Problems

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Solution:

1. \(\,\) \begin{align*} && \dot{x} &=12 \cos^2 t \sin t \\ && \dot{y} &= 12 \cos t - 12 \sin^2 t \cos t \\ && \frac{\d y}{\d x} &= \frac{12 \cos t - 12 \sin^2 t \cos t}{12 \cos^2 t \sin t} \\ &&&= \frac{1 - \sin^2 t}{\cos t \sin t} \\ &&&= \cot t \\ \\ && \frac{y - (12\sin\phi - 4\sin^3\phi)}{x - (-4 \cos^3 \phi)} &= - \tan \phi \\ && y &= -\tan \phi x -4 \cos^3 \phi \tan \phi + 12 \sin \phi -4\sin^3 \phi \\ &&&= -\tan \phi x -4 \cos^2 \phi \sin \phi + 12 \sin \phi -4\sin^3 \phi \\ &&&= -\tan \phi x - 4\sin \phi+12 \sin \phi \\ &&y&= -\tan \phi x + 8 \sin \phi \end{align*} Note that when \(x = 8\cos^3 \phi\) we have \(y =-8 \cos^2 \phi \sin \phi + 8 \sin \phi = 8 \sin^3 \phi\). So the point lies on the curve. Notice also that \((8\cos^3 \phi, 8 \sin^ 3\phi)\) is a parametrisation of \(x^{2/3} + y^{2/3} = 4\) and so we can use parametric differentiation to see the gradient is \(\frac{24\sin^2 \phi \cos \phi}{-24\cos^2 \phi\sin\phi} = - \tan \phi\) so it also has the same gradient as required.
   

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2. \(\,\) \begin{align*} && \dot{x} &= -\sin t + \sin t + t \cos t \\ &&&= t \cos t \\ && \dot{y} &= \cos t - \cos t + t \sin t \\ &&&= t \sin t \\ && \frac{\d y}{\d x} &= \frac{t \sin t}{t \cos t} = \tan t \\ \\ && \frac{y - (\sin \phi - \phi \cos \phi)}{x - (\cos \phi + \phi \sin \phi)} &= -\cot \phi \\ \Rightarrow && y &= -\cot \phi x + (\cos \phi + \phi \sin \phi) \cot \phi + \sin \phi - \phi \cos \phi \\ &&&= -\cot \phi x + \cos \phi \cot \phi + \phi \cos \phi + \sin \phi - \phi \cos \phi \\ &&&= -\cot \phi x + \frac{\cos^2 \phi + \sin^2 \phi}{\sin \phi} \\ &&&= -\cot \phi x + \cosec \phi \end{align*} The distance to the origin is \(\displaystyle \frac{|\cosec \phi|}{\sqrt{1 + \cot^2 \phi}} = 1\) so this normal is a tangent to \(x^2 + y^2 = 1\)
   

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This is an interesting question because many years ago this question of finding involutes and envelopes of questions would be considered extremely standard. (Particularing finding the involute of a circle). (It also seems to make sense mechanically imagine unwinding (or winding) a piece of string from a circle!)

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Solution:

1. \(\,\) \begin{align*} && \begin{pmatrix} 1 & -x & x \\ y & 1 & -y \\ -z & z & 1 \end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix} &= \begin{pmatrix} a-xb+xc \\ ay+b-yc \\ -za+zb+c \end{pmatrix} \\ &&&= \begin{pmatrix} a-x(b-c) \\ b-y(c-a) \\ c-z(a-b) \end{pmatrix} \\ &&&= \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \end{align*} Notice since \(a,b\) and \(c\) are distinct real numbers the vector \(\langle a,b,c \rangle\) cannot be the zero vector, so the determinant of the matrix is zero, ie \(0= 1(1+yz)+x(y-yz)+x(yz+z) = 1 +yz+yx+zx\). Notice also then that \begin{align*} && \frac{a^2}{(b-c)^2} + \frac{b^2}{(c-a)^2} + \frac{c^2}{(a-b)^2} &= x^2+y^2+z^2 \\ &&&= (x+y+z)^2 - 2(xy+yz+zx) \\ &&&= 2 + (x+y+z)^2 \geq 2 \end{align*}
2. \(\,\) \begin{align*} && \begin{pmatrix} 2 & -x & -x \\ -y & 2 & -y \\ -z & -z & 2 \end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix} &= \begin{pmatrix} 2a-xb-xc \\ -ay+2b-yc \\ -za-zb+2c \end{pmatrix} \\ &&&= \begin{pmatrix} 2a-x(b+c) \\ 2b-y(c+a) \\ 2c-z(a+b) \end{pmatrix} \\ &&&= \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \\ \Rightarrow && 0 &= \det \begin{pmatrix} 2 & -x & -x \\ -y & 2 & -y \\ -z & -z & 2 \end{pmatrix} \\ &&&= 2(4 -yz)+x(-2y-yz)-x(yz+2z) \\ &&&= 8 - 2yz-2yx-2xyz-2zx\\ \Rightarrow && 4 &= xyz+xy+yz+zx \end{align*} \begin{align*} && (2a + b + c)(a + 2b + c)(a + b + 2c) &> 5(b+c)(c+a)(a+b) \\ \Leftrightarrow && \left ( \frac{2a}{b+c}+1 \right)\left ( \frac{2b}{c+a}+1 \right)\left ( \frac{2c}{a+b}+1 \right) &> 5 \\ \Leftrightarrow && \left ( x+1 \right)\left ( y+1 \right)\left ( x+1 \right) &> 5 \\ \Leftrightarrow && xyz+xy+yz+zx+x+y+z+1 &> 5 \\ \Leftrightarrow && 5+x+y+z&> 5 \\ \end{align*} Which is clearly true since if \(a,b,c\) are positve real numbers so are \(x,y,z\). This final inequality is equivalent to showing \(x+y+z > 2\) ie \begin{align*} && x+y+z &> 2 \\ \Leftrightarrow && \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} &> 1 \\ \\ && \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} & > \frac{a}{a+b+c} + \frac{b}{a+b+c} + \frac{c}{a+b+c} = 1 \end{align*} So we're done.

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Solution: \begin{questionparts} \item \(\,\) \begin{align*} && I_n &= \int_0^{\beta} (\sec x + \tan x)^n \, \d x \\ && \tfrac12(I_{n+1}+I_{n-1}) &= \tfrac12\int_0^{\beta} \left ( (\sec x + \tan x)^{n+1}+(\sec x + \tan x)^{n-1}\right) \, \d x \\ && \tfrac12(I_{n+1}+I_{n-1}) &= \tfrac12\int_0^{\beta} (\sec x + \tan x)^{n-1}\left ( (\sec x + \tan x)^{2}+1\right) \, \d x \\ && \tfrac12(I_{n+1}+I_{n-1}) &= \tfrac12\int_0^{\beta} (\sec x + \tan x)^{n-1}\left ( \sec^2 x + \tan^2 x + 2\sec x \tan x + 1\right) \, \d x \\ && \tfrac12(I_{n+1}+I_{n-1}) &= \tfrac12\int_0^{\beta} (\sec x + \tan x)^{n-1}\left ( 2\sec x \tan x +2\sec^2 x \right) \, \d x \\ &&& = \left [\frac1n(\sec x + \tan x)^{n} \right]_0^{\beta} \\ &&&= \frac1n[(\sec \beta + \tan \beta)^n - 1] \end{align*} Notice that by AM-GM \(\tfrac12( ( (\sec x + \tan x)^{n+1}+(\sec x + \tan x)^{n-1}) \geq (\sec x + \tan x)^{n}\) with equality not holding most of the time. Integrating we obtain our result. \item \(\,\) \begin{align*} && J_n &= \int_0^{\beta} (\sec x \cos \beta + \tan x )^n \d x \\ && \tfrac12( J_{n+1} + J_{n-1}) &= \tfrac12 \int_0^{\beta} \left ( (\sec x \cos \beta + \tan x )^{n+1} +(\sec x \cos \beta + \tan x )^{n-1}\right ) \d x \\ && &= \tfrac12 \int_0^{\beta}(\sec x \cos \beta + \tan x )^{n-1} \left ( (\sec x \cos \beta + \tan x )^{2} + \right ) \d x \\ && &= \tfrac12 \int_0^{\beta}(\sec x \cos \beta + \tan x )^{n-1} \left ( \sec^2 x \cos^2 \beta + \tan^2 x+ 2\sec x \tan x \cos \beta +1 \right ) \d x \\ && &= \int_0^{\beta}(\sec x \cos \beta + \tan x )^{n-1} \left ( \sec x \tan x \cos \beta +\tfrac12(\cos^2 \beta +1)\sec^2 x \right ) \d x \\ && &< \int_0^{\beta}(\sec x \cos \beta + \tan x )^{n-1} \left ( \sec x \tan x \cos \beta +\sec^2 x \right ) \d x \\ &&&= \left [\frac1n (\sec x \cos \beta + \tan x)^{n} \right]_0^{\beta} \\ &&&= \frac1n[ (1 + \tan \beta)^n - \cos^n \beta] \end{align*} But notice we can use the same AM-GM argument from before to show that \(J_n < \tfrac12( J_{n+1} + J_{n-1}) < \frac1n[ (1 + \tan \beta)^n - \cos^n \beta]\)

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Solution:

1. \(\,\) \begin{align*} && \cos \angle MOB &= \frac{\mathbf{m} \cdot \mathbf{b}}{|\mathbf{m}||\mathbf{b}|} \\ &&&= \frac{\cos \theta + 1}{2\sqrt{\frac14(\mathbf{a}+\mathbf{b})\cdot(\mathbf{a}+\mathbf{b})}} \\ &&&= \frac{\cos \theta + 1}{\sqrt{1+1+2\cos \theta}} \\ &&&= \frac{1 + \cos \theta}{\sqrt{2(1+\cos \theta})} \\ &&&= \frac1{\sqrt{2}} \sqrt{1+\cos \theta} \\ &&&= \cos \tfrac{\theta}{2} \end{align*} Since \(0 < \theta < \pi\) we must have \(\angle MOB = \tfrac{\theta}{2}\) ie it is the angle bisector.
2. The plane through the origin perpendicular to \(\mathbf{c}\) has \(\mathbf{x} \cdot \mathbf{c} = 0\), so \begin{align*} && \mathbf{a}_1 \cdot \mathbf{c} &= (\mathbf{a} - (\mathbf{a} \cdot \mathbf{c}) \mathbf{c}) \cdot \mathbf{c} \\ &&&= \mathbf{a} \cdot \mathbf{c} - \mathbf{a} \cdot \mathbf{c} \\ &&&= 0 \\ \\ && |\mathbf{a}_1|^2 &= \mathbf{a}_1 \cdot \mathbf{a}_1 \\ &&&= (\mathbf{a} - (\mathbf{a} \cdot \mathbf{c}) \mathbf{c}) \cdot (\mathbf{a} - (\mathbf{a} \cdot \mathbf{c}) \mathbf{c}) \\ &&&= 1 - 2(\mathbf{a} \cdot \mathbf{c})^2 + \mathbf{a} \cdot \mathbf{c} \\ &&&= (1-\cos^2 \alpha) \\ &&&= \sin^2 \alpha \\ \Rightarrow && |\mathbf{a}_1| &= \sin \alpha \\ \Rightarrow && |\mathbf{b}_1| &= \sin \beta \tag{changing a and b} \\ \\ && \cos \phi &= \frac{\mathbf{a}_1 \cdot \mathbf{b}_1}{|\mathbf{a}_1||\mathbf{b}_1|} \\ &&&= \frac{(\mathbf{a} - (\mathbf{a} \cdot \mathbf{c}) \mathbf{c}) \cdot (\mathbf{b} - (\mathbf{b} \cdot \mathbf{c}) \mathbf{c})}{\sin \alpha \sin \beta} \\ &&&= \frac{\mathbf{a} \cdot \mathbf{b} - 2(\mathbf{a} \cdot \mathbf{c}) \cdot (\mathbf{b} \cdot \mathbf{c})+(\mathbf{a} \cdot \mathbf{c}) \cdot (\mathbf{b} \cdot \mathbf{c})}{\sin \alpha \sin \beta} \\ &&&= \frac{\cos \theta - \cos \alpha \cos \beta}{\sin \alpha \sin \beta} \end{align*}
3. Note that \(\mathbf{m}_1 = \tfrac12(\mathbf{a}_1 + \mathbf{b}_1)\) either by expanding or by noting that projection is linear \begin{align*} && \cos \angle M_1OB_1 &= \frac{\mathbf{m}_1 \cdot \mathbf{b}_1}{|\mathbf{m}_1||\mathbf{b}_1|} \\ &&&= \frac{(\mathbf{a}_1 + \mathbf{b}_1) \cdot \mathbf{b}_1}{2|\mathbf{m}_1||\mathbf{b}_1|} \\ &&&= \frac{\mathbf{a}_1 \cdot \mathbf{b}_1 + |\mathbf{b}_1|^2}{2|\mathbf{m}_1||\mathbf{b}_1|} \\ &&&= \frac{|\mathbf{a}_1 || \mathbf{b}_1| \cos \phi + |\mathbf{b}_1|^2}{2|\mathbf{m}_1||\mathbf{b}_1|} \\ &&&= \frac{|\mathbf{a}_1 |\cos \phi + |\mathbf{b}_1|}{2|\mathbf{m}_1|} \\ &&&= \frac{\sin \alpha \cos \phi + \sin \beta}{2\sin \frac{\theta}{2}} \\ &&&= \frac{\sin \alpha \frac{\cos \theta - \cos \alpha \cos \beta}{\sin \alpha \sin \beta} + \sin \beta}{2\sin \frac{\theta}{2}} \\ &&&= \frac{\cos \theta - \cos \alpha \cos \beta+ \sin^2 \beta}{2\sin \frac{\theta}{2} \sin \beta} \\ \Rightarrow && \cos \angle M_1OA_1 &= \frac{\cos \theta - \cos \beta \cos \alpha+ \sin^2 \alpha}{2\sin \frac{\theta}{2} \sin \alpha} \end{align*} \(M_1\) is a bisector iff these two cosines are equal, ie \begin{align*} && \cos \angle M_1OB_1 &= \cos \angle M_1OA_1 \\ \Leftrightarrow && \frac{\cos \theta - \cos \alpha \cos \beta+ \sin^2 \beta}{2\sin \frac{\theta}{2} \sin \beta} &= \frac{\cos \theta - \cos \beta \cos \alpha+ \sin^2 \alpha}{2\sin \frac{\theta}{2} \sin \alpha} \\ \Leftrightarrow && \cos \theta (\sin \alpha - \sin \beta) &= \cos \alpha \cos \beta(\sin \alpha - \sin \beta) + \sin \alpha \sin \beta (\sin \alpha - \sin \beta) \\ \Leftrightarrow &&0 &= (\sin \alpha - \sin \beta)( \cos \theta - (\cos \alpha \cos \beta + \sin \alpha \sin \beta)) \\ &&&= (\sin \alpha - \sin \beta) (\cos \theta - \cos (\alpha - \beta)) \end{align*} From which the result immediately follows

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Solution:

1. The curves meet when they have the same radius for a given \(\theta\) ie \begin{align*} && a + 2 \cos \theta &= 2 + \cos 2 \theta \\ &&&= 2 + 2\cos^2 \theta - 1 \\ \Rightarrow && 0 &= 2 \cos ^2 \theta - 2 \cos \theta + 1 - a \end{align*} The curves touch if this has a repeated root, ie \(0 = \Delta = 4 - 8(1-a) \Rightarrow a = \frac12\). The second way the curves can touch is if there is a single root, but it's at an extreme value of \(\cos \theta = \pm 1\) ie \(0 = 2 - 2\cdot(\pm1) + 1 - a \Rightarrow a = 3 \pm 2 = 1, 5\)
2. Suppose \(a = \frac12\) then the curves touch when \(0 = 2\cos^2 \theta - 2 \cos \theta + \frac12 = (2 \cos \theta-1 )(\cos \theta -\frac12) \Rightarrow \theta = \pm \frac{\pi}{3}\)
   

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3. \(a = 1\)
   

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   \(a = 5\)
   

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Solution:

1. \(\,\) \begin{align*} && f_\alpha(x) &= \tan^{-1}\!\left(\frac{x\tan\alpha + 1}{\tan\alpha - x}\right) \\ && f'_\alpha(x) &= \frac{1}{1 + \left(\frac{x\tan\alpha + 1}{\tan\alpha - x}\right) ^2} \cdot \frac{\tan \alpha \cdot (\tan \alpha - x) - (x \tan \alpha + 1) \cdot (-1)}{(\tan \alpha - x)^2} \\ &&&= \frac{\tan^2 \alpha -1}{(\tan \alpha - x)^2 + (x \tan \alpha +1)^2} \\ &&&= \frac{\tan^2 \alpha +1}{\tan^2 \alpha - 2x \tan \alpha + x^2 + x^2 \tan^2 \alpha + 2 x \tan \alpha + 1} \\ &&&= \frac{1+\tan^2 \alpha}{(1+\tan^2 \alpha(x^2 + 1)} = \frac{1}{1+x^2} \end{align*}
   

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2. Let \(g(x) = \tanh^{-1}(\sin x) - \sinh^{-1}(\tan x)\) then \begin{align*} && g'(x) &= \frac{1}{1-\sin^2 x} \cdot \cos x - \frac{1}{\sqrt{\tan^2 +1}} \cdot \sec^2 x \\ &&&= \sec x - \frac{\sec^2 x}{|\sec x|} \\ &&& = \begin{cases} 0 &\text{if } \sec x \geq 0 \\ 2 \sec x &\text{ otherwise} \end{cases} \end{align*} Therefore \(g'(x) = 2\sec x\) if \(\tfrac12 \pi < x < \tfrac32\pi\) Therefore $\displaystyle g(x) = \begin{cases} 0 & \text{if } x \in [0, \frac{\pi}{2}] \cup [\frac{3\pi}{2}, 2\pi] \\ \ln( (\tan x + \sec x)^2) + C &\text{otherwise} \end{cases}$
   

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Solution:

1. \(\,\) \begin{align*} && z &= \frac{e^{i \theta} + e^{i \phi}}{e^{i \theta} - e^{i \phi}} \\ &&&= \frac{e^{i\frac12(\theta +\phi)}(e^{i \frac12(\theta-\phi)} + e^{-i\frac12(\theta- \phi)})}{e^{i\frac12(\theta +\phi)}(e^{i \frac12(\theta-\phi)} - e^{-i\frac12(\theta- \phi)})} \\ &&&= \frac{(e^{i \frac12(\theta-\phi)} + e^{-i\frac12(\theta- \phi)})/2}{i(e^{i \frac12(\theta-\phi)} - e^{-i\frac12(\theta- \phi)})/2i} \\ &&&= \frac{\cos \frac12(\theta-\phi)}{i \sin \frac12(\theta-\phi)} \\ &&&= -i \cot \tfrac12(\theta-\phi) \end{align*} Therefore \(|z| = \cot \tfrac12(\theta-\phi)\) and \(\arg z = \frac{\pi}{2}\)
2. Since \(a,b\) lie on the unit circle, wlog \(a = e^{i \theta}, b = e^{i\phi}\). Not that the line \(OX\) has vector \(a+b\) and \(AB\) has vector \(b-a\) and not their ratio has argument \(\frac{\pi}{2}\) and hence they are perpendicular.
3. \(AH\) has vector \(h - a = (a+b+c) - a = b+c\) which we've already established is perpendicular to \(c-b\) which is the vector for \(BC\) (unless \(b+c = 0\) in which case \(h = a\)).
4. \(p = a +b+c, q = b+c+d\) etc. The midpoint of \(AQ = \frac12(a+b+c+d)\) which is the same as the midpoint of \(BR\), \(CS\) and \(DP\). Therefore we could say the transformation is reflection in the point \(\frac12(a+b+c+d)\)

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Solution:

1. Claim \(x_n \geqslant 2 + 4^{n-1}(a-2)\) Proof: (By induction) Base case: Note that when \(n = 1\), \(x_1 = a = 2 + 1 \cdot(a - 2)\). Inductive step, suppose true for some \(n\), then \begin{align*} && x_{n+1} &= x_n^2 - 2 \\ &&&\geq (2+4^{n-1}(a-2))^2 - 2 \\ &&&= 4 + 4^{2n-2}(a-2)^2 + 4^n(a-2) - 2 \\ &&&= 2 + 4^{n}(a-2) + 4^{2n-2}(a-2)^2 \\ &&&\geq 2 + 4^{n+1-1}(a-2) \end{align*} as required,
2. (\(\Leftarrow\)) Suppose \(a > 2\) then \(x_n \geq 2+4^{n-1}(a-2) \to \infty\) as required. Suppose \(a < -2\) then \(x_2 > 4 -2 = 2\) so the sequence starting from \(x_2\) clearly diverges for the same reason. (\(\Rightarrow\)) suppose \(|x_n| \leq 2\) then \(x_{n+1} = x_n^2 - 2 \leq 2\) so the sequence is bounded and cannot tend to \(\infty\).
3. Suppose \(y_n = \frac{Ax_1x_2 \cdots x_n}{x_{n+1}}\) and notice that \(x_{n+1}^2 - 4 = (x_n^2 -2)^2 - 4 = x_n^4 - 4x_n^2 = x_n^2(x_n^2-4)\). In particular, \(\frac{\sqrt{x_{n+1}^2-4}}{x_{n+1}} = \frac{x_n\sqrt{x_n^2-4}}{x_{n+1}} = \frac{x_n x_{n-1} \cdots x_1 \sqrt{x_1^2-4}}{x_{n+1}}\) Therefore if \(A = \sqrt{a^2-4}\) \(y_{n+1} = \frac{\sqrt{x_{n+1}^2-4}}{x_{n+1}}\). Notice that \begin{align*} && y_n &= \frac{\sqrt{x_{n+1}^2-4}}{x_{n+1}} \\ &&&= \sqrt{1 - \frac{4}{x_{n+1}^2}} \to 1 \end{align*}

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Solution:

\begin{align*} && \frac{x}{\sin \phi} &= \frac{PR}{\sin 60^{\circ}} \\ && \frac{a-x}{\sin (120^{\circ}-\theta)} &= \frac{QP}{\sin 60^{\circ}} \\ && \frac{PR}{\sin \theta} &= \frac{QP}{\sin(120^{\circ}-\phi)} \\ \\ \Rightarrow && PR &= \frac{\sqrt3}2 \frac{x}{\sin \phi} \\ && QP &= \frac{\sqrt3}2 \frac{a-x}{\sin(120^{\circ}-\theta)} \\ \Rightarrow && \frac{\sqrt3}2 \frac{x}{\sin \phi} \frac{1}{\sin \theta} &= \frac{\sqrt3}2 \frac{a-x}{\sin(120^{\circ}-\theta)} \frac{1}{\sin(120^{\circ}-\phi)} \\ \Rightarrow && a-x &= \frac{(\frac{\sqrt3}2 \cos \phi + \frac12 \sin \phi)(\frac{\sqrt3}2 \cos \theta + \frac12 \sin \theta)}{\sin \phi \sin \theta} x \\ \Rightarrow && 4(a-x) &= (\sqrt3 \cot \phi + 1)(\sqrt 3 \cot \theta + 1) x \end{align*} Notice that \(e = \frac{\tan 60^{\circ}}{\tan (120^{\circ}-\theta)} = \frac{\tan \phi}{\tan 60^{\circ}}\) or \begin{align*} && \tan \phi &= \sqrt 3 e \\ && \frac{\sqrt3}{e} &= \tan(120^{\circ} - \theta) \\ &&&= \frac{-\sqrt3 - \tan \theta}{1 - \sqrt3 \tan \theta} \\ &&&= \frac{\sqrt3 \cot \theta +1}{\sqrt3-\cot \theta} \\ \Rightarrow && \sqrt3(-e-1)\cot \theta &= e-3 \\ \Rightarrow && \cot\theta &= \frac{3-e}{\sqrt{3}(1+e)} \\ \\ \Rightarrow && 4(a-x) &= (\sqrt3 \cot \phi + 1)(\sqrt 3 \cot \theta + 1) x \\ &&&= \left (\sqrt3 \frac{1}{\sqrt3 e} + 1 \right) \left (\sqrt 3\frac{3-e}{\sqrt{3}(1+e)}+1 \right) x \\ &&&= \frac{1+e}{e}\frac{3-e+1+e}{1+e} x \\ \Rightarrow && (a-x) &= \frac{1}{e}x \\ \Rightarrow && a &= \frac{1+e}{e}x \\ \Rightarrow && x &= \frac{ae}{1+e} \end{align*} The ball will continue to move around \(PQR\) if \(e \tan(120^{\circ} - \phi) = \tan \theta\) ie \begin{align*} && e \frac{-\sqrt3-\tan \phi}{1-\sqrt3 \tan \phi} &= \tan \theta \\ \Rightarrow && e \frac{\sqrt3 + \sqrt3 e}{3e-1} &= \frac{\sqrt3(1+e)}{3-e} \\ \Rightarrow && \frac{e}{3e-1} &= \frac{1}{3-e} \tag{\(e \neq -1\)} \\ \Rightarrow && 3e-e^2 &= 3e-1 \\ \Rightarrow && e^2 &= 1 \\ \Rightarrow && e &= 1 \end{align*}

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Solution:

1. The line to the circle is tangent, and the point it meets the circle is \((a \cos \theta, a \sin \theta)\) and it will be a distance \(b - a \theta\) away, therefore it is at \((a \cos \theta - (b-a \theta) \sin \theta, a \sin \theta + (b-a \theta) \cos \theta)\)
2. The velocity will be \(\displaystyle \binom{-a \dot{\theta}\sin \theta-b \dot{\theta}\cos \theta + a \dot{\theta} \sin \theta + a \theta \dot{\theta} \cos \theta}{ a \dot{\theta} \cos \theta - b \dot{\theta} \sin \theta -a \dot{\theta} \cos \theta + a \theta \dot{\theta} \sin \theta}= \binom{-b \dot{\theta}\cos \theta + a \theta \dot{\theta} \cos \theta}{ - b \dot{\theta} \sin \theta + a \theta \dot{\theta} \sin \theta}\) Therefore the speed will be \(\dot{\theta}(b-a\theta)\)
3. Conservation of energy and the fact that the tension is perpendicular to the velocity means no work is being done on the particle and hence it's speed is unchanged. So \(u = \dot{\theta}(b-a\theta)\).
4. Note that the acceleration is \begin{align*} && \mathbf{a} &= \frac{\d}{\d t} \left (-\dot{\theta}(b-a\theta) \binom{\cos \theta}{\sin \theta} \right) \\ &&&=-u \dot{\theta}\binom{-\sin \theta}{\cos \theta} \\ \Rightarrow && T &= ma \\ &&&= \frac{mu^2}{b - a \theta} \end{align*} It would be valuable to have \(\theta\) in terms of \(t\), so we want to solve \begin{align*} &&\frac{\d \theta}{\d t} (b-a\theta) &= u \\ \Rightarrow && b \theta - a\frac{\theta^2}{2} + C &= ut \\ t = 0, \theta = 0: && C &= 0 \\ \Rightarrow && b\theta - \frac{a}{2} \theta^2 &= ut \\ \Rightarrow && \theta &= \frac{b \pm \sqrt{b^2-2aut}}{a} \end{align*} At \(t\) increases, \(\theta\) increases so \(a\theta = b -\sqrt{b^2-2aut}\) or \(b-a \theta = \sqrt{b^2-2aut}\) and the result follows

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Solution:

1. \(\,\) \begin{align*} && \mathbb{P}(Y = n) &= \mathbb{P}(X \in [n, n+1)) \\ &&&= \int_n^{n+1} \lambda e^{-\lambda x} \d x \\ &&&= \left [-e^{-\lambda x} \right]_n^{n+1} \\ &&&= e^{-\lambda n} - e^{-\lambda(n+1)} \\ &&&= e^{-\lambda n}(1- e^{-\lambda}) \end{align*}
2. \(,\) \begin{align*} && \mathbb{P}(Z < z) &= \sum_{i=0}^{\infty} \mathbb{P}(X \in (n, n+z)) \\ &&&= \sum_{i=0}^{\infty} \int_{n}^{n+z} \lambda e^{-\lambda x} \d x \\ &&&= \sum_{i=0}^{\infty} [-e^{-\lambda x}]_{n}^{n+z} \\ &&&= \sum_{i=0}^{\infty} (1-e^{-\lambda x})e^{-\lambda n} \\ &&&= \frac{1-e^{-\lambda x}}{1-e^{-\lambda}} \end{align*}
3. Give the cdf of \(Z\), we see that \(f_Z(z) = \frac{\lambda e^{-\lambda z}}{1-e^{-\lambda}}\) so \begin{align*} && \E[Z] &= \int_0^1 z \frac{\lambda e^{-\lambda z}}{1-e^{-\lambda}} \d z \\ &&&= \frac{\lambda}{1-e^{-\lambda}} \int_0^1 ze^{-\lambda z} \d z \\ &&&= \frac{\lambda}{1-e^{-\lambda}} \left ( \left [-\frac{1}{\lambda} ze^{-\lambda z} \right]_0^1+\int_0^1 \frac{1}{\lambda} e^{-\lambda z} \d z \right) \\ &&&= \frac{\lambda}{1-e^{-\lambda}} \left ( -\frac{e^{-\lambda}}{\lambda} + \frac{1-e^{-\lambda}}{\lambda^2} \right) \\ &&&= \frac{1-e^{-\lambda}(1+\lambda)}{\lambda (1-e^{-\lambda})} \end{align*}
4. \(\,\) \begin{align*} && \mathbb{P}(Y = n \text{ and }z_1 < Z < z_2)&= \mathbb{P}(X \in (n+z_1, n+z_2) ) \\ &&&= \int_{n+z_1}^{n+z_2} \lambda e^{-\lambda x} \d x \\ &&&= e^{-n\lambda}(e^{-\lambda z_1} - e^{-\lambda z_2}) \end{align*} Note that \(\mathbb{P}(z_1 < Z < z_2) = \mathbb{P}( Z < z_2) -\mathbb{P}(Z< z_1) =\frac{e^{-\lambda z_1} - e^{-\lambda z_2}}{1-e^{-\lambda}}\) Therefore \begin{align*} && \mathbb{P}(Y = n \text{ and }z_1 < Z < z_2) &= e^{-n\lambda}(e^{-\lambda z_1} - e^{-\lambda z_2}) \\ &&&= e^{-\lambda n}(1-e^{-\lambda}) \frac{e^{-\lambda z_1} - e^{-\lambda z_2}}{1-e^{-\lambda}} \\ &&&= \mathbb{P}(Y=n) \mathbb{P}(z_1 < Z < z_2) \end{align*} So they are independent, which is to be expected from the memorylessness property of the exponential distribution.

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Solution:

1. First note that \(\mathbb{P}(X_{ij} = 1) = \frac16\) since it doesn't matter what \(i\) rolls, it only matters that \(j\) rolls the same thing, which happens \(1/6\) of the time. \begin{align*} && \mathbb{P}(X_{12} = 1, X_{23} = 1) &= \mathbb{P}(1, 2\text{ and }3\text{ all roll the same})\\ &&&= \frac{6}{6^3}= \frac1{6^2} \\ &&&= \mathbb{P}(X_{12} = 1)\mathbb{P}(X_{23} = 1) \\ && \mathbb{P}(X_{12} = 1, X_{23} = 0) &= \mathbb{P}(1, 2\text{ roll the same and }3\text{ rolls different}) \\ &&&= \frac{6 \cdot 1 \cdot 5}{6^3} = \frac{5}{6^2} \\ &&&= \mathbb{P}(X_{12} = 1)\mathbb{P}(X_{23} = 0) \\ && \mathbb{P}(X_{12} = 0, X_{23} = 0) &= \mathbb{P}(2, 3 \text{ roll different to} 2)\\ &&&= \frac{6 \cdot 5 \cdot 5}{6^3}= \frac{5^2}{6^2} \\ &&&= \mathbb{P}(X_{12} = 0)\mathbb{P}(X_{23} = 0) \end{align*} Therefore they are independent (the final case is clear by symmetry from case 2). Note that the score is \(S = \sum_{i \neq j} X_{ij}\) so \begin{align*} && \E[S] &= \E \left [ \sum_{i \neq j} X_{ij} \right] \\ &&&= \sum_{i \neq j} \E \left [ X_{ij} \right] \\ &&&= \sum_{i \neq j} \frac16 \\ &&&= \binom{n}{2} \frac16 = \frac{n(n-1)}{12} \\ \\ && \var[S] &= \var \left [ \sum_{i \neq j} X_{ij} \right] \\ &&& \sum_{i \neq j} \var \left [X_{ij} \right] \tag{pairwise ind.} \\ &&&= \binom{n}{2} \frac{5}{36} = \frac{5n(n-1)}{72} \end{align*}
2. Note that \(\mathbb{P}(Z_{ij} = 1)=\mathbb{P}(Z_{ij} = -1) = \frac{3}{6^2} = \frac{1}{12}\) but that \(\mathbb{P}(Z_{12} = 1, Z_{23} = -1) = 0\). Notice that \(Z_{12}Z_{23}\) is either \(1\) or \(0\) (since \(2\) can't be both odd and even). \(\mathbb{P}(Z_{12}Z_{23} = 1) = \frac{1}{36}\). Notice that \(Z_{ij}, Z_{kl}\) are independent if \(i \neq j \neq k \neq l\) and so \begin{align*} && \E[T] &= \E \left [ \sum_{i \neq j} Z_{ij} \right] \\ &&&= \sum_{i \neq j}\E \left [ Z_{ij} \right] \\ &&&= 0 \\ \\ && \E[T^2] &= \E \left [ \left ( \sum_{i \neq j} Z_{ij} \right)^2 \right] \\ &&&= \E \left [ \sum_{i \neq j} Z_{ij}^2 + \sum_{i \neq j \neq k} Z_{ij}Z_{jk} + \sum_{i \neq j \neq k \neq l} Z_{ij}Z_{kl}\right] \\ &&&= \binom{n}{2} \frac{1}{6} + 2\frac{n(n-1)(n-2)}{2} \frac{1}{36} + 0 \\ &&&= \frac{n(n-1)}{12} + \frac{n(n-1)(n-2)}{6} \\ &&&= \frac{n(n-1)[3 + (n-2)]}{36} \\ &&&= \frac{n(n^2-1)}{36} \end{align*}