Problems

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Solution:

1. Let \(A = (0,0,0)\) and then \(B = 22b \mathbf{i}, D = 2d\mathbf{j}, C = 2b\mathbf{i}+2d\mathbf{j}\) and \(V = b \mathbf{i} + d\mathbf{j} + h\mathbf{k}\) We also have \begin{align*} && \tan \alpha &= \frac{h}{d}\\ && \tan \beta &= \frac{d}{b} \\ && \vec{AV} \times \vec{VB} &= \begin{pmatrix} b \\ d \\ h \end{pmatrix} \times \begin{pmatrix} -b \\ d \\ h \end{pmatrix} \\ &&&= \begin{pmatrix} 0 \\ -2bh \\ 2db \end{pmatrix} \\ &&&= 2b \begin{pmatrix} 0 \\ -d \tan \alpha \\ d \end{pmatrix} \\ &&&= k \begin{pmatrix} 0 \\ - \sin \alpha \\ \cos \alpha \end{pmatrix} \end{align*} similarly for the vector perpendicular to the other face it must be \(\begin{pmatrix}-\sin \beta \\ 0 \\ \cos \beta \end{pmatrix}\) Looking at the angle between these perpendicular (to find the angles between the faces we see: \begin{align*} \begin{pmatrix} 0 \\ - \sin \alpha \\ \cos \alpha \end{pmatrix} \cdot \begin{pmatrix}-\sin \beta \\ 0 \\ \cos \beta \end{pmatrix} &= \cos \alpha \cos \beta \end{align*} But this is also \(\pi -\) the angle between the planes, ie \(\cos \theta = \cos \alpha \cos \beta\)
2. \(\,\) \begin{align*} && \cot^2 \phi &= \frac{b^2+d^2}{h^2} \\ && \cot^2 \alpha &= \frac{d^2}{h^2} \\ && \cot^2 \beta &= \frac{b^2}{h^2} \\ \Rightarrow && cot^2 \phi &= \cot^2 \beta+\cot^2 \alpha \end{align*} \begin{align*} && \cos^2 \phi &= \frac{b^2+d^2}{b^2+d^2+h^2} \\ && \cos^2 \alpha &= \frac{d^2}{d^2+h^2} \\ && \cos^2 \beta &= \frac{b^2}{b^2+h^2} \\ && \frac{\cos^2 \alpha + \cos^2 \beta - 2 \cos^2 \theta}{1-\cos^2 \theta} &= \frac{\frac{d^2}{d^2+h^2}+\frac{b^2}{b^2+h^2}-2\cdot \frac{d^2}{d^2+h^2} \cdot \frac{b^2}{b^2+h^2}}{1 - \frac{d^2}{d^2+h^2} \cdot\frac{b^2}{b^2+h^2}} \\ &&&= \frac{d^2(b^2+h^2)+b^2(d^2+h^2)-2d^2b^2}{(d^2+h^2)(b^2+h^2)-d^2b^2} \\ &&&= \frac{h^2(b^2+d^2)}{h^2(b^2+d^2+h^2)} \\ &&&= \frac{b^2+d^2}{b^2+d^2+h^2} \\ &&&= \cos^2\phi \end{align*} Also notice that \begin{align*} && \cos^2 \alpha + \cos^2 \beta &\underbrace{\geq}_{AM-GM} 2 \cos \alpha \cos \beta \\ &&&= 2 \cos \theta \\ \Rightarrow && \frac{\cos^2 \alpha + \cos^2 \beta - 2 \cos^2 \theta}{1-\cos^2 \theta} &\geq \frac{2 \cos \theta - 2\cos^2 \theta}{1-\cos^2 \theta} \\ &&&= \frac{2\cos \theta}{1+\cos \theta} = \cos \theta \frac{2}{1+\cos \theta} \\ &&&> \cos^2 \theta \\ \Rightarrow && \phi &< \theta \end{align*}

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Solution: Note that \begin{align*} && \frac{\d a \sec \theta}{\d \theta} &= a \sec \theta \tan \theta \\ && \frac{\d b \tan \theta}{\d \theta} &= b \sec^2 \theta \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{b \sec^2 \theta}{a \sec \theta \tan \theta} \\ &&&= \frac{b}{a} \frac{1}{\sin \theta} \\ \Rightarrow && \frac{y-b \tan \theta}{x - a \sec \theta} &= \frac{b}{a} \frac{1}{\sin \theta} \\ \Rightarrow && a \sin \theta y - ab \tan \theta \sin \theta &= bx -ab \sec \theta \\ \Rightarrow && bx-ay\sin \theta &= ab \sec x (1 - \sin ^2 \theta) \\ &&&= ab \cos \theta \end{align*}

1. \begin{align*} S: &&& \begin{cases} bx-ay &= 0 \\ bx-ay \sin \theta &= ab \cos \theta \end{cases} \\ \Rightarrow && ay(1-\sin \theta) &= ab \cos \theta \\ \Rightarrow && y &= \frac{b \cos \theta}{1-\sin \theta} \\ &&x &=\frac{a\cos \theta}{1-\sin \theta} \\ T: &&& \begin{cases} bx+ay &= 0 \\ bx-ay \sin \theta &= ab \cos \theta \end{cases} \\ \Rightarrow && ay(1+\sin \theta) &= -ab \cos \theta \\ \Rightarrow && y &= \frac{-b \cos \theta}{1+\sin \theta} \\ &&x &=\frac{a\cos \theta}{1+\sin \theta} \\ M: && x &= \frac{a \cos \theta}{2} \frac{2}{1-\sin^2 \theta} \\ &&&= a \sec \theta \\ && y &= \frac{b \cos \theta}{2} \frac{2 \sin \theta}{1-\sin^2 \theta} \\ &&&= b \tan \theta \end{align*} The midpoint of \(ST\) is the same as \(P\).
2. The tangents are perpendicular, therefore \(\frac{b}{a} \cosec \theta = - \frac{a}{b} \sin \phi\), ie \(b^2 = -a^2 \sin \phi \sin \theta\) The will intersect at: \begin{align*} &&& \begin{cases} bx - ay \sin \theta &= ab \cos \theta \\ bx - ay \sin \phi &= ab \cos \phi \end{cases} \\ \Rightarrow && ay ( \sin \theta - \sin \phi) &= ab(\cos \phi - \cos \theta) \\ \Rightarrow && y &= \frac{b(\cos \phi - \cos \theta)}{(\sin \theta - \sin \phi)} \\ && y^2 &= \frac{-a^2 \sin \phi \sin \theta (\cos\phi - \cos \theta)^2}{(\sin \theta - \sin \phi)^2} \\ \Rightarrow && bx(\sin \phi - \sin \theta) &= ab(\cos \theta \sin \phi - \cos \phi \sin \theta) \\ \Rightarrow && x &= \frac{a(\cos \theta \sin \phi - \cos \phi \sin \theta)}{\sin \phi - \sin \theta} \\ &&&= \frac{a^2(\cos \theta \sin \phi - \cos \phi \sin \theta)^2}{(\sin \phi - \sin \theta)^2} \end{align*} Therefore \begin{align*} && x^2+y^2 &= \frac{a^2}{(\sin \phi - \sin \theta)^2} \l (\cos \theta \sin \phi- \cos \phi \sin \theta)^2 - \sin \phi \sin \theta (\cos\phi - \cos \theta)^2 \r \\ &&&= \frac{a^2}{(\sin \phi - \sin \theta)^2} \l (\sin \phi - \sin \theta)(\cos^2 \theta \sin \phi - \sin \theta \cos^2 \phi) \r \\ &&&=a^2-b^2 \end{align*}

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Solution: The line through \(A\) perpendicular to \(BC\) is \(\mathbf{a} + \lambda\mathbf{u}\). The line through \(B\) perpendicular to \(CA\) is \(\mathbf{b} + \mu \mathbf{v}\). They intersect when \(\mathbf{a} + \lambda\mathbf{u} = \mathbf{b} + \mu \mathbf{v}\). Since \(\mathbf{v}\) is perpendicular to \(CA\), we must have \begin{align*} && \mathbf{a} + \lambda\mathbf{u} &= \mathbf{b} + \mu \mathbf{v} \\ \Rightarrow && \mathbf{a}\cdot(\mathbf{c}-\mathbf{a}) + \lambda\mathbf{u}\cdot(\mathbf{c}-\mathbf{a}) &= \mathbf{b}\cdot(\mathbf{c}-\mathbf{a}) + \mu \mathbf{v}\cdot(\mathbf{c}-\mathbf{a}) \\ \\ \Rightarrow && \lambda &= \frac{\mathbf{b}\cdot(\mathbf{c}-\mathbf{a}) -\mathbf{a}\cdot(\mathbf{c}-\mathbf{a})}{\mathbf{u}\cdot(\mathbf{c}-\mathbf{a})} \\ &&&= \frac{(\mathbf{b}-\mathbf{a})\cdot(\mathbf{c}-\mathbf{a})}{\mathbf{u} \cdot(\mathbf{c}-\mathbf{a})} \end{align*} Therefore the point is \(\mathbf{a} + \frac{(\mathbf{b}-\mathbf{a})\cdot(\mathbf{c}-\mathbf{a})}{\mathbf{u} \cdot(\mathbf{c}-\mathbf{a})} \mathbf{u}\). The line \(CP\) is \(\mathbf{c} + \nu \left (\mathbf{p} - \mathbf{c} \right)\), to check this is perpendicular with \(AB\) we should dot \(\mathbf{p}-\mathbf{c}\) with \(\mathbf{a}-\mathbf{b}\), ie \begin{align*} && (\mathbf{p}-\mathbf{c}) \cdot (\mathbf{a}-\mathbf{b}) &= \left ( \mathbf{a} + \frac{(\mathbf{b}-\mathbf{a})\cdot(\mathbf{c}-\mathbf{a})}{\mathbf{u} \cdot(\mathbf{c}-\mathbf{a})} \mathbf{u} - \mathbf{c}\right) \cdot ( \mathbf{a}-\mathbf{b}) \\ &&&= \left ( \mathbf{a}- \mathbf{c} + \frac{(\mathbf{b}-\mathbf{a})\cdot(\mathbf{c}-\mathbf{a})}{\mathbf{u} \cdot(\mathbf{c}-\mathbf{a})} \mathbf{u} \right) \cdot ( \mathbf{a}-\mathbf{c}+(\mathbf{c}-\mathbf{b})) \\ &&&= (\mathbf{a}-\mathbf{c})\cdot(\mathbf{a}-\mathbf{c}) + \frac{(\mathbf{b}-\mathbf{a})\cdot(\mathbf{c}-\mathbf{a})}{\mathbf{u} \cdot(\mathbf{c}-\mathbf{a})}\mathbf{u} \cdot (\mathbf{a}-\mathbf{c}) + \\ &&&\quad (\mathbf{a}-\mathbf{c})\cdot(\mathbf{c}-\mathbf{b}) + \lambda \underbrace{\mathbf{u} \cdot (\mathbf{c}-\mathbf{b})}_{=0} \\ &&&=(\mathbf{a}-\mathbf{c})\cdot(\mathbf{a}-\mathbf{c}) -(\mathbf{b}-\mathbf{a})\cdot(\mathbf{c}-\mathbf{a})+ (\mathbf{a}-\mathbf{c})\cdot(\mathbf{c}-\mathbf{b}) \\ &&&= (\mathbf{a}-\mathbf{c})\cdot(\mathbf{a}-\mathbf{c}+\mathbf{b}-\mathbf{a}+\mathbf{c}-\mathbf{b}) \\ &&&= 0 \end{align*} as required.

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Solution: \(\angle POQ = 90^\circ\) means that if \(P(p^2,p^3)\) and \(Q(q^2,q^3)\) are our points then \(OP^2+OQ^2 = PQ^2\), so \begin{align*} && p^4+p^6+q^4+q^6 &= (p^2-q^2)^2+(p^3-q^3)^2 \\ &&&= p^4+q^4-2p^2q^2+p^6+q^6-2p^3q^3 \\ \Rightarrow && 0 &= 2p^2q^2(1+pq) \\ \Rightarrow && pq &= -1 \\ \\ && \frac{\d y}{ \d x} &= \frac{\frac{\d y }{\d t}}{\frac{\d x}{\d t}} \\ &&&= \frac{3t^2}{2t} = \tfrac32t \\ \Rightarrow && \frac{y-p^3}{x-p^2} &= \tfrac32p \\ \Rightarrow && 2(y-p^3) &=3p(x-p^2) \\ && 2(y-q^3) &=3q(x-q^2) \\ \Rightarrow && 2(q^3-p^3) &= (3p-3q)x+3(q^3-p^3) \\ && p^3-q^3 &= 3(p-q)x \\ \Rightarrow && x &= \tfrac13(p^2+q^2+pq) \\ && 2y &= 3p(\tfrac13(p^2+q^2+pq)-p^2)+2p^3 \\ &&&= p(p^2+q^2+pq)-p^3 \\ &&&= pq^2+p^2q \\ &&&= -p-q \\ &&y&= -\frac{p+q}{2} \\ \\ && 4y^2 &= p^2+q^2 \\ && 3x-1 &= p^2+q^2 \\ \end{align*} To check if they meet, try \(4t^6=3t^2 - 1\). Consider \(y = 4x^3-3x+1\) \(y(0) = 1\) and \(y' = 12x^2-3 = 3(4x^2-1)\) which has roots at \(\pm \tfrac12\), therefore we need to test \(y(\tfrac12) = \tfrac12-\tfrac32 + 1 = 0\), so there is a one intersection at \(x = \tfrac1{2}, y = \tfrac1{2\sqrt{2}}\)

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Solution: \begin{align*} && {\bf r} &=\left( \e^{t}\cos t \right) {\bf i}+ \left(\e^t \sin t\right) {\bf j} \\ \Rightarrow && \dot{\bf r} &= \left( \e^{t}\cos t -\e^t \sin t\right) {\bf i}+ \left(\e^t \sin t+\e^t \cos t\right) {\bf j} \\ \Rightarrow && \mathbf{r}\cdot\dot{ \mathbf{r}} &= e^{2t}(\cos^2 t - \sin t \cos t) + e^{2t}(\sin^2 t+ \sin t \cos t) \\ &&&= e^{2t} (\cos^2 t + \sin ^2 t)\\ &&&= e^{2t} \\ \\ && | {\bf r}| &= e^{t} \\ && |{\bf \dot{r}}| &= e^t \sqrt{(\cos t - \sin t)^2 + (\sin t + \cos t)^2} \\ &&&= e^t \sqrt{2 \cos^2 t + 2 \sin^2 t} \\ &&&= \sqrt{2} e^t \\ \\ \Rightarrow && \frac{\mathbf{r}\cdot\dot{ \mathbf{r}}}{ |{\bf {r}}| |{\bf \dot{r}}|} &= \frac{e^{2t}}{\sqrt{2}e^te^t} \\ &&&= \frac{1}{\sqrt{2}} \end{align*} Therefore the angle between the velocity and displacement is \(\frac{\pi}{4}\). \begin{align*} && \ddot{\bf{r}} &= \left( \e^{t}(\cos t - \sin t) - \e^t (\sin t + \cos t)\right) {\bf i}+ \left(\e^t (\sin t + \cos t) + \e^t(\cos t - \sin t)\right) {\bf j} \\ &&&= \left ( -2\e^{t} \sin t \right) {\bf i}+ \left ( 2\e^{t} \cos t \right) {\bf j} \\ \Rightarrow && {\bf r} \cdot \ddot{\bf{r}} &= 2e^{2t} \left ( -\sin t \cos t + \sin t \cos t \right) \\ &&&= 0 \end{align*} Therefore the acceleration is perpendicular.

\(Q\) has position $\mathbf{r}' = \left( \e^{t-T}\cos (t-T) \right) {\bf i}+ \left(\e^{t-T} \sin (t-T)\right) {\bf j}\( for \)t > T$. \begin{align*} && {\bf r' \cdot r} &= e^{2t-T} \left (\cos t \cos (t-T) + \sin t \sin(t - T) \right) \\ &&&= e^{2t-T} \cos (t - (t-T)) \\ &&&= e^{2t-T} \cos T \\ \\ && |{\bf r'}- {\bf r} |^2 &= |{\bf r}|^2 + |{\bf r}'|^2 - 2 {\bf r' \cdot r} \\ &&&= e^{2t} + e^{2(t-T)} - 2e^{2t-T} \cos T \\ &&&= e^{2t} \left (1 - 2e^{-T} \cos T + e^{-2T} \right) \\ \Rightarrow && |{\bf r'}- {\bf r} | &= e^{t} \sqrt{1 - 2e^{-T} \cos T + e^{-2T} } \end{align*} as required

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Solution:

Note that \(CP\) has equation \(\frac{y-0}{x-an} = \frac{a-0}{a-an} = \frac{1}{1-n} \Rightarrow y = \frac{x-an}{1-n}\) Therefore \(R = (0, -\frac{an}{1-n})\) Note that \(CQ\) has equation \(\frac{y-am}{x} = \frac{a-am}{a} = 1-m \Rightarrow y = (1-m)x + am\) Therefore \(S = (-\frac{am}{1-m}, 0)\) \(PQ\) has equation \(\frac{y}{x-an} = \frac{am-0}{0-an} \Rightarrow y = -\frac{m}{n}x +am\) \(SR\) has equation \(\frac{y}{x+\frac{am}{1-m}} = \frac{-\frac{an}{1-n}}{\frac{am}{1-m}} = -\frac{n(1-m)}{m(1-n)} \Rightarrow y =-\frac{n(1-m)}{m(1-n)} x -a\frac{n}{1-n}\) So \(PQ \cap SR\) has \begin{align*} && -\frac{m}{n}x +am &= -\frac{n(1-m)}{m(1-n)} x -a\frac{n}{1-n} \\ && x \left (\frac{n(1-m)}{m(1-n)} - \frac{m}{n} \right) &= -am - \frac{an}{1-n} \\ \Rightarrow && x \left ( \frac{n^2(1-m)-m^2(1-n)}{nm(1-n)} \right) &= -\frac{a(m(1-n)+n)}{1-n} \\ \Rightarrow && x \left ( \frac{(m-n)(mn-n-m)}{mn(1-n)} \right) &= \frac{a(mn-m-n)}{1-n} \\ \Rightarrow && x &= \frac{amn}{m-n} \\ && y &= -\frac{amn}{m-n} \end{align*} Therefore clearly \(TA\) is perpendicular to \(AC\) since they are the lines \(y = -x\) and \(y = x\) Given this method we can construct the perpendicular to the diagonal through the vertex. Doing this at \(A\) we can construct \(C'\) the reflection of \(C\) in \(AB\). We can do the same to find the reflection of \(A\) and so we have a square with sidelengths \(\sqrt{2}a\) and hence area \(2a^2\)

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Solution:

1. It is at the origin when both \(\sin 2t\) and \(\cos t = 0\), but this \(\sin 2t = 2 \sin t \cos t\) so this happens precisely when \(\cos t = 0\), ie when \(t = \frac{\pi}{2}, \frac{3\pi}{2}\)
2. \(\,\) \begin{align*} && \dot{\mathbf{r}} &= 2 \cos 2t \mathbf{i} - 2 \sin t \mathbf{j} \\ && \mathbf{r} \cdot \dot{\mathbf{r}} &= 2\cos 2t \sin 2t - 2 \sin t 2 \cos t \\ &&&= \sin 2t \left (2\cos 2t - 2 \right) \end{align*} Therefore they are perpendicular when \(\sin 2t = 0 \Rightarrow t = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi\) and when \(\cos 2t = 1 \Rightarrow 2t = 0, 2\pi, 4\pi \Rightarrow t = 0, \pi, 2\pi\), therefore all solutions are \( t = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi\)
3. For \(\mathbf{r}\) and \(\dot{\mathbf{r}}\) to be parallel, we would need \begin{align*} && \frac{2 \cos 2t}{\sin 2t} &= \frac{-2 \sin t}{2 \cos t}\\ && 2 \cos 2t \cos t &= - \sin t \sin 2t \\ && 0 &= 2\cos t (\cos 2t + \sin ^2 t) \\ &&&= 2 \cos t (\cos^2 t) \\ &&&= 2 \cos^3 t \end{align*} Therefore the only time we can be parallel is when \(\cos t = 0\), which is when we are at the origin.
4. \(\frac{\d }{\d t} (\mathbf{r} \cdot \mathbf{r}) = 2 \mathbf{r} \cdot \mathbf{\dot{r}}\) so we should check the values when velocity and displacement are perpendicular, ie \( t = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi\) which have values \(\mathbf{r} = \binom{0}{2}, \binom{0}{0}, \binom{0}{-2}, \binom{0}{0}, \binom{0}{2}\). Therefore the maximum distance is \(2\).
   

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Solution: Suppose \({\bf p} = \lambda {\bf a}\) and \({\bf p} + {\bf q} = {\bf a} + {\bf b}\) then \begin{align*} {\bf a} \cdot : && {\bf a} \cdot {\bf p} + {\bf a} \cdot {\bf p} &= {\bf a} \cdot {\bf a} + {\bf a} \cdot {\bf b} \\ && \lambda + 0 &= 1 + 3 = 4 \\ \Rightarrow && \mathbf{p} &= 4 \mathbf{a} \\ && \mathbf{q} &= \mathbf{b} - 3\mathbf{a} \\ \\ && \mathbf{P} &= 4p\mathbf{a} \\ && \mathbf{Q} &= q\mathbf{b} - 3q\mathbf{a} \\ \\ \mathbf{a} \cdot : && \mathbf{a} \cdot \mathbf{P} + \mathbf{a} \cdot \mathbf{Q} + \mathbf{a} \cdot \mathbf{R} &= \mathbf{a} \cdot \mathbf{a} + \mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c} \\ && 4p &= 1+3-2 \\ \Rightarrow && p &= \tfrac12 \\ \\ && {\bf P} + {\bf Q} + {\bf R} &= {\bf a}+{\bf b}+{\bf c} \\ \mathbf{b} \cdot : && \mathbf{b} \cdot \mathbf{P} + \mathbf{b} \cdot \mathbf{Q} + \mathbf{b} \cdot \mathbf{R} &= \mathbf{b} \cdot \mathbf{a} + \mathbf{b} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{c} \\ && 12p + 25q - 9q &= 3+25+2 \\ && 6+16q &= 30 \\ \Rightarrow && q &= \tfrac{3}{2}\\ && \\ && \mathbf{P} &= 2\mathbf{a} \\ && \mathbf{Q} &= \tfrac32 \mathbf{b} - \tfrac92 \mathbf{a} \\ && \mathbf{R} &= \tfrac72\mathbf{a} -\tfrac12 \mathbf{b} + \mathbf{c} \end{align*}

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Solution: The line \(OM\) is \(\lambda \begin{pmatrix} a \\ b \\ c \end{pmatrix}\), then we need \(1 = \lambda \begin{pmatrix} a \\ b \\ c \end{pmatrix} \cdot \begin{pmatrix} \frac1{a} \\ \frac1b \\ \frac1c \end{pmatrix} = 3 \lambda \Rightarrow \lambda = \frac13\). Therefore \(OM\) meets the plane at the centroid. The orthocentre is the point \(\mathbf{h}\) such that \((\mathbf{a}-\mathbf{b}) \cdot (\mathbf{c} - \mathbf{h}) = 0 \Leftrightarrow \begin{pmatrix} a \\ -b \\ 0 \end{pmatrix} \cdot \begin{pmatrix} -p \\ -q \\ c-r \end{pmatrix} \Leftrightarrow ap-bq = 0\) \((\mathbf{b}-\mathbf{c}) \cdot (\mathbf{a} - \mathbf{h}) = 0 \Leftrightarrow \begin{pmatrix} 0 \\ b \\ -c \end{pmatrix} \cdot \begin{pmatrix} a-p \\ -q \\ -r \end{pmatrix} \Leftrightarrow bq-cr = 0\) \((\mathbf{c}-\mathbf{a}) \cdot (\mathbf{b} - \mathbf{h}) = 0 \Leftrightarrow \begin{pmatrix} -a \\ 0 \\ c \end{pmatrix} \cdot \begin{pmatrix} -p \\ b-q \\ -r \end{pmatrix} \Leftrightarrow cr-ap = 0\) ie \(ap = bq = cr\) but this is clearly on the line \(\lambda \begin{pmatrix} \frac1{a} \\ \frac1b \\ \frac1c \end{pmatrix}\) therefore the orthocentre is on the perpendicular from \(O\) \(M-A = \begin{pmatrix} -a/2 \\ b/2 \\ c/2 \end{pmatrix}\) so \(|M-A|=|M-B|=|M-C|\) Also by pythagoras the point of intersection satisfies \(|M-P|^2 + |P-A|^2 = |M-A|^2\) so \(|P-A|^2 = |P-B|^2 = |P-C|^2\), therefore \(P\) is the circumcentre. Since all these points are in the same plane and \(OGM\) is a line, we have the points are in a line. Similar triangles gives the desired ratio

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Solution:

1. The line \(PO\) has gradient \(\frac{p^2}{p} = p\) and teh line \(QO\) has gradient \(q\), therefore we must have that \(pq = -1\). Therefore, \(R\) is the point \begin{align*} && R &= \left ( \frac{p-\frac{1}{p}}{2}, \frac{p^2+\frac{1}{p^2}}{2} \right) \\ &&&= \left ( \frac12\left ( p - \frac{1}{p} \right),2\left (\frac12 \left(p-\frac{1}{p}\right) \right)^2+1 \right) \\ &&&= \left ( t, 2t^2+1\right) \end{align*} So we are looking at another parabola.
   

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2. The tangents are \(y = 2px+c\), ie \(p^2 = 2p^2+c\), ie \(y = 2px -p^2\) so we have \begin{align*} && y - 2px &= -p^2 \\ && y - 2qx &= -q^2 \\ \Rightarrow && (2p-2q)x &= p^2-q^2 \\ \Rightarrow && x &= \frac12 (p+q)\\ && y &= p(p+q)-p^2 \\ && y &= pq = -1 \end{align*} Therefore \(x = \frac12(p - \frac1p), y= -1\), so we have the line \(y = -1\) (the directrix)
   

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Solution: The perpendicular bisector of the chords runs through the origin, therefore \(LM\) is perpendicular to \(PQ\) if and only if \(\frac{l+m}{2}\) is perpendicular to \(\frac{p+q}{2}\), ie \begin{align*} && (l+m) &= it (p+q) \\ \Leftrightarrow && \frac{l+m}{p+q} & \in i \mathbb{R} \\ \Leftrightarrow && 0 &= \frac{l+m}{p+q} + \frac{l^*+m^*}{p^*+q^*} \\ &&&= \frac{l+m}{p+q} + \frac{\frac{r^2}{l}+\frac{r^2}{m}}{\frac{r^2}{p}+\frac{r^2}{q}} \\ &&&=\frac{l+m}{p+q} + \frac{l+m}{p+q} \frac{pq}{lm} \\ &&&= \frac{l+m}{p+q} \left ( \frac{lm+pq}{lm} \right) \end{align*} Therefore as long as \(l+m, p+q \neq 0\) \(lm+pq = 0\) is equivalent to the chords being perpendicular. In the case where (say) \(l,m\) is a diameter, then the condition for the chords to be perpendicular is that \(p,q\) is also a diameter and at right angles, but clearly this is also equivalent to our condition. Suppose \(A_1, A_2, A_3\) are distinct points on \(S\), and \(A_1'\) is given and suppose \(a_i, a_i'\) are the corresponding complex numbers, then the conditions are: \begin{align*} A_1'A_2' \perp A_1A_2: && 0 &= a_1'a_2' + a_1a_2 \\ A_2'A_3' \perp A_2A_3: && 0 &= a_2'a_3' + a_2a_3 \\ A_3'A_1'' \perp A_3A_1: && 0 &= a_3'a_1'' + a_3a_1 \\ \\ \Rightarrow && a_2' &= -\frac{a_1a_2}{a_1'} \\ && a_3' &= -\frac{a_2a_3}{a_2'} \\ &&&= \frac{a_1'a_2a_3}{a_1a_2} \\ &&&= \frac{a_1'a_3}{a_1} \\ && a_1'' &= - \frac{a_3a_1}{a_3'} \\ &&&= \frac{a_3a_1a_1}{a_1'a_3} \\ &&&= \frac{a_1^2}{a_1'} \\ \Rightarrow && a_1'a_1'' &= a_1^2 \end{align*} Therefore \(a_1' = a_1''\) if \(a_1' = \pm a_1\) Suppose we have \(4\) points, then \begin{align*} A_1'A_2' \perp A_1A_2: && 0 &= a_1'a_2' + a_1a_2 \\ A_2'A_3' \perp A_2A_3: && 0 &= a_2'a_3' + a_2a_3 \\ A_3'A_4' \perp A_3A_4: && 0 &= a_3'a_4' + a_3a_4 \\ A_4'A_1'' \perp A_4A_1: && 0 &= a_4'a_1'' + a_4a_1 \\ \\ \Rightarrow && a_4' &= -\frac{a_3a_4}{a_3'} \\ &&&= -\frac{a_1a_3a_4}{a_1'a_3} \\ &&&= -\frac{a_1a_4}{a_1'} \\ \Rightarrow && a_1'' &= -\frac{a_4a_1}{a_4'} \\ &&&= \frac{a_4a_1a_1'}{a_1a_4} \\ &&&= a_1' \end{align*} So they coincide. For \(n\) points if there are an even number of points they coincide, an odd number and there are two points when they coincide.