Problems

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Solution: There are \(3\) possible numbers of points on the curved part of the perimeter. \(0\): The area is \(0\) \(1\):

The area of the triangle is \(\frac12 |x| \sin \theta\) Where \(X\) is the point along the diameter which is \(U[-1,1]\) and \(\theta \sim U(0, \pi)\) Therefore \begin{align*} \mathbb{E}(A|\text{one on diameter}) &= \int_{0}^\pi \frac{1}{\pi} \int_{-1}^1\frac{1}{2}\frac12 |x| \sin \theta \d x \d \theta \\ &= \frac{1}{2\pi}\frac12 \int_{0}^\pi \sin \theta \d \theta \cdot 2\int_{0}^1 x\d x \\ &=\frac{1}{2\pi}\cdot 2 \cdot \frac12 = \frac{1}{2\pi} \end{align*} \(2\): If both are on the curved section Then the area is \(\frac12 \sin \theta\) where \(\theta = |\theta_1 - \theta_2|\) and \(\theta_i \sim U[0, \pi]\) Therefore the area is \begin{align*} \mathbb{E}(A|\text{none on diameter}) &= \int_{0}^\pi\frac{1}{\pi} \int_{0}^\pi\frac{1}{\pi} \frac12 \sin |\theta_1 - \theta_2| \d \theta_1 \d \theta_2 \\ &= \frac{1}{\pi^2}\frac12 \int_{0}^\pi \left (\int_{0}^{\theta_2} \sin (\theta_2 - \theta_1) \d \theta_1-\int_{\theta_2}^{\pi} \sin (\theta_2 - \theta_1) \d \theta_1 \right)\d \theta_2 \\ &= \frac{1}{\pi^2}\frac12 \int_{0}^\pi \left [2\cos(\theta_2 - \theta_2)-\cos(\theta_2 - 0)-\cos(\theta_2 - \pi) \right]\d \theta_2 \\ &= \frac{1}{\pi} \end{align*} Therefore the expected area is: \begin{align*} \mathbb{E}(A ) &= \mathbb{E}(A|\text{one on diameter})\cdot \mathbb{P}(\text{one on diameter}) + \mathbb{E}(A|\text{none on diameter})\cdot \mathbb{P}(\text{none on diameter}) \\ &= \frac{1}{2\pi}\mathbb{P}(\text{one on diameter}) + \frac{1}{\pi}\cdot \mathbb{P}(\text{none on diameter}) \\ &= \frac{1}{2\pi} \cdot 2 \cdot \frac{\pi}{\pi + 2} \cdot \frac{2}{\pi + 2} + \frac1{\pi} \cdot \frac{\pi}{\pi + 2} \cdot \frac{\pi}{\pi+2} \\ &= \frac{2 + \pi}{(\pi+2)^2} \\ &= \frac{1}{\pi+2} \end{align*}

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Solution: The area for \(\alpha\) fixed is \(\frac12 r^2 \sin \alpha + \frac12 r^2 \sin \theta + \frac12 r^2 \sin (\pi - \theta - \alpha)\) So we wish to maximise \(V = \sin \theta + \sin(\pi - \theta-\alpha)\) \begin{align*} && V &= \sin \theta + \sin(\pi - \theta-\alpha)\\ &&&= 2\sin \l \frac{\pi-\alpha}2\r\cos \l \frac{2\theta + \alpha - \pi}{2}\r \end{align*} The largest \(\cos\) can be is \(1\) when \(\displaystyle 2\theta + \alpha - \pi = 0 \Rightarrow \theta = \frac{\pi-\alpha}2\). (ie we split the remaining area exactly in half). We are now trying to maximise \(W = \sin \alpha + 2 \sin \frac{\pi - \alpha}2\) ie \begin{align*} && W &= \sin \alpha + 2 \cos \frac{\alpha}{2} \\ \Rightarrow && \frac{\d W}{\d \alpha} &= \cos \alpha-\sin \frac{\alpha}{2} \\ &&&= 1-2 \sin^2 \frac{\alpha}{2} - \sin \frac{\alpha}{2} \\ &&&= (1+\sin \frac{\alpha}{2})(1-2\sin \frac{\alpha}{2}) \end{align*} Therefore \(\frac{\alpha}{2} = -\frac{3\pi}{2}, \frac{\alpha}{2} = \frac{\pi}{6}, \frac{5\pi}{6} \Rightarrow \alpha = -3\pi, \frac{\pi}{3}, \frac{5\pi}{3}\). The only turning point in our range is \(\frac{\pi}{3}\). This is obvious a a maximum by symmetry or checking the end points, but we could also check the second derivative \(\frac{\d^2 W}{\d \alpha^2} = -\sin \frac{\pi}{3}-\cos \frac{\pi}{3} < 0\) so we have a maximum. Therefore the largest possible area is: \(\displaystyle \frac{3\sqrt{3}}{4}r^2\)

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Solution: \(AG = r\), therefore the area is: \begin{align*} A &= [AHG] - 2*[ABC] + [CDE] \\ &= \frac12 \pi r^2 - \pi h^2 + \frac12 \pi (r-2h)^2 \\ &= \frac12 \pi \l r^2 - 2h^2 + r^2 -4rh+4h^2 \r \\ &= \frac12 \pi \l 2r^2 -4rh + 2h^2\r \\ &= \pi (r-h)^2 \end{align*} This is the same area as a circle radius \(r-h\) But \(HD = r + (r-2d) = 2(r-d)\), ie the circle with diameter \(HD\) has radius \(r-h\) as required. Suppose \(A = (-h, 0), C = (h, 0), B = (0, h)\) then our parabola is \(y = \frac1{h}(h^2-x^2)\)

The area of \(ABC\) would then be \(\int_{-h}^h \frac{1}{h}(h^2-x^2) \d x = \frac1{h} \left [ h^2x - \frac{x^3}{3} \right] = \frac1{h} \l 2h^3-2\frac{h^3}{3} \r = \frac{4}{3}h^2\) so we have: \begin{align*} A &= [AHG] - 2*[ABC] +[CDE] \\ &= \frac43 r^2-\frac83h^2+\frac43(r-2h)^2 \\ &= \frac43 \l r^2 -2h^2+r^2-4rh+4h^2) \\ &= \frac43 (r-h)^2 \end{align*}

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Solution:

1. 

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   \begin{align*} \mathbb{E}(A \mid \text{exactly one point on diameter}) &= \int_{-1}^1\int_0^\pi \frac12 (x-0)\cdot 1 \cdot \sin(\pi - \theta) \frac{1}{\pi} \d \theta \frac{1}{2} \d x \\ &= \int_{-1}^1\frac1{2\pi} x \d x \cdot \left [ -\cos \theta \right]_0^\pi \\ &= \frac{1}{2\pi} \end{align*}
2. 

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   \begin{align*} \mathbb{E}(A \mid \text{no point on diameter}) &= \int_0^{\pi} \frac12 \cdot 1 \cdot 1 \cdot \sin x \cdot 2(\pi - x)/\pi^2 \d x \\ &= \frac1{\pi^2} \int_0^\pi \sin x (\pi - x) \d x \\ &= \frac1{\pi^2} \int_0^\pi x\sin x \d x \\ &= \frac1{\pi^2} \left [ \sin x - x \cos x \right]_0^{\pi} \\ &= \frac{1}{\pi} \end{align*}

If both points lie on the diameter the area of the triangle is \(0\). Therefore: \begin{align*} \mathbb{E}(A) &= \frac{1}{2\pi} \mathbb{P}(\text{exactly one point on diameter}) + \frac{1}{\pi}\mathbb{P}(\text{no points on diameter}) \\ &= \frac1{2\pi} \cdot \left (2 \cdot \frac{2}{2+\pi} \cdot \frac{\pi}{2+\pi} \right) + \frac{1}{\pi} \cdot \left ( \frac{\pi}{2+\pi} \cdot \frac{\pi}{2+\pi}\right) \\ &= \frac{1}{\pi} \frac{2\pi + \pi^2}{(2+\pi)^2} \\ &= \frac{1}{2+\pi} \end{align*}

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Solution:

Splitting the arc up into strips of width \(\delta \theta\), then we must have \begin{align*} && \sum r\cos \theta (r \delta \theta) &= \bar{x}\sum (r \delta \theta) \\ \lim_{\delta \theta \to 0}: && \int_{-\theta}^{\theta} r^2 \cos \theta \d \theta &= \bar{x}2 \theta r \\ \Rightarrow && 2r^2 \sin \theta &= \bar{x} 2 \theta r \\ \Rightarrow && \bar{x} &= \frac{r\sin \theta}{\theta} \end{align*} The centre of mass will lie on the line of symmetry. It also must lie at the center of the base of the semi-circle (see diagram). Using a coordinate frame where that point is the origin we must have \begin{align*} && 0 &= -2r \cdot 2h - 4h \cdot h + \pi r \frac{r}{\frac{\pi}{2}} \\ &&&= -4rh-4h^2+2r^2\\ \Rightarrow && 0 &= r^2-2rh-h^2 \\ \Rightarrow && \frac{r}{h} &= 1 \pm \sqrt{3} \\ \Rightarrow && r &= (1+\sqrt{3})h \\ && h & = \frac12 (\sqrt{3}-1) r \end{align*}