HSC 2016 MX2 Marathon ADVANCED (archive) (1 Viewer)

Paradoxica · Jan 6, 2016

Re: HSC 2016 4U Marathon - Advanced Level

InteGrand said:
$\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.$

$I can't think of anything except Jensen's.$

glittergal96 · Jan 6, 2016

Re: HSC 2016 4U Marathon - Advanced Level

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

$\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.$

Pretty much yep

. There are a couple of subtle points that need to be addressed slightly differently but the ideas are there.

1. It is far from obvious that "flipping things to the outside" can transform an arbitrary n-gon into a convex n-gon. (The biggest difficulty here stems from the possibility of self-intersection after the flip.)

This technicality might be quite hard to deal with, I can't think of any easy way to do so.

However, there is at least one way I know of to avoid this technical issue rather cheaply.

2. I don't think this argument will really allow us to say that a circle is the UNIQUE optimal shape for enclosing area, just that any region bounded by a closed curve that can be well approximated by a polygon must satisfy A/L^2 =< 1/(4*pi).

(Of course, the circle is actually the unique rigorous solution to pretty much all formulations of the isoperimetric problem, but I don't believe that such elementary tools suffice to give us a rigorous solution to the full isoperimetric problem.)

Paradoxica · Jan 6, 2016

Re: HSC 2016 4U Marathon - Advanced Level

glittergal96 said:
Pretty much yep . There are a couple of subtle points that need to be addressed slightly differently but the ideas are there.

1. It is far from obvious that "flipping things to the outside" can transform an arbitrary n-gon into a convex n-gon. (The biggest difficulty here stems from the possibility of self-intersection after the flip.)

This technicality might be quite hard to deal with, I can't think of any easy way to do so.

However, there is at least one way I know of to avoid this technical issue rather cheaply.

2. I don't think this argument will really allow us to say that a circle is the UNIQUE optimal shape for enclosing area, just that any region bounded by a closed curve that can be well approximated by a polygon must satisfy A/L^2 =< 1/(4*pi).

(Of course, the circle is actually the unique rigorous solution to pretty much all formulations of the isoperimetric problem, but I don't believe that such elementary tools suffice to give us a rigorous solution to the full isoperimetric problem.)

I've seen topological solutions...

If we could somehow prove by elementary means that the nth ratio is strictly less than the n+1 th ratio, then we have an upper bound, as the sequence would be monotonically increasing.

glittergal96 · Jan 6, 2016

Re: HSC 2016 4U Marathon - Advanced Level

Paradoxica said:
I've seen topological solutions...

If we could somehow prove by elementary means that the nth ratio is strictly less than the n+1 th ratio, then we have an upper bound, as the sequence would be monotonically increasing.

Topological solutions to what?

An upper bound for what?

You are definitely onto something with the monotonicity property though, that is the key to the argument I mentioned in the first of my remarks (1).

Paradoxica · Jan 7, 2016

Re: HSC 2016 4U Marathon - Advanced Level

glittergal96 said:
Topological solutions to what? The isoperimetric inequality.

An upper bound for what? The nth ratio, which would imply the isoperimetric inequality.

You are definitely onto something with the monotonicity property though, that is the key to the argument I mentioned in the first of my remarks (1).

The inequality doesn't appear to be easily subject to elementary analysis. I'm looking at it already and my mind is going *nopenopenopegetmethefuckawayfromthisproblem*

seanieg89 · Jan 7, 2016

Re: HSC 2016 4U Marathon - Advanced Level

Paradoxica said:
The inequality doesn't appear to be easily subject to elementary analysis. I'm looking at it already and my mind is going *nopenopenopegetmethefuckawayfromthisproblem*

I know. As I said, the isoperimetric problem (ie showing that the optimal area-enclosing curves amongst curves of a certain regularity are precisely the circles, and the optimal value is 1/4pi) is not suitable as a high school problem. (It would be hard to make it instructive as well.)

The polygonal isoperimetric inequality is another matter though, especially is existence of a maximiser is already assumed, which is why I posed the question as I did.

As for b), that was to give a student an intuition for the isoperimetric inequality, without explictly looking at a rigorous definition of the class of curves we are optimising over.

seanieg89 · Jan 7, 2016

Re: HSC 2016 4U Marathon - Advanced Level

Paradoxica said:
The inequality doesn't appear to be easily subject to elementary analysis. I'm looking at it already and my mind is going *nopenopenopegetmethefuckawayfromthisproblem*

Also, by topological solutions of the isoperimetric inequality, do you mean:

-a proof by largely topological means that A/P^2 cannot exceed 1/4*pi, with equality only at circles. (quantifying over a set of curves of some regularity class.)
-a continuous function that is very rough (but still in some class of curves we can sensibly assign enclosed areas and arclengths to) that also satisfies the isoperimetric equality.

The first meaning seems much more likely to me now, but when I first read your post the context made it sound like you might mean the second so I thought I'd check.

seanieg89 · Jan 7, 2016

Re: HSC 2016 4U Marathon - Advanced Level

Oh lols, gave away my alt. Ah well, it's done now.

RealiseNothing · Jan 7, 2016

Re: HSC 2016 4U Marathon - Advanced Level

seanieg89 said:
Oh lols, gave away my alt. Ah well, it's done now.

Pretty sure it was obvious anyway

RealiseNothing · Jan 7, 2016

Re: HSC 2016 4U Marathon - Advanced Level

RealiseNothing said:
Pretty sure it was obvious anyway

At least for people who've been around awhile (me/sy/etc)

seanieg89 · Jan 7, 2016

Re: HSC 2016 4U Marathon - Advanced Level

RealiseNothing said:
Pretty sure it was obvious anyway

Yeah, after a while I just got lazier and lazier hahaha.

Paradoxica · Jan 7, 2016

Re: HSC 2016 4U Marathon - Advanced Level

seanieg89 said:
Also, by topological solutions of the isoperimetric inequality, do you mean:

-a proof by largely topological means that A/P^2 cannot exceed 1/4*pi, with equality only at circles. (quantifying over a set of curves of some regularity class.)
-a continuous function that is very rough (but still in some class of curves we can sensibly assign enclosed areas and arclengths to) that also satisfies the isoperimetric equality.

The first meaning seems much more likely to me now, but when I first read your post the context made it sound like you might mean the second so I thought I'd check.

I've seen paper abstracts with ideas in both, so both of those, I suppose...

Can anyone solve Integrand's problem? I still can't think of anything other than Jensen's Inequality.

seanieg89 · Jan 7, 2016

Re: HSC 2016 4U Marathon - Advanced Level

Paradoxica said:
I've seen paper abstracts with ideas in both, so both of those, I suppose...

Can anyone solve Integrand's problem? I still can't think of anything other than Jensen's Inequality.

Really? Could you possibly link me one of the ones corresponding to the second if you can find it? Genuinely interested, as I haven't encountered these particular pathological curves before.

The first would also interest me depending on what you meant by topological proof. It would be surprising to me if such a proof existed that did not involve much analysis.

For Integrand's problem, Jensens or Lagrange multipliers would be the most straightforward if things a bit outside of syllabus were allowed. Otherwise (if existence of a maximum was an allowable assumption, which I think it needs to be), you can employ a perturbation argument similar to your original post on the polygonal isoperimetric argument. i.e. showing that an n-gon which is not a regular n-gon (circumscribed on the unit circle) is not optimal.

Paradoxica · Jan 7, 2016

Re: HSC 2016 4U Marathon - Advanced Level

seanieg89 said:
Really? Could you possibly link me one of the ones corresponding to the second if you can find it? Genuinely interested, as I haven't encountered these particular pathological curves before.

The first would also interest me depending on what you meant by topological proof. It would be surprising to me if such a proof existed that did not involve much analysis.

For Integrand's problem, Jensens or Lagrange multipliers would be the most straightforward if things a bit outside of syllabus were allowed. Otherwise (if existence of a maximum was an allowable assumption, which I think it needs to be), you can employ a perturbation argument similar to your original post on the polygonal isoperimetric argument. i.e. showing that an n-gon which is not a regular n-gon (circumscribed on the unit circle) is not optimal.

aha, I only glanced briefly at those abstracts, and it was a while ago so I'm doubtful I'll be able to find them, as I only stumbled on them due to a corresponding google search on higher dimensional equivalents of the isoperimetric inequality.

I have no idea what a lagrange multiplier is.

I do not wish to go through that argument again, doing it once is enough for me.

I wonder if there's a geometric/graphical way to prove the inequality between subsequent ratios...

seanieg89 · Jan 8, 2016

Re: HSC 2016 4U Marathon - Advanced Level

Paradoxica said:
I wonder if there's a geometric/graphical way to prove the inequality between subsequent ratios...

Indeed there is.

Consider an n-gon that is optimal w.r.t. the n-gon isoperimetric inequality. Now construct an n+1 gon by severing two of the adjacent vertices of the n-gon and connecting them both to a new point Q, chosen in a way that avoids intersection. Keep a dotted line where the severed edge was. Now move this new point Q towards the dotted line from the outside or inside.

As you do this, your n+1 gon will have perimeter and enclosed area approaching that of the n-gon. So you can make n+1 gons with A/P^2 arbitrarily close to the optimal value of A/P^2 for n-gons.

I.e. n+1 gons cannot be "worse" at enclosing area than n-gons.

Paradoxica · Jan 8, 2016

Re: HSC 2016 4U Marathon - Advanced Level

seanieg89 said:
Indeed there is.

Consider an n-gon that is optimal w.r.t. the n-gon isoperimetric inequality. Now construct an n+1 gon by severing two of the adjacent vertices of the n-gon and connecting them both to a new point Q, chosen in a way that avoids intersection. Keep a dotted line where the severed edge was. Now move this new point Q towards the dotted line from the outside or inside.

As you do this, your n+1 gon will have perimeter and enclosed area approaching that of the n-gon. So you can make n+1 gons with A/P^2 arbitrarily close to the optimal value of A/P^2 for n-gons.

I.e. n+1 gons cannot be "worse" at enclosing area than n-gons.

This proves that each subsequent ratio is optimally better than the previous ratio. Since the infiniti-eth ratio is a finite value, this proves the two-dimensional isoperimetric inequality.

seanieg89 · Jan 8, 2016

Re: HSC 2016 4U Marathon - Advanced Level

Paradoxica said:
I have no idea what a lagrange multiplier is.

Nothing too fancy, just the standard tool for optimising functions w.r.t constraints. (e.g. Maximise e^(ye^x) subject to the constraint x^2+y^2=1).

Looking at stationary points and doing the higher dimensional analogues of the "derivative tests" to determine stationary point nature don't work in this setting. (unless you can nicely parametrise the constraint hypersurface).

seanieg89 · Jan 8, 2016

Re: HSC 2016 4U Marathon - Advanced Level

Paradoxica said:
Since the infiniti-eth ratio is a finite value, this proves the two-dimensional isoperimetric inequality.

Err, no. Although I am not sure if you are just joking by talking about infiniti-eth ratios.

KX · Jan 8, 2016

Re: HSC 2016 4U Marathon - Advanced Level

seanieg89 said:
Oh lols, gave away my alt. Ah well, it's done now.

....................

Wow lol

Glittergal was like an inspiration to the post-2014ers.....

To think that she was all fake.... Mind actually blown....

Actually tbh, it did cross my mind at least once that you and Glittergal were very very similar.

R.I.P Glittergal96 [25/07/14~08/01/16]

seanieg89 · Jan 8, 2016

Re: HSC 2016 4U Marathon - Advanced Level

KX said:
....................

Wow lol

Glittergal was like an inspiration to the post-2014ers.....

To think that she was all fake.... Mind actually blown....

Actually tbh, it did cross my mind at least once that you and Glittergal were very very similar.

R.I.P Glittergal96 [25/07/14~08/01/16]

Haha yeah, sorry about that. The start was just boredom / curiosity at how some of the more sexist people on this forum would react (as there are/were plenty of them). After a while people got to know the account though so I began to mostly use that one and just stuck to it.

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