Cambridge Admissions Exam Statistics 2006 The question I am trying to answer is



My work:
I understand that there are three cases. One where the two points are on the diameter, one where they are both on the circumference of the half circle, and one where one is on the diameter and one is on the circumference. The area in the first case is obviously $0$, however, I do not know how to proceed from here.
I have tried a couple of methods non of which seemed to work. The very first thing I tried was to assign angles to the two points, say $\theta$ and $\alpha$. The area is then given by
$$\frac{\sin (\theta - \alpha)}{2}$$
I thought of now defining a new angle $\phi=\theta - \alpha$ and then to take the expectancy of the area, however, this seems to go nowhere as I first don't know the distribution of this angle, and moreover, I am not sure you can take expectancy over $\sin$
.
Then I thought about using Heron's Formula for the area of a triangle and consider the distribution of the distance between the two points. Again, that led me to nowhere as I do not the distribution of the distance between them.
There are published answers for this question :
https://www.thestudentroom.co.uk/showthread.php?t=894936&page=4#post32939428
https://pmt.physicsandmathstutor.com/download/Maths/STEP/Solutions-and-Reports/2006%20Hints%20and%20Answers.pdf (page 28)
however I do not understand them. They assume that the points are in some regions and then they assume that the distance between them is just the distance between these regions. I am not following the logic of the whole argument. If anybody could help me understand this problem and help me answer it I would very greatful.
 A: First, we need to consider the probabilities of observing each of the three cases.  To do this, we consider the probability $p$ that a randomly selected point occurs on the diameter:  this is simply $$p = \frac{2}{2+\pi},$$ since the diameter has length $2$ and the semicircle's circumference has length $\pi$.  So the probability of the diameter-diameter ($DD$) case is $$\Pr[DD] = p^2 = \frac{4}{(2+\pi)^2};$$ the circumference-diameter ($CD$) case has probability $$\Pr[CD] = 2p(1-p) = \frac{4\pi}{(2+\pi)^2};$$ and the circumference-circumference ($CC$) case has probability $$\Pr[CC] = (1-p)^2 = \frac{\pi^2}{(2+\pi)^2}.$$
So the expected area is $$\operatorname{E}[A] = \operatorname{E}[A \mid DD]\Pr[DD] + \operatorname{E}[A \mid CD]\Pr[CD] + \operatorname{E}[A \mid CC]\Pr[CC],$$ and you noted already that $A \mid DD = 0$.  We want to figure out the other two conditional expectations.  Let's work with the circumference-diameter case.  In this situation, we are given that one point is on the diameter and the other is on the circumference.  So each point is placed independently and uniformly at random on their respective segments.  Therefore, if we place the semicircle on a coordinate plane with the midpoint of the diameter at the origin and the circumference corresponding to the curve $y = \sqrt{1-x^2}$, then the diameter point has some random $x$-coordinate, say $(X, 0)$ where $X \sim \operatorname{Uniform}(-1,1)$.  The point on the circumference needs to be parametrized by the counterclockwise angle it makes with the positive $x$-axis, i.e., $(\cos \theta, \sin \theta)$ for some $\theta \sim \operatorname{Uniform}(0, \pi)$.  Then the conditional area random variable $A \mid CD$ is a function of $X$ and $\theta$ as follows:
$$A \mid CD = \frac{1}{2} |X| \cdot \sin \theta.$$  The absolute value is needed since $|X|$ represents the absolute unsigned distance between $X$ and the origin.  Since $X$ and $\theta$ are independent, it follows that $$\operatorname{E}[A \mid CD] = \frac{1}{2} \operatorname{E}[|X|] \operatorname{E}[\sin \theta] = \frac{1}{2} \int_{x=-1}^1 |x| \cdot \frac{1}{2} \, dx \int_{\theta=0}^\pi \sin \theta \cdot \frac{1}{\pi} \, d\theta = \frac{1}{2\pi}.$$
Now we do the same thing for the $CC$ case.  Here, we have two angles, say $\theta$ and $\varphi$, where both are IID uniform on $(0,\pi)$.  The tricky part is that the conditional area uses a different formula, from $$|\triangle ABC| = \frac{1}{2}a b \sin C,$$ where $a = b = 1$ because the radius is $1$, and $C = |\theta - \varphi|$ is the magnitude of the difference in the two angles between the sides of unit length.  Consequently,
$$\operatorname{E}[A \mid CC] = \frac{1}{2} \int_{\theta = 0}^\pi \int_{\varphi = 0}^\pi \sin |\theta - \varphi| \frac{1}{\pi^2} \, d\varphi \, d\theta.$$  We can use a symmetry argument to write this double integral as twice an integral over half the domain:
$$\begin{align}
\operatorname{E}[A \mid CC] 
&= \frac{1}{\pi^2} \int_{\theta = 0}^\pi \int_{\varphi = 0}^\theta \sin (\theta - \varphi) \, d\varphi \, d\theta \\
&= \frac{1}{\pi^2} \int_{\theta = 0}^\pi \bigl[\cos (\theta - \varphi)\bigr]_{\varphi=0}^\theta \, d\theta \\
&= \frac{1}{\pi^2} \int_{\theta = 0}^\pi (1 - \cos \theta) \, d\theta \\
&= \frac{1}{\pi^2} \bigl[ \theta - \sin \theta \bigr]_{\theta = 0}^\pi \\
&= \frac{1}{\pi}.
\end{align}$$
Now we put everything together:
$$\operatorname{E}[A] = \frac{1}{2\pi} \cdot \frac{4\pi}{(2+\pi)^2} + \frac{1}{\pi} \cdot \frac{\pi^2}{(2+\pi)^2} = \frac{1}{2+\pi},$$ again remembering that the diameter-diameter case is zero.  This matches the claimed answer.

In case there is some confusion about the symmetry argument in the $CC$ case, note that we can explicitly write for any bounded function $f$
$$\begin{align}
\int_{\theta = 0}^\pi \int_{\varphi = 0}^\pi f(\theta, \varphi) \, d\varphi \, d\theta 
&= \int_{\theta = 0}^\pi \int_{\varphi = 0}^\theta f(\theta, \varphi) \, d\varphi \, d\theta + \int_{\theta = 0}^\pi \int_{\varphi = \theta}^\pi f(\theta, \varphi) \, d\varphi \, d\theta \\
&= \int_{\theta = 0}^\pi \int_{\varphi = 0}^\theta f(\theta, \varphi) \, d\varphi \, d\theta + \int_{\varphi = 0}^\pi \int_{\theta = 0}^\varphi f(\theta, \varphi) \, d\theta \, d\varphi.\end{align}$$  The double integral in the second term simply has its order of integration switched via Fubini's theorem.  Now we can exploit the aforementioned symmetry, with the choice $f(\theta, \varphi) = \sin |\theta - \varphi| = \sin |\varphi - \theta| = f(\varphi, \theta)$, from which it follows that the interchanged double integral is identical to the first term.
A: Not a full answer but I don't the the rights to comment or the time to give a full answer.  The points are presumably selected uniformly at random with the length of the perimeter the basis for two identical probability spaces.  This is a 2d-probability space with the positions of the two points the two dimensions.
You simply need to integrate across the range of point positions a function for the area of the triangle then divide by the total space.
We can assign the entire probability space $S=(2+\pi)^2$ in size.
An area of probability $\dfrac {2^2}S$ will return a triangle of size zero as you correctly observed, leaving $\dfrac{4\pi + \pi^2}S$ to be assessed.
To get the one point on diameter, one on perimeter area space, you need to parametrise the two points. The intuition they're testing is your ability to discern, is that you need to parametrize the point on the diameter based on its linear co-ordinate and the circumference point based upon its radial co-ordinate, because these parameters "move at constant speed" along the perimeter.
But bear in mind there are two such cases - $a$ on perimeter, $b$ on diameter and the converse so that gets doubled.  This is a double-integral, first between $-1,1$ and then between $-\pi$ and $0$
Then you need to parametrise both points being on the perimeter, this time, both radially, so a double integral with both going from $-\pi$ to $0$.
