Problems

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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:

Note that the sine rule gives us \begin{align*} && \frac{x}{\sin \alpha} &= \frac{AB}{\sin (180-3\alpha)} \\ \Rightarrow && AB &= x \frac{\sin 3\alpha}{\sin \alpha} \\ &&&= x \frac{\sin \alpha \cos 2\alpha + \cos \alpha \sin 2\alpha}{\sin \alpha} \\ &&&= x \left ( \frac{\sin \alpha (1-2\sin^2\alpha) + 2(1-\sin^2 \alpha)\sin \alpha}{\sin \alpha} \right) \\ &&&= x (3 - 4\sin^2 \alpha) \end{align*} Note that \(AD = (\tfrac32 - 2 \sin^2\alpha)x\) and \(AE = (3-4\sin^2\alpha-\cos2\alpha)x\) so \begin{align*} DE &= (3-4\sin^2\alpha-\cos2\alpha)x - (\tfrac32 - 2 \sin^2\alpha)x \\ &= (\tfrac32 - 2\sin^2 \alpha - (1-2\sin^2 \alpha))x \\ &= \tfrac12x \end{align*} Since \(DE = \frac12x\) if we drop the perpendicular from \(F\) to \(EC\) we have a \(30-60-90\) triangle. Therefore \(\angle FCE = 30^\circ\). Notice that \(\angle CEB = 90^{\circ} - 2\alpha\) and \(\angle ACB = 180^\circ - 3\alpha\), therefore \begin{align*} \angle ACF &= \angle ACB - \angle FCE - \angle ECB \\ &= (180^\circ - 3\alpha) - 30^\circ - (90^{\circ} - 2\alpha) \\ &= 60^\circ - \alpha \\ &= \frac13 \angle ACB \end{align*}

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Solution: The tangent at \((at^2, 2at)\) can be found \begin{align*} && \frac{\d y}{\d x} &= \frac{\dot{y}}{\dot{x}} \\ &&&= \frac{2a}{2at} = \frac1t \\ \Rightarrow && \frac{y-2at}{x-at^2} &= \frac1t \\ \Rightarrow && ty -x &= at^2 \\ \\ PT: && py - x &= ap^2 \\ QT: && qy - x &= aq^2 \\ \Rightarrow && (p-q)y &= a(p^2-q^2) \\ \Rightarrow && y &= a(p+q) \\ && x &= aq(p+q) - aq^2 \\ &&&= apq \end{align*} By the cosine rule: \begin{align*} && TP^2 &= FT^2 + FP^2 - 2 \cdot FT \cdot FP \cdot \cos \phi \\ && (apq - ap^2)^2 + (a(p+q)-2ap)^2 &= (a-apq)^2+(a(p+q))^2 + \\ &&&\quad + (a-ap^2) + (2ap)^2 - 2 \cdot FT \cdot FP \cdot \cos \phi \\ \Rightarrow && a^2p^2(q-p)^2 + a^2(q-p)^2 &= a^2(1-pq)^2+a^2(p+q)^2 + \\ &&&\quad + a^2(1-p^2)^2+4a^2p^2 - 2 \cdot FT \cdot FP \cdot \cos \phi \\ && a^2(p^2+1)(q-p)^2 &= a^2(1+p^2)(1+q^2) + a^2(1+p^2)^2 + \\ &&&\quad - 2 \cdot a^2(1+p^2)\sqrt{(1+p^2)(1+q^2)} \cos \phi \\ \Rightarrow && \cos \phi &= \frac{a^2(1+p^2)(2+q^2+p^2-(q-p)^2)}{2 a^2 (1+p^2)\sqrt{(1+p^2)(1+q^2)}} \\ &&&= \frac{1+pq}{\sqrt{(1+p^2)(1+q^2)}} \end{align*} As required. Notice that by symmetry, \(\cos \angle TFQ = \frac{1+qp}{\sqrt{(1+q^2)(1+p^2)}} = \cos \phi\). Therefore they have the same angle and \(FT\) bisects \(PFQ\)

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Solution:

\(\mathbf{p} = \lambda \mathbf{a} +(1- \lambda) \mathbf{b}\) Applying the Sine rule, we can see that: \begin{align*} && \frac{OA}{\sin \angle APO} = \frac{AP}{\sin \angle AOP} \\ && \frac{OB}{\sin \angle BPO} = \frac{BP}{\sin \angle BOP} \\ \end{align*} But \(\angle AOP = \angle BOP = \frac12 \angle AOP\) (since \(OP\) bisects the angle) and \(\angle APO = 180^{\circ} -\angle BPO\), so their sines are also equal. Therefore \begin{align*} && \frac{a}{AP} &= \frac{b}{BP} \\ \Rightarrow && \frac{a}{b} &= \frac{1-\lambda}{\lambda} \\ \Rightarrow && \lambda &= \frac{b}{a+b} \end{align*} Note that \(\mathbf{p} = \frac{b\mathbf{a} + a \mathbf{b}}{a+b} = \mathbf{a} + \frac{a}{a+b}(\mathbf{b}-\mathbf{a})\) and \(\mathbf{q} = \mathbf{b} +\frac{a}{a+b}(\mathbf{a}-\mathbf{b}) = \frac{a\mathbf{a} +b \mathbf{b}}{a+b}\) Therefore \begin{align*} && OQ^2 &= \frac{1}{(a+b)^2} \left (a\mathbf{a} + b \mathbf{b} \right)\cdot \left (a\mathbf{a} + b \mathbf{b} \right) \\ &&&= \frac{a^4+b^4+2ab\mathbf{a}\cdot\mathbf{b}}{(a+b)^2} \\ &&OP^2 &= \frac{1}{(a+b)^2} \left (b\mathbf{a} + a \mathbf{b} \right)\cdot \left (b\mathbf{a} + a \mathbf{b} \right) \\ &&&= \frac{2a^2b^2+2ab\mathbf{a}\cdot\mathbf{b}}{(a+b)^2} \\ \\ && OQ^2 - OP^2 &= \frac{a^4+b^4-2a^2b^2}{(a+b)^2} \\ &&&= \frac{(a^2-b^2)^2}{(a+b)^2} \\ &&&= (a-b)^2 \end{align*}

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Solution: \begin{align*} && 0 &= 3y^{2}-10xy+3x^{2}+16y-16x+15 \\ \Rightarrow && 0 &= 6y \frac{\d y}{\d x} - 10x \frac{\d y}{\d x} - 10y + 6x+ 16 \frac{\d y}{\d x } - 16 \\ &&&= \frac{\d y}{\d x} \left (6y - 10x +16 \right) - (10y-6x+16) \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{5y-3x+8}{3y-5x+8} \\ \Rightarrow && \frac{y-t}{x-s} &= \frac{5t-3s+8}{3t-5s+8} \\ && y &= \left(\frac{5t-3s+8}{3t-5s+8}\right)x -\left(\frac{5t-3s+8}{3t-5s+8}\right)s + t \\ &&&= \left(\frac{5t-3s+8}{3t-5s+8}\right)x - \frac{(5ts-3s^2+8s)-(3t^2-5st+8t)}{3t-5s+8} \\ &&&= \left(\frac{5t-3s+8}{3t-5s+8}\right)x - \frac{(5ts-3s^2+8s)-(3t^2-5st+8t)}{3t-5s+8} \\ &&&= \left(\frac{5t-3s+8}{3t-5s+8}\right)x - \frac{8s-8t-(3s^2+3t^2-10st)}{3t-5s+8} \\ &&&= \left(\frac{5t-3s+8}{3t-5s+8}\right)x - \frac{8s-8t-(16s-16t-15)}{3t-5s+8} \\ &&&= \left(\frac{5t-3s+8}{3t-5s+8}\right)x - \frac{8t-8s+15}{3t-5s+8} \\ \end{align*} While \(x \to \infty\) we still have \(3 \frac{y^2}{x^2} - 10 \frac{y}{x} + 3 + 16 \frac{y}{x^2} - 16\frac{1}{x} + 15 \frac{1}{x^2} = 0\), ie if \(\frac{y}{x} = k\), then \(3k^2 - 10k + 3 \to 0 \Rightarrow k \to 3, \frac13\). Therefore, as \(s \to \infty\) we can write \begin{align*} && y &= \left(\frac{5\frac{t}{s}-3+8\frac{1}{s}}{3\frac{t}{s}-5+8\frac1{s}}\right)x - \frac{8\frac{t}s-8+15\frac{1}{s}}{3\frac{t}{s}-5+8\frac{1}{s}} \\ k \to 3: &&& \to \left(\frac{15-3}{9-5}\right)x - \frac{24-8}{9-5} \\ &&&= 3x - 4 \\ k \to \frac13: && &\to \left(\frac{\frac53-3}{1-5}\right)x - \frac{\frac83-8}{1-5} \\ &&&= \frac13 x - \frac43 \end{align*} Therefore the equations are \(y = 3x-4\) and \(3y=x-4\) The lines are parallel to \(y = 3x\) and \(y = \frac13x\), so by considering the triangles formed with the origin and a point \(1\) along the \(x\) or \(y\) axis we can see the angles are identical. This means the line \(y = x\) is parallel to one axis and \(y = -x\) is parallel to the other. They must meet where our two lines meet which is \((1,-1)\), so our lines are \(y = x-2\) and \(y = -x\)

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Solution: \begin{align*} && \frac1{r} &= A \cos \theta + B \cos (\theta - \gamma) \\ &&&= A \cos \theta + B \cos \theta \cos \gamma + B \sin \theta \sin \gamma \\ &&&= (A+B \cos \gamma) \cos \theta + B \sin \gamma \sin \theta \\ \Longleftrightarrow && 1 &= (A+B \cos \gamma) x + B \sin \gamma y \end{align*} So if we choose \(B = \frac{m}{\sin \gamma}\) and \(A = l-m \cot \gamma\) we have the desired result. \begin{align*} && \frac{1 + e \cos (\gamma -\delta)}a &= A \cos (\gamma - \delta) + B \cos (\gamma - \delta - \gamma) \\ &&&= A \cos(\gamma-\delta) +B \cos \delta\\ && \frac{1 + e \cos (\gamma +\delta)}{a} &= A \cos (\gamma + \delta) + B \cos (\gamma + \delta - \gamma) \\ &&&= A \cos(\gamma + \delta) + B \cos \delta\\ \Rightarrow && \frac1{r} &= \frac{e}{a} \cos \theta + \frac{1}{a \cos \delta} \cos (\theta - \gamma) \\ \lim{\delta \to 0} &&\frac1{r} &= \frac{e}{a} \cos \theta+ \frac{1}{a} \cos (\theta - \gamma) \end{align*} Suppose we have have points \(L\) and \(M\) with \(\theta = \gamma_L, \gamma_M\) then our tangents are: \begin{align*} && \frac{a}{r} &= \cos \theta + \cos (\theta - \gamma_L) \\ && \frac{a}{r} &= \cos \theta + \cos (\theta - \gamma_M) \\ \Rightarrow && 0 &= \cos (\theta - \gamma_L) -\cos(\theta - \gamma_M) \\ &&&= - 2 \sin \frac{(\theta - \gamma_L)+(\theta - \gamma_M)}{2} \sin \frac{(\theta - \gamma_L)-(\theta - \gamma_M)}{2} \\ &&&= -2 \sin \left ( \theta - \frac{\gamma_L+\gamma_M}2 \right) \sin \left ( \frac{\gamma_M - \gamma_L}{2}\right) \\ \Rightarrow && \theta &= \frac{\gamma_L+\gamma_M}{2} \end{align*} Therefore clearly \(ST\) bisects \(LSM\). The line \(LM\) can be seen as the chord from the points \(\frac{\gamma_L+\gamma_M}{2} \pm \frac{\gamma_L-\gamma_M}{2}\), so the line is: \begin{align*} && \frac{a}{r} &= e \cos \theta + \frac{1}{\cos \left ( \frac{\gamma_L-\gamma_M}{2}\right)} \cos \left (\theta - \frac{\gamma_L+\gamma_M}{2} \right) \end{align*} and we want the point on the line where \(\theta =\frac{\gamma_L+\gamma_M}{2}\) so \begin{align*} && \frac{a}{r} &= e \cos \left ( \frac{\gamma_L+\gamma_M}{2} \right) + \frac{1}{\cos \left ( \frac{\gamma_L-\gamma_M}{2}\right)} \\ \Rightarrow && r &= \frac{a}{e \cos \left ( \frac{\gamma_L+\gamma_M}{2} \right) + \frac{1}{\cos \left ( \frac{\gamma_L-\gamma_M}{2}\right)}} \end{align*}