove them

InteGrand · Jan 4, 2016

Re: HSC 2016 4U Marathon - Advanced Level

JusticeTackle said:
Also, can someone clarify what realisenothing means when he says that m is the highest integer power but contradicts themselves in the next line when they say that m doesn't necessarily have to be an integer?

I and another user were also a bit confused by this so we asked some questions about it on the thread where it was first posted, so you might want to read those posts: http://community.boredofstudies.org...4u-marathon-advanced-level-9.html#post6965906

(But spoiler alert, that question is solved on the next page on from that link by the user glittergal96 .)

JusticeTackle · Jan 4, 2016

Re: HSC 2016 4U Marathon - Advanced Level

So, the only problems left for the 2016ers are Integrand's, Heroic Pandas, Lita1000's, and Dan964's.

dan964 · Jan 5, 2016

Re: HSC 2016 4U Marathon - Advanced Level

InteGrand said:
$Also, not sure about dan964's Q. If $f(nx)$ and $g(nx)$ are identical degree $n$ polynomials, they are equal for more than $n$ input values, so they are identical polynomials. Then if $f(x)^2 + g(x)^2=1$ (by this I assume he meant identically 1), we have $f(x)^2 + f(x)^2 \equiv 1\Rightarrow f(x) \equiv \pm \frac{1}{\sqrt{2}}$. This doesn't have anything to do with $n\geq 5$.$

aah this problem just wait, I will explain.

InteGrand · Jan 6, 2016

Re: HSC 2016 4U Marathon - Advanced Level

$\textbf{ANOTHER QUESTION}$$

$\noindent Let $A = \sin \theta _1 + \sin \theta _2 + \cdots + \sin \theta_n$, where $n$ is a fixed positive integer with $n\geqslant 3$, and each $\theta _i >0$ and $\theta_1 + \theta _2 + \cdots + \theta _n = 2\pi$. Show that the quantity $A$ is maximised if and only if $\theta_1 = \theta_2 = \cdots = \theta_n$ $\Big{(}$so each $\theta_i$ is equal to $\frac{2\pi}{n}\Big{)}$.$

$\noindent \textbf{Note.} This proves that the maximal-area $n$-sided polygon inscribed in a given circle is precisely the regular $n$-gon inscribed in that circle.$

glittergal96 · Jan 6, 2016

Re: HSC 2016 4U Marathon - Advanced Level

Related (harder): Find with proof the maximum possible value of A/P^2 where the variables denote area and perimeter of an n-sided polygon. (Find this maximum for each fixed n).

(You may assume without proof that such a maximum exists.)

b) Use this to say something about the problem of enclosing as large an area as possible with a string of unit length.

Paradoxica · Jan 6, 2016

Re: HSC 2016 4U Marathon - Advanced Level

glittergal96 said:
$\noindent Related (harder): Find with proof the maximum possible value of $\frac{A}{P^2}$ where the variables denote area and perimeter of an $n$-sided polygon.$\\$Find this maximum for each fixed $n.\\$You may assume without proof that such a maximum exists.$\\ $b) Use this to say something about the problem of enclosing as large an area as possible with a string of unit length.$

$\noindent Let the perimeter of the polygon be fixed. It now suffices to maximise the area of the polygon with a fixed perimeter. Suppose that the polygon with maximum area is concave. But this polygon is concave, so we can take the part that is concave, and flip it to the other such that the number of sides and perimeter is unchanged, while the area of the polygon has increased. This contradicts the first supposition, and hence the polygon cannot be concave if the area is to be a maximum. Suppose that the polygon is convex and the sides and interior angles arbitrary, subject to restrictions. Take a pair of adjacent sides and construct a diagonal at their unique ends to form a sub-triangle in the polygon. The length of the diagonal is fixed. Let the two sides be variable, but only up to the point that their sum remains the same. Construct a circumcircle on the three vertices that define this sub-triangle.$

$\noindent The angle subtended by a chord (in this case, the diagonal) on the circumference of a circle is always the same, provided it does not cross either of the points of intersection of the chord and the circle. As we can see, if this were to happen the polygon would self-intersect, which breaks the convexity of the polygon. Thus, this situation cannot arise, and we can safely ignore it. Let the variable sides be $x$ and $y$, and we have $x+y=z$. Let the constant side be $w$ and the angle at the circumference be $\theta$. The area of the triangle is thus $\frac{xy}{2}\sin{\theta}$. By the AM-GM inequality, $z^2=(x+y)^2 \geq 4xy$. This places an upper bound on the area of the triangle, as the left hand side is fixed.$

$$\noindent Hence, the maximum possible area of the triangle occurs only when $x=y$, which means that the sides must be equal in order for the area of this sub-triangle to achieve maximum. Iterating this situation across all pairs of adjacent sides, we find that all the interior angles must be equal to one another, due to all the sub-triangles being congruent, and that all the sides are equal to each other. This situation is the very definition of a regular polygon, and thus, in order for a polygon with a fixed perimeter to have maximum area, it must be a regular polygon. This proves that our ratio is a maximum if and only if the polygon is a regular polygon.\\ The perimeter/area of a regular $n$-gon can be partitioned into $n$ identical line segments/triangles, respectively. The angles at the centre are all $\frac{2\pi}{n}.$

$$\noindent Construct a circumcircle with radius $R$ that fully encompasses the regular $n$-gon. The length of each perimeter segment, by the chord function crd$(\theta)$ is $2R\sin{\frac{\pi}{n}}.$ Thus the perimeter is $2Rn\sin{\frac{\pi}{n}}.$ The area of each triangle, by the Sine Formula for area of a triangle, is $\frac{R^2}{2}\sin{\frac{2\pi}{n}}.\\$ Thus, the total area of the $n$-gon is $\frac{n}{2}R^2\sin{\frac{2\pi}{n}}.\\$ Putting this together, we have $\frac{A}{P^2} = \frac{nR^2\sin{\frac{2\pi}{n}}}{8R^2n^2 \sin^2{\frac{\pi}{n}}}=\frac{\cot{\frac{\pi}{n}}}{4n}.\\$ Now we have a ratio that is strictly dependent on $n$. Due to the proof of the statements above, we know that this is the maximum possible ratio for an $n$-sided polygon.$

$\noindent As the number of sides of a regular polygon increases to infinity, the polygon's area and circumference converge to the area and perimeter of a circle, respectively. Taking limits, we have:$\\ \lim_{n \rightarrow \infty} \frac{\cot{\frac{\pi}{n}}}{4n}=\lim_{m \rightarrow 0} \frac{m\cos{(\pi m)}}{4\sin{(\pi m)}}=\lim_{m \rightarrow 0} \frac{\cos{(\pi m)}}{4\pi \cdot \frac{\sin{\pi m}}{\pi m}}=\frac{\cos0}{4\pi \cdot 1}=\frac{1}{4\pi}=\frac{\pi r^2}{(2 \pi r)^2}=\frac{A}{P^2}\\$Conclusion: The optimal shape to enclose the maximum area with a fixed perimeter is a circle.$

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