Problems

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Solution:

1. \(A\), \(Q\), \(C\) lie on a straight line if \(q = \lambda a + (1-\lambda)c\) for some \(\lambda \in \mathbb{R}\), \begin{align*} && q &= \lambda a + (1-\lambda)c \\ \Leftrightarrow && q - a &= (\lambda - 1)a + (1-\lambda)c \\ \Leftrightarrow && q - a &= (\lambda - 1)(a-c) \\ \Leftrightarrow && \frac{q - a}{c-a} &= 1-\lambda \\ \end{align*} therefore \(\frac{q-a}{c-a} \in \mathbb{R}\) \begin{align*} && \frac{q-a}{c-a} & \in \mathbb{R} \\ \Leftrightarrow && \left (\frac{q-a}{c-a} \right)^* &= \frac{q-a}{c-a} \\ \Leftrightarrow && (q^*-a^*)(c-a) &= (q-a)(c^*-a^*) \\ \end{align*} Given \(aa^* = cc^* = 1\), \begin{align*} && (q^*-a^*)(c-a) &= (q-a)(c^*-a^*) \\ \Leftrightarrow && q^*(c-a) - \frac{c}{a}+1 &= q \frac{a-c}{ca} - \frac{a}{c}+1 \\ \Leftrightarrow && (c-a)\l q^* +\frac{q}{ca}\r &= \frac{c}{a} - \frac{a}{c} \\ &&&= \frac{c^2-a^2}{ac} \\ \Leftrightarrow && q^* +\frac{q}{ca} &= \frac{c+a}{ac} \\ \Leftrightarrow && q^*ac +q &= a+c \end{align*}
2. Since \(Q\) lies on \(AC\) and \(BD\) we must have \begin{align*} &&& \begin{cases} q^*ac +q &= a+c \\ q^*bd +q &= b+d \\ \end{cases} \\ \Rightarrow && q^*(ac-bd) &= (a+c)-(b+d) \\ \Rightarrow && q(ac-bd) &= (b+d)ac-(a+c)bd \\ \Rightarrow && (q+q^*)(ac-bd) &= (a+c)(1-bd)+(b+d)(ac-1) \\ &&&=a-abd+c-bcd+abc-b+acd-d \\ &&&= a(1+cd)-b(1+cd)+c(1+ab)-d(1+ab) \\ &&&= (a-b)(1+cd)+(c-d)(1+ab) \end{align*}
3. If \(AB\) and \(CD\) meet at \(p\) we must have \(p^*ab + p = a+b\), ie \(p(1+ab) = a+b\) amd \(p(1+cd) = c+d\), so \begin{align*} && (q+q^*)(ac-bd) &= (a-b) \frac{c+d}{p} + (c-d) \frac{a+b}{p} \\ \Leftrightarrow && p(q+q^*)(ac-bd) &= (a-b)(c+d)+(c-d)(a+b) \\ &&&= ac+ad-bc-bd+ac+bc-ad-bd \\ &&&= 2(ac-bd) \\ \Leftrightarrow && p(q+q^*) &= 2 \end{align*}

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Solution:

Notice that \(\mathbf{d} = \frac{1}{r+1} \mathbf{b}\) and \(\mathbf{e} = \frac{s}{s+1}\mathbf{a}\). We must also have that the line \(AD\) is \(\mathbf{a} + \lambda \left (\mathbf{a} - \frac{1}{r+1} \mathbf{b}\right)\) and \(BE\) is \(\mathbf{b} + \mu \left (\mathbf{b} - \frac{s}{s+1} \mathbf{a}\right)\) at their point of intersection we must have \begin{align*} && \mathbf{a} + \lambda \left (\mathbf{a} - \frac{1}{r+1} \mathbf{b}\right) &= \mathbf{b} + \mu \left (\mathbf{b} - \frac{s}{s+1} \mathbf{a}\right) \\ [\mathbf{a}]: && 1 + \lambda &= -\frac{\mu s}{s+1} \\ [\mathbf{b}]: && -\frac{\lambda}{r+1} &= 1 + \mu \\ \Rightarrow && \lambda &= -\frac{1+s+\mu s}{s+1} \\ \Rightarrow && \mu &= \frac{1+s+\mu s}{(1+r)(1+s)} - 1 \\ \Rightarrow && (1+r+rs)\mu &= 1+s - 1 - r - s - rs \\ \Rightarrow && \mu &= -\frac{r+rs}{1+r+rs} \\ \Rightarrow && \mathbf{g} &= \mathbf{b} -\frac{r+rs}{1+r+rs}\left (\mathbf{b} - \frac{s}{s+1} \mathbf{a}\right) \\ &&&= \frac{rs}{1+r+rs} \, \mathbf{a} + \frac 1 {1+r+rs} \, \mathbf{b} \end{align*} \item The line \(OG\) is \(\lambda \mathbf{g}\). The line \(AB\) is \(\mathbf{a} + \mu(\mathbf{b}-\mathbf{a})\), so we need \begin{align*} && \lambda \mathbf{g} &= \mathbf{a} + \mu(\mathbf{b}-\mathbf{a}) \\ [\mathbf{a}]: && \lambda \frac{rs}{1+r+rs} &= 1-\mu \\ [\mathbf{b}]: && \lambda \frac{1}{1+r+rs} &= \mu \\ \Rightarrow && \lambda \frac{1+rs}{1+r+rs} &= 1 \\ \Rightarrow && \lambda &= \frac{1+r+rs}{1+rs} \\ \Rightarrow && \mu &= \frac{1}{1+rs} \end{align*} Therefore the line is divided in the ratio \(rs : 1\), and therefore we have proven Ceva's Theorem.

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Solution: Find the gradient of the tangent of the ellipse at \(P\): \begin{align*} && \frac{2x}{a^2} + \frac{2y}{b^2} \frac{\d y}{\d x} &= 0 \\ \Rightarrow && \frac{\d y}{\d x} &= - \frac{2xb^2}{2ya^2} \\ &&&=- \frac{a \cos \theta b^2}{b \sin \theta a^2} \\ &&&=-\frac{b}{a} \cot \theta \end{align*} Therefore the gradient of \(ON\) is \(\frac{a}{b} \tan \theta\). \begin{align*} && y &= \frac{a}{b} \tan \theta x \\ && \frac{y-0}{x-(-ea)} &= \frac{b\sin \theta-0}{a\cos \theta -(-ea)} \\ && y &= \frac{b \sin \theta}{a(e+\cos \theta)}(x+ea) \\ \Rightarrow && y &= \frac{b \sin \theta}{a(\cos \theta+e)}\frac{b}{a} \cot \theta y+ \frac{eb \sin \theta}{\cos \theta + e} \\ &&&= \frac{b^2 \cos \theta}{a^2(\cos \theta +e)}y + \frac{eb \sin \theta}{\cos \theta + e} \\ \Rightarrow && (\cos \theta+e)y &= (1-e^2)\cos \theta y +eb \sin \theta\\ && e(1+e\cos \theta)y &= eb \sin \theta \\ \Rightarrow && y &= \frac{b \sin \theta}{1+e\cos \theta} \\ && x &= \frac{b \sin \theta}{1+e\cos \theta} \frac{b}{a} \cot \theta \\ &&&= \frac{b^2 \cos \theta}{a(1+e\cos \theta)} \end{align*} Therefore \(\displaystyle T\left (\frac{b^2 \cos \theta}{a(1+e\cos \theta)}, \frac{b \sin \theta}{1+e\cos \theta} \right)\). Finally, we can look at the distance of \(T\) from \(S\) \begin{align*} && d^2 &= \left (\frac{b^2 \cos \theta}{a(1+e\cos \theta)}-(-ea) \right)^2 + \left (\frac{b \sin \theta}{1+e\cos \theta} -0\right)^2 \\ &&&= \frac{\left (b^2 \cos \theta+ea^2(1+e\cos\theta)\right)^2 + \left ( ab \sin \theta\right)^2}{a^2(1+e\cos \theta)^2} \\ &&&= \frac{b^4\cos^2\theta+e^2a^4(1+e\cos\theta)^2+2ea^2b^2(1+e\cos\theta)+a^2b^2\sin^2\theta}{a^2(1+e\cos\theta)^2} \\ &&&= \frac{a^4(1-e^2)^2\cos^2\theta+e^2a^4(1+e\cos\theta)^2+2ea^2a^2(1-e^2)(1+e\cos\theta)+a^4(1-e^2)\sin^2\theta}{a^2(1+e\cos\theta)^2} \\ &&&= a^2 \left ( \frac{(1-e^2)^2\cos^2\theta+e^2(1+e\cos\theta)^2+2e(1-e^2)(1+e\cos\theta)+(1-e^2)(1-\cos^2\theta)}{(1+e\cos\theta)^2} \right) \\ &&&= a^2 \left ( \frac{e^2(1+e\cos\theta)^2+(1-e^2)((1-e^2)\cos^2\theta+2e(1+e\cos\theta)+(1-\cos^2\theta))}{(1+e\cos\theta)^2} \right) \\ &&&= a^2 \left ( \frac{e^2(1+e\cos\theta)^2+(1-e^2)(1+e\cos\theta)^2}{(1+e\cos\theta)^2} \right) \\ &&&= a^2 \end{align*} Therefore a circle radius \(a\) centre \(S\).