\[ \text{Intuition suggests that the values }\alpha,\,\beta\text{ and }\gamma\text{ should somehow}\\ \text{correspond to the same ones given in the previous part.}\\ \text{Else it does not make sense.} \]
\[ \text{How the answers start off is basically relying on part i) for some simplifications.}\\ \text{In part i), it is clear that the angles all meet at a point, which gives the}\\ \alpha+\beta+\gamma = 2\pi\text{ bit.} \]
Which is just 3 times the expression for the \(x\)-coordinate of the centroid. So the centroid must have \(x\)-coordinate \( \frac{2\pi}{3} \).
\[ \text{But on one hand, the area of }\triangle TUV\text{ was }\frac12 r^2 (\sin\alpha+\sin\beta+\sin\gamma)\\ \text{whilst the }y\text{-coordinate of the centroid is }\frac{\sin\alpha+\sin\beta+\sin\gamma}{3}. \]
\[ \text{So pretty much a bit of rearranging is what's needed here to show that}\\ \text{the centroid has }y\text{-coordinate }\frac{2\times \operatorname{Area}_{\triangle TUV}}{3r^2}. \]
The answers write the area as \(|\triangle TUV|\) instead. This isn't really common in the HSC but it does get used at times in the real world, so I'll use the absolute value brackets too.
Of course, that was the easy mark to get. By considering the link between the two parts we've more or less reinterpreted the problem. What we wished to do is to maximise \( |\triangle TUV| \), but we see that the \(y\)-coordinate of the centroid of \(\triangle ABC\) happens to be a scalar multiple of it. (Note of course that \(r\) is constant.)
So if we maximise the \(y\)-coordinate of said centroid, we achieve the same goal. This is a much harder mark to get. Naively, we'd be tempted to go back to maximising \(\sin\alpha+\sin\beta+\sin\gamma\), but this problem is still not viable because we only have the one constraint \(\alpha+\beta+\gamma = 2\pi\). This is where we get the intuition that something about the curve \(y=\sin x\) must be useful - we want to do this maximisation with a completely different argument, not via calculus in optimisation.
To my knowledge, concavity hasn't been examined much at all in the HSC - in fact, I only know of one question in 2018 or 2017 that did. This is the idea.
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\[ \text{Given two points }x_1\text{ and }x_2,\text{ we say that a curve is concave down for all }x_1 \leq x \leq x_2\\ \text{if for every }0\leq t \leq 1,\text{ it is true that}\\ f(tx_1 + (1-t)x_2) \geq tf(x_1) + (1-t)f(x_2).\]
\[ \text{This definition is awkward to work with, so we clarify what is going on.}\\ \text{Essentially, if your curve is concave down over a region, what you can do is draw a straight line segment}\\ \text{from one endpoint of the region, to the other.} \]
\[ \text{A curve is concave }\textbf{down}\text{ if that line segment}\\ \text{is guaranteed to lie }\textbf{below}\text{ the curve about that region.} \]
Note similarly that for a curve concave up over a certain interval, the line segment drawn between the endpoints should lie above the curve.
\[ \text{This is illustrated in the diagram. For example, between the points }A\text{ and }C\\ \text{the curve is concave down.}\\ \text{The interval }AC\text{ is also, in fact, under the curve.} \]
Similarly, between the points \(A\) and \(B\) the curve is concave down, and the line segment \(AB\) is below the curve. Same goes for \(BC\).
The answers claim that this will always be the case, regardless of what \(A\), \(B\) and \(C\) are. This is because the \(x\)-coordinates of these points are respectively \(\alpha\), \(\beta\) and \(\gamma\), which in part (i) we're told are always between \(0\) and \(\pi\). For a MX2 Q16 problem, it should be safe to just assume without proof that \(y=\sin x\) is always concave down for \(0 < x< \pi\) - this can easily be proven using the second derivative, if a graph is not adequately convincing.
(Aside: The second derivative is used as a theorem to prove concavity. The definition, however, is the clunky thing I mentioned earlier.)
\[ \text{So because we know that }AB, \, AC\textbf{ and }BC\text{ lie under the curve }y=\sin x,\\ \text{it follows that }\triangle ABC\text{ also lies under it too.} \]
And after all, since the centroid is just the intersection of the triangles' medians, that is consequently forced under the curve as well.
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\[ \text{However, the crucial thing to note now is that }\textbf{only}\text{ the }y\text{-coordinate}\\ \text{influences }|\triangle TUV|. \text{ Recall that the }x\text{-coordinate}\\ \text{of the centroid is }\textbf{constant}.\text{ It is always equal to }\frac{2\pi}{3}. \]
This is where advanced intuition becomes necessary. The idea is that we know the centroid can never float above \(y=\sin x\). Rather, the \(y\)-coordinate of the centroid is bounded above by the curve \(y=\sin x\).
So the natural question to ask her is whether or not this bound is attainable. That is, is there a configuration of points \(A\), \(B\) and \(C\) that would ensure that the centroid must lie on the curve as well. This is useful, because if we know it can lie on the curve, and also know that it can't go above it, then we've deduced that the maximum possible \(y\)-coordinate IS when it is on the curve.
\[ \text{The answer to these questions is usually yes,}\\ \text{when we allow for a thing called }\textbf{degeneration}. \]
\[ \text{The idea is to }\textit{collapse }\triangle ABC\text{ somehow.}\\ \text{Without loss of generality, let's start by collapsing }BC. \]
\[ \text{What we are going to do is let }B\text{ and }C\text{ become the same point.}\\ \text{That is, }C=B.\\ \text{Then the concept of }\triangle ABC\text{ no longer exists.} \]
\[ \text{However what happens is that if we gradually bring }C\text{ closer to }B,\\ \text{what happens geometrically is that }\triangle ABC\\ \text{will gradually turn into just another line segment, namely the interval }AB! \]
Once we're left with the interval \(AB\), the "centroid" of \(\triangle ABC\) will be nothing more than a point on said interval. (To be somewhat specific, it will be the point that divides \(AB\) internally in the ratio 1:2, I believe. It could be 2:1 though.)
But the point is this is not enough. We know that \(AB\) is still under the curve \(y=\sin x\), so if the centroid lies on \(AB\), it is also under the curve.
\[ \text{This is why we need to think about the most }\textit{borderline}\text{ case.}\\ \text{This is when we make all }\textbf{three}\text{ points coincide.}\\ \text{That is, set }A = B = C\text{, as the answers have stated.} \]
The intuition behind this is a general rule of mathematics - borderline cases tend to influence optimality (be it minimality or maximality). This can prove useful at times, but trying to recall this in an exam is hard.
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\[ \text{And the thing is, we see that this is good!}\\ \text{When }A=B=C\text{, we're saying that all 3 points are now the same point}\\ \text{and hence }\triangle ABC\text{ has also collapsed down into a single point.} \]
\[ \text{Consequently, the centroid has also been collapsed to a single point.}\\ \text{But the key idea is, the centroid now has nowhere to run.}\\ \text{Since }A,\, B\text{ and }C\text{ are the same point, the centroid is forced to being the same point as well.} \]
\[ \text{The reason why this is optimal is because we know that }A,\, B\text{ and }C\text{ are on the curve.}\\ \text{Hence the centroid is }\textbf{also}\text{ on the curve }y=\sin x\text{ now!} \]
Thus, we've shown that the centroid both cannot lie above the curve, but can also lie ON the curve. This must therefore give the maximum \(y\)-coordinate.
Basically, the argument was pretty much fully geometric. But in a specific way.
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\[ \text{Once this bit of the puzzle is cracked, everything is easy.}\\ \text{We know that }A = B = C\text{ automatically imposes the condition}\\ \text{that their corresponding }x\text{-coordinates are equal.} \]
\[ \text{Therefore }\alpha = \beta = \gamma.\\ \text{So by subbing back into }\alpha+\beta+\gamma = 2\pi\\ \text{we can conclude that in fact, }\\ \alpha = \beta = \gamma = \frac{2\pi}{3}.\\ \text{Which, of course, can now be plugged back into the area formula.} \]
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